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From the outset of our study of energy, we recognize that we are always dealing with energy changes. Even when we write $E=E\left(P,T,h\right)$ to indicate that energy is a function of $P$, $T$, and $h$, we recognize that $E$ represents the energy difference between the state of the system characterized by $P$, $T$, and $h$ and the state of the system when the independent variables correspond to a reference state in which, by definition, $E=0$. As we observe in §6-2, we can sort thermodynamic variables into two classes. Some, like $P$, $V$, and $T$, can be measured only for a system. Others, like $q$, $w$, $E$, $S$, $H$, $G$, and $A$, can be measured only for a process. To say that the volume of a system is one cubic meter has absolute significance. To say that the energy of a system is one joule means nothing unless we know the reference state.
When we intend to specify that the reference state for energy is the particular state specified by $P{=P}_0$, $T{=T}_0$, and $h{=h}_0$, we write “$E\left(P_0,T_0,h_0\right)=0$.” Otherwise, when we write “$E=E\left(P,T,h\right)$,” we could equally well write “$\Delta E=E\left(P,T,h\right)$.” We intend either of these formulations to mean the same thing as “$E=E\left(P,T,h\right)-E\left(P_0,T_0,h_0\right)$ and $E\left(P_0,T_0,h_0\right)=0$.”
Whether we write $E$ or $\Delta E$, the quantity represented is the difference in energy between some initial and some final state. When we focus on very small changes, we can write $dE$ or $d\left(\Delta E\right)$. If our perspective is that we are describing a process, we may prefer to write “$E$”; if our perspective is that we are describing a change in the system, we may prefer to write “$\Delta E$.” In practice, our choice depends primarily on what we have grown accustomed to in the context at hand. In the discussion above, we write $E=q+w$. We could equally well write $\Delta E=q+w$. The meaning is the same. We can make similar statements about most thermodynamic functions. Often there is no particular reason to prefer $X$ over $\Delta X$, or vice versa.
However, there are circumstances in which the delta notation serves particular purposes. If a system undergoes a change in which some thermodynamic variables remain constant, the delta notation provides a convenient way to indicate that a particular variable is not constant. For example, if the volume of a system changes while the applied pressure remains constant, we write $w=-P_{applied}\Delta V$.
Similarly, we often want to describe processes in which some state functions are different in the final state than they are in the initial state, while other state functions are the same in both states, but not necessarily constant throughout the process. In the next few chapters, we develop properties of the state functions entropy, $S$, enthalpy, $H$, and Gibbs free energy, $G$. We define the Gibbs free energy by the relationship $G=H-TS$. To specify the relationship among the changes in these state functions when the final temperature is the same as the initial temperature, we write $\Delta G=\Delta H-T\Delta S$. Here too, we often say that this relationship relates the changes in $\Delta G$, $\Delta H$, and $\Delta S$ when “the temperature is constant.” This is another useful, but potentially misleading, figure of speech. It is important to remember that the equation is valid for any path between the same two states, even if the temperature varies wildly along that path, so long as the initial and final states are at the same temperature.
Finally, we find it convenient to use subscripted versions of the delta notation to specify particular kinds of processes. For a process in which one mole of a pure substance vaporizes to its gas at a particular temperature, we write ${\Delta }_{vap}H$ and ${\Delta }_{vap}G$ to denote the changes in enthalpy and Gibbs free energy, respectively. (We can write ${\Delta }_{vap}E$ to denote the change in the energy; however, ${\Delta }_{vap}E$ is not a quantity that we find useful very often.) Similarly for the fusion and sublimation of one mole of a pure substance at a particular temperature, we write ${\Delta }_{fus}G$, ${\Delta }_{sub}G$, ${\Delta }_{fus}H$, and ${\Delta }_{sub}H$. We also find it convenient to write ${\Delta }_rH$ and ${\Delta }_rG$ to denote the changes in these quantities when a chemical reaction occurs. When we do so, it is essential that we specify the corresponding stoichiometric equation.
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If we heat or do work on any gas—real or ideal—the energy change is $E=q+w$. When we investigate the energy change that accompanies a temperature change, we can obtain reproducible results by holding either the pressure or the volume constant. With volume held constant, we measure $C_V$. With pressure held constant, the energy change we measure depends on both $C_P$ and the relationship among the pressure, volume, and temperature of the gas. If we know an equation of state for the gas and the values of both $C_V$ and $C_P$, we can find the energy change between any two states of the gas, because the same change of state can be achieved in two steps, one at constant pressure and one at constant volume.
To see this, we recognize that the state of any pure gas is completely specified by specifying its pressure, temperature, and volume. Any change of state necessarily involves changing at least two of these state functions. Any change of state that changes all three of them can be achieved in an alternate way that involves two changes, each of which occurs with one variable held constant. For example, the change $\left(P_1,V_1,T_1\right)\to \left(P_2,V_2,T_2\right) \nonumber$ can be achieved by the constant-pressure sequence $\left(P_1,V_1,T_1\right)\to \left(P_1,V_2,T_i\right) \nonumber$ followed by the constant-volume sequence $\left(P_1,V_2,T_i\right)\to \left(P_2,V_2,T_2\right) \nonumber$ where $T_i$ is some intermediate temperature. Note that this sequence has to be possible: with $P$ held constant, specifying a change in $T$ is sufficient to determine the change in $V$; with $V$ held constant, specifying a change in $T$ is sufficient to determine the change in $P$.
Let us consider how the energy of one mole of any pure substance changes with temperature at constant volume. The rate of change of $E$ with $T$ is
${\left(\frac{\partial E}{\partial T}\right)}_V={\left(\frac{\partial q}{\partial T}\right)}_V+{\left(\frac{\partial w}{\partial T}\right)}_V=C_V+{\left(\frac{\partial w}{\partial T}\right)}_V \nonumber$
where we use the definition of $C_V$. For any system, and hence for any substance, the pressure–volume work is zero for any process in which the volume remains constant throughout; therefore, we have ${\left({\partial w}/{\partial T}\right)}_V=0$ and
${\left(\frac{\partial E}{\partial T}\right)}_V=C_V \nonumber$
(one mole of any substance, only PV work possible)
When we develop the properties of ideal gases by treating them as point mass molecules, we find that their average translational kinetic energy is ${3RT}/{2}$ per mole or ${3kT}/{2}$ per molecule, which clearly depends only on temperature. Translational kinetic energy is the only form of energy available to a point-mass molecule, so these relationships describe all of the energy of any point-mass molecule. In particular, they describe all of the energy of a monatomic ideal gas. Since the energy of a monatomic ideal gas is independent of pressure and volume, the temperature derivative must be independent of pressure and volume. The ordinary derivative and the partial derivatives at constant pressure and constant volume all describe the same thing, which, we have just seen, is $C_V$.
$\frac{dE}{dT}={\left(\frac{\partial E}{\partial T}\right)}_P={\left(\frac{\partial E}{\partial T}\right)}_V=C_V=\frac{3}{2}R \nonumber$
(one mole of a monatomic ideal gas)
It is useful to extend the idea of an ideal gas to molecules that are not monatomic. When we do so, we have in mind molecules that do not interact significantly with one another. Another way of saying this is that the energy of the collection of molecules is not affected by any interactions among the molecules; we can get the energy of the collection by adding up the energies that the individual molecules would have if they were isolated from one another. In our development of statistical thermodynamics, we find that the energy of a collection of non-interacting molecules depends only on the molecules’ energy levels and the temperature. The molecules’ energy levels are fixed. This means that if we extend our idea of ideal gases to include non-interacting polyatomic compounds, the energies of such gases still depend only on temperature. For any ideal gas, we have
$\frac{dE}{dT}={\left(\frac{\partial E}{\partial T}\right)}_P={\left(\frac{\partial E}{\partial T}\right)}_V=C_V \nonumber$ (one mole of any ideal gas)
However, for polyatomic molecules it will no longer be true that $C_V={3R}/{2}$. Let us see why. Recall that we construct our absolute temperature scale by extrapolating the Charles’ law graph of volume versus temperature to zero volume. (Figure 2-2.) By experiment, we find that this graph is the same for one mole of a polyatomic ideal gas as it is for one mole of a monatomic ideal gas. Evidently, our definition of temperature depends only on the translational energy of ideal gas molecules and vice-versa. At a fixed temperature, the average translational kinetic energy is the same for any ideal gas; it is independent of the mass of the molecule and of the kinds of atoms in it. To increase the temperature by one degree requires that the translational kinetic energy increase by ${3R}/{2}$, and vice versa.
Consider what happens when we add energy to a polyatomic ideal gas. Polyatomic gas molecules have energy in rotational and vibrational modes of motion. When we add energy to such molecules, some of the added energy goes into these rotational and vibrational modes. To achieve the same increase in translational kinetic energy, the total amount of energy added must be greater. We find that we need a larger $\Delta E$ to achieve the same $\Delta T$, which means that the heat capacity (either $C_V$ or $C_P$) of the polyatomic ideal gas is greater than that of a monatomic ideal gas.
Now let us consider the rate of change of $E$ with $T$ at constant pressure. For one mole of any substance, we have
${\left(\frac{\partial E}{\partial T}\right)}_P={\left(\frac{\partial q}{\partial T}\right)}_P+{\left(\frac{\partial w}{\partial T}\right)}_P=C_P+{\left(\frac{\partial w}{\partial T}\right)}_P \nonumber$
This equation is as far as we can go, unless we can focus on a particular situation for which we know how work varies with temperature at constant pressure.
For one mole of an ideal gas, we have this information. From $PV=RT$ at constant $P$, we have $PdV=RdT$. If reversible work is done on the ideal gas,
$w=\int{-P_{applied}dV=\int{-PdV}}$ and
${\left(\frac{\partial w}{\partial T}\right)}_P={\left[\frac{\partial }{\partial T}\int{-PdV}\right]}_P={\left[\frac{\partial }{\partial T}\int{-RdT}\right]}_P=-R \nonumber$
(any ideal gas)
That is, when enough heat is added to increase the temperature of one mole of ideal gas by one degree kelvin at constant pressure, $-R$ units of work are done on the gas. This is the energy change that occurs because of the increase in volume that accompanies the one-degree temperature increase. Since, for any ideal gas,
$C_V={\left(\frac{\partial E}{\partial T}\right)}_P={\left(\frac{\partial q}{\partial T}\right)}_P+{\left(\frac{\partial w}{\partial T}\right)}_P=C_P-R \nonumber$
we have $C_P=C_V+R \nonumber$
(one mole of any ideal gas)
For a monatomic ideal gas,
$C_P=C_V+R=\frac{3}{2}R+R=\frac{5}{2}R \nonumber$ (one mole of a monatomic ideal gas)
The heat capacity functions have a pivotal role in thermodynamics. We consider many of their properties further in the next section and in later chapters (particularly §10-9 and §10-10.) Because we want to use these properties before we get around to justifying them all, let us summarize them now:
1. For monatomic ideal gases, $C_V$ and $C_P$ are independent of temperature.
2. For polyatomic gases, real or ideal, $C_V$ and $C_P$ are functions of temperature.
3. $C_P$ is always greater than $C_V$, but as the temperature decreases, their values converge, and both vanish at absolute zero.
4. At ordinary temperatures, $C_V$ and $C_P$ increase only slowly as temperature increases. For many purposes they can be taken to be constant over rather wide temperature ranges.
5. For real substances, $C_V$ is a weak function of volume, and $C_P$ is a weak function of pressure. These dependencies are so small that they can be neglected for many purposes.
6. For ideal gases, $C_V$ is independent of volume, and $C_P$ is independent of pressure.
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It is easy to maintain a constant pressure on a solid while varying its temperature. To keep its volume rigorously constant over a range of temperatures is difficult. Because the direct measurement of $C_P$ is straightforward, most heat-capacity experiments on solids measure $C_P$. In Section 10.9, we derive a general relationship between $C_P$, $C_V$, and other measurable properties of a substance. This relationship makes it possible to evaluate $C_V$ indirectly. For a solid, this relationship shows that $C_P$ and $C_V$ are usually about the same.
Heat capacities of solids have been investigated over wide temperature ranges. For most solids, $C_P$ is approximately constant at room temperature and above. For any of the heavier elements, this constant has about the same value. This observation was first made in 1819. It is called the law of Dulong and Petit, in honor of the discoverers. It played an important role in the establishment of correct atomic weights for the elements. The value of the constant found by Dulong and Petit is about $3R$. Remarkably, the law can be extended to polyatomic molecules containing only the heavier elements. Often the solid-state heat capacity of such molecules is about $\mathrm{3}R$ per mole of atoms in the molecule. Correlations that are more detailed have been developed. These relate the heat capacity of a mole of a molecular solid to its molecular formula. In such correlations, the heat capacity per mole increases by a fixed increment for each atom of, say, carbon in the molecule; by a different fixed increment for each atom of nitrogen in the molecule; etc. For the lighter elements, the increments are less than $\mathrm{3}R$. For the heavier elements, the increment is approximately $\mathrm{3}R$, as observed by Dulong and Petit.
As the temperature of any solid decreases, its heat capacity eventually begins to decrease. At temperatures near absolute zero, the heat capacity approaches zero. The graph in Figure 6 shows the shape of the heat capacity versus temperature curve for solid mercury. The shape of this curve can be predicted from a very simple model for the energy modes available to the atoms in a solid. Albert Einstein developed this model in 1907. Einstein’s model for the heat capacity of a solid was an important milestone in the development of quantum mechanics. Since then, the basic ideas have been extended and refined to create more detailed theories that achieve good quantitative agreement with the experimental results for particular substances. We discuss Einstein’s treatment in Section 22.6.
7.15: Defining Enthalpy H
Any mathematical expression that involves only state functions must itself be a state function. We could define $\mathit{\Gamma}=E^2P^2VT$, and $\mathit{\Gamma}$ would be a state function. However, it is not a useful state function. We can define several state functions that have the units of energy and that turn out to be particularly useful. One of them is named enthalpy and is customarily represented by the symbol $H$. We define enthalpy:
$H = E + PV. \nonumber$
One reason that enthalpy is a useful state function emerges if we examine the change in $H$ when the system pressure is equal to the applied pressure, and both are constant. (When these conditions are satisfied, we usually denote the heat accepted by the system as “$q_P$.”) If all of the work is pressure–volume work, we have
\begin{align*} \Delta H &= \Delta E+P\Delta V \[4pt] &=q_P+w+P\Delta V \[4pt] &= q_P-P_{applied}\int{dV}+P\Delta V \[4pt] &= q_P-\left(P_{applied}-P\right)\Delta V \[4pt] &=q_P \end{align*}
If these conditions are satisfied, the enthalpy change is the same thing as the heat added to the system. When we want to express the requirement that the system and applied pressures are equal and constant, we often just say that the process “occurs at a constant pressure.” This is another convenient figure of speech. It also reflects our expectation that the system pressure and the applied pressure will equilibrate rapidly in most circumstances.
For an ideal gas, the molar energy depends only on temperature. Since $PV=RT$ for an ideal gas, $PV$ depends only on temperature. Hence, the molar enthalpy of an ideal gas also depends only on temperature. For an ideal gas, we have the parallel relationships:
${\left(\frac{\partial E}{\partial T}\right)}_V={\left(\frac{\partial E}{\partial T}\right)}_P={\left(\frac{\partial q}{\partial T}\right)}_V=C_V \nonumber$ and ${\left(\frac{\partial H}{\partial T}\right)}_V={\left(\frac{\partial H}{\partial T}\right)}_P={\left(\frac{\partial q}{\partial T}\right)}_P=C_p \nonumber$
Earlier we asserted that, while energy is a state function, heat and work are not. Hess’s law, as originally formulated in 1840, says that the heat changes for a series of chemical reactions can be summed to get the heat change for the overall process described by the sum of the chemical reactions. This amounts to saying that heat is a state function. As it stands, this is a contradiction. The resolution is, of course, that Hess’s law was formulated for a series of chemical reactions that occur at the same constant pressure. Then the heat involved in each step is the enthalpy change for that step, and since enthalpy is a state function, there is no contradiction. Modern statements of Hess’s law frequently forego historical accuracy in favor of scientific accuracy to assert that the enthalpy change for a series of reactions can be summed to get the enthalpy change for the overall process. Thus revised, Hess’s law ceases to be a seminal but imperfect conjecture and becomes merely a special case of the principle that enthalpy is a state function.
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If a system is in thermal contact with its surroundings, a reversible change can involve the exchange of heat between the system and the surroundings. In Chapter 6, we make a number of important observations about the nature of any heat transfer that occurs during a reversible process. Let us review these ideas.
A system can undergo a change in which it accepts (or liberates) heat while its temperature remains constant. If we boil a liquid at constant pressure, a thermometer immersed in the liquid continues to show the same temperature even though we add more and more heat energy. The added heat is used within the system to convert the liquid to its vapor. If the liquid is stirred well, any localized temperature excursions away from the equilibrium temperature are small; it is a good approximation to say that the temperature of the system is homogenous throughout the system and that it has a constant value.
Nevertheless, for a finite boiling rate we recognize that the idea of an isothermal process is indeed an approximation. For heat transfer to occur from the surroundings to the system, the surroundings must be at a higher temperature than the system. The portion of the system in immediate contact with the wall of the vessel must be at a higher temperature than the portion in the interior of the vessel.
When we think about a constant-temperature system undergoing a reversible change while in thermal contact with its surroundings, we imagine that heat can be transferred in either direction with equal facility. If the system is taking up heat as the process proceeds, we imagine that we can reverse the direction of the change simply by changing the direction of heat transfer. Heat will flow from the system to the surroundings, and the process will run backwards. We can reverse the direction of heat flow by changing the temperature of the surroundings. Initially the surroundings must be hotter than the system. To reverse the direction of heat flow, we must make the temperature of the surroundings less than that of the system. Since a reversible process is one whose direction can be reversed by an arbitrarily small change in some state function, the original temperatures must be arbitrarily close to one another.
For a system that exchanges heat with its surroundings, a process can be reversible only if the temperatures of the system and the surroundings are arbitrarily close to one another. In a reversible process, net heat transfer occurs between two entities—the system and its surroundings—that are arbitrarily close to thermal equilibrium. Such a process is an idealization. As we have noted several times, a reversible process is a creature of theory that is merely approximated in real systems. A reversible process does not have to be a constant-temperature process. If the temperatures of system and surroundings change simultaneously, they can remain arbitrarily close to one another throughout the process. Nor must a system undergoing reversible change be in thermal contact with its surroundings. A system can undergo a reversible change adiabatically.
Finally, we have noted that the term “isothermal process” is often intended to mean a constant-temperature thermally-reversible process. However, the same words are frequently intended to indicate only that the final temperature of the system is the same as the initial temperature. This is the case whenever the “isothermal process” is a spontaneous process. The intended meaning is usually clear from the context.
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To develop the theory of thermodynamics, we must be able to model the thermodynamic properties of gases as functions of pressure, temperature, and volume. To do so, we consider processes in which the volume of a gas changes. For the expansion (or compression) of a gas to be a reproducible process, the exchange of heat between the system and its surroundings must be controlled. There are two straightforward ways to do this. We can immerse the system in a constant temperature bath whose temperature is the same as that of the system; in this case, $\Delta T=0$, and we can say that the process is isothermal.
Alternatively, we can isolate the system so that it cannot exchange heat with the surroundings; in this case $q=0$, and the process is said to be adiabatic. In §7 we find that the work done on a system when its volume changes by $dV$ under the influence of an applied pressure, $P_{applied}$, is
$dw=-P_{applied}dV.\nonumber$
Any expansion of a system in which the applied pressure is less than the system pressure can be called a free expansion. In Section 10.14 we consider the adiabatic expansion of a real gas against a constant applied pressure—a process known as a Joule-Thomson expansion. We find that we must introduce a new parameter—the Joule-Thomson coefficient—in order to describe the behavior of a real gas in a free expansion. The Joule-Thomson coefficient varies with pressure and temperature.
Literally, an isothermal process is one in which the temperature of the system remains the same throughout the process. However, we often use the term to mean merely that the process occurs while the system is in thermal contact with constant-temperature surroundings. The free expansion of a gas is an irreversible process; in principle, the temperature of a gas undergoing a free expansion is not a meaningful quantity. When we talk about an isothermal free expansion of a gas, we mean that the final temperature is the same as the initial temperature.
Here we consider the behavior of ideal gases, and we begin by considering the limiting case of a free expansion in which the applied pressure is zero. Physically, this corresponds to the expansion of a system into a (very large) evacuated container. Under this condition, $dw=0$, and the energy change is $dE=dq$. For one mole of any substance, $C_V={\left({\partial E}/{\partial T}\right)}_V$. If only pressure–volume work is possible and the applied pressure is zero, we have $dE=dq=C_VdT$, and
$\Delta E=q=\int^{T_2}_{T_1}{C_VdT}\nonumber$
where $T_1$ and $T_2$ are the temperatures of the substance before and after the expansion, respectively.
At ordinary temperatures, $C_V$ changes only slowly as the temperature changes. Over a short temperature range, it is usually a good approximation to assume that $C_V$ is constant. We have
$\Delta E=q=C_V\left(T_2-T_1\right)\nonumber$ (one mole of any gas or other substance)
For a monatomic ideal gas, the energy change is exactly $\Delta E=q=\frac{3}{2}R\left(T_2-T_1\right)\nonumber$ (one mole of a monatomic ideal gas)
The enthalpy change for any process is $\Delta H=\Delta E+\Delta \left(PV\right)$. If the system is one mole of an ideal gas, we have, because $\Delta \left(PV\right)=R\Delta T=R\left(T_2-T_1\right)$,
$\Delta H=C_V\left(T_2-T_1\right)+R\left(T_2-T_1\right)=\left(C_V+R\right)\left(T_2-T_1\right)=C_P\left(T_2-T_1\right)\nonumber$ (one mole of any ideal gas)
For an isothermal free expansion against an applied pressure of zero, we have $\Delta T=0$, and so neither the energy nor the enthalpy of the gas changes. Since also $dw=0$, there can be no exchange of heat with the surroundings. We have
$w=q=\Delta T=\Delta E=\Delta H=0\nonumber$ (free expansion, ideal gas)
For an adiabatic free expansion, we have $dq=0$ and $dw=0$, and it follows again that $w=q=\Delta T=\Delta E=\Delta H=0$. We see that the isothermal and adiabatic expansions of an ideal gas into a vacuum are equivalent processes. If the expansion is opposed by a non-zero applied pressure, the two processes cease to be equivalent.
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In Section 7.16 we consider heat transfer in reversible processes. Similar considerations apply to the exchange of work between a system and its surroundings. When we use a piston to compress a gas in a cylinder, we must apply sufficient inward force on the piston to overcome the outward force applied by the gas. In any real system, it is necessary also to overcome the force of friction in order to slide the piston into the cylinder. We ignore friction, imagining that we can make its effects arbitrarily small.
The gas can be compressed only if the applied pressure exceeds the gas pressure. If the applied pressure equals the gas pressure, the piston remains stationary. If the applied pressure is greater than the gas pressure by any ever-so-small amount, the gas will be compressed. Conversely, if the applied pressure is infinitesimally less than the gas pressure, the gas will expand. The work done under such conditions is reversible work; an arbitrarily small change in the relative pressures can reverse the direction in which the piston moves. We summarize these conditions by saying that reversible pressure–volume work can occur only if the system and its surroundings are at mechanical equilibrium.
Now, let us think about calculating the reversible work for isothermally compressing a gas by sliding a piston into a cylinder. In any real experiment, we must have $P_{applied}>P_{gas}$, and any real experiment is necessarily irreversible. In a reversible experiment, we have $P_{applied}=P_{gas}=P$, and the reversible work, $w^{rev}$, is
\begin{align*} w^{rev} &=\int^{V_2}_{V_1}{-P_{applied}dV} \[4pt] &=\int^{V_2}_{V_1}{-P_{gas}dV} \end{align*}\nonumber
For one mole of an ideal gas, we have $P={RT}/{V}$. Since the temperature is constant, the reversible isothermal work becomes
$w^{rev}=\int^{V_2}_{V_1}{-\frac{RT}{V}dV}=-RT{ \ln \frac{V_2}{V_1}\ }\nonumber$
where $V_1$ and $V_2$ are the initial and final volumes of the gas, respectively. This has a straightforward graphical interpretation. For an ideal gas at constant temperature, $-P_{gas}$ is inversely proportional to $V_{gas}$. As sketched in Figure 7, the reversible work corresponds to the area between this curve and the abscissa and between the initial, $V_1$, and the final, $V_2$, gas volumes.
In contrast, an irreversible expansion corresponds to movement of the piston when $P_{gas}>P_{applied}$, or equivalently, $-P_{gas}<-P_{applied}$. Therefore, the work done on the gas is less in the reversible case than it is in the irreversible case. (Both work terms are less than zero. The absolute value of the reversible work is greater than the absolute value of the irreversible work.) From our definitions of reversible and irreversible pressure–volume work, we have ${dw}^{rev}<{dw}^{irrev}$ and$\ w^{rev}<w^{irrev}$, so long as the initial and final states are the same in the irreversible process as they are in the reversible constant-temperature process. The shaded area in Figure 7 represents the work done on the gas when the applied pressure is instantaneously decreased to the final pressure, $P_a$, attained by the gas in the reversible process.
For the reversible process, the pressure–volume curve accurately depicts the state of the gas as the volume increase takes place. The temperature of the gas is constant along this curve. While we can trace a similar line of pressure–volume points for the irreversible expansion, this line does not define a set of intermediate states that the system occupies during the irreversible expansion. The state of the gas is well defined only in the equilibrium state that precedes the irreversible pressure drop and in the equilibrium state that the system ultimately attains. It is convenient to describe these two processes as a reversible process and a spontaneous process that “take the system from the same initial state to the same final state.” However, this language obscures a significant point. In the initial state for the reversible process, we have
$P_{gas}={RT}/{V_1=P_{applied}}\nonumber$
In the initial state for the spontaneous process, we have
$P_{gas}=\dfrac{RT}{V_1}\nonumber$
and
$P_{applied}=\frac{RT}{V_2}\nonumber$
What we mean, of course, is that the values of all of the state functions for the hypothetical initial state of the spontaneous process are the same as those for the equilibrium initial state of the reversible process.
So long as we can say that the process takes the system from the same initial state to the same final state, a similar argument can be made for reversible and irreversible work of any kind. Whatever the force, the isothermal reversible work done on the system is always less than the irreversible work for taking the system between the same two states. This is an important result. In Chapter 9, we find that it is a logical consequence of the second law of thermodynamics.
Finally, let us consider a reversible process in which a system completes a pressure–volume cycle. The system traverses a closed path in the pressure–volume plane. Such a path is depicted in Figure 8. We let the smallest and largest volumes reached during the cycle be $V_{\ell }$ and $V_h$, respectively. The closed path is composed of a high-pressure segment and a low-pressure segment that meet at $V_{\ell }$ and $V_h$. On each of these segments, the pressure is a function of volume. We let pressures on the high- and low-pressure segments be $P_h\left(V\right)$ and $P_{\ell }\left(V\right)$, respectively. (In the interval $V_{\ell }<v_h$>, we have $P_h\left(V\right)>P_{\ell }\left(V\right)>0$. At the limiting volumes, we have $P_h\left(V_h\right)=P_{\ell }\left(V_h\right)$ and $P_h\left(V_{\ell }\right)=P_{\ell }\left(V_{\ell }\right)$.) The system temperature varies continuously around the closed path. The work done on the system as it traverses the high-pressure segment from $V_{\ell }$ to $V_h$ is represented in Figure 8 by area $A_h$.
We have
$w_h\left(V_{\ell }\mathrm{\to }V_h\right) = \int^{V_h}_{V_{\ell }}{{ + P}_h\left(V\right)dV} = A_h\mathrm{<0}\nonumber$
The work done on the system as it traverses the low-pressure segment from $V_{\ell }$ to $V_h$ is represented by area $A_{\ell }$. We have
$w_{\ell }\left(V_{\ell }\to V_h\right)=\int^{V_h}_{V_{\ell }}{{-P}_{\ell }\left(V\right)dV}=A_{\ell }<0\nonumber$
When the low-pressure segment is traversed in the opposite direction, we have $w_{\ell }\left(V_h\to V_{\ell }\right)=-A_{\ell }>0$. When the system traverses the cycle in the counterclockwise direction, the net work done on the system is
\begin{align*} w_{net} &=w_h\left(V_{\ell}\to V_h\right)+w_{\ell}\left(V_h\to V_{\ell}\right) \[4pt] &=\int^{V_h}_{V_{\ell}}{-P_h\left(V\right)dV}+\int^{V_{\ell}}_{V_h}{{-P}_{\ell }\left(V\right)dV} \[4pt] &=\oint{-P\left(V\right)dV} \[4pt] &=A_h-A_{\ell }<0 \end{align*}\nonumber
Thus the net work done on the system is represented on the graph by the area $A_h-A_{\ell }$, which is just the (negative) area in the pressure–volume plane that is bounded by the closed path.
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For an isothermal reversible expansion of an ideal gas, we have by definition that $\Delta T=0$. Since the energy of an ideal gas depends only on the temperature, a constant temperature implies constant energy, so that $\Delta E=0=q^{rev}+w^{rev}$. Using the equation we find for $w^{rev}$ in the previous section, we have
$-q^{rev}=w^{rev}=-RT{ \ln \frac{V_2}{V_1}\ } \nonumber$ (ideal gas, isothermal reversible expansion)
where $V_1$ and $V_2$ are the initial and final volumes, respectively. Since enthalpy is defined as $H=E+PV$, we have $\Delta H=\Delta E+\Delta \left(PV\right)=\Delta E+\Delta \left(RT\right)=0$.
For the spontaneous isothermal expansion of an ideal gas from $V_1$ to $V_2$ against a constant applied pressure, we again have $\Delta T=\Delta E=\Delta H=0$. These are state functions, and the amounts by which they change in this spontaneous process must be the same as those for the reversible process between the same two states. The heat and work exchanged in the spontaneous process are different, demonstrating that heat and work are not state functions. We have $-q=w=\int^{V_2}_{V_1}{-P_{applied}dV}=-P_{applied}\left(V_2-V_1\right)=RT\left(\frac{P_{applied}}{P_1}-1\right) \nonumber$ (one mole ideal gas, isothermal free expansion, $P_{applied}>0$)
7.20: Adiabatic Expansions of An Ideal Gas
Consider an ideal gas that undergoes a reversible adiabatic expansion from an initial state, specified by known values $V_1$ and $T_1$, to a new state in which the value of the volume, $V_2$, is known but the value of the temperature, $T_2$, is not known. For an adiabatic reversible process, $q=0$, and $w=\Delta E$. Since ${\left({\partial E}/{\partial T}\right)}_V=C_V$, we have $dE=C_VdT$, so that
$w=\Delta E=\int^{T_2}_{T_1}{C_V}dT \nonumber$
For any gas, we can assume that $C_V$ is approximately constant over a small temperature range. Taking $C_V$ to be constant in the interval $T_1<t_2$>, we have $w=\Delta E=C_V\left(T_2-T_1\right)$. We obtain the enthalpy change from
$\Delta H=\Delta E+\Delta \left(PV\right)=\Delta E+\Delta \left(RT\right)=C_V\left(T_2-T_1\right)+R\left(T_2-T_1\right)=C_P\left(T_2-T_1\right) \nonumber$
where we use our ideal-gas result from Section 7.16, $C_P=C_V+R$.
While these relationships yield the values of the various thermodynamic quantities in terms of the temperature difference, $T_2-T_1$, we have yet to find the final temperature, $T_2$. To find $T_2$, we return to the first law: $dE=dq+dw$. Substituting for $dE$, $dq$, and $dw$, and making use of the ideal gas equation, we have
$C_VdT=-PdV=-\frac{RT}{V}dV \nonumber$
from which, by separation of variables, we have
$\int^{T_2}_{T_1}{C_V}\frac{dT}{T}=-R\int^{V_2}_{V_1}{\frac{dV}{V}}$ (one mole ideal gas, reversible adiabatic expansion)
If we know $C_V$ as a function of temperature, we can integrate to find a relationship among $T_1$, $T_2$, $V_1$, and $V_2$. Given any three of these quantities, we can use this relationship to find the fourth. If $C_V$ is independent of temperature, as it is for a monatomic ideal gas, we have
$\ln \frac{T_2}{T_1} =-\frac{R}{C_V} \ln \frac{V_2}{V_1} =\frac{R}{C_V} \ln \frac{V_1}{V_2} = \ln \left(\frac{V_1}{V_2}\right)^{R/C_V} \nonumber$
so that
$\frac{T_2}{T_1}= \left(\frac{V_1}{V_2}\right)^{R/C_V} \nonumber$ (monatomic ideal gas, reversible adiabatic expansion)
For the spontaneous adiabatic expansion of an ideal gas against a constant applied pressure, we have $dq=0$, so that $dE=dw$, and $C_VdT=-P_{applied}dV$. Given the initial conditions, we can find the final temperature from
$\int^{T_2}_{T_1}{C_VdT}=\int^{V_2}_{V_1}{-P_{applied}dV}=-P_{applied}\left(\frac{RT_2}{P_{applied}}-\frac{RT_1}{P_1}\right)=R\left(\frac{P_{applied}T_1}{P_1}-T_2\right) \nonumber$ (spontaneous adiabatic process)
The changes in the remaining state functions can then be calculated from the relationships above. In this spontaneous adiabatic process, all of the other thermodynamic quantities are different from those of a reversible adiabatic process that reaches the same final volume.
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1. Which of the following differential expressions are exact?
a. $df=ydx+xdy$
b. $df=2xy^2dx+2x^2ydy$
c. $df=2xydx+2x^2ydy$
d. $df=\left[\left(1-xy\right)e^{-xy}\right]dx-\left[x^2e^{-xy}\right]dy$
e. $df=\left({\mathrm{cos} x\ }{\mathrm{cos} y\ }\right)dx-\left({\mathrm{sin} x{\mathrm{sin} y\ }\ }\right)dy$
f. $df=\left({\mathrm{cos} x\ }{\mathrm{cos} y\ }\right)dx-\left({\mathrm{sin} y\ }\right)dy$
2. Show that $df=e^{-y}dx-xe^{-y}dy$ is exact. Find $f\left(x,y\right)$ by integrating the $dx$ term. Find $f\left(x,y\right)$ by integrating the $dy$ term.
3. A marble of mass $m$ is free to move on a surface whose height above the $x,y$-plane is $h=ax^2+by^2$.
a. What is the gravitational potential energy of the marble expressed as a function of $x$ and $y$, $E\left(x,y\right)$?
b. The force experienced by the marble due to gravity is the vector function $\mathop{f}\limits^{\rightharpoonup}\left(x,y\right)=-\mathrm{\nabla }E\left(x,y\right)=-{\left(\frac{\partial E}{\partial x}\right)}_y\mathop{i}\limits^{\rightharpoonup}-{\left(\frac{\partial E}{\partial y}\right)}_x\mathop{j}\limits^{\rightharpoonup} \nonumber$ What is $\mathop{f}\limits^{\rightharpoonup}\left(x,y\right)$ on this surface?
c. What is the differential of $E$? Is $dE$ exact or inexact?
d. The vector description of a general path, $\left\{\left(x,y\right)\right\}$, is the position vector, $\mathop{r}\limits^{\rightharpoonup}=x\mathop{i}\limits^{\rightharpoonup}+y\mathop{j}\limits^{\rightharpoonup}$, and so $d\mathop{r}\limits^{\rightharpoonup}=dx\mathop{\ i}\limits^{\rightharpoonup}+dy\ \mathop{j}\limits^{\rightharpoonup}$. If we push the marble up the surface from point $\left(0,0,E\left(0,0\right)\right)$ to point $\left(2,2,E\left(2,2\right)\right)$ along the path $y=x$, express $d\mathop{r}\limits^{\rightharpoonup}$ as a vector function of $dx$.
e. If we push the marble along the path in part d with a force just large enough to overcome the force of gravity, what is the increment of work, $dw$, associated with an increment of motion, $d\mathop{r}\limits^{\rightharpoonup}$?
f. How much work must we do if we are to move the marble from $\left(0,0,E\left(0,0\right)\right)$ to point $\left(2,2,E\left(2,2\right)\right)$ along the path in part d, using the force in part e? What is the relationship between this amount of work and the change in the energy of the marble during this process?
g. Suppose that we push the marble up the surface from point $\left(0,0,E\left(0,0\right)\right)$ to point $\left(2,2,E\left(2,2\right)\right)$ along the path $y={x^2}/{2}$. What is the vector description of this path?
h. How much work must we do if we are to move the marble from point $\left(0,0,E\left(0,0\right)\right)$ to point $\left(2,2,E\left(2,2\right)\right)$ along the path in part g using the force in part b? Compare this result to your result in part f. Explain.
4. Consider the plane, $f\left(x,y\right)=1-2x-3y$. What is $df$ for this surface? Evaluate $\Delta f=f\left(1,1\right)-f\left(-1,-1\right)$by integrating $df$ along each of the following paths:
a. $y=x$
b. $y=x^3$
c. $y=1+x-x^2$
d. $y={\mathrm{sin} \left({\pi x}/{2}\right)\ }$
5. A $\mathrm{2.00}$ mole sample of a monatomic ideal gas is expanded reversibly and isothermally at $\mathrm{350}$ K from $\mathrm{5.82}$ L to $\mathrm{58.20}$ L. How much work is done on the gas? What are $q$, $\Delta E$, and $\Delta H$ for the gas in this process?
6. A $\mathrm{2.00}$ mole sample of a monatomic ideal gas is expanded irreversibly from $\mathrm{5.82}$ L to $\mathrm{58.20}$ L at a constant applied pressure equal to the final pressure of the gas. The initial and final temperatures are $\mathrm{350}$ K. How much work is done on the gas? What are $q$, $\Delta E$, and $\Delta H$ for the gas in this process? Compare $w$,$\ q$, $\Delta E$, and $\Delta H$ for this process to the corresponding quantities for the process in problem 5. Compare the initial and final states of the gas to the corresponding states in problem 5.
7. A $\mathrm{2.00}$ mole sample of a monatomic ideal gas is expanded reversibly and adiabatically from $\mathrm{5.82}$ L to $\mathrm{58.20}$ L. The initial temperature is $\mathrm{350}$ K. What is the final temperature? What are the initial and final pressures? How much work is done on the gas? What are $q$, $\Delta E$, and $\Delta H$ for the gas in this process?
8. The equation of state for a “hard-sphere gas” is $P\left(V-nb\right)=nRT$, where $n$ is the number of moles and $b$ is the molar volume of the hard spheres. How much work is done on this gas when n moles of it expand reversibly and isothermally from $V_1$ to $V_2$?
9. Strictly speaking, can the spontaneous expansion of a real gas be isothermal? Can it be free? Can it be adiabatic? Can the reversible expansion of a gas be isothermal? Can it be free? Can it be adiabatic?
10. Consider a machine that operates in a cycle and converts heat into a greater amount of work. What would happen to the energy of the universe if this machine could be operated in reverse?
11. Show that the product of pressure and volume has the units of energy.
12. Give a counter-example to prove that each of the following propositions is false:
a. If $X$ is a state function,$\ X$ is conserved.
b. If $X$ is an extensive quantity that satisfies $X+\hat{X}=0$, $X$ is a state function.
Notes
$^{1}$ Since the temperature of the water increases and the process is to be reversible, we must keep the temperature of the thermal reservoir just $dT$ greater than that of the water throughout the process. We can accomplish this by using a quantity of ideal gas as the heat reservoir. By reversibly compressing the ideal gas, we can reversibly deliver the required heat while maintaining the required temperature. We consider this operation further in Section 12.5.
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• 8.1: Enthalpy
• 8.2: Using Thermochemical Cycles to Find Enthalpy Changes
• 8.3: How Enthalpy Depends on Pressure
• 8.4: Standard States and Enthalpies of Formation
• 8.5: The Ideal Gas Standard State
The ideal gas standard state is a useful invention, which has additional advantages that emerge as our development proceeds. For permanent gases—gases whose behavior is approximately ideal anyway—there is a negligible difference between the enthalpy in the ideal gas state and the enthalpy at 1 bar.
• 8.6: Standard Enthalpies of Reaction
If we have the enthalpies of formation, we can compute the enthalpy change for the reaction. We can demonstrate this by writing out the chemical equations corresponding to the formation of A, B, C, and D from their elements. When we multiply these chemical equations by the appropriately signed stoichiometric coefficient and add them, we obtain the chemical equation for the indicated reaction of A and B to give C and D. Because enthalpy is a state function.
• 8.7: Standard State Heat Capacities
We have observed that CV depends on volume and temperature, while CP depends on pressure and temperature. Compilations of heat capacity data usually give values for CP , rather than CV . When the temperature-dependence of CP is known, such compilations usually express it as an empirical polynomial function of temperature.
• 8.8: How The Enthalpy Change for a Reaction Depends on Temperature
We use tabulated enthalpies of formation to calculate the enthalpy change for a particular chemical reaction. Such tables typically give enthalpies of formation at a number of different temperatures, so that the enthalpy change for a given reaction can also be calculated at these different temperatures; it is just a matter of repeating the same calculation at each temperature.
• 8.9: Calorimetry
Calorimetry is the experimental science of measuring the heat changes that accompany chemical or physical changes. The accurate measurement of small amounts of heat is experimentally challenging. Nevertheless, calorimetry is an area in which great experimental sophistication has been achieved and remarkably accurate measurements can be made. Numerous devices have been developed to measure heat changes.
• 8.10: Problems
08: Enthalpy and Thermochemical Cycles
In Chapter 7, we introduce the enthalpy function, which we define as
$H\ =\ E\ +\ PV \nonumber$
When the only form of work possible is pressure–volume work and a system change occurs at constant pressure, the enthalpy change is synonymous with the heat added to the system.
$H=q_P$ (only PV work)
Since we define $C_P={\left({\partial q}/{\partial T}\right)}_P$, it follows that ${\left({\partial H}/{\partial T}\right)}_P=C_P$. Recalling our earlier discovery that ${\left({\partial E}/{\partial T}\right)}_V=C_V$, we have the important parallel relationships: $C_P={\left(\frac{\partial q}{\partial T}\right)}_P={\left(\frac{\partial H}{\partial T}\right)}_P \nonumber$ and
$C_V={\left(\frac{\partial q}{\partial T}\right)}_V={\left(\frac{\partial E}{\partial T}\right)}_V \nonumber$
We can find the enthalpy change for heating a substance at constant pressure by integrating its constant-pressure heat capacity, $C_P$, over the change in temperature. That is, $C_P={\left({\partial H}/{\partial T}\right)}_P$ implies that
$\Delta H=\int^{T_2}_{T_1}{C_PdT} \nonumber$ Similarly, we have $\Delta E=\int^{T_2}_{T_1}{C_VdT} \nonumber$
for a process in which a substance is heated at constant volume.
One reason that the enthalpy function is useful in chemistry is that many processes are carried out at conditions (constant pressure, only $PV$ work) where the enthalpy change is synonymous with the heat exchanged. The heat exchanged in a process is frequently an important consideration. If we want to carry out an endothermic process, we must provide means to add sufficient
heat. If we want to carry out an exothermic process, we may have to make special arrangements to safely transfer the heat evolved from the system to its surroundings.
One of our principal objectives is to predict whether a given process can occur spontaneously. We will see that the heat evolved in a process is not a generally valid predictor of whether or not the process can occur spontaneously; however, it is true that a very exothermic process is usually one that can occur spontaneously. (We will see that $\Delta H<0$ is a rigorous criterion for whether the process can occur spontaneously if and only if the process is one for which both the entropy and the pressure remain constant.)
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Because enthalpy is a state function, the enthalpy change in going between any two states of a system is independent of the path. For a series of changes that restore a system to its original state, the sum of all the enthalpy changes must be zero. This fact enables us to find the enthalpy changes for many processes for which it is difficult to measure heat and work directly. It is easiest to see what is involved by considering a specific example. Figure 1 shows a cyclic path, A$\mathrm{\to }$A*$\mathrm{\to }$B$\mathrm{\to }$$\mathrm{\to }$A, superimposed on a not-to-scale presentation of the phase diagram for water. Let us look at the sublimation of ice at the melting point of pure water. The sublimation of ice is the conversion of pure ice to pure water vapor. (The melting point of pure water is the temperature at which pure ice is at equilibrium with pure liquid water at a pressure of one atmosphere; it is represented by points A and A* on the diagram. We want to find the enthalpy of sublimation at the temperature and pressure represented by points D and D*.)
Points A, A*, D, and D* are all at the same temperature; this temperature is about 273.153 K or 0.003 C. (This temperature is very slightly greater than 273.15 K or 0 C—which is the temperature at which ice and water are at equilibrium in the presence of air at a total pressure of one atmosphere.) We want to calculate the enthalpy change for the equilibrium conversion of one mole of ice to gaseous water at the pressure where the solid–gas equilibrium line intersects the line $T=\mathrm{273.153\ K}\approx \mathrm{0\ C}$.
On the diagram, this sublimation pressure is represented as $P_{sub}$ and the sublimation process is represented as the transition from D* to D. $P_{sub}$ is less than the triple-point pressure of $\mathrm{611\ Pa}$ or $6.03\times {10}^{-3}\ \mathrm{atm}$. However, the difference is less than $1.4\times {10}^{-5}\ \mathrm{atm}$ or $\mathrm{1.4\ Pa}$. In equation form, the successive states traversed in this cycle are:
A (ice at 0 C and 1 atm) $\mathrm{\to }$
A* (water at 0 C and 1 atm) $\mathrm{\to }$
B (water at 100 C and 1 atm) $\mathrm{\to }$
B* (water vapor at 100 C and 1 atm) $\mathrm{\to }$
C (water vapor at 100 C and $P_{sub}$) $\mathrm{\to }$
D (water vapor at 0 C and $P_{sub}$) $\mathrm{\to }$
D*(ice at 0 C and $P_{sub}$) $\mathrm{\to }$
A (ice at 0 C and 1 atm)
We select these steps because it is experimentally straightforward to find the enthalpy change for all of them except the sublimation step (D*$\mathrm{\to }$D). All of these steps can be carried out reversibly. This strategy is useful in general. We make extensive use of reversible cycles to find thermodynamic information for chemical systems. The enthalpy changes for these steps are
$H_2O$ (s, 0 C, 1 atm) $\mathrm{\to }$ $H_2O$ (liq, 0 C, 1 atm) $\Delta H\left(\mathrm{A}\mathrm{\to }{\mathrm{A}}^{\mathrm{*}}\right)={\Delta }_{fus}H \nonumber$
$H_2O$ (liq, 0 C, 1 atm) $\mathrm{\to }$ $H_2O$ (liq, 100 C, 1 atm) $\Delta H\left({\mathrm{A}}^{\mathrm{*}}\mathrm{\to }\mathrm{B}\right)=\int^{372.15\ \mathrm{K}}_{273.15\ \mathrm{K}}{C_P\left(H_2O,\ \mathrm{liq}\right)\ dT} \nonumber$
$H_2O$ (liq, 100 C, 1 atm) $\mathrm{\to }$ $H_2O$ (g, 100 C, 1 atm) $\Delta H\left(\mathrm{B}\mathrm{\to }{\mathrm{B}}^{\mathrm{*}}\right)={\Delta }_{vap}H \nonumber$
$H_2O$ (g, 100 C, 1 atm) $\mathrm{\to }$ $H_2O$ (g, 100 C, $P_{sub}$) $\Delta H\left({\mathrm{B}}^{\mathrm{*}}\mathrm{\to }\mathrm{C}\right)=\int^{P=P_{sub}}_{P=1}{{\left(\frac{\partial H\left(H_2O,\ \mathrm{g}\right)}{\partial P}\right)}_T\ dP\approx 0} \nonumber$
$H_2O$ (g, 100 C, $P_{sub}$) $\mathrm{\to }$$H_2O$ (g, 0 C, $P_{sub}$) $\Delta H\left(\mathrm{C}\mathrm{\to }\mathrm{D}\right)=\int^{272.15\ \mathrm{K}}_{373.15\ \mathrm{K}}{C_P\left(H_2O,\ \mathrm{g}\right)\ dT} \nonumber$
$H_2O$ (g, 0 C, $P_{sub}$) $\mathrm{\to }$ $H_2O$ (s, 0 C, $P_{sub}$) $\Delta H\left(\mathrm{D}\mathrm{\to }{\mathrm{D}}^{\mathrm{*}}\right)={-\Delta }_{sub}H \nonumber$
$H_2O$ (s, 0 C, $P_{sub}$) $\mathrm{\to }$$H_2O$ (s, 0 C, 1 atm)
$\Delta H\left({\mathrm{D}}^{\mathrm{*}}\mathrm{\to }\mathrm{A}\right)=\int^{P=1}_{P=P_{sub}}{{\left(\frac{\partial H\left(H_2O,\ \mathrm{s}\right)}{\partial P}\right)}_T\ dP\approx 0} \nonumber$ Summing the enthalpy changes around the cycle gives
$0={\Delta }_{fus}H+\int^{372.15\ \mathrm{K}}_{273.15\ \mathrm{K}}{C_P\left(H_2O,\ \mathrm{liq}\right)\ dT}+{\Delta }_{vap}H \nonumber$ $+\Delta H\left({\mathrm{B}}^{\mathrm{*}}\mathrm{\to }\mathrm{C}\right)+\int^{272.15\ \mathrm{K}}_{373.15\ \mathrm{K}}{C_P\left(H_2O,\ \mathrm{g}\right)\ dT}{-\Delta }_{sub}H \nonumber$ $+\Delta H\left({\mathrm{D}}^{\mathrm{*}}\mathrm{\to }\mathrm{A}\right) \nonumber$
Using results that we find in the next section, $H\left({\mathrm{B}}^{\mathrm{*}}\mathrm{\to }\mathrm{C}\right)\approx 0$ and $\Delta H\left({\mathrm{D}}^{\mathrm{*}}\mathrm{\to }\mathrm{A}\right)\approx 0$, we have
$0={\Delta }_{fus}H+\int^{372.15\ \mathrm{K}}_{273.15\ \mathrm{K}}{C_P\left(H_2O,\ \mathrm{liq}\right)\ dT}+{\Delta }_{vap}H \nonumber$ $\ \ +\int^{272.15\ \mathrm{K}}_{373.15\ \mathrm{K}}{C_P\left(H_2O,\ \mathrm{g}\right)\ dT}{-\Delta }_{sub}H \nonumber$
The enthalpy of fusion, the enthalpy of vaporization, and the heat capacities are measurable in straightforward experiments. Their values are given in standard compilations, so we are now able to evaluate ${\Delta }_{sub}H$, a quantity that is not susceptible to direct measurement, from other thermodynamic quantities that are. (See Problem 8.)
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Let us look briefly at the approximations $\Delta H\left({\mathrm{B}}^{\mathrm{*}}\mathrm{\to }\mathrm{C}\right)\approx 0$ and $\Delta H\left({\mathrm{D}}^{\mathrm{*}}\mathrm{\to }\mathrm{A}\right)\approx 0$ that we used in Section 8.2. In these steps, the pressure changes while the temperature remains constant. In Chapter 10, we find a general relationship for the pressure-dependence of a system’s enthalpy: ${\left(\frac{\partial H}{\partial P}\right)}_T=-T{\left(\frac{\partial V}{\partial T}\right)}_P+V \nonumber$
This evaluates to zero for an ideal gas and to a negligible quantity for many other systems.
For liquids and solids, information on the variation of volume with temperature is collected in tables as the coefficient of thermal expansion, $\alpha$, where
$\alpha =\frac{1}{V}{\left(\frac{\partial V}{\partial T}\right)}_P \nonumber$
Consequently, the dependence of enthalpy on pressure is given by ${\left(\frac{\partial H}{\partial P}\right)}_T=V\left(1-\alpha T\right) \nonumber$
For ice, $\alpha \approx 50\times {10}^{-6}\ {\mathrm{K}}^{-1}$ and the molar volume near 0 C is $\mathrm{19.65}\ {\mathrm{cm}}^3\ {\mathrm{mol}}^{-1}$. The enthalpy change for compressing one mole of ice from the sublimation pressure to 1 atm is $\Delta H\left({\mathrm{D}}^{\mathrm{*}}\mathrm{\to }\mathrm{A}\right)=2\ \mathrm{J}\mathrm{\ }{\mathrm{mol}}^{-1}$.
To find the enthalpy change for expanding one mole of water vapor at 100 C from 1 atm to the sublimation pressure, we use the virial equation and tabulated coefficients for water vapor to calculate ${\left({\partial H}/{\partial P}\right)}_{\mathrm{398\ K}}$. We find $\Delta H\left({\mathrm{B}}^{\mathrm{*}}\mathrm{\to }\mathrm{C}\right)=220\ \mathrm{J}\ {\mathrm{mol}}^{-1}$. (See problem 9.)
8.04: Standard States and Enthalpies of Formation
A useful convention makes it possible to tabulate enthalpy data for individual compounds in such a way that the enthalpy change for any chemical reaction can be calculated from the tabulated information for the reaction’s reactants and products. The convention comprises the following rules:
I. At any particular temperature, we define the standard state of any liquid or solid substance to be the most stable form of that substance at a pressure of one bar. For example, for water at $-10$ C, the standard state is ice at a pressure of one bar; at $+10$ C, it is liquid water at a pressure of one bar.
II. At any particular temperature, we define the standard state of a gas to be the ideal gas standard state at that temperature. By the ideal gas standard state, we mean a finite low pressure at which the real gas behaves as an ideal gas. We know that it is possible to find such a pressure, because any gas behaves as an ideal gas at a sufficiently low pressure. Since the enthalpy of an ideal gas is independent of pressure, we can also think of a substance in its ideal gas standard state as a hypothetical substance whose pressure is one bar but whose molar enthalpy is that of the real gas at an arbitrarily low pressure.
III. For any substance at any particular temperature, we define the standard enthalpy of formation as the enthalpy change for a reaction in which the product is one mole of the substance and the reactants are the compound’s constituent elements in their standard states.
For water at –10 C, this reaction is
$H_2\left(\mathrm{g},-10\ \mathrm{C},\ 1\ \mathrm{bar}\right)+\ \frac{1}{2}\ O_2\left(\mathrm{g},-10\ \mathrm{C},\ 1\ \mathrm{bar}\right) \ \mathrm{\to } H_2O\left(\mathrm{s},-10\ \mathrm{C},\ 1\ \mathrm{bar}\right) \nonumber$
For water at +10 C, it is
$H_2\left(\mathrm{g},+10\ \mathrm{C},\ 1\ \mathrm{bar}\right)+\ \frac{1}{2}\ O_2\left(\mathrm{g},+10\ \mathrm{C},\ 1\ \mathrm{bar}\right) \ \mathrm{\to } H_2O\left(\mathrm{liq},+10\ \mathrm{C},\ 1\ \mathrm{bar}\right) \nonumber$
For water at +110 C, it is
$H_2\left(g,+110\ \mathrm{C},\ 1\ \mathrm{bar}\right)+\ \frac{1}{2}\ O_2\left(\mathrm{g},+110\ \mathrm{C},\ 1\ \mathrm{bar}\right) \ \mathrm{\to } H_2O\left(\mathrm{g},+110\ \mathrm{C},\ 1\ \mathrm{bar}\right) \nonumber$
IV. The standard enthalpy of formation is given the symbol $\boldsymbol{\Delta }_{\boldsymbol{f}} \boldsymbol{H}^{\boldsymbol{o}}$, where the superscript degree sign indicates that the reactants and products are all in their standard states. The subscript, $\boldsymbol{f}$, indicates that the enthalpy change is for the formation of the indicated compound from its elements. Frequently, the compound and other conditions are specified in parentheses following the symbol. The solid, liquid, and gas states are usually indicated by the letters “s”, “$\ell$” (or “liq”), and “g”, respectively. The letter “c” is sometimes used to indicate that the substance is in a crystalline state. In this context, specification of the gas state normally means the ideal gas standard state.
Thermochemical-data tables that include standard enthalpies of formation can be found in a number of publications or on the internet. For some substances, values are available at a number of temperatures. For substances for which less data is available, these tables usually give the value of the standard enthalpy of formation at 298.15 K. (In this context, 298.15 K is frequently abbreviated to 298 K.)
V. For any element at any particular temperature, we define the standard enthalpy of formation to be zero. When we define standard enthalpies of formation, we choose the elements in their standard states as a common reference state for the enthalpies of all substances at a given temperature. While we could choose any arbitrary value for the enthalpy of an element in its standard state, choosing it to be zero is particularly convenient.
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The ideal gas standard state is a useful invention, which has additional advantages that emerge as our development proceeds. For permanent gases—gases whose behavior is approximately ideal anyway—there is a negligible difference between the enthalpy in the ideal gas state and the enthalpy at 1 bar.
For volatile substances that are normally liquid or solid at 1 bar, the ideal gas standard state becomes a second standard state. For such substances, data tables frequently give the standard enthalpy of formation for both the condensed phase (designated ${\Delta }_fH^o\left(\mathrm{liq}\right)$ or ${\Delta }_fH^o\left(\mathrm{s}\right)$) and the ideal gas standard state (designated ${\Delta }_fH^o\left(\mathrm{g}\right)$). For example, the CODATA${}^{1}$ values for the standard enthalpies of formation for liquid and ideal-gas methanol are $-\mathrm{239.2}$ and $-201.0\ \mathrm{k}\mathrm{J}\ {\mathrm{mol}}^{-1}$, respectively, at 298.15 K. The difference between these values is the enthalpy change in vaporizing one mole of liquid methanol to its ideal gas standard state at 298.15 K:
$CH_3OH\left(\mathrm{liq},\ 298.15\mathrm{\ K},\ 1\ \mathrm{bar}\right) \mathrm{\to } CH_3OH\left(\mathrm{ideal\ gas},\ 298.15\mathrm{\ K},\ \sim 0\ \mathrm{bar}\right) \nonumber$
Since this is the difference between the enthalpy of methanol in its standard state as an ideal gas and methanol in its standard state as a liquid, we can call this difference the standard enthalpy of vaporization for methanol:
${\Delta }_{vap}H^o={\Delta }_fH^o\left(\mathrm{g,\ 298.15\ K,\ }\sim 0\mathrm{\ bar}\right) \ -{\Delta }_fH^o\left(\mathrm{g,\ 298.15\ K,\ }1\mathrm{\ bar}\right) \ =37.40\ \mathrm{k}\mathrm{J}\ {\mathrm{mol}}^{-1} \nonumber$
This is not a reversible process, because liquid methanol at 1 bar is not at equilibrium with its vapor at an arbitrarily low pressure at 298.15 K.
Note that ${\Delta }_{vap}H^o$ is not the same as the ordinary enthalpy of vaporization, ${\Delta }_{vap}H$. The ordinary enthalpy of vaporization is the enthalpy change for the reversible vaporization of liquid methanol to real methanol vapor at a pressure of 1 atm and the normal boiling temperature. We write it without the superscript degree sign because methanol vapor is not produced in its standard state. For methanol, the normal boiling point and enthalpy of vaporization${}^{2}$ are $337.8\ \mathrm{K}$ and $35.21\ \mathrm{k}\mathrm{J}\ {\mathrm{mol}}^{-1}$, respectively.
We can devise a cycle that relates these two vaporization processes to one another: Summing the steps below yields the process for vaporizing liquid methanol in its standard state to methanol vapor in its standard state.
1. $CH_3OH$ (liq, 298.15 K, 1 bar) $\mathrm{\to }$ $CH_3OH$ (liq, 337.8 K, 1 bar) ${\Delta }_{\left(1\right)}H$
2. $CH_3OH$ (liq, 337.8 K, 1 bar) $\mathrm{\to }$ $CH_3OH$ (liq, 337.8 K, 1 atm) ${\Delta }_{\left(2\right)}H$
3. $CH_3OH$ (liq, 337.8 K, 1 atm) $\mathrm{\to }$ $CH_3OH$ (g, 337.8 K, 1 atm) ${\Delta }_{\left(3\right)}H={\Delta }_{vap}H \nonumber$
4. $CH_3OH$ (g, 337.8 K, 1 atm) $\mathrm{\to }$ $CH_3OH$ (g, 337.8 K, $\mathrm{\sim}$0 bar) ${\Delta }_{\left(4\right)}H$
5. $CH_3OH$ (g, 337.8 K, $\mathrm{\sim}$0 bar) $\mathrm{\to }$ $CH_3OH$ (g, 298.15 K, $\mathrm{\sim}$0 bar) ${\Delta }_{\left(5\right)}H$
Thus, we have
${\Delta }_{vap}H^o={\Delta }_{\left(1\right)}H+{\Delta }_{\left(2\right)}H+{\Delta }_{vap}H+{\Delta }_{\left(4\right)}H+{\Delta }_{\left(5\right)}H \nonumber$
${\Delta }_{\left(1\right)}H$ and ${\Delta }_{\left(5\right)}H$ can be evaluated by integrating the heat capacities for the liquid and gas, respectively. ${\Delta }_{\left(2\right)}H$ and ${\Delta }_{\left(4\right)}H$ can be evaluated by integrating ${\left({\partial H}/{\partial P}\right)}_T$ for the liquid and gas, respectively.$\ {\Delta }_{\left(2\right)}H$ is negligible. (For the evaluation of these quantities, see problem 10.)
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The benefit of these conventions is that, at any particular temperature, the standard enthalpy change for a reaction
$aA+bB+\dots \ \to \ cC+dD+\dots\nonumber$
which we designate as $\Delta_rH^o$, is given by
$\Delta_rH^o=\underbrace{c{\Delta}_fH^o\left(C\right)+d{\Delta}_fH^o\left(D\right)+\dots}_{\text{product enthalpies}} - \underbrace{a{\Delta}_fH^o\left(A\right)-b\Delta_fH^o\left(B\right)-\dots}_{\text{reactant enthalpies}}\nonumber$
If we have the enthalpies of formation, we can compute the enthalpy change for the reaction. We can demonstrate this by writing out the chemical equations corresponding to the formation of A, B, C, and D from their elements. When we multiply these chemical equations by the appropriately signed stoichiometric coefficient and add them, we obtain the chemical equation for the indicated reaction of A and B to give C and D. (See below.) Because enthalpy is a state function, the enthalpy change that we calculate this way will be valid for any process that converts the specified reactants into the specified products.
The oxidation of methane to methanol is a reaction that illustrates the value of this approach. The normal products in the oxidation of methane are, of course, carbon dioxide and water. If the reaction is done with an excess of methane, a portion of the carbon-containing product will be carbon monoxide rather than carbon dioxide. In any circumstance, methanol is, at best, a trace product. Nevertheless, it would be very desirable to devise a catalyst that quantitatively—or nearly quantitatively—converted methane to methanol according to the equation
$\ce{CH4 + 1/2O_2 -> CH3OH}\nonumber$
(This is frequently called a selective oxidation, to distinguish it from the non-selective oxidation that produces carbon dioxide and water.)
If the catalyst were not inordinately expensive or short-lived, and the operating pressure were sufficiently low, this would be an economical method for the manufacture of methanol. (Methanol is currently manufactured from methane. However, the process involves two steps and requires a substantial capital investment.) If the cost of manufacturing methanol could be decreased sufficiently, it would become economically feasible to convert natural gas, which cannot be transported economically unless it is feasible to build a pipeline for the purpose, into liquid methanol, which is readily transported by ship. (At present, the economic feasibility of marine transport of liquefied natural gas, LNG, is marginal, but it appears to be improving.) This technology would make it possible to utilize the fuel value of known natural gas resources that are presently useless because they are located too far from population centers.
When we contemplate trying to develop a catalyst and a manufacturing plant to carry out this reaction, we soon discover reasons for wanting to know the enthalpy change. One is that the oxidative manufacture of methanol will be exothermic, so burning the methanol produced will yield less heat than would be produced by burning the methane from which it was produced. We want to know how much heat energy is lost in this way.
Another reason is that a manufacturing plant will have to control the temperature of the oxidation reaction in order to maintain optimal performance. (If the temperature is too low, the reaction rate will be too slow. If the temperature is too high, the catalyst may be deactivated in a short time, and the production of carbon oxides will probably be excessive.) A chemical engineer designing a plant will need to know how much heat is produced so that he can provide adequate cooling equipment.
Because we do not know how to carry out this reaction, we cannot measure its enthalpy change directly. However, if we have the enthalpies of formation for methane and methanol, we can compute this enthalpy change:
$\ce{C(s) + 2H_2 (g) + 1/2O2 (g) -> CH3OH (g) })\nonumber$
$\Delta H=\Delta_fH^o\left(CH_3OH,\ g\right)\nonumber$
$CH_4\left(g\right)\to C\left(\mathrm{s}\right)+2\ H_2\left(g\right)\nonumber$
$\Delta H={-\Delta }_fH^o\left(CH_4,g\right)\nonumber$
$1/2 O_2\left(g\right)\to 1/2 O_2\left(g\right)\nonumber$
$\Delta H=-1/2 \Delta_fH^o\left(O_2,g\right)=0\nonumber$
Summing the reactions gives
$\ce{ CH4 (g) + 1/2O2 (g) \to CH3OH (g)}\nonumber$
$\Delta H=\Delta_rH^o\nonumber$
and summing the enthalpy changes gives
$\Delta_rH^o=\Delta_fH^o (CH_3OH, g) -\Delta_f H^o (CH_4, g)- 1/2 \Delta_fH^o\left(O_2,g\right)\nonumber$
The diagram in Figure 2 shows how these conventions, and the fact that enthalpy is a state function, work together to produce, for the reaction $aA+bB+\dots \to cC+dD+\dots$, the result that the standard reaction enthalpy is given by
$\Delta_rH^o={c\ \Delta }_fH^o\left(C\right)+{d \Delta }_fH^o\left(D\right)+\dots -{a \Delta }_fH^o\left(A\right)-{b\ \Delta }_fH^o\left(B\right)-\dots\nonumber$
This cycle highlights another aspect of the conventions that we have developed. Note that $\Delta_rH^o$ is the difference between the enthalpies of formation of the separated products and the enthalpies of formation of the separated reactants. We often talk about $\Delta_rH^o$ as if it were the enthalpy change that would occur if we mixed $a$ moles of $A$ with $b$ moles of $B$ and the reaction proceeded quantitatively to yield a mixture containing $c$ moles of $C$ and $d$ moles of $D$. This is usually a good approximation. However, to relate rigorously the standard enthalpy of reaction to the enthalpy change that would occur in a real system in which this reaction took place, it is necessary to recognize that there can be enthalpy changes associated with the pressure–volume changes and with the processes of mixing the reactants and separating the products.
Let us suppose that the reactants and products are gases in their hypothetical ideal-gas states at 1 bar, and that we carry out the reaction by mixing the reactants in a sealed pressure vessel. We suppose that the reaction is then initiated and that the products are formed rapidly, reaching some new pressure and an elevated temperature. (To be specific, we could imagine the reaction be the combustion of methane. We would mix known amounts of methane and oxygen in a pressure vessel and initiate the reaction using an electrical spark.) We allow the temperature to return to the original temperature of the reactants; there is an accompanying pressure change.
Experimentally, we measure the heat evolved as the mixed reactants are converted to the mixed products, at the original temperature. To complete the process corresponding to the standard enthalpy change, however, we must also separate the products and bring them to a pressure of 1 bar. That is, the standard enthalpy of reaction and the enthalpy change we would measure are related by the following sequence of changes, where the middle equation corresponds to the process whose enthalpy change we actually measure.
${\left(aA+bB\right)}_{\mathrm{separate\ reactants\ at\ }P = 1 \text{ bar}}\to {\left(aA+bB\right)}_{\mathrm{homogenous\ mixture\ at\ }P}\nonumber$ $\Delta H_{\mathrm{compression}}\nonumber$
${\left(aA+bB\right)}_{\mathrm{separate\ reactants\ at\ }P}\to {\left(aA+bB\right)}_{\mathrm{homogeneous\ mixture\ at\ }P}\nonumber$ $\Delta H_{\mathrm{mixing}}\nonumber$ ${\left(aA+bB\right)}_{\mathrm{homogeneous\ mixture\ at\ }P}\to {\left(cC+dD\right)}_{\mathrm{homogeneous\ mixture\ at\ }P^*}\nonumber$ $\Delta H_{\mathrm{measured}}\nonumber$ ${\left(cC+dD\right)}_{\mathrm{homogeneous\ mixture\ at\ }P^*}\to {\left(cC+dD\right)}_{\mathrm{separate\ products\ at\ }P^*}\nonumber$ $\Delta H_{\mathrm{separation}}\nonumber$ ${\left(cC+dD\right)}_{\mathrm{separate\ products\ at\ }P^*}\to {\left(cC+dD\right)}_{\mathrm{separate\ products\ at\ }P=1\mathrm{\ bar\ }}\nonumber$ $\Delta H_{\mathrm{expansion}}\nonumber$
Summing the reaction equations gives
${\left(aA+bB\right)}_{\mathrm{separate\ reactants\ at\ }P=1\ \mathrm{bar}}\to {\left(cC+dD\right)}_{\mathrm{separate\ products\ at\ }P=1\mathrm{\ bar\ }}\nonumber$ $\Delta_rH^o\nonumber$
and summing the enthalpy changes for the series of steps gives the standard enthalpy change for the reaction:
$\Delta_rH^o=\Delta H_{\mathrm{compression}}+\Delta H_{\mathrm{mixing}}+\Delta H_{\mathrm{measured}} + \Delta H_{\mathrm{separation}}+\Delta H_{\mathrm{expansion}}\nonumber$
It turns out that the enthalpy changes for the compression, mixing, separation, and expansion processes are usually small compared to $\Delta_rH^o$. This is the principal justification for our frequent failure to consider them explicitly. For ideal gases, these enthalpy changes are identically zero. (In Chapter 13, we see that the entropy changes for the mixing and separation processes are important.)
When we call $\Delta_rH^o$ the standard enthalpy change “for the reaction,” we are indulging in a degree of poetic license. Since $\Delta_rH^o$ is a computed difference between the enthalpies of the pure products and those of the pure reactants, the corresponding “reaction” is a purely formal change, which is a distinctly different thing from the real-world process that actually occurs.
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We have observed that $C_V$ depends on volume and temperature, while $C_P$ depends on pressure and temperature. Compilations of heat capacity data usually give values for $C_P$, rather than $C_V$. When the temperature-dependence of $C_P$ is known, such compilations usually express it as an empirical polynomial function of temperature. In Chapter 10, we find an explicit function for the dependence of $C_P$ on pressure:
${\left(\frac{\partial C_P}{\partial P}\right)}_T=-T{\left(\frac{{\partial }^2V}{\partial T^2}\right)}_P \nonumber$
If we have an equation of state for a substance, we can find this pressure dependence immediately. It is usually negligible. For ideal gases, it is zero, and $C_P$ is independent of pressure.
Compilations often give data for the standard state heat capacity, $C^o_P$, at a specified temperature. For condensed phases, this is the heat capacity for the substance at one bar. For gases, this is the heat capacity of the substance in its ideal gas standard state.
300 K 400 K $\boldsymbol{a}\left(\mathrm{J}\right)$ $b\left(\mathrm{J}\boldsymbol{\ }{\mathrm{K}}^{-\boldsymbol{1}}\right)$
C_s 0 0 -1.482 0.03364
${\boldsymbol{H}}_{\boldsymbol{2}}\left(\boldsymbol{g}\right)$ 0 0 27.853 0.00332
${\boldsymbol{O}}_{\boldsymbol{2}}\left(\boldsymbol{g}\right)$ 0 0 27.221 0.00722
${\boldsymbol{CH}}_{\boldsymbol{4}}\left(\boldsymbol{g}\right)$ -74.656 -77.703 21.167 0.04866
${\boldsymbol{CH}}_{\boldsymbol{3}}\boldsymbol{OH}\left(\boldsymbol{g}\right)$ -201.068 -204.622 21.737 0.07494
8.08: How The Enthalpy Change for a Reaction Depends on Temperature
In Section 8.6, we see how to use tabulated enthalpies of formation to calculate the enthalpy change for a particular chemical reaction. Such tables typically give enthalpies of formation at a number of different temperatures, so that the enthalpy change for a given reaction can also be calculated at these different temperatures; it is just a matter of repeating the same calculation at each temperature.
We often need to find the enthalpy change associated with increasing the temperature of a substance at constant pressure. As we observe in §1, this enthalpy change is readily calculated by integrating the heat capacity over the temperature change. We may want to know, for example, the enthalpy change for increasing the temperature of one mole of methane from 300 K to 400 K, with the pressure held constant at one bar. In Table 1, we find
$\Delta_fH^o\left(CH_4 ,g,300\, K\right) =-74.656\ \mathrm{k}\mathrm{J}\ \mathrm{mol}^{-1} \nonumber$
$\Delta_fH^o\left(CH_4\mathrm{,g,400\ K}\right) = -77.703\ \mathrm{k}\mathrm{J}\ \mathrm{mol}^{-1} \nonumber$
We might be tempted to think that the difference represents the enthalpy change associated with heating the methane. This is not so! The reason becomes immediately apparent if we consider a cycle in which we go from the elements to a compound at two different temperatures. For methane, this cycle is shown in Figure 3.
The difference between the standard enthalpies of formation of methane at 300 K and 400 K reflects the enthalpy change for increasing the temperatures of all of the reactants and products from 300 K to 400 K. That is,
$\Delta_fH^o\left(CH_4\mathrm{,g,400\ K}\right)-\Delta_fH^o\left(CH_4\mathrm{,g,300\ K}\right) \nonumber$ $=\int^{400}_{300}{C_P\left(CH_4\mathrm{,g}\right)dT}-\int^{400}_{300}{C_P\left(C\mathrm{,s}\right)dT} -2\int^{400}_{300}{C_P\left(H_2\mathrm{,g}\right)dT} \nonumber$
Over the temperature range from 300 K to 400 K, the heat capacities of carbon, hydrogen, and methane are approximated by $C_P=a+bT$, with values of $a$ and $b$ given in Table 1. From this information, we calculate the enthalpy change for increasing the temperature of one mole of each substance from 300 K to 400 K at 1 bar: $\Delta H\left(C\right)=1,029\ \mathrm{J}\ {\mathrm{mol}}^{-1}$, $\Delta H\left(H_2\right)=2,902\ \mathrm{J}\ {\mathrm{mol}}^{-1}$, and $\Delta H\left(CH_4\right)=3,819\ \mathrm{J}\ {\mathrm{mol}}^{-1}$. Thus, from the cycle, we calculate:
$\Delta_fH^o\left(CH_4\mathrm{,g,400\ K}\right)=-74,656+3,819-1,029-2\left(2,902\right)\ \mathrm{J}\ {\mathrm{mol}}^{-1}=\ -77,670\ \mathrm{J}\ {\mathrm{mol}}^{-1} \nonumber$
The tabulated value is $-77,703\ \mathrm{J}\ {\mathrm{mol}}^{-1}$. The two values differ by $33\ \mathrm{J}\ {\mathrm{mol}}^{-1}$, or about 0.04%. This difference arises from the limitations of the two-parameter heat-capacity equations.
As another example of a thermochemical cycle, let us consider the selective oxidation of methane to methanol at 300 K and 400 K. From the enthalpies of formation in Table 1, we calculate the enthalpies for the reaction to be $\Delta_rH^o\left(3\mathrm{00\ K}\right)=-126.412\ \mathrm{k}\mathrm{J}\ {\mathrm{mol}}^{-1}$ and $\Delta_rH^o\left(4\mathrm{00\ K}\right)=-126.919\ \mathrm{k}\mathrm{J}\ {\mathrm{mol}}^{-1}$. As in the previous example, we use the tabulated heat-capacity parameters to calculate the enthalpy change for increasing the temperature of one mole of each of these gases from 300 K to 400 K at 1 bar. We find: $\Delta H\left(CH_3OH\right)=4,797\ \mathrm{J}\ {\mathrm{mol}}^{-1}$, $\Delta H\left(CH_4\right)=3,819\ \mathrm{J}\ {\mathrm{mol}}^{-1}$, and $\Delta H\left(O_2\right)=2,975\ \mathrm{J}\ {\mathrm{mol}}^{-1}$.
The cycle is shown in Figure 4. Inspecting this cycle, we see that we can calculate the enthalpy change for warming one mole of methanol from 300 K to 400 K by summing the enthalpy changes around the bottom, left side, and top of the cycle; that is,
$\Delta H\left(CH_3OH\right)=126,412+3,819+\left(\frac{1}{2}\right)2,975-126,919\ \mathrm{J}\ {\mathrm{mol}}^{-1}=4,800\ \mathrm{J}\ {\mathrm{mol}}^{-1} \nonumber$
This is 3 J or about 0.06 % larger than the value obtained $\left(4,797\ \mathrm{J}\right)$ by integrating the heat capacity for methanol.
8.09: Calorimetry
Calorimetry is the experimental science of measuring the heat changes that accompany chemical or physical changes. The accurate measurement of small amounts of heat is experimentally challenging. Nevertheless, calorimetry is an area in which great experimental sophistication has been achieved and remarkably accurate measurements can be made. Numerous devices have been developed to measure heat changes.
Some of these devices measure a (usually small) temperature change. Such devices are calibrated by measuring how much their temperature increases when a known amount of heat is introduced. This is usually accomplished by passing a known electric current through a known resistance for a known time. Other calorimeters measure the amount of some substance that undergoes a phase change. The ice calorimeter is an important example of the latter method. In an ice calorimeter, the heat of the process is transferred to a mixture of ice and water. The amount of ice that melts is a direct measure of the amount of heat released by the process. The amount of ice melted can be determined either by direct measurement of the increase in the amount of water present or by measuring the change in the volume of the ice–water mixture. (Since ice occupies a greater volume than the same mass of water, melting is accompanied by a decrease in the total volume occupied by the mixture of ice and water.)
The processes that can be investigated accurately using calorimetry are limited by two important considerations. One is that the process must go to completion within a relatively short time. No matter how carefully it is constructed, any calorimeter will exchange thermal energy with its environment at some rate. If this rate is not negligibly small compared to the rate at which the process evolves heat, the accuracy of the measurement is degraded. The second limitation is that the process must involve complete conversion of the system from a known initial state to a known final state. When the processes of interest are chemical reactions, these considerations mean that the reactions must be quantitative and fast.
Combustion reactions and catalytic hydrogenation reactions usually satisfy these requirements, and they are the most commonly investigated. However, even in these cases, there can be complications. For a compound containing only carbon, hydrogen, and oxygen, combustion using excess oxygen produces only carbon dioxide and water. For compounds containing heteroatoms like nitrogen, sulfur, or phosphorus, there may be more than one heteroatom-containing product. For example, combustion of an organosulfur compound might produce both sulfur dioxide and sulfur trioxide. To utilize the thermochemical data obtained in such an experiments, a chemical analysis must be done to determine the amount of each oxide present.
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1. One mole of an ideal gas reversibly traverses Cycle I above. Step a is isothermal. Step b is isochoric (constant volume). Step c is isobaric (constant pressure). Assume $C_V$ and $C_P$ are constant. Find $q$, $w$, $\Delta E$, and $\Delta H$ for each step and for the cycle. Prove $C_P=C_V+R$.
2. One mole of an ideal gas reversibly traverses Cycle II below. Step a is the same isothermal process as in problem 1. Step d is adiabatic. Step e is isobaric. Assume $C_V$ and $C_P$ are constant. Find $q$, $w$, $\Delta E$, and $\Delta H$ for each step and for the cycle.
3. One mole of an ideal gas reversibly traverses Cycle III below. Step a is the same isothermal process as in
problem 1. Step f is adiabatic. Step g is isochoric. Assume $C_V$ and $C_P$ are constant. Find $q$, $w$, $\Delta E$, and $\Delta H$ for each step and for the cycle.
4. One mole of an ideal gas reversibly traverses Cycle IV. Step h is isobaric. Step f is the same adiabatic process as in problem 3. Step i is isochoric. Assume $C_V$ and $C_P$ are constant. Find $q$, $w$, $\Delta E$, and $\Delta H$ for each step and for the cycle.
5. Prove that the work done on the system is positive when the system traverses Cycle I. Note that Cycle I traverses the region of the $PV$ plane that it encloses in a counter-clockwise direction. Hint: Note that $T_2<t_1$. Show that ${V_2}/{V_1}={T_2}/{T_1}$.
6. Cycles III and IV share a common adiabatic step. Express the work done in each of these cycles in terms of $V_1$, $V_2$, and $T_1$. Prove that the work done in Cycle IV is greater than the work done in Cycle III.
7. Cycles I, II, and III share a common first step, a. Express $V_3$, $T_3$, and $T_4$ in terms of $V_1$, $V_2$, and $T_1$. For $V_1=10\ \mathrm{L}$, $V_2=2\ \mathrm{L}$, and $T_1=400\ \mathrm{K}$, show that the work done decreases in the order Cycle I $\mathrm{>}$ Cycle III $\mathrm{>}$ Cycle II.
8. For water, the enthalpies of fusion and vaporization are $6.009$ and $40.657\ \mathrm{k}\mathrm{J}\ {\mathrm{mol}}^{-1}$, respectively. The heat capacity of liquid water varies only weakly with temperature and can be taken as $\mathrm{75.49\ }\mathrm{J}\ {\mathrm{mol}}^{-1}\ {\mathrm{K}}^{-1}$. The heat capacity of water vapor varies with temperature: $C_P\left(H_2O\mathrm{,\ g}\right)=30.51+\left(1.03\times {10}^{-2}\right)T \nonumber$ where $T$ is in degrees K and the heat capacity is in $\mathrm{J}\ {\mathrm{mol}}^{-1}\ {\mathrm{K}}^{-1}$. Estimate the enthalpy of sublimation of water.
9. If we truncate the virial equation $\left(Z=1+B^*\left(T\right)P+\dots \right)$ and make use of $B\left(T\right)=RTB^*\left(T\right)$, where$\ B\left(T\right)$ is the “second virial coefficient” most often given in data tables, the molar volume is $\overline{V}=\frac{RT}{P}+B\left(T\right) \nonumber$ Show that ${\left(\frac{\partial H}{\partial P}\right)}_T=B\left(T\right)-T\left(\frac{dB}{dT}\right) \nonumber$
The Handbook of Chemistry and Physics (CRC Press, 79${}^{th}$ Ed., 1999, p. 6–25) gives the temperature dependence of $B$ for water vapor as
$B=-1158-5157t-10301t^2-10597t^3-4415t^4 \nonumber$
where $t=\left({298.15}/{T}\right)-1$, $T$ is in degrees kelvin, and the units of $B$ are ${\mathrm{cm}}^{-3\ }{\mathrm{mol}}^{-1}$. Estimate the enthalpy change when one mole of water vapor at 1 atm and 100 C is expanded to the equilibrium sublimation pressure, which for this purpose we can approximate as the triple-point pressure, $610\ \mathrm{Pa}$. How does this value compare to the result of problem 8?
10. The heat capacities of methanol liquid and gas are $81.1$ and $44.1\ \mathrm{J}\ {\mathrm{mol}}^{-1}\ {\mathrm{K}}^{-1}$, respectively. The second virial coefficient for methanol vapor is
$B=-1752-4694t \nonumber$
where $t=\left({298.15}/{T}\right)-1$, $T$ is in degrees kelvin, and the units of $B$ are ${\mathrm{cm}}^{-3\ }{\mathrm{mol}}^{-1}$. Referring to the discussion of methanol vaporization in §5, calculate ${\Delta }_{\left(1\right)}H$, ${\Delta }_{\left(4\right)}H$, ${\Delta }_{\left(5\right)}H$, ${\Delta }_{\left(vap\right)}H^o$. Compare this value of ${\Delta }_{\left(vap\right)}H^o$ to the value given in the text. [Data from the Handbook of Chemistry and Physics, CRC Press, 79${}^{th}$ Ed., 1999, p. 5-27 and p. 6-31.]
Molecular formula Name ${\Delta }_fH^o$$\left(\mathrm{k}\mathrm{J}\ {\mathrm{mol}}^{-1}\right)$
$H_2O\ \left(\mathrm{liq}\right)$ Water $-285.8$
$CO\ \left(\mathrm{g}\right)$ Carbon monoxide $-110.5$
$CO_2\ \left(\mathrm{g}\right)$ Carbon dioxide $-393.5$
$CH_4\left(\mathrm{g}\right)$ Methane $-74.6$
$C_2H_4\left(\mathrm{g}\right)$ Ethylene $52.4$
$C_2H_6\left(\mathrm{g}\right)$ Ethane $-84.0$
$CH_3CH_2OH\ \left(\mathrm{liq}\right)$ Ethanol $-277.6$
$CH_3CHO\ \left(\mathrm{liq}\right)$ Acetaldehyde $-192.2$
$CH_3CO_2H\ \left(\mathrm{liq}\right)$ Acetic acid $-484.3$
$CH_3CH_2CHO\ \left(\mathrm{liq}\right)$ Propanal $-215.6$
$C_6H_6\ \left(\mathrm{liq}\right)$ Benzene $49.1$
$C_6H_5CO_2H\ \left(\mathrm{s}\right)$ Benzoic acid $-385.2$
11. Using data from the table above, find the enthalpy change for each of the following reactions at 298 K.
(a) $C_2H_6\left(\mathrm{g}\right)+ \ O_2\left(\mathrm{g}\right)\to CH_3CH_2OH\left(\mathrm{liq}\right)$
(b) $C_2H_4\left(\mathrm{g}\right)+ \ O_2\left(\mathrm{g}\right)\to CH_3CHO\left(\mathrm{liq}\right)$
(c) $C_2H_6\left(\mathrm{g}\right)+ \ O_2\left(\mathrm{g}\right)\to CH_3CHO\left(\mathrm{liq}\right)+H_2O\left(\mathrm{liq}\right)$
(d) $C_6H_6\left(\mathrm{liq}\right)+\ CO_2\left(\mathrm{g}\right)\to C_6H_5CO_2H\left(\mathrm{s}\right)$
(e) $CH_3CHO\left(\mathrm{liq}\right)+ \ O_2\left(\mathrm{g}\right)\to CH_3CO_2H\left(\mathrm{liq}\right)$
(f) $CH_4\left(\mathrm{g}\right)+H_2O\left(\mathrm{liq}\right)\to CO\left(\mathrm{g}\right)+3\ H_2\left(\mathrm{g}\right)$
(g) $CH_4\left(\mathrm{g}\right)+H_2O\left(\mathrm{liq}\right)+ \ O_2\left(\mathrm{g}\right)\to CO_2\left(\mathrm{g}\right)+3\ H_2\left(\mathrm{g}\right)$
(h) $C_2H_4\left(\mathrm{g}\right)+CO\left(\mathrm{g}\right)+\ H_2\left(\mathrm{g}\right)\to CH_3CH_2CHO\left(\mathrm{liq}\right)$
Notes
$^{1}$ Data compiled by The Committee on Data for Science and Technology (CODATA) and reprinted in D. R. Linde, Editor, The Handbook of Chemistry and Physics, 79${}^{th}$ Edition (1998-1999), CRC Press, Section 5.
${}^{2}$ D. R. Linde, op. cit., p. 6-104.
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The first law of thermodynamics is concerned with energy and its properties. As we saw in Chapter 7, the first law arose from the observation that the dissipation of mechanical work through friction creates heat. In a synthesis that was partly definition and partly a generalization from experience, it was proposed that mechanical energy and heat are manifestations of a common quantity, energy. Later, by further definition and generalization, the concept was expanded to include other forms of energy. The energy concept evolved into the prescript that there exists a quantity (state function) that is conserved through any manner of change whatsoever.
The element of definition arises from the fact that we recognize new forms of energy whenever necessary in order to ensure that the conservation condition is satisfied. The element of experience arises from the fact that this prescript has resulted in a body of theory and a body of experimental results that are mutually compatible. When we define and measure energy “correctly” we do indeed find that energy is a state function and that it is conserved.
The theory of relativity introduced a significant expansion of the energy concept. For chemical processes, we can view mass and energy conservation as independent postulates. For processes in which fundamental particles undergo changes and for systems moving at velocities near that of light, we cannot. Relativity asserts that the energy of a particle is given by Einstein’s equation,
$E^2=p^2c^2+m^2_0c^4. \nonumber$
In this equation, $E$ is the particle energy, $p$ is its momentum, $m_0$ is its rest mass, and $c$ is the speed of light. In transformations of fundamental particles in which the sum of the rest masses of the product particles is less than that of the reactant particles, conservation of energy requires that the sum of the momenta of the product particles exceed that of the reactant particles. The momentum increase means that the product particles have high velocities, corresponding to a high temperature for the product system. The most famous expression of this result is that $E=m_0c^2$, meaning that we can associate this quantity of energy with the mass, $m_0$, of a stationary particle, for which $p=0$.
The situation with respect to the second law is similar. From experience with devices that convert heat into work, the idea evolved that such devices must have particular properties. Consideration of these properties led to the discovery of a new state function, which we call entropy, and to which we customarily assign the symbol “$S$”. We introduce the laws of thermodynamics in §6-13. We repeat our statement of the second law here:
The Second Law of Thermodynamics
In a reversible process in which a closed system accepts an increment of heat, $\boldsymbol{d}\boldsymbol{q}^\boldsymbol{rev}$, from its surroundings, the change in the entropy of the system, $\boldsymbol{dS}$, is $\boldsymbol{dS}\boldsymbol{=}\boldsymbol{dq}^\boldsymbol{rev}/\boldsymbol{T}$. Entropy is a state function. For any reversible process, $\boldsymbol{dS}_\boldsymbol{universe}\boldsymbol{=}\boldsymbol{0}$, and conversely. For any spontaneous process, $\boldsymbol{dS}_\boldsymbol{universe}\boldsymbol{>}\boldsymbol{0}$, and conversely.
If a spontaneous process takes a system from state A to state B, state B may or may not be an equilibrium state. State A cannot be an equilibrium state. Since we cannot use the defining equation to find the entropy change for a spontaneous process, we must use some other method if we are to estimate the value of the entropy change. This means that we must have either an empirical mathematical model from which we can estimate the entropy of a non-equilibrium state or an equilibrium system that is a good model for the initial state of the spontaneous process.
We can usually find an equilibrium system that is a good model for the initial state of a spontaneous process. Typically, some alteration of an equilibrium system makes the spontaneous change possible. The change-enabled state is the initial state for a spontaneous process, but its thermodynamic state functions are essentially identical to those of the pre-alteration equilibrium state. For example, suppose that a solution contains the reactants and products for some reaction that occurs only in the presence of a catalyst. In this case, the solution can be effectively at equilibrium even when the composition does not correspond to an equilibrium position of the reaction. (In an effort to be more precise, we can term this a quasi-equilibrium state, by which we mean that the system is unchanging even though a spontaneous change is possible.) If we introduce a very small quantity of catalyst, and consider the state of the system before any reaction occurs, all of the state functions that characterize the system must be essentially unchanged. Nevertheless, as soon as the catalyst is introduced, the system can no longer be considered to be in an equilibrium state. The spontaneous reaction proceeds until it reaches equilibrium. We can find the entropy change for the spontaneous process by finding the entropy change for a reversible process that takes the initial, pre-catalyst, quasi-equilibrium state to the final, post-catalyst, equilibrium state.
Our statement of the second law establishes the properties of entropy by postulate. While this approach is rigorously logical, it does not help us understand the ideas involved. Like the first law, the second law can be stated several ways. To develop our understanding of entropy and its properties, it is useful to again consider a more traditional statement of the second law:
A Traditional statement of the second law
It is impossible to construct a machine that operates in a cycle, exchanges heat with its surroundings at only one temperature, and produces work in the surroundings.
When we introduce the qualification that the machine “exchanges heat with its surroundings at only one temperature,” we mean that the temperature of the surroundings has a particular value whenever the machine and surroundings exchange heat. The statement does not place any conditions on the temperature of the machine at any time.
In this chapter, we have frequent occasion to refer to each of these statements. To avoid confusing them, we will refer to our statement of the second law as the entropy-based statement. We will refer to the statement above as the machine-based statement of the second law.
By “a machine”, we mean a heat engine—a device that accepts heat and produces mechanical work. This statement asserts that a “perpetual motion machine of the second kind” cannot exist. Such a machine accepts heat energy and converts all of it into work, while itself returning to the same state at the end of each cycle. (In §7-11, we note that a “perpetual motion machine of the first kind” is one whose operation violates the principle of conservation of energy.) Normally, we view this statement as a postulate. We consider that we infer it from experience. Unlike our statements about entropy, which are entirely abstract, this statement makes an assertion about real machines of the sort that we encounter in daily life. We can understand the assertion that it makes in concrete terms: A machine that could convert heat from a constant-temperature source into work could extract heat from ice water, producing ice cubes in the water and an equivalent amount of work elsewhere in the surroundings. This machine would not exchange heat with any other heat reservoir. Our machine-based statement of the second law postulates that no such machine can exist.
Our entropy-based statement of the second law arose from thinking about the properties of machines that do convert heat into work. We trace this thinking to see how our entropy-based statement of the second law was developed. Understanding this development gives us a better appreciation for the meaning of entropy. We find that we must supplement the machine-based statement of the second law with additional assumptions in order to arrive at all of the properties of the entropy function that are asserted in the entropy-based statement.
However, before we undertake to develop the entropy-based statement of the second law from the machine-based statement, let us develop the converse; that is, let us show that the machine-based statement is a logical consequence of the entropy-based statement. To do so, we assume that a perpetual motion machine of the second kind is possible. To help keep our argument clear, let proposition $\mathrm{MSL}$ be the machine-based statement. We are assuming that proposition $\mathrm{MSL}$ is false, so that proposition $\sim \mathrm{MSL}$ is true. We let $\mathrm{SL}$ be the entropy-based statement of the second law.
The sketch in Figure 1 describes the interaction of this perpetual motion machine, $\mathrm{PPM}$, with its surroundings. From our entropy-based statement of the second law, we can assert some important facts about the entropy changes that accompany operation of the machine. Since entropy is a state function, $\Delta S=0$ for one cycle of the machine. If the machine works (that is, $\sim \mathrm{MSL}$ is true), then the entropy-based statement requires that $\Delta S_{universe}=\Delta S+\Delta \hat{S}\ge 0$. Since $\Delta S=0$, it follows that $\Delta \hat{S}\ge 0$. We can make this more explicit by writing: $\left(\mathrm{SL\ and}\ \sim \mathrm{MSL}\right)\Rightarrow \Delta \hat{S}\ge 0$.
The machine-based statement of the second law also enables us to determine the entropy change in the surroundings from our second-law definition of entropy. In one cycle, this machine (system) delivers net work, $\hat{w}>0$, to the surroundings; it accepts a net quantity of heat, $q>0$, from the surroundings, which are at temperature, $\hat{T}$. Simultaneously, the surroundings surrender a quantity of heat, $\hat{q}$, where $\hat{q}=-q$, and $\hat{q}<0$. The change that occurs in one cycle of the machine need not be reversible. However, whether the change is reversible or not, the entire thermal change in the surroundings consists in the exchange of an amount of heat, $\hat{q}<0$, by a constant temperature reservoir at $\hat{T}$. We can effect identically the same change in the surroundings using some other process to reversibly extract this amount of heat. The entropy change in the surroundings in this reversible process will be ${\hat{q}}/{\hat{T}}$, and this will be the same as the entropy change for the surroundings in one cycle of the machine. (We consider this conclusion further in §15.) It follows that $\Delta \hat{S}={\hat{q}}/{\hat{T}}$, and since $\hat{q}<0$, while $\hat{T}>0$, we have $\Delta \hat{S}<0$. We can write this conclusion more explicitly: $\left(\mathrm{SL\ and}\ \sim \mathrm{MSL}\right)\Rightarrow \Delta \hat{S}<0$.
By assuming a perpetual motion machine of the second kind is possible—that is, by assuming $\sim \mathrm{MSL}$ is true—we derive the contradiction that both $\Delta \hat{S}\ge 0$ and $\Delta \hat{S}<0$. Therefore, proposition $\sim \mathrm{MSL}$ must be false. Proposition $\mathrm{MSL}$ must be true. The entropy-based second law of thermodynamics implies that a perpetual motion machine of the second kind is not possible. That is, the entropy-based statement of the second law implies the machine-based statement. (We prove that $\sim \left(\mathrm{SL\ and}\ \sim \mathrm{MSL}\right)$; it follows that $\mathrm{SL\ }\mathrm{\Rightarrow }\mathrm{MSL}$. For a more detailed argument, see problem 2.)
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Historically, the steam engine was the first machine for converting heat into work that could be exploited on a large scale. The steam engine played a major role in the industrial revolution and thus in the development of today’s technology-intensive economy. It was important also in the development of the basic concepts of thermodynamics. A steam engine produces work when hot steam under pressure is introduced into a cylinder, driving a piston outward. A shaft connects the piston to a flywheel. When the connecting shaft reaches its greatest extension, the spent steam is vented to the atmosphere. Thereafter the flywheel drives the piston inward.
The economic viability of the steam engine derives, in part, from the fact that the spent steam can be vented to the atmosphere at the end of each cycle. However, this is not a necessary feature of heat engines. We can devise engines that alternately heat and cool a captive working fluid to convert heat energy into mechanical work. Stirling engines are practical devices of this type. A Carnot engine is a conceptual engine that exploits the response of a closed system to temperature changes. A Carnot engine extracts heat from one reservoir at a fixed high temperature and discharges a lesser amount of heat into a second reservoir at a fixed lower temperature. An amount of energy equal to the difference between these increments of heat energy appears in the surroundings as work.
For one cycle of the Carnot engine, let the heat transferred to the system from the hot and cold reservoirs be $q_h$ and $q_{\ell }$ respectively. We have $q_h>0$ and $q_{\ell }<0$. Let the net work done on the system be $w_{net}$ and the net work that appears in the surroundings be ${\hat{w}}_{net}$. We have
${\hat{w}}_{net}>0$, ${\hat{w}}_{net}=-w_{net}$, and $w_{net}<0$. For one cycle of the engine, $\Delta E=0$, and since
$\Delta E=q_h+q_{\ell }+w_{net}=q_h+q_{\ell }-{\hat{w}}_{net}, \nonumber$
it follows that ${\hat{w}}_{net}=q_h+q_{\ell }$. The energy input to the Carnot engine is $q_h$, and the useful work that appears in the surroundings is ${\hat{w}}_{net}$. (The heat accepted by the low-temperature reservoir, ${\hat{q}}_{\ell }=-q_{\ell }>0$, is a waste product, in the sense that it represents energy that cannot be converted to mechanical work using this cycle. All feasible heat engines share this feature of the Carnot engine. In contrast, a perpetual motion machine of the second kind converts its entire heat intake to work; no portion of its heat intake goes unused.) The efficiency, $\epsilon$, with which the Carnot engine converts the input energy, $q_h$, to useful output energy, ${\hat{w}}_{net}$, is therefore,
$\epsilon =\frac{\hat{w}_{net}}{q_h}=\frac{q_h+q_{\ell}}{q_h}=1+\frac{q_{\ell }}{q_h} \nonumber$
We can generalize our consideration of heat engines to include any series of changes in which a closed system exchanges heat with its surroundings at more than one temperature, delivers a positive quantity of work to the surroundings, and returns to its original state. We use the Carnot cycle and the machine-based statement of the second law to analyze systems that deliver pressure–volume work to the surroundings. We consider both reversible and irreversible systems. We begin by considering reversible Carnot cycles. If any system reversibly traverses any closed path on a pressure–volume diagram, the area enclosed by the path represents the pressure–volume work exchanged between the system and its surroundings. If the area is not zero, the system temperature changes during the cycle. If the cycle is reversible, all of the heat transfers that occur must occur reversibly. We can apply our reasoning about reversible cycles to any closed system containing any collection of chemical substances, so long as any phase changes or chemical reactions that occur do so reversibly. This means that all phase and chemical changes that occur in the system must adjust rapidly to the new equilibrium positions that are imposed on them as a system traverses a Carnot cycle reversibly.
In Figure 2, we describe the operation of a reversible Carnot engine in which the working fluid is an ideal gas. We designate the system’s initial pressure, volume, and temperature by $P_1$, $V_1$, and $T_h$. From this initial state, we cause the ideal gas to undergo a reversible isothermal expansion in which it absorbs a quantity of heat, $q_h$, from a high-temperature heat reservoir at ${\hat{T}}_h$. We designate the pressure, volume, and temperature at the end of this isothermal expansion as $P_2$, $V_2$, and $T_h$. In a second step, we reversibly and adiabatically expand the ideal gas until its temperature falls to that of the second, low-temperature, heat reservoir. We designate the pressure, volume, and temperature at the end of this adiabatic expansion as $P_3$, $V_3$, and $T_{\ell }$. We begin the return portion of the cycle by reversibly and isothermally compressing the ideal gas at the temperature of the cold reservoir. We continue this reversible isothermal compression until the ideal gas reaches the pressure and volume from which an adiabatic compression will just return it to the initial state. We designate the pressure, volume, and temperature at the end of this isothermal compression by $P_4$, $V_4$, and $T_{\ell }$. During this step, the ideal gas gives up a quantity of heat, $q_{\ell }<0$, to the low-temperature reservoir. Finally, we reversibly and adiabatically compress the ideal gas to its original pressure, volume, and temperature.
For the high-temperature isothermal step, we have $-q_h=w_h=-RT_h \ln \left(\frac{V_2}{V_1}\right) \nonumber$
and for the low-temperature isothermal step, we have
$-q_{\ell }=w_{\ell }=-RT_{\ell } \ln \left(\frac{V_4}{V_3}\right) \nonumber$
For the adiabatic expansion and compression, we have $q_{exp}=q_{comp}=0 \nonumber$
The corresponding energy and work terms are
${\Delta }_{exp}E=w_{exp}=\int^{T_{\ell }}_{T_h}{C_VdT} \nonumber$
for the adiabatic expansion and
${\Delta }_{comp}E=w_{comp}=\int^{T_h}_{T_{\ell }}{C_VdT} \nonumber$
for the adiabatic compression. The heat-capacity integrals are the same except for the direction of integration; they sum to zero, and we have $w_{exp}+w_{comp}=0$. The net work done on the system is the sum of the work for these four steps, $w_{net}=w_h+w_{exp}+w_{\ell }+w_{comp}=w_h+w_{\ell }$. The heat input occurs at the high-temperature reservoir, so that $q_h>0$. The heat discharge occurs at the low-temperature reservoir, so that $q_{\ell }<0$.
For one cycle of the reversible, ideal-gas Carnot engine,
$\epsilon =1+\frac{q_{\ell}}{q_h}=1+\frac{RT_{\ell } \ln \left({V_4}/{V_3}\right)}{RT_h \ln \left(\frac{V_2}{V_1}\right)} \nonumber$
Because the two adiabatic steps involve the same limiting temperatures, the energy of an ideal gas depends only on temperature, and $dE=dw$ for both steps, we see from Section 9.7-9.20 that
$\int^{T_{\ell }}_{T_h}{\frac{C_V}{T}}dT=-\int^{V_3}_{V_2}{\frac{R}{V}}dV=-R{ \ln \left(\frac{V_3}{V_2}\right)\ } \nonumber$
and
$\int^{T_h}_{T_{\ell }}{\frac{C_V}{T}}dT=-\int^{V_1}_{V_4}{\frac{R}{V}}dV=-R{ \ln \left(\frac{V_1}{V_4}\right)\ } \nonumber$
The integrals over $T$ are the same except for the direction of integration. They sum to zero, so that $-R{ \ln \left({V_3}/{V_2}\right)\ }-R{ \ln \left({V_1}/{V_4}\right)\ }=0$ and
$\frac{V_2}{V_1}=\frac{V_3}{V_4} \nonumber$
Using this result, the second equation for the reversible Carnot engine efficiency becomes
$\epsilon =1-\frac{T_{\ell }}{T_h} \nonumber$
Equating our expressions for the efficiency of the reversible Carnot engine, we find
$\epsilon =1+\frac{q_{\ell }}{q_h}=1-\frac{T_{\ell }}{T_h} \nonumber$ from which we have
$\frac{q_h}{T_h}+\frac{q_{\ell }}{T_{\ell }}=0 \nonumber$
Since there is no heat transfer in the adiabatic steps, $q_{exp}=q_{comp}=0,$ and we can write this sum as
$\sum_{cycle}{\frac{q_i}{T_i}}=0 \nonumber$
If we divide the path around the cycle into a large number of very short segments, the limit of this sum as the $q_i$ become very small is
$\oint{\frac{dq^{rev}}{T}}=0 \nonumber$
where the superscript “$rev$” serves as a reminder that the cycle must be traversed reversibly. Now, we can define a new function, $S$, by the differential expression
$dS=\frac{dq^{rev}}{T} \nonumber$
In this expression, $dS$ is the incremental change in $S$ that occurs when the system reversibly absorbs a small of increment of heat, ${dq}^{rev}$, at a particular temperature, $T$. For an ideal gas traversing a Carnot cycle, we have shown that
$\Delta S=\oint{dS}=\oint{\frac{dq^{rev}}{T}}=0 \nonumber$
$S$ is, of course, the entropy function described in our entropy-based statement of the second law.
We now want to see what the machine-based statement of the second law enables us to deduce about the properties of $S$. Since the change in $S$ is zero when an ideal gas goes around a complete Carnot cycle, we can conjecture that $S$ is a state function. Of course, the fact that $\Delta S=0$ around one particular cycle does not prove that $S$ is a state function. If $S$ is a state function, it must be true that $\Delta S=0$ around any cycle whatsoever. We now prove this for any reversible cycle.
The proof has two steps. In the first, we show that $\oint{dq^{rev}/T}=0$ for a machine that uses any reversible system operating between two constant-temperature heat reservoirs to convert heat to work. In the second step, we show that $\oint{dq^{rev}/T}=0$ for any system that reversibly traverses any closed path.
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To show that $\oint dq^{rev}/T=0$ for any reversible system taken around a Carnot cycle, we first observe that the Carnot cycle can be traversed in the opposite direction. In this case, work is delivered to the engine and a quantity of heat is transferred from the low-temperature reservoir to the high-temperature reservoir. Operated in reverse, the Carnot engine is a refrigerator. Suppose that we have two identical ideal-gas Carnot machines, one of which we operate as an engine while we operate the other as a refrigerator. If we configure them so that the work output of the engine drives the refrigerator, the effects of operating them together cancel completely. The refrigerator exactly consumes the work output of the engine. The heat transfers to and from the heat reservoirs offset exactly.
Now, let us consider an ideal-gas Carnot engine and any other reversible engine that extracts heat from a high-temperature reservoir and rejects a portion of it to a low-temperature reservoir. Let us call these engines A and B. We suppose that one is operated to produce work in its surroundings ($w<0$); the other is operated to consume this work and transfer net heat energy from the low-temperature to the high-temperature reservoir. Let the net work done in one cycle on machines A and B be $w_{netA}$ and $w_{netB}$, respectively. We can choose to make these engines any size that we please. Let us size them so that one complete cycle of either engine exchanges the same quantity of heat with the high-temperature reservoir. That is, if the high-temperature reservoir delivers heat $q_{hA}$ to engine A, then it delivers heat $q_{hB}=q_{hA}$ to engine B. Figure 3 diagrams these engines. With one operating as an engine and the other operating as a refrigerator, we have $q_{hA}+q_{hB}=0$. When both engine and refrigerator have completed a cycle, the high temperature reservoir has returned to its original state.
We can create a combined device that consists of A running as an engine, B running as a refrigerator, and the high-temperature reservoir. Figure 3 also diagrams this combination. When it executes one complete cycle, the initial condition of the combined device is restored. Therefore, since E is a state function, we have
\begin{align} \Delta E &= w_{netA}+q_{hA}+q_{\ell A}+w_{netB}+q_{hB}+q_{\ell B} \[4pt] &=w_{netA}+w_{netB}+q_{\ell A}+q_{\ell B} \[4pt] &=0. \end{align} \nonumber
where we use the constraint $q_{hA}+q_{hB}=0$. Let us consider the possibility that $w_{netA}+w_{netB}<0$; that is, the combined device does net work on the surroundings. Then, $\Delta E=0$ implies that $q_{\ell A}+q_{\ell B}>0$.
In this cyclic process, the combined device takes up a positive quantity of heat from a constant-temperature reservoir and delivers a positive quantity of work to the surroundings. There is no other change in either the system or the surroundings. This violates the machine-based statement of the second law. Evidently, it is not possible for the combined device to operate in the manner we have hypothesized. We conclude that any such machine must always operate such that $w_{netA}+w_{netB}\ge 0$; that is, the net work done on the combined machine during any complete cycle must be either zero or some positive quantity.
In concluding that $w_{netA}+w_{netB}\ge 0$, we specify that the combined machine has A running as a heat engine and B running as a refrigerator. Now, suppose that we reverse their roles, and let $w^*_{netA}$ and $w^*_{netB}$ represent the net work for the reversed combination. Applying the same argument as previously, we conclude that $w^*_{netA}+w^*_{netB}\ge 0$. But, since the direction of operation is reversed for both machines, we must also have $w^*_{netA}=-w_{netA}$ and $w^*_{netB}=-w_{netB}$. Hence we have $-w_{netA}-w_{netB}\ge 0$ or $w_{netA}+w_{netB}\le 0$. We conclude, therefore, that
$w_{netA}+w_{netB}=0 \nonumber$
for any two, matched, reversible engines operating around a Carnot cycle.
This conclusion can be restated as a condition on the efficiencies of the two machines. The individual efficiencies are ${\epsilon }_A=-w_{netA}/q_{hA}$ and ${\epsilon }_B=-w_{netB}/q_{hB}$.
(The efficiency equation is unaffected by the direction of operation, because changing the direction changes the sign of every energy term in the cycle. Changing the direction of operation is equivalent to multiplying both the numerator and denominator by minus one.) Then, from $w_{netA}+w_{netB}=0$, it follows that
${\epsilon }_Aq_{hA}+{\epsilon }_Bq_{hB}=0 \nonumber$
Since we sized A and B so that $q_{hA}+q_{hB}=0$, we have
$\epsilon_Aq_{hA}-{\epsilon }_Bq_{hA}=0 \nonumber$ so that ${\epsilon }_A={\epsilon }_B \nonumber$
for any reversible Carnot engines A and B operating between the same two heat reservoirs.
For the ideal gas engine, we found $\epsilon =1-T_{\ell }/{T_h}$. For any reversible Carnot engine, we have $\Delta E=0=w_{net}+q_h+q_{\ell }$, so that $-w_{net}=q_h+q_{\ell }$, and
$\epsilon =\frac{-w_{net}}{q_h}=1+\frac{q_{\ell }}{q_h} \nonumber$
This means that the efficiency relationship
$\epsilon =1-\frac{T_{\ell }}{T_h}=1+\frac{q_{\ell }}{q_h} \nonumber$
applies to any reversible Carnot engine. It follows that the integral of $dq^{rev}/T$ around a Carnot cycle is zero for any reversible system.
The validity of these conclusions is independent of type of work that the engine produces; if engine A is an ideal-gas engine, engine B can be comprised of any system and can produce any kind of work. In obtaining this result from the machine-based statement of the second law, we make the additional assumption that pressure–volume work can be converted entirely to any other form of work, and vice versa. That is, we assume that the work produced by engine A can reversibly drive engine B as a refrigerator, whether engines A and B produce the same or different kinds of work.
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Any system reversibly traversing any closed curve on a pressure–volume diagram exchanges work with its surroundings, and the area enclosed by the curve represents the amount of this work. In the previous section, we found $\oint{dq^{rev}/T}=0$ for any system that traverses a Carnot cycle reversibly. We now show that this is true for any system that traverses any closed path reversibly. This establishes that $\Delta S$ is zero for any system traversing any closed path reversibly and proves that $S$, defined by $dS=dq^{rev}/T$, is a state function.
To do so, we introduce an experience-based theorem: The pressure–volume diagram for any reversible system can be tiled by intersecting lines that represent isothermal and adiabatic paths. These lines can be packed as densely as we please, so that the tiling of the pressure–volume diagram can be made as closely spaced as we please. The perimeter of any one of the resulting tiles corresponds to a path around a Carnot cycle. Given any arbitrary closed curve on the pressure–volume diagram, we can select a set of tiles that just encloses it. See Figure 4. The perimeter of this set of tiles approximates the path of the arbitrary curve. Since the tiling can be made as fine as we please, the perimeter of the set of tiles can be made to approximate the path of the arbitrary curve as closely as we please.
Suppose that we traverse the perimeter of each of the individual tiles in a clockwise direction, adding up $q^{rev}/T$ as we go. Segments of these perimeters fall into two groups. One group consists of segments that are on the perimeter of the enclosing set of tiles. The other group consists of segments that are common to two tiles. When we traverse both of these tiles in a clockwise direction, the shared segment is traversed once in one direction and once in the other. When we add up $q^{rev}/T$ for these two traverses of the same segment, we find that the sum is zero, because we have $q^{rev}/T$ in one direction and $-q^{rev}/T$ in the other. This means that the sum of $q^{rev}/T$ around all of the tiles will just be equal to the sum of $q^{rev}/T$ around those segments that lie on the perimeter of the enclosing set. That is, we have
$\sum_{ \begin{array}{c} \mathrm{cycle} \ \mathrm{perimeter} \end{array}} \frac{q^{rev}}{T}+\sum_{ \begin{array}{c} \mathrm{interior} \ \mathrm{seqments} \end{array}} \frac{q^{rev}}{T}=\sum_{ \begin{array}{c} \mathrm{all} \ \mathrm{tiles} \end{array}} \left\{\sum_{ \begin{array}{c} \mathrm{tile} \ \mathrm{perimeter} \end{array}} \frac{q^{rev}}{T}\right\} \nonumber$
where $\sum_{ \begin{array}{c} \mathrm{interior} \ \mathrm{seqments} \end{array} } \frac{q^{rev}}{T}=0 \nonumber$
because each interior segment is traversed twice, and the two contributions cancel exactly.
This set of tiles has another important property. Since each individual tile represents a reversible Carnot cycle, we know that
$\sum_{ \begin{array}{c} \mathrm{tile} \ \mathrm{perimeter} \end{array} }{\frac{q^{rev}}{T}}=0 \nonumber$
around each individual tile. Since the sum around each tile is zero, the sum of all these sums is zero. It follows that the sum of ${q^{rev}}/{T}$ around the perimeter of the enclosing set is zero:
$\sum_{ \begin{array}{c} \mathrm{cycle} \ \mathrm{perimeter} \end{array} }{\frac{q^{rev}}{T}}=0 \nonumber$
By tiling the pressure–volume plane as densely as necessary, we can make the perimeter of the enclosing set as close as we like to any closed curve. The heat increments become arbitrarily small, and
$\mathop{\mathrm{lim}}_{q^{rev}\to {dq}^{rev}} \left[\sum_{ \begin{array}{c} cycle \ perimeter \end{array} } \frac{q^{rev}}{T}\right]\ =\oint{\frac{dq^{rev}}{T}}=0 \nonumber$ For any reversible engine producing pressure–volume work, we have $\oint{dS=0}$ around any cycle.
We can extend this analysis to reach the same conclusion for a reversible engine that produces any form of work. To see this, let us consider the tiling theorem more carefully. When we say that the adiabats and isotherms tile the pressure–volume plane, we mean that each point in the pressure–volume plane is intersected by one and only one adiabat and by one and only one isotherm. When only pressure–volume work is possible, every point in the pressure–volume plane represents a unique state of the system. Therefore, the tiling theorem asserts that every state of the variable-pressure system can be reached along one and only one adiabat and one and only one isotherm.
From experience, we infer that this statement remains true for any form of work. That is, every state of any reversible system can be reached by one and only one isotherm and by one and only one adiabat when any form of work is done. If more than one form of work is possible, there is an adiabat for each form of work. If changing ${\theta }_1$ and changing ${\theta }_2$ change the energy of the system, the effects on the energy of the system are not necessarily the same. In general, ${\mathit{\Phi}}_1$ is not the same as ${\mathit{\Phi}}_2$, where
${\mathit{\Phi}}_i={\left(\frac{\partial E}{\partial {\theta }_i}\right)}_{V,{\theta }_{m\neq i}} \nonumber$
From §3, we know that a reversible Carnot engine doing any form of work can be matched with a reversible ideal-gas Carnot engine in such a way that the engines complete the successive isothermal and adiabatic steps in parallel. At each step, each engine experiences the same heat, work, energy, and entropy changes as the other. Just as we can plot the reversible ideal-gas Carnot cycle as a closed path in pressure–volume space, we can plot a Carnot cycle producing any other form of work as a closed path with successive isothermal and adiabatic steps in ${\mathit{\Phi}}_i{--\theta }_i$ space. Just as any closed path in pressure–volume space can be tiled (or built up from) arbitrarily small reversible Carnot cycles, so any closed path in ${\mathit{\Phi}}_i{--\theta }_i$ space can be tiled by such cycles. Therefore, the argument we use to show that $\oint{dS=0}$ for any closed reversible cycle in pressure–volume space applies equally well to a closed reversible cycle in which heat is used to produce any other form of work.
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We view the tiling theorem as a generalization from experience, just as the machine-based statement of the second law is such a generalization. Let us consider the kinds of familiar observations from which we infer that every equilibrium state of any system is intersected by one and only one adiabat and by one and only one isotherm.
When only pressure–volume work is possible, each pressure–volume point specifies a unique equilibrium state of the system. Since temperature is a state function, the temperature of this state has one and only one value. When another form of work is possible, every ${\mathit{\Phi}}_i{-\theta }_i$ point specifies a unique state for which the temperature has one and only one value. From experience, we know that we can produce a new state of the system, at the same temperature, by exchanging heat and work with it in a concerted fashion. We can make this change of state arbitrarily small, so that successive equilibrium states with the same temperature are arbitrarily close to one another. This succession of arbitrarily close equilibrium states is an isotherm. Therefore, at least one isotherm intersects any equilibrium state. There cannot be two such isotherms. If there were two isotherms, the system would have two temperatures, violating the principle that temperature is a state function.
In an adiabatic process, the system exchanges energy as work but not as heat. From experience, we know that we can effect such a change with any reversible system. The result is a new equilibrium state. When we make the increment of work arbitrarily small, the new equilibrium state is arbitrarily close to the original state. Successive exchanges of arbitrarily small work increments produce successive equilibrium states that are arbitrarily close to one another. This succession of arbitrarily close equilibrium states is an adiabat.
If the same state of a system could be reached by two reversible adiabats involving the same form of work, the effect of doing a given amount of this work on an equilibrium system would not be unique. From the same initial state, two reversible adiabatic experiments could do the same amount of the same kind of work and reach different final states of the system. For example, in two different experiments, we could raise a weight reversibly from the same initial elevation, do the same amount of work in each experiment, and find that the final elevation of the weight is different. Any such outcome conflicts with the observations that underlie our ideas about reversible processes.
More specifically, the existence of two adiabats through a given point, in any ${\mathit{\Phi}}_i{-\theta }_i$ space, violates the machine-based statement of the second law. Two such adiabats would necessarily intersect a common isotherm. A path along one adiabat, the isotherm, and the second adiabat would be a cycle that restored the system to its original state. This path would enclose a finite area. Traversed in the appropriate direction, the cycle would produce work in the surroundings. By the first law, the system would then accept heat as it traverses the isotherm. The system would exchange heat with surroundings at a single temperature and produce positive work in the surroundings, thus violating the machine-based statement.
If an adiabatic process that connects two states A and B is reversible, we see that the system follows the same path, in opposite directions, when it does work going from A to B as it does when work is done on it as it goes from B to A.
From another perspective, we can say that the tiling theorem is a consequence of our assumptions about reversible processes. Our conception of a reversible process is that the energy, pressure, temperature, and volume are continuous functions of state, with continuous derivatives. That there is one and only one isotherm for every state is equivalent to the assumption that temperature is a continuous (single-valued) function of the state of the system. That there is one and only one adiabat for every state is equivalent to the assumption that ${\left({\partial E}/{\partial V}\right)}_{T,{\theta }_1}$, or generally, ${\left({\partial E}/{\partial {\theta }_i}\right)}_{T,V,{\theta }_{m\neq i}}$, is a continuous, single-valued function of the state of the system.
With these ideas in mind, let us now observe that any reversible cycle can be described by a closed path in a space whose coordinates are $T$ and ${q^{rev}}/{T}$ (entropy). In Figure 5, we sketch this space with ${q^{rev}}/{T}$ on the abscissa; then an isotherm is a horizontal line, and line of constant entropy (an isentrope) is vertical. A reversible Carnot cycle is a closed rectangle, and the area of this rectangle corresponds to the reversible work done by the system on its surroundings in one cycle. Any equilibrium state of the system corresponds to a particular point in this space. Any closed path can be tiled arbitrarily densely by isotherms and isentropes. Any reversible cycle involving any form of work is represented by a closed path in this space. Figure 5 is an alternative illustration of the argument that we make in Section 9.4. The path in this space is independent of the kind of work done, reinforcing the conclusion that $\oint{dq^{rev}/T=0}$ for a reversible Carnot cycle producing any form of work. The fact that a cyclic process corresponds to a closed path in this space is equivalent to the fact that entropy is a state function.
To appreciate this aspect of the path of a cyclic process in $T-q^{rev}/{T}$ space, let us describe the path of the same process in a space whose coordinates are $T$ and $q^{rev}$. With $q^{rev}$ on the abscissa, isotherms are again horizontal lines and adiabats are vertical lines. In this space, a reversible Carnot cycle does not begin and end at the same point. The path is not closed. Similarly, the representation of an arbitrary reversible cycle is not a closed figure. See Figure 6. The difference between the representations of a reversible cyclic process in these two spaces illustrates graphically the fact that entropy is a state function while heat is not.
9.06: Entropy Changes for A Reversible Process
Let us consider a closed system that undergoes a reversible change while in contact with its surroundings. Since the change is reversible, the portion of the surroundings that exchanges heat with the system is at the same temperature as the system: $T=\hat{T}$. From $q^{rev}=- \hat{q}^{rev}$ and the definition, $dS=dq^{rev}/T$, the entropy changes are
$\Delta S={q^{rev}}/{T} \nonumber$
and
$\Delta \hat{S}=\hat{q}^{rev}/T=- q^{rev}/T=-\Delta S \nonumber$
Evidently, for any reversible process, we have
$\Delta S_{universe}=\Delta S+\Delta \hat{S}=0 \nonumber$
Note that these ideas are not sufficient to prove that the converse is true. From only these ideas, we cannot prove that $\Delta S_{universe}=0$ for a process means that the process is reversible; it remains possible that there could be a spontaneous process for which $\Delta S_{universe}=0$. However, our entropy-based statement of the second law does assert that the converse is true, that $\Delta S_{universe}=0$ is necessary and sufficient for a process to be reversible.
In the next section, we use the machine-based statement of the second law to show that $\Delta S\ge 0$ for any spontaneous process in an isolated system. We introduce heuristic arguments to infer that $\Delta S=0$ is not possible for a spontaneous process in an isolated system. From this, we show that $\Delta S_{universe}>0$ for any spontaneous process and hence that $\Delta S_{universe}=0$ is not possible for any spontaneous process. We conclude that $\Delta S_{universe}=0$ is sufficient to establish that the corresponding process is reversible.
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In Section 9.6, we consider the entropy changes in a system and its surroundings when the process is reversible. We consider now the diametrically opposite situation in which an isolated system undergoes a spontaneous change. From the entropy-based statement of the second law, we know how the entropy of this system and its surroundings change. Since the system is isolated, no change occurs in the surroundings. Thus, $\Delta \hat{S}=0$; and since $\Delta S+\Delta \hat{S}>0$, we have $\Delta S>0$.
Let us attempt to develop these conclusions from the machine-based statement of the second law. Since the process occurs irreversibly, we cannot use the heat of the process to find the entropy change for the system. We can calculate the entropy change for a process from the defining equation only if the process is reversible. However, entropy is a state function; using the figure of speech that we introduce in Section 7.21, we can find the entropy change for the spontaneous process by evaluating $\Delta S$ along a second and reversible path that connects the same initial and final states.
In Figure 7, these paths are diagrammed in temperature–entropy space. The transition from state A to state B occurs irreversibly, and therefore it does not necessarily correspond to a path that we can specify on this diagram. The dashed line drawn for this transition is supposed to remind us of this fact. We can readily devise a reversible path from B back to A. First, we reversibly and adiabatically return the temperature of the system to its original value $\hat{T}$. In this step, the system does work on the surroundings, or vice versa. The system reaches point C on the diagram. Then we reversibly and isothermally add or remove heat from the system to return to the original state at point A. For the transfer of heat to be reversible, we must have $T=\hat{T}$ for this step. Hence, the final (and original) temperature of the system at point A is equal to the temperature of the surroundings. The reversible path $B\to C\to A$ must exist, because the tiling theorem asserts that adiabats (vertical lines) and isotherms (horizontal lines) tile the $T--S$-plane arbitrarily densely.
Taken literally, this description of state A is inconsistent. We suppose that the initial state A is capable of spontaneous change; therefore, it cannot be an equilibrium state. We suppose that the final state A is reached by a reversible process; therefore, it must be an equilibrium state. We bridge this contradiction by refining our definition of the initial state. The final state A is an equilibrium state with well-defined state functions. What we have in mind is that these final equilibrium-state values also characterize the initial non-equilibrium state. Evidently, the initial state A that we have in mind is a hypothetical state. This hypothetical state approximates the state of a real system that undergoes spontaneous change. By invoking this hypothetical initial state, we eliminate the contradiction between our descriptions of initial state A and final state A. Given a real system that undergoes spontaneous change, we must find approximate values for the real system’s state functions by finding an equilibrium—or quasi-equilibrium—system that adequately models the initial state of the spontaneously changing system.
In the development below, we place no constraints on the nature of the system or the spontaneous process. We assume that the state functions of any hypothetical initial state A can be adequately approximated by some equilibrium-state model. However, before we consider the general argument, let us show how these conditions can be met for another specific system. Consider a vessel whose interior is divided by a partition. The real gas of a pure substance occupies the space on one side of the partition. The space on the other side of the partition is evacuated. We suppose that this vessel is isolated. The real gas is at equilibrium. We can measure its state functions, including its pressure, volume, and temperature. Now suppose that we puncture the partition. As soon as we do so, the gas expands spontaneously to fill the entire vessel, reaching a new equilibrium position, at a new pressure, volume, and temperature. The gas undergoes a free expansion, as defined in Section 7.17.
At the instant the partition is punctured, the system becomes able to undergo spontaneous change. In this hypothetical initial state, before any significant quantity of gas passes through the opening, neither the actual condition of the gas nor the values of its state functions have changed. After the expansion to the new equilibrium state, the original state can be restored by reversible processes of adiabatic compression and isothermal volume adjustment. (Problems 13 and 14 in Chapter 10 deal with the energy and entropy changes for ideal and real gases around a cycle in which spontaneous expansion in an isolated system is followed by reversible restoration of the initial state.)
Returning to the general cycle depicted in Figure 7, we see that there are some important conditions on the heat and work terms in the individual steps. Since the system is isolated while it undergoes the transition from A to B, it exchanges no heat or work with the surroundings in this step: $q^{spon}_{AB}=w^{spon}_{AB}=0$. For the reversible adiabatic transition from B to C, $d_{BC}q^{rev}=0$ in every incremental part of the path. The transition from C to A occurs reversibly and isothermally; letting the heat of this step be $q^{rev}_{CA}$, the entropy changes for these reversible steps are, from the defining equation,
${\Delta }_{BC}S=\int^{T_C}_{T_B}{\frac{d_{BC}q^{rev}}{T}}=0 \nonumber$ and ${\Delta }_{CA}S=\frac{q^{rev}_{CA}}{\hat{T}} \nonumber$
The energy and entropy changes around this cycle must be zero, whether the individual steps occur reversibly or irreversibly. We have
$\Delta E=q^{spon}_{AB}+w^{spon}_{AB}+q^{rev}_{BC}+w^{rev}_{BC}+q^{rev}_{CA}+w^{rev}_{CA}=q^{rev}_{CA}+w^{rev}_{BC}+w^{rev}_{CA}=0 \nonumber$ and
$\Delta S={\Delta }_{AB}S+{\Delta }_{BC}S+{\Delta }_{CA}S ={\Delta }_{AB}S+{q^{rev}_{CA}}/{\hat{T}}=0 \nonumber$
We want to analyze this cycle using the machine-based statement of the second law. We have $w^{rev}_{BC}=-{\hat{w}}^{rev}_{BC}$, $w^{rev}_{CA}=-{\hat{w}}^{rev}_{CA}$, and $q^{rev}_{CA}=-{\hat{q}}^{rev}_{CA}$. Let us assume that the system does net work on the surroundings as this cycle is traversed so that ${\hat{w}}^{rev}_{BC}+{\hat{w}}^{rev}_{CA}>0$. Then,
$-\left({\hat{w}}^{rev}_{BC}+{\hat{w}}^{rev}_{CA}\right)=w^{rev}_{BC}+w^{rev}_{CA}<0 \nonumber$
and it follows that $q^{rev}_{CA}>0$. The system exchanges heat with the surroundings in only one step of this process. In this step, the system extracts a quantity of heat from a reservoir in the surroundings. The temperature of this reservoir remains constant at $\hat{T}$ throughout the process. The heat extracted by the system is converted entirely into work. This result contradicts the machine-based statement of the second law. Hence, $w^{rev}_{BC}+w^{rev}_{CA}<0$ is false; it follows that
$w^{rev}_{BC}+w^{rev}_{CA}\ge 0 \nonumber$
and that $q^{rev}_{CA}\le 0 \nonumber$
For the entropy change in the spontaneous process in the isolated system, we have
${\Delta }_{AB}S=-{q^{rev}_{CA}}/{\hat{T}}\ge 0 \nonumber$
Now, we introduce the premise that $q^{rev}_{CA}\neq 0$. If this is true, the entropy change in the spontaneous process in the isolated system becomes
${\Delta }_{AB}S>0 \nonumber$
(The converse is also true; that is, ${\Delta }_{AB}S>0$ implies that $q^{rev}_{CA}\neq 0$.) The premise that $q^{rev}_{CA}\neq 0$ is independent of the machine-based statement of the second law, which requires only that $q^{rev}_{CA}\le 0$, as we just demonstrated. It is also independent of the first law, which requires only that $q^{rev}_{CA}=-w^{rev}_{BC}-w^{rev}_{CA}$. If $q^{rev}_{CA}\neq 0$, we can conclude that, for a spontaneous process in an isolated system, we must have $w_{BC}+w_{CA}>0$ and $q^{rev}_{CA}<0$. These conditions correspond to doing work on the system and finding that heat is liberated by the system. There is no objection to this; it is possible to convert mechanical energy into heat quantitatively. The conclusions that $q^{rev}_{CA}<0$ and ${\Delta }_{AB}S>0$ have important consequences; we consider them below. First, however, we consider a line of thought that leads us to infer that $q^{rev}_{CA}\neq 0$ and hence that $q^{rev}_{CA}<0$ must be true.
Because $q_{AB}=0$ and $w_{AB}=0$, we have ${\ E}_A=E_B$. The system can be taken from state A to state B by the reversible process A$\to C\to B$. Above we see that if $q^{rev}_{CA}=0$, we have ${\ S}_A=S_B$. In §6-10, we introduce Duhem’s theorem, which asserts that two thermodynamic variables are sufficient to specify the state of a closed reversible system in which only pressure–volume work is possible. We gave a proof of Duhem’s theorem when the two variables are chosen from among the pressure, temperature, and composition variables that describe the system. We avoided specifying whether other pairs of variables can be used. If we assume now that specifying the variables energy and entropy is always sufficient to specify the state of such a system, it follows that states A and B must in fact be the same state. (In §14, and in greater detail in Chapter 10, we see that the first law and our entropy-based statement of the second law do indeed imply that specifying the energy and entropy specifies the state of a closed reversible system in which only pressure–volume work is possible.)
If state A and state B are the same state; that is, if the state functions of state A are the same as those of state B, it is meaningless to say that there is a spontaneous process that converts state A to state B. Therefore, if A can be converted to B in a spontaneous process in an isolated system, it must be that $q^{rev}_{CA}\neq 0$. That is,
$\left[\left(q^{rev}_{CA}=0\right)\Rightarrow \sim \left(\mathrm{A\ can\ go\ to\ B\ spontaneously}\right)\right] \nonumber$ $\Rightarrow \left[\left(A\ \mathrm{can\ go\ to\ B\ spontaneously}\right)\Rightarrow \left(q^{rev}_{CA}\neq 0\right)\right] \nonumber$
From the machine-based statement of the second law, we find ${\Delta }_{AB}S=-{q^{rev}_{CA}}/{\hat{T}}\ge 0$. When we supplement this conclusion with our Duhem’s theorem-based inference that $q^{rev}_{CA}\neq 0$, we can conclude that $\Delta S>0$ for any spontaneous process in any isolated system. Because the system is isolated, we have $\hat{q}=0$, and $\Delta \hat{S}=0$. For any spontaneous process in any isolated system we have
$\Delta S_{universe}=\Delta S+\Delta \hat{S}>0. \nonumber$
We can also conclude that the converse is true; that is, if ${\Delta }_{AB}S=S_B-S_A>0$ for a process in which an isolated system goes from state A to state B, the process must be spontaneous. Since any process that occurs in an isolated system must be a spontaneous process, it is only necessary to show that ${\Delta }_{AB}S>0$ implies that state B is different from state A. This is trivial. Because entropy is a state function, $S_B-S_A>0$ requires that state B be different from state A.
None of our arguments depends on the magnitude of the change that occurs. Evidently, the same inequality must describe every incremental portion of any spontaneous process; otherwise, we could define an incremental spontaneous change for which the machine-based statement of the second law would be violated. For every incremental part of any spontaneous change in any isolated system we have $dS>0$ and
$dS_{universe}=dS+d\hat{S}>0. \nonumber$
These are pivotally important results; we explore their ramifications below. Before doing so, however, let us again consider a system in which only pressure–volume work is possible. There is an alternative way to express the idea that such a system is isolated. Since an isolated system cannot interact with its surroundings in any way, it cannot exchange energy with its surroundings. Its energy must be constant. Since it cannot exchange pressure–volume work, its volume must be constant. Hence, isolation implies constant $E$ and $V$. If only pressure–volume work is possible, the converse must be true; that is, if only pressure–volume work is possible, constant energy and volume imply that there are no interactions between the system and its surroundings. Therefore, constant $E$ and $V$ imply that the system is isolated, and it must be true that $\Delta \hat{S}=0$. In this case, a spontaneous process in which $E$ and $V$ are constant must be accompanied by an increase in the entropy of the system. (If $V$ is constant and only pressure–volume work is possible, the process involves no work.) We have a criterion for spontaneous change:
${\left(\Delta S\right)}_{EV}>0 \nonumber$ (spontaneous process, only pressure–volume work)
where the subscripts indicate that the energy and volume of the system are constant. (In Section 9.21, we arrive at this conclusion by a different argument.)
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In Section 9.7, we conclude that the entropy change is positive for any spontaneous change in an isolated system. Since we can consider the universe to be an isolated system, it follows that $\Delta S_{universe}>0$ for any spontaneous process.
To reach this conclusion by a more detailed argument, let us consider an arbitrary system that is in contact with its surroundings. We can subdivide these surroundings into subsystems. As diagrammed in Figure 8, we define a surroundings subsystem (Surroundings 1) that interacts with the system and a more remote surroundings subsystem (Surroundings 2) that does not. That is, we assume that we can define Surroundings 2 so that it is unaffected by the process. Then we define an augmented system consisting of the original system plus Surroundings 1. The augmented system is isolated from the remote portion of the surroundings, so that the entropy change for the augmented system is positive by the argument in the previous section. Denoting entropy changes for the system, Surroundings 1, Surroundings 2, and the augmented system by $\Delta S$, $\Delta {\hat{S}}_1$, $\Delta {\hat{S}}_2$, and $\Delta S_{augmented}$, respectively, we have $\Delta S_{augmented}=\Delta S+\Delta {\hat{S}}_1>0$, and
$\Delta \hat{S}=\Delta {\hat{S}}_1+\Delta {\hat{S}}_2>0$. Since the remote portion of the surroundings is unaffected by the change, we have $\Delta {\hat{S}}_2=0$. For any spontaneous change, whether the system is isolated or not, we have
$\Delta S_{universe}=\Delta S+\Delta {\hat{S}}_1+\Delta {\hat{S}}_2=\Delta S_{augmented}+\Delta {\hat{S}}_2=\Delta S_{augmented}>0 \nonumber$ (any spontaneous change)
This statement is an essential part of the entropy-based statement of the second law. We have now developed it from the machine-based statement of the second law by convincing, but not entirely rigorous arguments. In Section 9.6 we find that $\Delta S_{universe}=\Delta S+\Delta \hat{S}=0$ for any reversible process. Thus, for any possible process, we have
$\Delta S_{universe}=\Delta S+\Delta \hat{S}\ge 0 \nonumber$ The equality applies when the process is reversible; the inequality applies when it is spontaneous.
Because entropy is a state function, $\Delta S$ and $\Delta \hat{S}$ change sign when the direction of a process is reversed. We say that a process for which $\Delta S+\Delta \hat{S}<0$ is an impossible process. Our definitions mean that these classifications—reversible, spontaneous, and impossible—are exhaustive and mutually exclusive. We conclude that $\Delta S_{universe}=\Delta S+\Delta \hat{S}=0$ is necessary and sufficient for a process to be reversible; $\Delta S_{universe}=\Delta S+\Delta \hat{S}>0$ is necessary and sufficient for a process to be spontaneous. (See problem 19.)
9.09: The Significance of The Machine-based Statement of The Second
Our entropy-based statement of the second law asserts the definition and basic properties of entropy that we need in order to make predictions about natural processes. The ultimate justification for these assertions is that the predictions they make agree with experimental observations. We have devoted considerable attention to arguments that develop the definition and properties of entropy from the machine-based statement of the second law. These arguments parallel those that were made historically as these concepts were developed. Understanding these arguments greatly enhances our appreciation for the relationship between the properties of the entropy function and the changes that can occur in various physical systems.
While these arguments demonstrate that our machine-based statement implies the entropy-based statement, we introduce additional postulates in order to make them. These include: the premise that the pressure, temperature, volume, and energy of a reversible system are continuous functions of one another; Duhem’s theorem; the tiling theorem; and the presumption that the conclusions we develop for pressure–volume work are valid for any form of work. We can sum up this situation by saying that our machine-based statement serves a valuable heuristic purpose. The entropy-based statement of the second law is a postulate that we infer by reasoning about the consequences of the machine-based statement. When we want to apply the second law to physical systems, the entropy-based statement and other statements that we introduce below are much more useful.
Finally, we note that our machine-based statement of the second law is not the only statement of this type. Other similar statements have been given. The logical relationships among them are interesting, and they can be used to develop the entropy-based statement of the second law by arguments similar to those we make in Section 9.2 to Section 9.8.
9.10: A Slightly Philosophical Digression on Energy and Entropy
The content of the first law of thermodynamics is that there is a state function, which we call energy, which has the property that \(\Delta E_{universe}=0\) for any process that can occur. The content of the second law is that there is a state function, which we call entropy, which has the property that \(\Delta S_{universe}>0\) for any spontaneous process.
These two state functions exhaust the range of independent possibilities: Suppose that we aspire to find a new and independent state function, call it \(B\), which further characterizes the possibilities open to the universe. What other condition could B impose on the universe—or vice versa? The only available candidate might appear to be \(\Delta B_{universe}<0\). However, this does not represent an independent condition, since its role is already filled by the quantity \(-\Delta S_{universe}\).
Of course, we can imagine a state function, \(B\), which is not simply a function of \(S\), but for which
\(\Delta B_{universe}>0\), \(\Delta B_{universe}=0\), or \(\Delta B_{universe}<0\), according as the process is spontaneous, reversible, or impossible, respectively. For any given change, \(\Delta B\) would not be the same as \(\Delta S\); however, \(\Delta B\) and \(\Delta S\) would make exactly the same predictions. If \(\Delta B_{universe}\) were more easily evaluated than \(\Delta S_{universe}\), we would prefer to use \(B\) rather than \(S\). Nevertheless, if there were such a function \(B\), its role in our description of nature would duplicate the role played by \(S\).
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Let us consider another frequently cited alternative statement of the second law, which, for easy reference, we call the temperature-based statement of the second law:
The temperature-based statement of the second law
The spontaneous transfer of heat from a colder body to a warmer body is impossible.
In the discussion below, we refer to this statement as proposition $\mathrm{TSL}$. By “body”, we simply mean any system or object. By the “spontaneous transfer of heat,” we mean that the transfer of heat energy can be initiated by bringing the two bodies into contact with one another or by enabling the transmission of radiant energy between them. The surroundings do no work and exchange no heat with either reservoir; there is no change of any sort in the surroundings.
We can show that the entropy-based statement and the temperature-based statement of the second law are equivalent: Given the definition of entropy, one implies the other.
Let us begin by showing that the entropy-based statement implies the temperature-based statement of the second law. That is, we prove $\mathrm{SL}\Rightarrow \mathrm{TSL}$. To do so, we prove $\sim \mathrm{TSL}\Rightarrow \sim \mathrm{SL}$. That is, we assume that spontaneous transfer of heat from a colder to a warmer body is possible and show that this leads to a contradiction of the entropy-based statement of the second law. Let the quantity of heat received by the warmer body be $dq_{warmer}>0$, and let the temperatures of the warmer and colder bodies be $T_{warmer}$ and $T_{colder}$, respectively. We have $T_{warmer}-T_{colder}>$0. The colder body receives heat
$dq_{colder}=-dq_{warmer}<0$. We make the heat increment so small that there is no significant change in the temperature of either body. No other changes occur. The two bodies are the only portions of the universe that are affected. Let the entropy changes for the warmer and colder bodies be $dS_{colder}$ and $dS_{warmer}$, respectively.
To find $dS_{colder}$ and $dS_{warmer}$ we must find a reversible path to effect the same changes. This is straightforward. We can effect identically the same change in the warmer body by transferring heat, $q_{warmer}>0$, to it through contact with some third body, whose temperature is infinitesimally greater than $T_{warmer}$. This process is reversible, and the entropy change is $dS_{warmer}={dq_{warmer}}/{T_{warmer}}$. Similarly, the entropy change for the colder body is $dS_{colder}={dq_{colder}}/{T_{colder}=-{dq_{warmer}}/{T_{colder}}}$. It follows that
\begin{aligned} dS_{universe} & =dS_{warmer}+dS_{colder} \ ~ & =\frac{dq_{warmer}}{T_{warmer}}-\frac{dq_{warmer}}{T_{colder}} \ ~ & =-dq_{warmer}\left(\frac{T_{warmer}-T_{colder}}{T_{warmer}T_{colder}}\right) \ ~ & <0 \end{aligned} \nonumber
However, if ${dS}_{universe}<0$ for a spontaneous process, the second law ($\mathrm{SL}$) must be false. We have shown that a violation of the temperature-based statement implies a violation of the entropy-based statement of the second law: $\sim \mathrm{TSL}\Rightarrow \sim \mathrm{SL}$, so that $\mathrm{SL}\Rightarrow \mathrm{TSL}$.
It is equally easy to show that the temperature-based statement implies the entropy-based statement of the second law. To do so, we assume that the entropy-based statement is false and show that this implies that the temperature-based statement must be false. By the arguments above, the entropy change that the universe experiences during the exchange of the heat increment is
${dS}_{universe}=-dq_{warmer}\left(\frac{T_{warmer}-T_{colder}}{T_{warmer}T_{colder}}\right) \nonumber$
If the entropy-based statement of the second law is false, then ${dS}_{universe}<0$. It follows that $dq_{warmer}>0$; that is, the spontaneous process transfers heat from the colder to the warmer body. This contradicts the temperature-based statement. That is, $\sim \mathrm{SL}\Rightarrow \sim \mathrm{TSL}$, so that $\mathrm{TSL}\Rightarrow \mathrm{SL}$.
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The entropy-based criteria that we develop in Section 9.2 through Section 9.8 are of central importance. If we are able to evaluate the change in the entropy of the universe for a prospective process and find that it is greater than zero, we can conclude that the process can occur spontaneously. The reverse of a spontaneous process cannot occur; it is an impossible process and the change in the entropy of the universe for such a process must be less than zero. Since an equilibrium process is a reversible process, the entropy of the universe must remain unchanged when a system goes from an initial state to a final state along a path whose every point is an equilibrium state. Using another figure of speech, we often say that a change that occurs along a reversible path is a change that “occurs at equilibrium.”
These conclusions are what make the entropy function useful: If we can calculate ${\Delta S}_{universe}$ for a prospective process, we know whether the system is at equilibrium with respect to that process; whether the process is possible; or whether the process cannot occur. If we find ${\Delta S}_{universe}>0$ for a process, we can conclude that the process is possible; however, we cannot conclude that the process will occur. Indeed, many processes can occur spontaneously but do not do so. For example, hydrocarbons can react spontaneously with oxygen; most do so only at elevated temperatures or in the presence of a catalyst.
The criteria $\Delta S_{universe}=\Delta S+\Delta \hat{S}\ge 0$ are completely general. They apply to any process occurring under any conditions. To apply them we must determine both $\Delta S$ and $\Delta \hat{S}$. By definition, the system comprises the part of the universe that is of interest to us; the need to determine $\Delta \hat{S}$ would appear to be a nuisance. This proves not to be the case. So long as the surroundings have a well-defined temperature, we can develop additional criteria for equilibrium and spontaneous change in which $\Delta \hat{S}$ does not occur explicitly. In §14, we develop criteria that apply to reversible processes. In §15, we find a general relationship for $\Delta \hat{S}$ that enables us to develop criteria for spontaneous processes.
To develop the criteria for spontaneous change, we must define what we mean by spontaneous change more precisely. To define a spontaneous process in an isolated system as one that can take place on its own is reasonably unambiguous. However, when a system is in contact with its surroundings, the properties of the surroundings affect the change that occurs in the system. To specify a particular spontaneous process we must specify some properties of the surroundings or—more precisely—properties of the system that the surroundings act to establish. The ideas that we develop in §15 lead to criteria for changes that occur while one or more thermodynamic functions remain constant. These criteria supplement the second-law criteria $\Delta S+\Delta \hat{S}\ge 0$. In using these criteria, we can say that the change occurs subject to one or more constraints.
Some of these criteria depend on the magnitudes of $\Delta E$ and $\Delta H$ in the prospective process. We also find criteria that are expressed using new state functions that we call the Helmholtz and Gibbs free energies. In the next section, we introduce these functions.
9.13: Defining the Helmholtz and Gibbs Free Energies
The first and second laws of thermodynamics define energy and entropy. Energy and entropy are fundamental state functions that we use to define other state functions. In Chapter 8, we use the energy function to define enthalpy. We use the energy and entropy functions to define two more state functions that also prove to have useful properties. These are the Helmholtz and Gibbs free energies. The Helmholtz free energy is usually given the symbol $A$, and the Gibbs free energy is usually given the symbol $G$. We define them by
$\begin{array}{c c} A=E-TS ~ & ~ \text{(Helmholtz free energy)} \end{array} \nonumber$
and
$\begin{array}{c c} G=H-TS ~ & ~ \text{(Gibbs free energy)} \end{array} \nonumber$
Note that $PV$, $TS$, $H$, $A$, and $G$ all have the units of energy, $E$.
The sense of the name “free energy” is that a constant-temperature process in which a system experiences an entropy increase ($\Delta S>0$) is one in which the system’s ability to do work in the surroundings is increased by an energy increment $T\Delta S$. Then, adding $T\Delta S$ to the internal energy lost by the system yields the amount of energy that the process actually has available (energy that is “free”) to do work in the surroundings. When we consider how $\Delta A$ and $\Delta G$ depend on the conditions under which system changes, we find that this idea leads to useful results.
The rest of this chapter develops important equations for $\Delta E$, $\Delta H$, $\Delta S$, $\Delta A$, and $\Delta G$ that result when we require that a system change occur under particular sets of conditions.
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To begin exploring the possibilities for stating the criteria for change using only the properties of the system, let us consider how some thermodynamic functions change when a process is reversible. We consider a closed system and focus on making incremental changes in the state of the system. For a reversible process, we have ${dq}^{rev}=TdS$. The reversible pressure–volume work is ${dw}^{rev}_{PV}=-PdV$. If non-pressure–volume work is also possible, the reversible work becomes ${dw}^{rev}=-PdV+{dw}^{rev}_{NPV}$, where ${dw}^{rev}_{NPV}$ is the increment of reversible, non-pressure–volume work. The energy change is
$dE=dq^{rev}+dw^{rev}=TdS-PdV+dw^{rev}_{NPV} \nonumber$ (any reversible process)
This equation is of central importance. It is sometimes called the combined first and second laws of thermodynamics or the fundamental equation. It applies to any closed system that is undergoing reversible change. It specifies a relationship among the changes in energy, entropy, and volume that must occur if the system is to remain at equilibrium while an increment of non-pressure–volume work, ${dw}^{rev}_{NPV}$, is done on it. The burden of our entire development is that any reversible process must satisfy this equation. Conversely, any process that satisfies this equation must be reversible.
For a reversible process at constant entropy, we have $dS=0$, so that ${\left(dE\right)}_S=-PdV+{dw}^{rev}_{NPV}$. Since $-PdV$ is the reversible pressure–volume work, ${dw}^{rev}_{PV}$, and the sum ${dw}^{rev}_{net}=-PdV+{dw}^{rev}_{NPV}$ is the net work, we have
${{\left(dE\right)}_S=dw}^{rev}_{net} \nonumber$ (reversible process, constant S)
where the subscript “$S$” specifies that the entropy is constant. For a reversible process in which all of the work is pressure–volume work, we have ${dw}^{rev}_{NPV}=0$, and the fundamental equation becomes
$dE=TdS-PdV \nonumber$ (reversible process, only pressure–volume work)
For a reversible process in which only pressure–volume work is possible, this equation gives the amount, $dE$, by which the energy must change when the entropy changes by $dS$ and the volume changes by $dV$.
Now, let us apply the fundamental equation to an arbitrary process that occurs reversibly and at constant entropy and constant volume. Under these conditions, $dS=0$ and $dV=0$. Therefore, at constant entropy and volume, a necessary and sufficient condition for the process to be reversible—and hence to be continuously in an equilibrium state as the process takes place—is that
${{\left(dE\right)}_{SV}=dw}^{rev}_{NPV} \nonumber$ (reversible process)
and if only pressure–volume work is possible,
${\left(dE\right)}_{SV}=0 \nonumber$ (reversible process, only pressure–volume work)
where the subscripts indicate that entropy and volume are constant.
If we consider an arbitrary reversible process that occurs at constant energy and volume, we have $dE=0$ and $dV=0$, and the fundamental equation reduces to
${\left(dS\right)}_{EV}=-\frac{dw^{rev}_{NPV}}{T} \nonumber$ (reversible process)
and if only pressure–volume work is possible,
${\left(dS\right)}_{EV}=0 \nonumber$ (reversible process, only pressure–volume work)
In this case, as noted in §7, the system is isolated. In §1-6, we note that an isolated system in an equilibrium state can undergo no further change. Thus, the condition ${\left(dS\right)}_{EV}=0$ defines a unique or primitive equilibrium state.
If a closed system behaves reversibly, any composition changes that occur in the system must be reversible. For chemical applications, composition changes are of paramount importance. We return to these considerations in Chapter 14, where we relate the properties of chemical substances—their chemical potentials—to the behavior of systems undergoing both reversible and spontaneous composition changes.
If a closed system behaves reversibly and only pressure–volume work is possible, we see from the fundamental equation that specifying the changes in any two of the three variables, $E$, $S$, and $V$, is sufficient to specify the change in the system. In particular, if energy and entropy are constant, $dE=dS=0$, the volume is also constant, $dV=0$, and the system is isolated. Thus, the state of an equilibrium system whose energy and entropy are fixed is unique; $dE=dS=0$ specifies a primitive equilibrium state. We see that the internal consistency of our model passes a significant test: From the entropy-based statement of the second law, we deduce the same proposition that we introduce in §7 as a heuristic conjecture. In Chapter 10, we expand on this idea.
Starting from the fundamental equation, we can find similar sets of relationships for enthalpy, the Helmholtz free energy, and the Gibbs free energy. We define $H\ =\ E\ +\ PV$. For an incremental change in a system we, have $dH=dE+PdV+VdP \nonumber$
Using the fundamental equation to substitute for dE, this becomes
$dH=TdS-PdV+dw^{rev}_{NPV}+PdV+VdP=TdS+VdP+dw^{rev}_{NPV} \nonumber$
For a reversible process in which all of the work is pressure–volume work, we have
$dH=TdS+VdP \nonumber$ (reversible process, only pressure–volume work)
For a reversible process in which only pressure–volume work is possible, this equation gives the amount, $dH$, by which the enthalpy must change when the entropy changes by $dS$ and the pressure changes by $dP$. If a reversible process occurs at constant entropy and pressure, then $dS=0$ and $dP=0$. At constant entropy and pressure, the process is reversible if and only if
${\left(dH\right)}_{SP}=dw^{rev}_{NPV} \nonumber$ (reversible process)
If only pressure–volume work is possible,
${\left(dH\right)}_{SP}=0 \nonumber$ (reversible process, only pressure–volume work)
where the subscripts indicate that entropy and pressure are constant.
If we consider an arbitrary reversible process that occurs at constant enthalpy and pressure, we have $dH=0$ and $dP=0$, and the total differential for $dH$ reduces to
${\left(dS\right)}_{HP}=-\frac{dw^{rev}_{NPV}}{T}$(reversible process)
and if only pressure–volume work is possible,
${\left(dS\right)}_{HP}=0 \nonumber$ (reversible process, only pressure–volume work)
From $A=E-TS$, we have $dA=dE-TdS-SdT$. Using the fundamental equation to substitute for $dE$, we have
$dA=TdS-PdV+dw^{rev}_{NPV}-TdS-SdT=-PdV-SdT+dw^{rev}_{NPV} \nonumber$
For a reversible process in which all of the work is pressure–volume work,
$dA=-SdT-PdV \nonumber$ (reversible process, only pressure–volume work)
For a reversible process in which only pressure–volume work is possible, this equation gives the amount, $dA$, by which the Helmholtz free energy must change when the temperature changes by$\ dT$ and the volume changes by $dV$. For a reversible isothermal process, we have $dT=0$, and from $dA=-PdV-SdT+dw^{rev}_{NPV}$ we have
${\left(dA\right)}_T=-PdV+dw^{rev}_{NPV}=dw^{rev}_{PV}+dw^{rev}_{NPV}=dw^{rev}_{net} \nonumber$ (reversible isothermal process)
where we recognize that the reversible pressure–volume work is $dw^{rev}_{PV}=-PdV$, and the work of all kinds is $dw^{rev}_{net}=dw^{rev}_{PV}+dw^{rev}_{NPV}$. We see that ${\left(dA\right)}_T$ is the total of all the work done on the system in a reversible process at constant temperature. This is the reason that “$A$” is used as the symbol for the Helmholtz free energy: “$A$” is the initial letter in “Arbeit,” a German noun whose meaning is equivalent to that of the English noun “work.”
If a reversible process occurs at constant temperature and volume, we have $dT=0$ and $dV=0$. At constant temperature and volume, a process is reversible if and only if
${\left(dA\right)}_{TV}=dw^{rev}_{NPV} \nonumber$ (reversible process)
If only pressure–volume work is possible,
${\left(dA\right)}_{TV}=0 \nonumber$ (reversible process, only pressure–volume work)
where the subscripts indicate that volume and temperature are constant. (Of course, these conditions exclude all work, because constant volume implies that there is no pressure–volume work.)
From $G=H-TS=E+PV-TS \nonumber$
and the fundamental equation, we have $dG=dE+PdV+VdP-SdT-TdS=\ \ TdS-PdV+dw^{rev}_{NPV}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -TdS+PdV+VdP-SdT=-SdT+VdP+dw^{rev}_{NPV} \nonumber$
For a reversible process in which all of the work is pressure–volume work,
$dG=-SdT+VdP \nonumber$ (reversible process, only pressure–volume work)
For a reversible process in which only pressure–volume work is possible, this equation gives the amount, $dG$, by which the Gibbs free energy must change when the temperature changes by $dT$ and the pressure changes by $dP$. For a reversible process that occurs at constant temperature and pressure, $dT=0$ and $dP=0$. At constant temperature and pressure, the process will be reversible if and only if
${\left(dG\right)}_{TP}=dw^{rev}_{NPV} \nonumber$ (any reversible process)
If only pressure–volume work is possible,
${\left(dG\right)}_{TP}=0 \nonumber$ (reversible process, only pressure–volume work)
where the subscripts indicate that temperature and pressure are constant.
In this section, we develop several criteria for reversible change, stating these criteria as differential expressions. Since each of these expressions applies to every incremental part of a reversible change that falls within its scope, corresponding expressions apply to finite changes. For example, we find ${\left(dE\right)}_S=dw^{rev}_{net}$ for every incremental part of a reversible process in which the entropy has a constant value. Since we can find the energy change for a finite amount of the process by summing up the energy changes in every incremental portion, it follows that
${\left(\Delta E\right)}_S=w^{rev}_{net} \nonumber$ (reversible process)
Each of the other differential-expression criteria for reversible change also gives rise to a corresponding criterion for a finite reversible change. These criteria are summarized in §25.
In developing the criteria in this section, we stipulate that various combinations of the thermodynamic functions that characterize the system are constant. We develop these criteria for systems undergoing reversible change; consequently, the requirements imposed by reversibility must be satisfied also. In particular, the system must be composed of homogeneous phases and its temperature must be the same as that of the surroundings. The pressure of the system must be equal to the pressure applied to it by the surroundings. When we specify that a reversible process occurs at constant temperature, we mean that $T=\hat{T}=\mathrm{constant}$. When we specify that a reversible process occur at constant pressure, we mean that $P=P_{applied}=\mathrm{constant}$.
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In a reversible process, the changes that occur in the system are imposed by the surroundings; reversible change occurs only because the system responds to changes in the conditions imposed on it by its surroundings. A reversible process is driven by the surroundings. In contrast, a spontaneous process is driven by the system. Nevertheless, when a spontaneous process occurs under some specific set of imposed conditions (specific values of the temperature and pressure, for example) the system’s equilibrium state depends on these conditions. To specify a particular spontaneous change, we must specify enough constraints to fix the final state of the system.
To see these points from a slightly different perspective, let us consider a closed reversible system in which only pressure–volume work is possible. Duhem’s theorem asserts that a change in the state of this system can be specified by specifying the changes in some pair of state functions, say $X$ and $Y$. If the imposed values of $X$ and $Y$ are constant at their eventual equilibrium values, but the system is changing, the system cannot be on a Gibbsian equilibrium manifold. We say that the system is undergoing a spontaneous change at constant$\ X$ and $Y$.
This description is a figure of speech in that the system’s $X$ and $Y$ values do not necessarily attain the imposed values and become constant until equilibrium is reached. An example is in order: A system whose original pressure and temperature are $P_i$ and $T_i$ can undergo a spontaneous change while the surroundings impose a constant pressure, $P_{applied}=P_f$, and the system is immersed in constant temperature bath at $T=T_f$. The pressure and temperature of the system may be indeterminate as the process occurs, but the equilibrium pressure and temperature must be $P_f$ and $T_f$.
If the surroundings operate to impose particular values of $\boldsymbol{X}$ and $\boldsymbol{Y}$ on the system, then the position at which the system eventually reaches equilibrium is determined by these values. The same equilibrium state is reached for any choice of surroundings that imposes the same values of $\boldsymbol{X}$ and $\boldsymbol{Y}$ on the system at the time that the system reaches equilibrium. For every additional form of non-pressure–volume work that affects the system, we must specify the value of one additional variable in order to specify a unique equilibrium state.
The entropy changes that occur in the system and its surroundings during a spontaneous process have predictive value. However, our definitions do not enable us to find the entropy change for a spontaneous process, and the temperature of the system may not have a meaningful value. On the other hand, we can always carry out the process so that the temperature of the surroundings is known at every point in the process. Indeed, if the system is in thermal contact with its surroundings as the process occurs, we cannot specify the conditions under which the process occurs without specifying the temperature of the surroundings along this path.
Figure 9 describes a spontaneous process whose path can be specified by the values of thermodynamic variable $Y$ and the temperature of the surroundings, $\hat{T}$, as a function of time, $t$. Let us denote the curve that describes this path as $C$. We can divide this path into short intervals. Let $C_k$ denote a short segment of this path along which the temperature of the surroundings is approximately constant. For our present purposes, the temperature of the system, $T$, is irrelevant; since the process is spontaneous, the temperature of the system may have no meaningful value within the interval $C_k$. As the system traverses segment $C_k$, it accepts a quantity of heat, $q_k$, from the surroundings, which are at temperature ${\hat{T}}_k$. The heat exchanged by the surroundings within $C_k$ is ${\hat{q}}_k=-q_k$. Below, we show that it is always possible to carry out the process in such a way that the change in the surroundings occurs reversibly. Then
$\Delta {\hat{S}}_k=\frac{{\hat{q}}_k}{{\hat{T}}_k}=-\frac{q_k}{{\hat{T}}_k} \nonumber$
and since $\Delta S_k+\Delta {\hat{S}}_k>0$, it follows that
$\Delta S_k>\frac{q_k}{{\hat{T}}_k} \nonumber$
This is the Clausius inequality. It plays a central role in the thermodynamics of spontaneous processes. When we make the intervals $C_k$ arbitrarily short, we have
$dS_k>\frac{{dq}_k}{{\hat{T}}_k} \nonumber$
To demonstrate that we can measure the entropy change in the surroundings during a spontaneous process, let us use a conceptual device to transfer the heat, $q_k$, that must be exchanged from the surroundings at temperature, $\hat{T}_k$ to the system. As sketched in Figure 10, we imagine a very small, reversible, ideal-gas Carnot engine, whose high-temperature reservoir is also very small. We suppose that the Carnot engine delivers a very small heat increment $\delta q$ to the high temperature reservoir in every cycle. While the system is within $C_k$, we maintain the Carnot engine’s high temperature reservoir at $\hat{T}_k$, and allow heat $q_k$ to pass from the high temperature reservoir to the system. The high temperature reservoir is the only part of the surroundings that is in thermal contact with the system; $q_k$ is the only heat exchanged by the system while it is within $C_k$.
To maintain the high temperature reservoir at ${\hat{T}}_k$ we operate the Carnot engine for a large integral number of cycles, $n$, such that $q_k\approx n\times \delta q$, and do so at a rate that just matches the rate at which heat passes from the high-temperature reservoir to the system. When the system passes from path-segment $C_k$ to path-segment $C_{k+1}$, we alter the steps in the reversible Carnot cycle to maintain the high-temperature reservoir at the new surroundings temperature, ${\hat{T}}_{k+1}$. The low-temperature heat reservoir for this Carnot engine is always at the constant temperature ${\hat{T}}_{\ell }$. Let the heat delivered from the high-temperature reservoir to the Carnot engine within $C_k$ be ${\left({\hat{q}}_k\right)}_h$. We have $q_k=-{\left({\hat{q}}_k\right)}_h$. Let the heat delivered from the low-temperature reservoir to the Carnot engine within $C_k$ be ${\left({\hat{q}}_k\right)}_{\ell }$. Let the heat delivered to the low-temperature reservoir within $C_k$ be $q_{\ell }$. We have $q_{\ell }=-{\left({\hat{q}}_k\right)}_{\ell }$. Since the Carnot engine is reversible, we have
$\frac{\left(\hat{q}_k\right)_h}{\hat{T}_k}+\frac{\left( \hat{q}_k\right)_{\ell }}{\hat{T}_{\ell }}=0 \nonumber$
and
$-\frac{q_k}{\hat{T}_k}-\frac{q_{\ell }}{\hat{T}_{\ell }}=0 \nonumber$
so that
$\frac{q_{\ell }}{\hat{T}_{\ell }}=-\frac{q_k}{\hat{T}_k} \nonumber$
While the system is within $C_k$, it receives an increment of heat $q_k$ from the high temperature reservoir. Simultaneously, three components in the surroundings also exchange heat. Let the entropy changes in the high-temperature reservoir, the Carnot engine, and the low-temperature reservoir be ${\left(\Delta {\hat{S}}_{HT}\right)}_k$, ${\left(\Delta {\hat{S}}_{CE}\right)}_k$, and ${\left(\Delta {\hat{S}}_{LT}\right)}_k$, respectively. The high temperature reservoir receives heat $q_k$ from the Carnot engine and delivers the same quantity of heat to the system. The net heat accepted by the high temperature reservoir is zero. No change occurs in the high-temperature reservoir. We have ${\left(\Delta {\hat{S}}_{HT}\right)}_k=0$. The reversible Carnot engine completes an integral number of cycles, so that ${\left(\Delta {\hat{S}}_{CE}\right)}_k=0$. The low temperature reservoir accepts heat ${-\left({\hat{q}}_k\right)}_{\ell }=q_{\ell }$, at the fixed temperature ${\hat{T}}_{\ell }$, during the reversible operation of the Carnot engine, so that
$\left(\Delta \hat{S}_{LT}\right)_k=\frac{q_{\ell }}{\hat{T}_{\ell }}=-\frac{q_k}{\hat{T}_k} \nonumber$
The entropy change in the surroundings as the system passes through $C_k$ is
$\Delta \hat{S}_{k}= \left(\Delta \hat{S}_{HT}\right)_k+ \left(\Delta \hat{S}_{CE}\right)_k+\left(\Delta \hat{S}_{LT}\right)_k=\frac{q_{\ell }}{\hat{T}_{\ell }}=-\frac{q_k}{\hat{T}_k} \nonumber$
so that, as we observed above,
$\Delta S_k >- \Delta {\hat{S}}_k =\frac{q_k}{\mathrm{\ }{\hat{T}}_k} \nonumber$
Since $C_k$ can be any part of path C, and $C_k$ can be made arbitrarily short, we have for every increment of any spontaneous process occurring in a closed system that can exchange heat with its surroundings, $d\hat{S}=-{dq}/{\hat{T}}$, and
$dS>\frac{dq}{\hat{T}} \nonumber$
If the temperature of the surroundings is constant between any two points A and B on curve C, we can integrate over this interval to obtain $\mathrm{\Delta }_{\mathrm{AB}}\widehat{\mathrm{S}}\mathrm{=-} \mathrm{q}_{\mathrm{AB}}/\widehat{\mathrm{T}}$ and
$\mathrm{\Delta }_{\mathrm{AB}}\mathrm{S>}\frac{\mathrm{q}_{\mathrm{AB}}}{\widehat{\mathrm{T}}} \nonumber$
For an adiabatic process, $q=0$. For any arbitrarily small increment of an adiabatic process, $dq=0$. It follows that $\Delta S>0$ and $dS>0$ for any spontaneous adiabatic process.
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For every incremental part of any process, we have $dS+d\hat{S}\ge 0$. Let us define a new quantity, the external entropy change, as $d_eS=-d\hat{S}$. The change criteria become $dS-d_eS\ge 0$. Now, let us define the internal entropy ${}^{1}$ change as $d_iS=dS-d_eS$. The entropy change for a system is the sum of its internal and external entropy changes, $dS=d_iS+d_eS$. We use $d_iS$ and $d_eS$ to represent incremental changes. To represent macroscopic changes, we use ${\Delta }_iS$ and ${\Delta }_eS$. Since two processes can effect different changes in the surroundings while the change that occurs in the system is the same, $\Delta \hat{S}$ and ${\Delta }_eS$ are not completely determined by the change in the state of the system. Neither the internal nor the external entropy change depends solely on the change in the state of the system. Nevertheless, we see that $d_iS\ge 0$ or ${\Delta }_iS\ge 0$ is an alternative expression of the thermodynamic criteria.
The external entropy change is that part of the entropy change that results from the interaction between the system and its surroundings. The internal entropy is that part of the entropy change that results from processes occurring entirely within the system. (We also use the term “internal energy.” The fact that the word “internal” appears in both of these terms does not reflect any underlying relationship of material significance.) The criterion $d_iS>0$ makes it explicit that a process is spontaneous if and only if the events occurring within the system act to increase the entropy of the system. In one common figure of speech, we say “entropy is produced” in the system in a spontaneous process. (It is, of course, possible for a spontaneous process to have $d_iS>0$ while $d_eS<0$, and $dS<0$.)
In Section 14.1, we introduce a quantity,
$\sum^{\omega }_{j=1}{{\mu }_j}{dn}_j \nonumber$
that we can think of as a change in the chemical potential energy of a system. The internal entropy change is closely related to this quantity: We find
$d_iS=-\frac{1}{T}\sum^{\omega }_{j=1}{{\mu }_j}{dn}_j \nonumber$
As required by the properties of $d_iS$, we find that $\sum^{\omega }_{j=1}{{\mu }_j}{dn}_j\le 0$ is an expression of the thermodynamic criteria for change. Internal entropy is a useful concept that is applied to particular advantage in the analysis of many different kinds of spontaneous processes in non-homogeneous systems.
9.17: Notation and Terminology- Conventions for Spontaneous Processe
We now want to consider criteria for a spontaneous process in which a closed system passes from state A to state B. State B can be an equilibrium state, but state A is not. We can denote the energy change for this process as \({\Delta }_{AB}E\), and we can find it by measuring the heat and work exchanged with the surroundings as the process takes place, \({\Delta }_{AB}E=q+w\), or for a process in which the increments of heat and work are arbitrarily small, \(d_{AB}E=dq+dw\). Likewise, we can denote the entropy change for the spontaneous process as \({\Delta }_{AB}S\) or \(d_{AB}S\), but we cannot find the entropy change by measuring \(q^{spon}\) or \(dq^{spon}\). If we cannot find the entropy change, we cannot find the Helmholtz or Gibbs free energy changes from their defining relationships, \(A=E-TS\) and \(G=H-TS\). Moreover, intensive variables—pressure, temperature, and concentrations—may not have well-defined values in a spontaneously changing system.
When we say that a reversible process occurs with some thermodynamic variable held constant, we mean what we say: The thermodynamic variable has the same value at every point along the path of reversible change. In the remainder of this chapter, we develop criteria for spontaneous change. These criteria are statements about the values of \(\Delta E\), \(\Delta H\), \(\Delta A\), and \(\Delta G\) for a system that can undergo spontaneous change under particular conditions. In stating some of these criteria, we specify the conditions by saying that the pressure or the temperature is constant. As we develop these criteria, we will see that these stipulations have specific meanings. When we say that a process occurs “at constant volume” (isochorically), we mean that the volume of the system remains the same throughout the process. When we say that a spontaneous process occurs “at constant pressure” (isobarically or isopiestically), we mean that the pressure applied to the system by the surroundings is constant throughout the spontaneous process and that the system pressure is equal to the applied pressure, \(P=P_{applied}\), at all times. When we say that a spontaneous process occurs “at constant temperature”, we may mean only that
1. the system is continuously in thermal contact with its surroundings
2. the temperature of the surroundings is constant
3. in the initial and final states, the system temperature is equal to the surroundings temperature.
9.18: The Heat Exchanged by A Spontaneous Process at Constant Entrop
To continue our effort to find change criteria that use only properties of the system, let us consider a spontaneous process, during which the system is in contact with its surroundings and the entropy of the system is constant. For every incremental part of this process, we have $dS=0$ and $dS+d\hat{S}>0$. Hence, $d\hat{S}>0$. It follows that $\Delta S=0$, $\Delta S+\Delta \hat{S}>0$, and $\Delta \hat{S}>0$. (Earlier, we found that the entropy changes for a spontaneous process in an isolated system are $\Delta S>0$ and $\Delta \hat{S}=0$. The present system is not isolated.) Since the change that occurs in the system is irreversible, $dS=0$ does not mean that $dq=0$. The requirement that $dS=0$ places no constraints on the temperature of the system or of the surroundings at any time before, during, or after the process occurs.
In Section 9.15, we find $dS>dq^{spon}/\hat{T}$ for any spontaneous process in a closed system. If the entropy of the system is constant, we have
${dq}^{spon}<0 \nonumber$ (spontaneous process, constant entropy)
for every incremental part of the process. For any finite change, it follows that the overall heat must satisfy the same inequality:
$q^{spon}<0 \nonumber$ (spontaneous process, constant entropy)
For a spontaneous process that occurs with the system in contact with its surroundings, but in which the entropy of the system is constant, the system must give up heat to the surroundings. $dq<0$ and $q<0$ are criteria for spontaneous change at constant system entropy.
In Section 9.14, we develop criteria for reversible processes. The criteria relate changes in the system’s state functions to the reversible non-pressure–volume work that is done on the system during the process. Now we can develop parallel criteria for spontaneous processes.
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From the fundamental equation, $dE-TdS+PdV=dw^{rev}_{NPV}$ for a reversible process. We find that the criterion for reversible change at constant entropy is ${\left(dE\right)}_S=dw^{rev}_{net}$. For a reversible process at constant entropy and volume, we find ${\left(dE\right)}_{SV}=dw^{rev}_{NPV}$
To consider the energy change for a spontaneous process, we begin with $dE=dq+dw$, which is independent of whether the change is spontaneous or reversible. For a spontaneous process in which both pressure–volume, $dw^{spon}_{PV}$, and non-pressure–volume work, $dw^{spon}_{NPV}$, are possible, we have $dE=dq^{spon}+dw^{spon}_{PV}+dw^{spon}_{NPV}$, which we can rearrange to
$dE-dw^{spon}_{PV}-dw^{spon}_{NPV}=dq^{spon} \nonumber$
For a spontaneous, constant-entropy change that occurs while the system is in contact with its surroundings, we have ${dq}^{spon}<0$. Hence, we have ${\left(dE\right)}_S-{dw}^{spon}_{PV}-{dw}^{spon}_{NPV}<0$. Lettting ${dw}^{spon}_{net}={dw}^{spon}_{PV}+{dw}^{spon}_{NPV}$, we can express this as
${\left(dE\right)}_S<{dw}^{spon}_{net} \nonumber$ (spontaneous process, constant S)
If we introduce the further condition that the spontaneous process occurs while the volume of the system remains constant, we have ${dw}^{spon}_{PV}=0$. Making this substitution and repeating our earlier result for a reversible process, we have the parallel relationships ${\left(dE\right)}_{SV}<{dw}^{spon}_{NPV} \nonumber$ (spontaneous process, constant S and V)
${\left(dE\right)}_{SV}={dw}^{rev}_{NPV} \nonumber$ (reversible process, constant S and V)
If we introduce the still further requirement that only pressure–volume work is possible, we have ${dw}^{spon}_{NPV}=0$. The parallel relationships become
${\left(dE\right)}_{SV}<0 \nonumber$ (spontaneous process, constant S and V, only PV work)
${\left(dE\right)}_{SV}=0 \nonumber$ (reversible process, constant S and V, only PV work)
These equations state the criteria for change under conditions in which the entropy and volume of the system remain constant. If the process is reversible, the energy change must be equal to the non-pressure–volume work. If the process is spontaneous, the energy change must be less than the non-pressure volume work. If only pressure–volume work is possible, the energy of the system must decrease in a spontaneous process and remain constant in a reversible process. Each of these differential-expression criteria applies to every incremental part of a change that falls within its scope. In consequence, corresponding criteria apply to finite spontaneous changes. These criteria are listed in the summary in Section 9.25.
Now the question arises: What sort of system can undergo a change at constant entropy? If the process is reversible and involves no heat, the entropy change will be zero. If we have a system consisting of a collection of solid objects at rest, we can rearrange the objects without transferring heat between the objects and their surroundings. For such a process, the change in the energy of the system is equal to the net work done on the system. Evidently, reversible changes in mechanical systems occur at constant entropy and satisfy the criterion
${\left(dE\right)}_S={dw}^{rev}_{net} \nonumber$
For a change that occurs reversibly and in which the entropy of the system is constant, the energy change is equal to the net work (of all kinds) done on the system. A spontaneous change in a mechanical system dissipates mechanical energy as heat by friction. If this heat appears in the surroundings and the thermal state of the system remains unchanged, such a spontaneous processes satisfies the criterion ${\left(dE\right)}_S<{dw}^{spon}_{net} \nonumber$
We have arrived at the criterion for change that we are accustomed to using when we deal with a change in the potential energy of a constant-temperature mechanical system: A spontaneous change can occur in such a system if and only if the change in the system’s energy is less than the net work done on it. The excess work is degraded to heat that appears in the surroundings. This convergence notwithstanding, the principles of mechanics and those of thermodynamics, while consistent with one another, are substantially independent. We address this issue briefly in Section 12.2.
In the next section, we develop spontaneous-change criteria based on the enthalpy change for a constant-entropy process. In subsequent sections, we consider other constraints and find other criteria. We find that the Helmholtz and Gibbs free energy functions are useful because they provide criteria for spontaneous change when the process is constrained to occur isothermally.
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From $H=E+PV$, we have $dH=dE+d\left(PV\right)$. For a spontaneous process in which both pressure–volume and non-pressure–volume work are possible, we can write this as $dH={dq}^{spon}+{dw}^{spon}_{PV}+{dw}^{spon}_{NPV}+d\left(PV\right)$, which we can rearrange to $dH-{dw}^{spon}_{PV}-{dw}^{spon}_{NPV}-d\left(PV\right)={dq}^{spon}$. For a spontaneous constant-entropy change that occurs while the system is in contact with its surroundings, we have ${dq}^{spon}$$<0$, so that ${\left(dH\right)}_S-{dw}^{spon}_{PV}-{dw}^{spon}_{NPV}-d\left(PV\right)<0. \nonumber$
Now, let us introduce the additional constraint that the system is subjected to a constant applied pressure, $P_{applied}$, throughout the process. Thus $P_{applied}$ is a well-defined property that can be measured at any stage of the process. The incremental pressure–volume work done by the surroundings on the system is ${dw}^{spon}_{PV}=-P_{applied}dV$. In principle, the system can undergo spontaneous change so rapidly that there can be a transitory difference between the system pressure and the applied pressure. In practice, pressure adjustments occur very rapidly. Except in extreme cases, we find that $P=P_{applied}$ is a good approximation at all times. Then the change in the pressure volume product is $d\left(PV\right)=P_{applied}dV$. Making these substitutions, the enthalpy inequality becomes
${\left(dH\right)}_{SP}<{dw}^{spon}_{NPV} \nonumber$ (spontaneous process, constant S and $P_{applied}$)
From our earlier discussion of reversible processes, we have the parallel relationship
${\left(dH\right)}_{SP}={dw}^{rev}_{NPV} \nonumber$ (any reversible process, constant S and $P_{applied}$)
If we introduce the still further requirement that only pressure–volume work is possible, we have ${dw}_{NPV}=0$. The parallel relationships become
${\left(dH\right)}_{SP}<0 \nonumber$ (spontaneous process, constant S and P, only PV work)
${\left(dH\right)}_{SP}=0 \nonumber$ (reversible process, constant S and P, only PV work)
These equations state the criteria for change under conditions in which the entropy and pressure of the system remain constant. If the process is reversible, the enthalpy change must be equal to the non-pressure–volume work. If the process is spontaneous, the enthalpy change must be less than the non-pressure–volume work. If only pressure–volume work is possible, the enthalpy of the system must decrease in a spontaneous process and remain constant in a reversible process. Since each of these differential criteria applies to every incremental part of a reversible change that falls within its scope, corresponding criteria apply to finite spontaneous changes. These criteria are listed in the summary in Section 9.25.
9.21: The Entropy Change for A Spontaneous Process at Constant E and
For any spontaneous process, we have $dE={dq}^{spon}$+${dw}^{spon}$, which we can rearrange to ${dq}^{spon}=dE-{dw}^{spon}$. Substituting our result from Section 9.15, we have
$\hat{T}dS>dE-dw^{spon} \nonumber$ (spontaneous process)
If the energy of the system is constant throughout the process, we have $dE=0$ and
$\hat{T}{\left(dS\right)}_E>-dw^{spon} \nonumber$ (spontaneous process, constant energy)
The spontaneous work is the sum of the pressure–volume work and the non-pressure–volume work, $\ dw^{spon}={dw}^{spon}_{PV}+{dw}^{spon}_{NPV}$. If we introduce the further condition that the spontaneous process occurs while the volume of the system remains constant, we have ${dw}^{spon}_{PV}=0$. Making this substitution and repeating our earlier result for a reversible process, we have the parallel relationships
${\left(dS\right)}_{EV}>\frac{-dw^{spon}_{NPV}}{\hat{T}} \nonumber$ (spontaneous process, constant $E$ and $V$)
${\left(dS\right)}_{EV}=\frac{-dw^{spon}_{NPV}}{\hat{T}} \nonumber$ (reversible process, constant $E$ and $V$)
(For a reversible process, $T=\hat{T}$.) If the spontaneous process occurs while $\hat{T}$ is constant, summing the incremental contributions to a finite change of state produces the parallel relationships
${\left(\Delta S\right)}_{EV}>\frac{-w^{spon}_{NPV}}{\hat{T}} \nonumber$ (spontaneous process, constant $E$, $V$, and $\hat{T}$)
${\left(\Delta S\right)}_{EV}=\frac{-w^{spon}_{NPV}}{\hat{T}} \nonumber$ (reversible process, constant $E$, $V$, and $\hat{T}$)
Constant $\hat{T}$ corresponds to the common situation in chemical experimentation in which we place a reaction vessel in a constant-temperature bath. If we introduce the further condition that only pressure–volume work is possible, we have $dw^{spon}_{NPV}=0$. The parallel relationships become
${\left(dS\right)}_{EV}>0 \nonumber$ (spontaneous process, constant $E$ and $V$, only $PV$ work)
${\left(dS\right)}_{EV}=0 \nonumber$ (reversible process, constant $E$ and $V$, only$\ PV$ work)
If the energy and volume are constant for a system in which only pressure–volume work is possible, the system is isolated. The conditions we have just derived are entirely equivalent to our earlier conclusions that $dS=0$ and $dS>0$ for an isolated system that is at equilibrium or undergoing a spontaneous change, respectively. Summing the incremental contributions to a finite change of state produces the parallel relationships
${\left(\Delta S\right)}_{EV}>0 \nonumber$ (spontaneous process, only $PV$ work)
${\left(\Delta S\right)}_{EV}=0 \nonumber$ (reversible process, only $PV$ work)
The validity of these expressions is independent of any variation in either $T$ or $\hat{T}$.
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For any spontaneous process, we have
$dH=dE+PdV+VdP=dq^{spon}-P_{applied}dV+dw^{spon}_{NPV}+PdV+VdP \nonumber$
If the pressure is constant ($P=P_{applied}=\mathrm{constant}$), this becomes $dq^{spon}=dH-dw^{spon}_{NPV}$. Substituting our result from Section 9.15, we have
$\hat{T}{\left(dS\right)}_P>dH-dw^{spon}_{NPV} \nonumber$ (spontaneous process, constant $P$)
If the enthalpy of the system is also constant throughout the process, we have
$\hat{T}{\left(dS\right)}_{HP}>-dw^{spon}_{NPV} \nonumber$ (spontaneous process, constant $H$ and $P$)
Dividing by $\hat{T}$ and repeating our earlier result for a reversible process, we have the parallel relationships
${\left(dS\right)}_{HP}>\frac{-dw^{spon}_{NPV}}{\hat{T}} \nonumber$ (spontaneous process, constant $H$ and $P$) ${\left(dS\right)}_{HP}=\frac{-dw^{rev}_{NPV}}{\hat{T}} \nonumber$ (reversible process, constant $H$ and $P$)
If it is also true that the temperature of the surroundings is constant, summing the incremental contributions to a finite change of state produces the parallel relationships
${\left(\Delta S\right)}_{HP}>\frac{-w^{spon}_{NPV}}{\hat{T}} \nonumber$ (spontaneous process, constant $H$, $P$, and $\hat{T}$)
${\left(\Delta S\right)}_{HP}>\frac{-w^{rev}_{NPV}}{\hat{T}} \nonumber$ (reversible process, constant $H$, $P$, and $\hat{T}=T$)
If only pressure–volume work is possible, we have $dw^{spon}_{NPV}=0$, and
${\left(dS\right)}_{HP}>0 \nonumber$ (spontaneous process, constant $H$, $P$, only $PV$ work)
${\left(dS\right)}_{HP}=0 \nonumber$ (reversible process, constant $H$ and $P$, only $PV$ work)
and for a finite change of state,
${\left(\Delta S\right)}_{HP}>0 \nonumber$ (spontaneous process, only $PV$ work)
${\left(\Delta S\right)}_{HP}=0 \nonumber$ (reversible process, only $PV$ work)
In this and earlier sections, we develop criteria for spontaneous change that are based on $dE$ and $dH$. We are now able to develop similar criteria for a spontaneous change in a system that is in thermal contact with constant-temperature surroundings. These criteria are based on $dA$ and $dG$. However, before doing so, we develop a general relationship between the isothermal work in a spontaneous process and the isothermal work in a reversible process, when these processes take a system from a common initial state to a common final state.
9.23: The Reversible Work is the Minimum Work at Constant T
The Clausius inequality leads to an important constraint on the work that can be done on a system during a spontaneous process in which the temperature of the surroundings is constant. As we discuss in Section 9.7, the initial state of the spontaneous process cannot be a true equilibrium state. In our present considerations, we assume that the initial values of all the state functions of the spontaneously changing system are the same as those of a true equilibrium system. Likewise, we assume that the final state of the spontaneously changing system is either a true equilibrium state or a state whose thermodynamic functions have the same values as those of a true equilibrium system.
From the first law applied to any spontaneous process in a closed system, we have ${\Delta E}^{rev}={\Delta E}^{spon}$ and $q^{rev}+w^{rev}=q^{spon}+w^{spon}$. Since the temperature of the system and its surroundings are equal and constant for the reversible process, we have $q^{rev}=T\Delta S=\hat{T}\Delta S$. So long as the temperature of the surroundings is constant, we have $q^{spon}<\hat{T}\Delta S$ for the spontaneous process. It follows that
$\hat{T}\Delta S+w^{rev}-w^{spon}=q^{spon}<\hat{T}\Delta S \nonumber$
so that $w^{rev}<w^{spon} \nonumber$ ($\hat{T}$ constant)
A given isothermal process does the minimum possible amount of work on the system when it is carried out reversibly. (In Section 7.20, we find this result for the special case in which the only work is the exchange of pressure–volume work between an ideal gas and its surroundings.) Equivalently, a given isothermal process produces the maximum amount of work in the surroundings when it is carried out reversibly: Since $w^{rev}=-{\hat{w}}^{rev}$ and $w^{spon}=-{\hat{w}}^{spon}$, we have $-{\hat{w}}^{rev}<-{\hat{w}}^{spon}$ or
${\hat{w}}^{rev}>{\hat{w}}^{spon} \nonumber$
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Now let us consider the change in the Helmholtz free energy when a system undergoes a spontaneous change while in thermal contact with surroundings whose temperature remains constant at $\hat{T}$. We begin by considering an arbitrarily small increment of change in a process in which the temperature of the system remains constant at $T=\hat{T}$. The change in the Helmholtz free energy for this process is ${\left(dA\right)}_T=dE-TdS$. Substituting $dE=dq^{spon}+dw^{spon}$ gives
${\left(dA\right)}_T=dq^{spon}+dw^{spon}-TdS \nonumber$ (spontaneous process, constant $T$)
Rearranging, we have ${\left(dA\right)}_T-dw^{spon}+TdS=dq^{spon}$. Using the inequality $dq^{spon}<\hat{T}dS$, we have
${\left(dA\right)}_T-dw^{spon}+TdS<\hat{T}dS \nonumber$
When we stipulate that $T=\hat{T}=\mathrm{constant}$, this becomes
${\left(dA\right)}_T<dw^{spon} \nonumber$ (spontaneous process, constant $T$)
where $dw^{spon}$ is all of the work of any kind done on the system during a small increment of the spontaneous process. If we introduce the still further requirement that the volume is constant, we have $dw^{spon}_{PV}=0$ and $dw^{spon}=dw^{spon}_{NPV}$. Then
${\left(dA\right)}_{TV}<dw^{spon}_{npv} \nonumber$ (spontaneous process, constant $T$ and $V$)
and if only pressure–volume work is possible,
${\left(dA\right)}_{TV}<0 \nonumber$ (spontaneous process, constant $T$ and $V$, only $PV$ work)
From our earlier discussion of reversible processes, we have the parallel relationships
${\left(dA\right)}_T=dw^{rev}_{net} \nonumber$ (reversible isothermal process)
${\left(dA\right)}_{TV}=dw^{rev}_{NPV} \nonumber$ (reversible process at constant $T$ and $V$)
${\left(dA\right)}_{TV}=0 \nonumber$ (reversible process at constant $T$ and $V$, only $PV$ work)
Similarly, under these conditions, the change in the Gibbs free energy for a spontaneous isothermal process is
\begin{aligned} \left(dG\right)_T & =dH-TdS \ ~& =dE+d\left(PV\right)-TdS \ ~ & =dq^{spon}+dw^{spon}_{PV}+dw^{spon}_{NPV}+d\left(PV\right)-TdS \end{aligned} \nonumber
Rearranging, we have
${\left(dG\right)}_T-dw^{spon}_{PV}-dw^{spon}_{NPV}-d\left(PV\right)+TdS ={dq}^{spon}<\hat{T}dS \nonumber$
and since $T=\hat{T}=\mathrm{constant}$,
${\left(dG\right)}_T<dw^{spon}_{pv}+dw^{spon}_{npv}+d\left(pv\right) \nonumber$ (spontaneous process, constant $T$)
As we did when considering the enthalpy change for a spontaneous process, we introduce the additional constraints that the system is subjected to a constant applied pressure, $P_{applied}$, and that ${P=P}_{applied}$ throughout the process. The irreversible pressure–volume work done by the surroundings on the system becomes $dw^{spon}_{PV}={-P}_{applied}dV$, and the change in the pressure volume product becomes ${d\left(PV\right)=P}_{applied}dV$. The Gibbs free energy inequality becomes
${\left(dG\right)}_{TP}<dw^{spon}_{npv} \nonumber$ (spontaneous process, constant $P_{applied}$ and $T$)
If only pressure–volume work is possible, this becomes
${\left(dG\right)}_{TP}<0 \nonumber$ (spontaneous process, constant $P_{applied}$ and $T$, only $PV$ work)
From our earlier discussion of reversible processes, we have the parallel relationships
${\left(dG\right)}_{TP}=dw^{rev}_{NPV} \nonumber$ (reversible process, constant $P$ and $T$) ${\left(dG\right)}_{TP}=0 \nonumber$ (reversible process, constant $P$ and $T$,
only $PV$ work)
Since each of these differential-expression criteria applies to every incremental part of a reversible change that falls within its scope, we have the following criteria for finite spontaneous changes when the temperature of the system is constant:
${\left(\Delta A\right)}_T<w^{spon} \nonumber$ (spontaneous process, constant $T$)
${\left(\Delta A\right)}_{TV}<w^{spon}_{npv} \nonumber$ (spontaneous process, constant $T$ and $V$)
${\left(\Delta A\right)}_{TV}<0 \nonumber$ (spontaneous process, constant $T$ and $V$, only $PV$ work)
${\left(\Delta G\right)}_{TP}<w^{spon}_{npv} \nonumber$ (spontaneous process, constant $P_{applied}$ and $T$)
${\left(\Delta G\right)}_{TP}=0 \nonumber$ (spontaneous process, constant $P_{applied}$ and $T$, only $PV$ work)
While the development we have just made assumes that the system temperature is strictly constant, the validity of these finite-change inequalities is not restricted to the condition of strictly constant system temperature. We can derive these finite-change inequalities by essentially the same argument from less restrictive conditions.
Let us consider a spontaneous process in which a system goes from state B to state C while in contact with surroundings whose temperature remains constant at $\hat{T}$. We suppose that in both state B and state C the system temperature is equal to the surroundings temperature; that is, $T_B=T_C=\hat{T}=\mathrm{constant}$. However, at any intermediate point in the process, the system can have any temperature whatsoever. In states B and C, the Helmholtz free energies are $A_B=E_B-\hat{T}S_B$ and $A_C=E_C-\hat{T}S_C$. The change in the Helmholtz free energy is $\left(A_C-A_B\right)=\left(E_C{-E}_B\right)-\hat{T}\left(S_C-S_B\right)$ or ${\left(\Delta A\right)}_{\hat{T}}=\Delta E-\hat{T}\Delta S=q^{spon}+w^{spon}-\hat{T}\Delta S$. Rearranging, and using $q^{spon}<\hat{T}\Delta S$, we have
${\left(\Delta A\right)}_{\hat{T}}-w^{spon}+\hat{T}\Delta S=q^{spon}<\hat{T}\Delta S$, so that
${\left(\Delta A\right)}_{\hat{T}}<w^{spon} \nonumber$ (spontaneous process, constant $\hat{T}$)
If we require further that the system volume remain constant, there is no pressure–volume work, and we have
${\left(\Delta A\right)}_{V\hat{T}}<w^{spon}_{npv} \nonumber$ (spontaneous process, constant $\hat{T}$ and $V$)
If only pressure–volume work is possible, $w^{spon}_{NPV}=0$, and ${\left(\Delta A\right)}_{V\hat{T}}<0 \nonumber$ (spontaneous process, constant $\hat{T}$ and $V$, only $PV$ work)
Under the same temperature assumptions, and assuming that $P_B=P_C=P_{applied}=\mathrm{constant}$, the Gibbs free energies are $G_B=E_B+P_{applied}V_B-\hat{T}S_B$ and $G_c=E_C+P_{applied}V_C-\hat{T}S_C$. So that $\left(G_C-G_B\right)=\left(E_C{-E}_B\right)+P_{applied}\left(V_C-V_B\right)-\hat{T}\left(S_C-S_B\right)$ or
\begin{aligned} {\left(\Delta G\right)}_{P\hat{T}} & =\Delta E+P_{applied}\Delta V-\hat{T}\Delta S \ ~ & =q^{spon}+w^{spon}_{PV}+w^{spon}_{NPV}+P_{applied}\Delta V-\hat{T}\Delta S \end{aligned} \nonumber
The pressure–volume work is $w^{spon}_{PV}=-P_{applied}\Delta V$. Cancelling and rearranging, we have
${\left(\Delta G\right)}_{P\hat{T}}-w^{spon}_{NPV}+\hat{T}\Delta S=q^{spon}<\hat{T}\Delta S \nonumber$
and
${\left(\Delta G\right)}_{P\hat{T}}<w^{spon}_{npv} \nonumber$ (spontaneous process, constant $\hat{T}$ and $P$)
If only pressure–volume work is possible,
${\left(\Delta G\right)}_{P\hat{T}}<0 \nonumber$ (spontaneous process, constant $\hat{T}$ and $P$, only $PV$ work)
We find ${\left(\Delta G\right)}_{P\hat{T}}<w^{spon}_{npv}$> for any spontaneous process that occurs at constant pressure, while the system is in contact with surroundings at the constant temperature $\hat{T}$, and in which the initial and final system temperatures are equal to $\hat{T}$. These are the most common conditions for carrying out a chemical reaction. Consider the situation after we mix non-volatile reactants in an open vessel in a constant-temperature bath. We suppose that the initial temperature of the mixture is the same as that of the bath. The atmosphere applies a constant pressure to the system. The reaction is an irreversible process. It proceeds spontaneously until its equilibrium position is reached. Until equilibrium is reached, the reaction cannot be reversed by an arbitrarily small change in the applied pressure or the temperature of the surroundings. ${\left(\Delta G\right)}_{P\hat{T}}<w^{spon}_{npv}$ and ${\left(\Delta G\right)}_{P\hat{T}}<0$ are criteria for spontaneous change that apply to this situation whatever the temperature of the system might be during any intermediate part of the process.
9.25: Summary- Thermodynamic Functions as Criteria for Change
For a spontaneous process, we conclude that the entropy change of the system must satisfy the inequality $\Delta S+\Delta \hat{S}>$$0$. For any process that occurs reversibly, we conclude that $\Delta S+\Delta \hat{S}=0$. For every incremental part of a reversible process that occurs in a closed system, we have the following relationships: $dE=TdS-PdV+dw^{rev}_{NPV} \nonumber$ $dH=TdS+VdP+dw^{rev}_{NPV} \nonumber$ $dA=-SdT-PdV+dw^{rev}_{NPV} \nonumber$ $dG=-SdT+VdP+dw^{rev}_{NPV} \nonumber$
At constant entropy, the energy relationship becomes:
${\left(dE\right)}_S=dw^{rev}_{net} \nonumber$ ${\left(\Delta E\right)}_S=w^{rev}_{net} \nonumber$
At constant temperature, the Helmholtz free energy relationship becomes:
${\left(dA\right)}_T=dw^{rev}_{net} \nonumber$ ${\left(\Delta A\right)}_T=w^{rev}_{net} \nonumber$
For reversible processes in which all work is pressure–volume work:
$dE=TdS-PdV \nonumber$ $dH=TdS+VdP \nonumber$ $dA=-SdT-PdV \nonumber$ $dG=-SdT+VdP \nonumber$
From these general equations, we find the following relationships for reversible processes when various pairs of variables are held constant:
${\left(dS\right)}_{EV}={-dw^{rev}_{NPV}}/{T} {\left(\Delta S\right)}_{EV}={-w^{rev}_{NPV}}/{T} \nonumber$ ${\left(dS\right)}_{HP}={-dw^{rev}_{NPV}}/{T} {\left(\Delta S\right)}_{HP}={-w^{rev}_{NPV}}/{T} \nonumber$ ${\left(dE\right)}_{SV}=dw^{rev}_{NPV} {\left(\Delta E\right)}_{SV}=w^{rev}_{NPV} \nonumber$ ${\left(dH\right)}_{SP}=dw^{rev}_{NPV} {\left(\Delta H\right)}_{SP}=w^{rev}_{NPV} \nonumber$ ${\left(dA\right)}_{TV}=dw^{rev}_{NPV} {\left(\Delta A\right)}_{TV}=w^{rev}_{NPV} \nonumber$ ${\left(dG\right)}_{TP}=dw^{rev}_{NPV} {\left(\Delta G\right)}_{TP}=w^{rev}_{NPV} \nonumber$
If the only work is pressure–volume work, then $dw^{rev}_{NPV}=0$, $w^{rev}_{NPV}=0$, and these relationships become:
${\left(dS\right)}_{EV}=0 {\left(\Delta S\right)}_{EV}=0 \nonumber$ ${\left(dS\right)}_{HP}=0 {\left(\Delta S\right)}_{HP}=0 \nonumber$ ${\left(dE\right)}_{SV}=0 {\left(\Delta E\right)}_{SV}=0 \nonumber$ ${\left(dH\right)}_{SP}=0 {\left(\Delta H\right)}_{SP}=0 \nonumber$ ${\left(dA\right)}_{TV}=0 {\left(\Delta A\right)}_{TV}=0 \nonumber$ ${\left(dG\right)}_{TP}=0 {\left(\Delta G\right)}_{TP}=0 \nonumber$
For every incremental part of an irreversible process that occurs in a closed system at constant entropy:
${dq}^{spon}<0 \nonumber$
and
${\left(dE\right)}_S<{dw}^{spon}_{net} \nonumber$
and
$q^{spon}<0 \nonumber$
and
${\left(\Delta E\right)}_S<w^{spon}_{net} \nonumber$
For an irreversible process at constant temperature:
${dq}^{spon}<\hat{T}dS \nonumber$
and
${\left(dA\right)}_{\hat{T}}<{dw}^{spon}_{net} \nonumber$
and
$q^{spon}<\hat{T}\Delta S \nonumber$
and
${\left(\Delta A\right)}_{\hat{T}}<w^{spon}_{net} \nonumber$
When an irreversible process occurs with various pairs of variables held constant, we find:
${\left(dS\right)}_{EV}>{-dw^{spon}_{NPV}}/{\hat{T}} {\left(\Delta S\right)}_{EV}={-w^{spon}_{NPV}}/{\hat{T}} \nonumber$
${\left(dS\right)}_{HP}>{-dw^{spon}_{NPV}}/{\hat{T}} {\left(\Delta S\right)}_{HP}>{-w^{spon}_{NPV}}/{\hat{T}} \nonumber$
${\left(dE\right)}_{SV} \nonumber$
${\left(dH\right)}_{SP} \nonumber$
${\left(dA\right)}_{\hat{T}V} \nonumber$
${\left(dG\right)}_{\hat{T}P} \nonumber$
For irreversible processes in which the only work is pressure–volume work, these inequalities become:
${\left(dS\right)}_{EV}>0 {\left(\Delta S\right)}_{EV}>0 \nonumber$ ${\left(dS\right)}_{HP}>0 {\left(\Delta S\right)}_{HP}>0 \nonumber$ ${\left(dE\right)}_{SV}<0 {\left(\Delta E\right)}_{SV}<0 \nonumber$ ${\left(dH\right)}_{SP}<0 {\left(\Delta H\right)}_{SP}<0 \nonumber$ ${\left(dA\right)}_{\hat{T}V}<0 {\left(\Delta A\right)}_{\hat{T}V}<0 \nonumber$ ${\left(dG\right)}_{\hat{T}P}<0 {\left(\Delta G\right)}_{\hat{T}P}<0 \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/09%3A_The_Second_Law_-_Entropy_and_Spontaneous_Change/9.24%3A_The_Free_Energy_Changes_for_A_Spontaneous_Process_at_Constant_.txt
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Problems
1. Does a perpetual motion machine of the second kind violate the principle of conservation of energy?
2. What is the contrapositive of $\left(\mathrm{SL\ and}\ \sim \mathrm{MSL}\right)\Rightarrow \left(\Delta \hat{S}<0\right)$? It is a theorem of logic that $\sim (\mathrm{B\ and\ C})\Rightarrow (\sim B{\mathrm{and}}/{\mathrm{or}}\sim \mathrm{C})$. Interpret this theorem. Given that SL is true and that $\sim \left(\mathrm{SL\ and}\ \sim \mathrm{MSL}\right)$ is true, prove that $\sim \mathrm{MSL}$ is true.
3. Max Planck introduced the following statement of the second law:
“It is impossible to construct an engine which will work in a complete cycle, and produce no effect except the raising of a weight and the cooling of a heat-reservoir.”
(M. Planck, Treatise on Thermodynamics, 3rd Edition, translated from the seventh German Edition, Dover Publications, Inc., p 89.) Since we take “raising a weight” to be equivalent to “produces work in the surroundings,” the Planck statement differs from our machine-based statement only in that it allows the temperature of the heat source to decrease as the production of work proceeds. We can now ask whether this difference has any material consequences. In particular, can we prove that the Planck statement implies our machine-based statement, or vice versa? (Suggestion: Suppose that we have identical Planck-type machines, each with its own heat reservoir. We dissipate by friction the work produced by one machine in the heat reservoir of the other.)
4. Our statements of the first and second laws have a common format: Assertion that a state function exists; operational definition by which the state function can be measured; statement of a property exhibited by this state function. Express the zero-th law of thermodynamics (Chapter 1) in this format.
5. A 0.400 mol sample of $N_2$ is compressed from 5.00 L to 2.00 L, while the temperature is maintained constant at 350 K. Assume that $N_2$ is an ideal gas. Calculate the change in the Helmholtz free energy, $\Delta A$.
6. Show that $\Delta G=\Delta A$ when an ideal gas undergoes a change at constant temperature.
7. Calculate $\Delta E$, $\Delta H$, and $\Delta G$ for the process in problem 5.
8. A sample of 0.200 mol of an ideal gas, initially at 5.00 bar, expands reversibly and isothermally from 1.00 L to 10.00 L. Calculate $\Delta E$, $\Delta H$, and $\Delta G$ for this process.
9. A 100.0 g sample of carbon tetrachloride is compressed from 1.00 bar to 10.00 bar at a constant temperature of 20 C. At 20 C, carbon tetrachloride is a liquid whose density is $1.5940\ \mathrm{g}\ \mathrm{m}{\mathrm{L}}^{-1}$. Assume that the density does not change significantly with pressure. What is $\Delta G$ for this process?
10. Calculate the Helmholtz free energy change ($\Delta A$) in problem 9.
11. If $C_V$ is constant, show that the initial and final temperatures and volumes for an adiabatic ideal-gas expansion are related by the equation $\left(\frac{T_f}{T_i}\right)=\left(\frac{V_i}{V_f}\right)^{R/C_V} \nonumber$
12. At 25 C, the initial volume of a monatomic ideal gas is 5 L at 10 bar. This gas expands to 20 L against a constant applied pressure of 1 bar.
(a) Is this process impossible, spontaneous, or reversible?
(b) What is the final temperature?
(c) Find $q$, $w$, $\Delta E$, and $\Delta H$ for this process.
13. The same change of state experience by the monatomic ideal gas in problem 12 can be effected in two steps. Let step A be the reversible cooling of the gas to its final temperature while the pressure is maintained constant at 10 bar. Let step B be the reversible isothermal expansion of the resulting gas to a pressure of 1 bar.
(a) Find $q$, $w$, $\Delta E$, and $\Delta H$ for step A.
(b) Find $q$, $w$, $\Delta E$, and $\Delta H$ for step B.
(c) From your results in (a) and (b), find$\ q$, $w$, $\Delta E$, and $\Delta H$ for the overall process of step A followed by step B.
(d) Compare the values of $q$, $w$, $\Delta E$, and $\Delta H$ that you find in (c) to the values for the same overall process that you found in problem 12.
(e) Find $\Delta S$ and $\Delta \hat{S}$ for step A.
(f) Find $\Delta S$ and $\Delta \hat{S}$ for step B.
(g) Find $\Delta S$, $\Delta \hat{S}$, and $\Delta S_{universe}$ for the overall process.
14. Assume that the process in problem 12 occurs while the gas is in thermal contact with its surroundings and that the temperature of the surroundings is always equal to the final temperature of the gas. Find $\Delta \hat{S}$ and $\Delta S_{universe}$ for this process.
15. At 25 C, the initial volume of a monatomic ideal gas is 5 L at 10 bar. The gas expands to 20 L while in thermal contact with surroundings at 125 C. During the expansion, the applied pressure is constant and equal to the equilibrium pressure at the final volume and temperature.
(a) Is this process impossible, spontaneous, or reversible?
(b) Find $q$, $w$, $\Delta E$, $\Delta H$, and $\Delta \hat{S}$ for this process.
(c) Find $\Delta S$ and $\Delta S_{universe}$ for this process. To find $\Delta S$, it is necessary to find a reversible alternative path that effects the same change in the system’s state functions.
16. At 60 C, the density of water is $\mathrm{0.98320\ g}\ {\mathrm{cm}}^{-3}$, the vapor pressure is 19,932 Pa, and the enthalpy of vaporization is $42,482\mathrm{\ J}\ {\mathrm{mol}}^{-1}$. Assume that gaseous water behaves as an ideal gas. A vessel containing liquid and gaseous water is placed in a constant 60 C bath, and the applied pressure is maintained at 19,932 Pa while 100 g of water vaporizes.
(a) Is this process impossible, spontaneous, or reversible?
(b) Find $q$, $w$, $\Delta E$, $\Delta H$, $\Delta S$, $\Delta A$, and $\Delta G$ for this process.
(c) Is ${\left(\Delta S\right)}_{EV}=0$ a criterion for equilibrium that applies to this system? Why or why not? ${\left(\Delta H\right)}_{SP}=0$? ${\left(\Delta A\right)}_{VT}=0$? ${\left(\Delta G\right)}_{PT}=0$?
17. This problem compares the efficiency and $\sum{q/T}$ for one mole of a monatomic ideal gas taken around a reversible Carnot cycle to the same quantities for the same gas taken around an irreversible cycle using the same two heat reservoirs.
(i) Let the successive step of the reversible Carnot cycle be a, b, c, and d. Isothermal step a begins with the gas occupying 5.00 L at 600 K and ends with the gas occupying 20.00 L. Adiabatic expansion step b ends with the gas at 300 K. After the isothermal compression step c, the gas is adiabatically compressed in step d to the original state. Find $P$, $V$, and $T$ for the gas at the end of each step of this reversible cycle. Find $\sum q/\hat{T}$, $\Delta S$, and $\Delta \hat{S}$ for the cycle a, b, c, d. What is the efficiency of this cycle?
(ii) Suppose that following step b, the ideal gas is warmed at constant volume to 400 K by exchanging heat with the 600 K reservoir. Call this step e. Following step e, the gas is cooled at constant pressure to 300 K by contact with the 300 K reservoir. Call this step f. Following step f, the gas is isothermally and reversibly compressed at 300 K to the same $P$, $V$, and $T$ as the gas reaches at the end of step c. Call this step g. Find $P$, $V$, and $T$ for the gas at the ends of steps e, f, and g. Although steps e and f are not reversible, the same changes can be effected reversibly by keeping $T=\hat{T}$ as the gas is warmed at constant volume (step e) and cooled at constant pressure (step f). (We discuss this point in further in Section 12.4.) Consequently, ${\Delta }_eS=\int^{400\ K}_{300\ K} \frac{C_V}{T}dT$ and ${\Delta }_fS=\int^{400\ K}_{300\ K} \frac{C_P}{T}dT$. Find $\sum q/\hat{T}$, $\Delta S$, and $\Delta \hat{S}$, and $\Delta S_{universe}$ for the cycle a, b, e, f, g, d. What is the efficiency of this cycle?
(iii) Compare the value of $\sum q/ \hat{T}$ that you obtained in part (ii) to value of $\sum q/ \hat{T}$ that you obtained in part (i).
(iv) Clausius’ theorem states that $\sum q/ \hat{T}=0$ for a cycle traversed reversibly, and $\sum q/ \hat{T}<0$ for a cycle traversed spontaneously. Comment.
18. For a spontaneous cycle traversed while the temperature changes continuously, Clausius’ theorem asserts that $\oint{dq/\hat{T}}<0$. Show that this inequality follows from the result, $dS>dq/ \hat{T}$, that we obtained in Section 9.15 for any spontaneous process in a closed system.
19. In Sections 9.6 through 9.8, we conclude that $\Delta S+\Delta \hat{S}=0$ is necessary for a reversible process, $\Delta S+\Delta \hat{S}>0$ is necessary for a spontaneous process, and $\Delta S+\Delta \hat{S}<0$ is necessary for an impossible process. That is: $(\mathrm{Process\ is\ reversible})\ \ \ \Rightarrow \ \left(\Delta S+\Delta \hat{S}=0\right) \nonumber$ ($\mathrm{Process\ is\ spontaneous}$)$\ \ \Rightarrow \left(\Delta S+\Delta \hat{S}>0\right)$, and ($\mathrm{Process\ is\ impossible}$) $\ \ \Rightarrow \left(\Delta S+\Delta \hat{S}<0\right)$.
Since we have defined the categories reversible, spontaneous, and impossible so that they are exhaustive and mutually exclusive, the following proposition is true:
$\sim \left(\mathrm{Process\ is\ spontaneous}\right)\ \mathrm{and}\ \sim \left(\mathrm{Process\ is\ impossible}\right) \nonumber$ $\Rightarrow \left(\mathrm{Process\ is\ reversible}\right) \nonumber$
(a) Prove that $\Delta S+\Delta \hat{S}=0$ is sufficient for the process to be reversible; that is, prove: $\left(\Delta S+\Delta \hat{S}=0\right)\ \ \Rightarrow \ \ \left(\mathrm{Process\ is\ reversible}\right) \nonumber$
(b) Prove that $\Delta S+\Delta \hat{S}>0$ is sufficient for the process to be spontaneous; that is, prove: $\left(\Delta S+\Delta \hat{S}>0\ \right)\ \Rightarrow \ \ \left(\mathrm{Process\ is\ spontaneous}\right) \nonumber$
(c) Prove that $\Delta S+\Delta \hat{S}<0$ is sufficient for the process to be impossible; that is, prove: $\left(\Delta S+\Delta \hat{S}<0\right)\ \ \Rightarrow \ \ \left(\mathrm{Process\ is\ impossible}\right) \nonumber$
20. Label the successive steps in a reversible Carnot cycle A, B, C, and D, where A is the point at which the pressure is greatest.
(a) Sketch the path ABCD in $P$–$V$ space.
(b) Sketch the path ABCD in $T$–$dq^{rev}/T$ space.
(c) Sketch the path ABCE in $T$– $q^{rev}$ space.
(d) Sketch the path BCDA in $T$– $q^{rev}$ space.
(e) Sketch the path CDAB in $T$– $q^{rev}$ space.
(d) Sketch the path DABC in $T$– $q^{rev}$ space.
21. Assume that the earth’s atmosphere is pure nitrogen and that it behaves as an ideal gas. Assume that the molar energy of this nitrogen is constant and that its molar entropy changes are adequately modeled by $d\overline{S}=\left({C_V}/{T}\right)dT+\left({R}/{\overline{V}}\right)d\overline{V}$. For this atmosphere, show that ${\left(\frac{\partial T}{\partial h}\right)}_E=\frac{-\overline{M}g}{C_V} \nonumber$ where $h$ is the height above the earth’s surface, $\overline{M}$ is the molar mass of dinitrogen ($0.0280\ \ \mathrm{kg}\ {\mathrm{mol}}^{-1}$), $g$ is the acceleration due to gravity ($\mathrm{9.80}\ \mathrm{m}\ {\mathrm{s}}^{-1}$), and $C_V$ is the constant-volume heat capacity ($20.8\ \mathrm{J}\ {\mathrm{K}}^{-1}\ {\mathrm{mol}}^{-1}$). [Suggestion: Write the total differential for $\overline{E}=\overline{E}\left(\overline{S},\overline{V},h\right)$. What are ${\left({\partial \overline{E}}/{\partial \overline{S}}\right)}_{\overline{V},h}$, ${\left({\partial \overline{E}}/{\partial \overline{V}}\right)}_{\overline{S},h}$, and ${\left({\partial \overline{E}}/{\partial h}\right)}_{\overline{S},\overline{V}}$?]
If the temperature at sea level is 300 K, what is the temperature on the top of a 3000 m mountain?
22. Assume that the earth’s atmosphere is pure nitrogen and that it behaves as an ideal gas. Assume that the molar enthalpy of this nitrogen is constant and that its molar entropy changes are adequately modeled by $d\overline{S}=\left({C_P}/{T}\right)dT-\left({R}/{P}\right)dP$. For this atmosphere, show that ${\left(\frac{\partial T}{\partial h}\right)}_S=\frac{-\overline{M}g}{C_P} \nonumber$ where $h$ is the height above the earth’s surface, $\overline{M}$ is the molar mass of dinitrogen ($0.0280\ \ \mathrm{kg}\ {\mathrm{mol}}^{-1}$), $g$ is the acceleration due to gravity ($\mathrm{9.80}\ \mathrm{m}\ {\mathrm{s}}^{-1}$), and $C_P$ is the constant-pressure heat capacity ($29.1\ \mathrm{J}\ {\mathrm{K}}^{-1}\ {\mathrm{mol}}^{-1}$). [Suggestion: Write the total differential for$\ \overline{H}=\overline{H}\left(\overline{S},P,h\right)$. What are ${\left({\partial \overline{H}}/{\partial \overline{S}}\right)}_{P,h}$, ${\left({\partial \overline{H}}/{\partial P}\right)}_{\overline{V},h}$, and ${\left({\partial \overline{H}}/{\partial h}\right)}_{\overline{S},\overline{V}}$?]
Use this approximation to calculate the temperature on the top of a 3000 m mountain when the temperature at sea level is 300 K.
23. Hikers often say that, as a rule-of-thumb, the temperature on a mountain decreases by 1 C for every 100 m increase in elevation. Is this rule in accord with the relationships developed in problems 21 and 22? In these problems, we assume that the temperature of an ideal-gas atmosphere varies with altitude but that the molar energy or enthalpy does not. Does this assumption contradict the principle that the energy and enthalpy of an ideal gas depend only on temperature?
24. Derive the barometric formula (Section 2.10) from the assumptions that the earth’s atmosphere is an ideal gas whose molar mass is $\overline{\boldsymbol{M}}$ and whose temperature and Gibbs free energy are independent of altitude.
25. Run in reverse, a Carnot engine consumes work $\left(w>0\right)$ and transfers heat $\left(q_{\ell }>0\right)$ from a low-temperature reservoir to a high temperature reservoir $\left(q_h<0\right)$. The work done by the machine is converted to heat that is discharged to the high-temperature reservoir. In one cycle of the machine, $\Delta E=q_{\ell }+q_h=0$. For a refrigerator—or for a heat pump operating in air-conditioning mode—we are interested in the quantity of heat removed $\left(q_{\ell }\right)$ per unit of energy expended $\left(w\right)$. We define the coefficient of performance as $COP\left(cooling\right)={q_{\ell }}/{w}$. This is at a maximum for the reversible Carnot engine. Show that the theoretical maximum is $COP\left(cooling\right)=\frac{1-\epsilon }{\epsilon }=\frac{T_{\ell }}{T_h-T_{\ell }} \nonumber$ where $\epsilon$ is the reversible Carnot-engine efficiency, $\epsilon =1-\frac{T_{\ell }}{T_h} \nonumber$ 26. For a heat pump operating in heating mode—as a “furnace”—we are interested in the quantity of heat delivered to the space being heated $\left(-q_h\right)$ per unit of energy expended $\left(w\right)$. We define the coefficient of performance as $COP\left(heating\right)=-{q_h}/{w}$. Show that the theoretical maximum is $COP\left(heating\right)=\frac{1}{\epsilon }=\frac{T_h}{T_h-T_{\ell }} \nonumber$
27. For $T_{\ell }=300\ \mathrm{K}$ and $T_h=500\ \mathrm{K}$, calculate the theoretical maxima for $COP\left(cooling\right)$ and $COP\left(heating\right)$.
28. Find the theoretical maximum $COP\left(cooling\right)$ for a refrigerator at $40\ \mathrm{F}$ in a room at $72\ \mathrm{F}$.
29. Find the theoretical maximum $COP\left(cooling\right)$ for a heat pump that keeps a room at $72\ \mathrm{F}$ when the outside temperature is $100\ \mathrm{F}$.
30. Find the theoretical maximum $COP\left(heating\right)$ for a heat pump that keeps a room at $72\ \mathrm{F}$ when the outside temperature is $32\ \mathrm{F}$.
Notes
${}^{1}$ For an introduction to the concept of internal entropy and its applications, see Ilya Prigogine, Introduction to the Thermodynamics of Irreversible Processes, Interscience Publishers, 1961.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/09%3A_The_Second_Law_-_Entropy_and_Spontaneous_Change/9.26%3A_Problems.txt
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In Chapter 9, we substitute $dq^{rev} = TdS$, from the second law, into
$dE = dq + PdV \nonumber$
from the first law, to obtain, for any closed system undergoing a reversible change in which the only work is pressure–volume work, the fundamental equation, $dE = TdS + PdV$. In view of the mathematical properties of state functions that we develop in Chapter 7, this result means that we can express the energy of the system as a function of entropy and volume, $E = E\left(S,V\right)$. With this choice of independent variables, the total differential of $E$ is
$(dE = {\left({\partial E}/{\partial S}\right)}_VdS + {\left({\partial E}/{\partial V}\right)}_SdV. \nonumber$
Equating these expressions for $dE$, we find
$\left[{\left(\frac{\partial E}{\partial S}\right)}_V + T\right]dS + \left[{\left(\frac{\partial E}{\partial V}\right)}_S + P\right]dV\mathrm{=0} \nonumber$
for any such system. Since $S$ and $V$ are independent variables, this equation can be true for any arbitrary state of the system only if the coefficients of$\mathrm{\ }dS$ and $dV$ are each identically equal to zero. It follows that
${\left(\frac{\partial E}{\partial S}\right)}_V = T \nonumber$
and
${\left(\frac{\partial E}{\partial V}\right)}_S\mathrm{=-}P \nonumber$
Moreover, because dE is an exact differential, we have
$\frac{\partial }{\partial V}{\left(\frac{\partial E}{\partial S}\right)}_V = \frac{\partial }{\partial S}{\left(\frac{\partial E}{\partial V}\right)}_S \nonumber$
so that
${\left(\frac{\partial T}{\partial V}\right)}_S\mathrm{=-}{\left(\frac{\partial P}{\partial S}\right)}_V \nonumber$
Using the result $dH = TdS + VdP$, parallel arguments show that enthalpy can be expressed as a function of entropy and pressure, $H = H\left(S,P\right)$, so that
${\left(\frac{\partial H}{\partial S}\right)}_P = T \nonumber$
and
${\left(\frac{\partial H}{\partial P}\right)}_S = V \nonumber$ and ${\left(\frac{\partial T}{\partial P}\right)}_S = {\left(\frac{\partial V}{\partial S}\right)}_P \nonumber$
Since $dA\mathrm{=-}SdT + PdV$, the Helmholtz free energy must be a function of temperature and volume, $A = A\left(T,V\right)$, and we have
${\left(\frac{\partial A}{\partial T}\right)}_V\mathrm{=-}S \nonumber$ and ${\left(\frac{\partial A}{\partial V}\right)}_T\mathrm{=-}P \nonumber$ and ${\left(\frac{\partial S}{\partial V}\right)}_T = {\left(\frac{\partial P}{\partial T}\right)}_V \nonumber$
Likewise, $dG\mathrm{=-}SdT + VdP$ implies that the Gibbs free energy is a function of temperature and pressure, $G = G\left(P,T\right)$, so that
${\left(\frac{\partial G}{\partial T}\right)}_P\mathrm{=-}S \nonumber$
and
${\left(\frac{\partial G}{\partial P}\right)}_T = V \nonumber$ and ${\left(\frac{\partial V}{\partial T}\right)}_P\mathrm{=-}{\left(\frac{\partial S}{\partial P}\right)}_T \nonumber$
10.02: dE TdS - PdV and Internal consistency
In Chapter 1, we observe that the business of science is the creation of models that are internally consistent and that accurately describe reality. Logical deduction from tentative hypotheses is a valuable tool in our effort to create new models. Such logical arguments often take the form of “Gedanken (thought) experiments,” as exemplified by our various arguments about the properties of hypothetical, friction-free, piston engines. Nevertheless, the route by which we arrive at a theory is irrelevant; what counts are the theory’s internal consistency and predictive capability. Let us pause, therefore, to note that we have arrived at mathematical expressions of ideas that we initially introduced as principles inferred from experience.
In Chapter 6, we prove Duhem’s theorem when the variables are chosen from the set pressure, temperature, volume, and component concentrations. However, the theorem is more general. It asserts that two variables are sufficient to specify changes in the state of a closed, reversible system, in which only pressure–volume work is possible. Our derivations have now led us to the conclusion that the energy of such a system can be expressed as a function of entropy and volume. Given the entropy, the volume, and the function $E = E\left(S,V\right)$, the relationships developed above mean that we know $S$, $V$, $E$, $P$, and $T$ for the system. Given these, we can calculate $H$, $A$, and $G$. That is, specifying the changes in the two variables $S$ and $V$ is sufficient to specify the change in the state of the system. Moreover, we can rearrange the fundamental equation to
$dV = \left(\frac{T}{P}\right)dS + \left(\frac{\mathrm{1}}{P}\right)dE \nonumber$
so that the volume can be expressed as a function of entropy and energy. Given $S$, $E$, and the function $V = V\left(S,E\right)$, we can find $P$ and $T$. Specifying the changes in $S$ and $E$ is sufficient to specify the change in the system. Finally, we can rearrange the fundamental equation to
$dS = \left(\frac{\mathrm{1}}{T}\right)dE + \left(\frac{P}{T}\right)dV \nonumber$
so that $S = S\left(V,E\right)$ and specifying changes in $E$ and $V$ is sufficient to specify the change in the system.
Now, let us return to our discussion in Section 9.7 of the entropy change for an isolated system undergoing a spontaneous change. That discussion explores the use of the machine-based statement of the second law to establish that the entropy of an isolated system must increase during any spontaneous process. To infer that the system’s entropy must increase in such a process, we consider the special case in which only pressure–volume work is possible and argue that a change in which $\mathrm{\Delta }E = \mathrm{\Delta }S\mathrm{=0}$ is no change at all. That is, we assume that specifying the change in $E$ and $S$ is sufficient to specify the change in the state of such a system. It is, therefore, a significant check on the internal consistency of our thermodynamic model to see that $dE = TdS + PdV$ implies that $E$ and $S$ are indeed a sufficient pair.
Finally, let us consider the relationship of a spontaneous process in a closed system to the surface that describes reversible processes in the same system. The energy of the system undergoing reversible change is expressed as $E = E\left(S,V\right)$. An energy surface in $V-S-E$-space is sketched in Figure 1. At any point on this surface, the system is at equilibrium. The point $\left(V_0,S_0,E_0\right)$ is such a point. The tangent to this surface at $\left(V_0,S_0,E_0\right)$ and in the plane $V = V_0$ is the partial derivative $T = {\left({\partial E}/{\partial S}\right)}_V$. The tangent to the surface at $\left(V_0,S_0,E_0\right)$ and in the plane $S = S_0$ is the partial derivative $P\mathrm{=-}{\left({\partial E}/{\partial V}\right)}_S$.
No point in $V-S-E$-space that is off the $E = E\left(S,V\right)$ surface can describe an equilibrium state of the closed system. As a practical matter, some of these points may represent states that the system can attain. If so, they are transient states of a spontaneously changing system. Let us suppose, for example, that we are able to
maintain $S = S_0$ and $V = V_0$ while a slow chemical reaction occurs in the system. At every instant, such a state must have a non-equilibrium composition. It must have an energy, and this energy must exceed $E_0$; since we must have ${\left(\mathrm{\Delta }E\right)}_{SV}\mathrm{<0}$ for a spontaneous process, a point $\left(V_0,S_0,E_{\mathrm{1}}\right)$ can represent a state of the system during a spontaneous change only if $E_{\mathrm{1}}\mathrm{>}E_0$. If $E_{\mathrm{2}}\mathrm{<}E_0$, the point $\left(V_0,S_0,E_{\mathrm{2}}\right)$ cannot represent a state of the system that can spontaneously go to equilibrium at $\left(V_0,S_0,E_0\right)$. Similarly, if $S_{\mathrm{1}}\mathrm{<}S_0$ and the system is isolated with $E = E_0$ and $V = V_0$, the point $\left(V_0,S_{\mathrm{1}},E_0\right)$ represents a state of the system that can go to equilibrium at $\left(V_0,S_0,E_0\right)$ spontaneously. This process would satisfy the entropy criterion, ${\left(\mathrm{\Delta }S\right)}_{EV}\mathrm{>0}$.
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We have found simple differential expressions for $E$, $S$, $H$, A, and $G\mathrm{\ }$that apply to closed, reversible systems in which only pressure–volume work is possible. From $dE = TdS + PdV$, we infer that $E = E\left(S,V\right)$. From $dH = TdS + VdP$, we infer that $H = H\left(S,P\right)$. An argument parallel to that above leads us to the conclusion that specifying the changes in $S$ and $P$ is sufficient to specify the change in the state of the system. Similarly, from $dA\mathrm{=-}SdT + PdV$ and $dG\mathrm{=-}SdT + VdP$, we see that it is sufficient to specify the changes in either $V$ and $T$ or $P$ and $T$. These total differentials show that that specifying the change in two state functions is sufficient to specify the change that occurs in the state of a closed system, when the change is reversible and all of the work is pressure–volume work. We have now found seven pairs of state functions that are sufficient; they are $\{S,V\}$, $\{S,E\}$, $\{E,V\}$, and the four pairs in which we choose one variable from the set $\{S\mathrm{,\ }T\}$ and one from the set $\{P\mathrm{,\ }V\}$. However, each of the equations we have obtained so far uses a different pair of independent variables.
Evidently, we should be able to express any thermodynamic function using various pairs of state functions. We can do this by transforming the equations that we have already derived. We are particularly interested in $P$, $V$, and $T$ as independent variables, because these quantities are readily measured for most systems. In the sections below, we find exact differentials for $dS$, $dE$, $dH$, $dA$, and $dG$ with $V$ and $T$ and with $P$ and $T$ as the independent variables.
While specifying the change in some pair of variables is always sufficient to specify the change in the state of a closed reversible system, we should note that it is not always necessary. If the system has only one degree of freedom, specifying some single variable is sufficient. For example, so long as both phases remain present, the change in the state of a pure substance at liquid–vapor equilibrium can be specified by specifying the change in the temperature, the pressure, the volume, or the number of moles of either phase. We discuss this further in Section 10.7.
At present, we are developing relationships among state functions that are valid for any closed reversible system in which all work is pressure–volume work. The next several chapters explore the implications of these results. If the composition of the closed reversible system changes during these processes, this composition change does not affect the relationships we develop here. Of course, any composition change that occurs during a reversible process must be reversible; if the components of the system can react, this reaction must be at equilibrium throughout the process. In Chapter 14, we extend the relationships that we develop here to explicitly include molar compositions as independent variables. This enables us to express our theory for equilibrium using composition variables.
In Section 6.10, we assume—infer from experience—that specifying the changes in $P$ and $T$ is sufficient to specify a change in the state of a closed equilibrium system whose phase composition is fixed and in which only pressure–volume work is possible. We use this assumption to give a partial proof of Duhem’s theorem. In Section 10.5, we see that this assumption is also a consequence of the theory we have developed. This is another check on the internal consistency of the theory.
Finally, it is time to consider a question we have thus far avoided: Is any arbitrary pair of state functions a sufficient set? The answer is no. In Section 10.8, we find that neither $\{P,V\}$ nor $\{S,T\}$ is a sufficient pair in all cases.
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If we choose $V$ and $T$ as the independent variables, we can express the differential of $E$ as a function of $V$ and $T$. We also have the differential relationship $dE=TdS-PdV$. These expressions for $dE$ must be equal:
$dE = {\left(\frac{\partial E}{\partial V}\right)}_TdV + {\left(\frac{\partial E}{\partial T}\right)}_VdT = TdS + PdV \nonumber$ Rearranging, we find a total differential for $dS$ with $V$ and $T$ as the independent variables:
$dS = \frac{\mathrm{1}}{T}{\left(\frac{\partial E}{\partial T}\right)}_VdT + \frac{\mathrm{1}}{T}\left[{\left(\frac{\partial E}{\partial V}\right)}_T + P\right]dV \nonumber$
From the coefficient of $dT$, we have
${\left(\frac{\partial S}{\partial T}\right)}_V = \frac{\mathrm{1}}{T}{\left(\frac{\partial E}{\partial T}\right)}_V = \frac{C_V}{T} \nonumber$
where we use the definition ${\left({\partial E}/{\partial T}\right)}_V=C_V$. (When we write “$C_V$,” we usually think of it as a property of a pure substance. The relationship above is valid for any reversible system. When we are describing a system that is not a pure substance, $C_V$ is just an abbreviation for ${\left({\partial E}/{\partial T}\right)}_V$.) From the coefficient of $dV$, we have
${\left(\frac{\partial S}{\partial V}\right)}_T = \frac{\mathrm{1}}{T}\left[{\left(\frac{\partial E}{\partial V}\right)}_T + P\right] = {\left(\frac{\partial P}{\partial T}\right)}_V \nonumber$
where we use the relationship ${\left({\partial S}/{\partial V}\right)}_T={\left({\partial P}/{\partial T}\right)}_V$ that we find in §1. Substituting into the expression for $dS$, we find $dS = \frac{C_V}{T}dT + {\left(\frac{\partial P}{\partial T}\right)}_VdV \nonumber$
Now, from $dE=TdS-PdV$, we have
$dE = C_VdT + \left[T{\left(\frac{\partial P}{\partial T}\right)}_V + P\right]dV \nonumber$
From $H=E+PV$, we have
\begin{align*} dH &= dE + d\left(PV\right) \[4pt] &= dE + {\left(\frac{\partial \left(PV\right)}{\partial T}\right)}_VdT + {\left(\frac{\partial \left(PV\right)}{\partial V}\right)}_TdV \[4pt]&= dE + V{\left(\frac{\partial P}{\partial T}\right)}_VdT + \left[P + V{\left(\frac{\partial P}{\partial V}\right)}_T\right]dV \[4pt]&= \left[C_V + V{\left(\frac{\partial P}{\partial T}\right)}_V\right]dT + \left[T{\left(\frac{\partial P}{\partial T}\right)}_V + V{\left(\frac{\partial P}{\partial V}\right)}_T\right]dV \end{align*}
Of course, we already have
$dA\mathrm{=-}SdT + PdV \nonumber$
From $G=H-TS$, by an argument that parallels the above derivation of $dH$, we obtain
$dG = \left[V{\left(\frac{\partial P}{\partial T}\right)}_V + S\right]dT + V{\left(\frac{\partial P}{\partial V}\right)}_TdV \nonumber$
Finally, we can write $P=P\left(T,V\right)$ to find
$dP = {\left(\frac{\partial P}{\partial T}\right)}_VdT + {\left(\frac{\partial P}{\partial V}\right)}_TdV \nonumber$
$P$, $T$, $V$, $C_V$, ${\left({\partial P}/{\partial T}\right)}_V$, and ${\left({\partial P}/{\partial V}\right)}_T$ are all experimentally accessible for any reversible system. If we have this information for a system that undergoes a change from a state specified by $T_1$ and $V_1$ to a second state specified by $T_2$ and $V_2$, we can use these relationships to calculate $\Delta E$, $\Delta S$, and $\Delta H$. To do so, we calculate the appropriate line integral along a reversible path. One such path is an isothermal reversible change, at $T_1$, from $V_1$ to $V_2$, followed by a constant-volume change, at $V_2$, from $T_1$ to $T_2$. In principle, the same procedure can used to calculate $\Delta A$ and $\Delta G$. However, because $S$ appears in the differentials$\ dA$ and $dG$, this requires that we first find $S$ as a function of $V$ and$\ T$.
If the system is a pure substance for which we have an equation of state, we can find ${\left({\partial P}/{\partial T}\right)}_V$, and ${\left({\partial P}/{\partial V}\right)}_T$ by straightforward differentiation. When the substance is a gas, an equation of state may be available in the literature. When the substance is a liquid or a solid, these partial derivatives can still be related to experimentally accessible quantities. The compressibility of a substance is the change in its volume that results from a change in the applied pressure, at a constant temperature. The thermal expansion of a substance is the change in its volume that results from a change in its temperature, at a constant applied pressure. It is convenient to convert measurements of these properties into intensive functions of the state of the substance by expressing the volume change as a fraction of the original volume. That is, we define the coefficient of thermal expansion:
$\alpha = \frac{\mathrm{1}}{V}{\left(\frac{\partial V}{\partial T}\right)}_P \nonumber$
and the coefficient of isothermal compressibility:
$\beta \mathrm{=-}\frac{\mathrm{1}}{V}{\left(\frac{\partial V}{\partial P}\right)}_T \nonumber$
Coefficients of thermal expansion and isothermal compressibility are available in compilations of thermodynamic data for many liquids and solids. In general, both coefficients are weak functions of temperature. We have
$\left(\frac{\partial P}{\partial V}\right)_T=-\frac{1}{\beta V} \nonumber$
and
$\left(\frac{\partial P}{\partial T}\right)_V=-\left(\frac{\partial V}{\partial T}\right)_P/ \left(\frac{\partial V}{\partial P}\right)_T=\frac{\alpha }{\beta } \nonumber$
Using these coefficients, we can estimate a pressure change, for example, as a line integral of
$dP=\left(\frac{\alpha }{\beta }\right)dT-\left(\frac{1}{\beta V}\right)dV \nonumber$
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We can follow a parallel development to express these thermodynamic functions with $P$ and $T$ as the independent variables. We have the differential relationship $dH=TdS+VdP$. We expand $\ dH$ with $P$ and $T$ as the independent variables. Equating these, we obtain
\begin{align} dH &= \left(\frac{\partial H}{\partial P}\right)_TdP + \left(\frac{\partial H}{\partial T}\right)_PdT \[4pt] &= TdS + VdP \end{align} \nonumber
so that we have
$dS = \frac{\mathrm{1}}{T}{\left(\frac{\partial H}{\partial T}\right)}_PdT + \frac{\mathrm{1}}{T}\left[{\left(\frac{\partial H}{\partial P}\right)}_T - V\right]dP \nonumber$
From the coefficient of $dT$ and the definition ${\left({\partial H}/{\partial T}\right)}_P=C_P$, we have
$\left(\frac{\partial S}{\partial T}\right)_P = \frac{\mathrm{1}}{T}{\left(\frac{\partial H}{\partial T}\right)}_P = \frac{C_P}{T} \nonumber$
(When we are describing a reversible system that is not a pure substance, $C_P$ is just an abbreviation for ${\left({\partial H}/{\partial T}\right)}_P$.) From the coefficient of $dP$ and the relationship ${\left({\partial S}/{\partial P}\right)}_T=-{\left({\partial V}/{\partial T}\right)}_P$ that we find in Section 10.1, we have
${\left(\frac{\partial S}{\partial P}\right)}_T = \frac{\mathrm{1}}{T}\left[{\left(\frac{\partial H}{\partial P}\right)}_T - V\right] = -{\left(\frac{\partial V}{\partial T}\right)}_P \nonumber$
Substituting into the expression for $dS$, we find
$dS = \frac{C_P}{T}dT - {\left(\frac{\partial V}{\partial T}\right)}_PdP \nonumber$
Using the same approach as in the previous section, we can now obtain
\begin{align} dE &= \left[C_P - P{\left(\frac{\partial V}{\partial T}\right)}_P\right]dT - \left[{P\left(\frac{\partial V}{\partial P}\right)}_T + T{\left(\frac{\partial V}{\partial T}\right)}_P\right]dP \[4pt] dH &= C_PdT + \left[V - T{\left(\frac{\partial V}{\partial T}\right)}_P\right]dP \[4pt] dA &= -\left[S + P{\left(\frac{\partial V}{\partial T}\right)}_P\right]dT - {P\left(\frac{\partial V}{\partial P}\right)}_TdP \end{align} \nonumber
and, we already have
$dG = VdP - SdT \nonumber$
Finally, we can write $V=V\left(P,T\right)$ to find
${dV = \left(\frac{\partial V}{\partial P}\right)}_TdP + {\left(\frac{\partial V}{\partial T}\right)}_PdT \nonumber$
so that we have total differentials for all of the principal thermodynamic functions when they are expressed as functions of $P$ and $T$. If an equation of state is not known but the coefficients of thermal expansion and isothermal compressibility are available, we have ${\left({\partial V}/{\partial T}\right)}_P=\alpha V$ and ${\left({\partial V}/{\partial P}\right)}_T=-\beta V$. Then we can estimate a volume change, for example, as a line integral of
$dV=\alpha VdT-\beta VdP \nonumber$
10.06: The Transformation of Thermodynamic Variables in G
Let us suppose that $M$, $Q$, $R$, $X$, and $Y$ are state functions and that we know the total differentials
$dM={\left(\dfrac{\partial M}{\partial X}\right)}_YdX+{\left(\dfrac{\partial M}{\partial Y}\right)}_XdY \nonumber$
$dQ={\left(\dfrac{\partial Q}{\partial X}\right)}_YdX+{\left(\dfrac{\partial Q}{\partial Y}\right)}_XdY \nonumber$
$dR={\left(\dfrac{\partial R}{\partial X}\right)}_YdX+{\left(\dfrac{\partial R}{\partial Y}\right)}_XdY \nonumber$
To find the total differential of $M\left(Q,R\right)$,
$dM={\left(\dfrac{\partial M}{\partial Q}\right)}_RdQ+{\left(\dfrac{\partial M}{\partial R}\right)}_QdR \nonumber$
we solve the total differentials of $Q\left(X,Y\right)$ and $R\left(X,Y\right)$ to find $dX$ and $dY$ in terms of $dQ$ and $dR$. Since $dQ$ and $dR$ are simultaneous equations in the variables $dX$ and $dY$, we can apply Cramer’s rule to obtain
$dX=\dfrac{\left| \begin{array}{cc} dQ & {\left({\partial Q}/{\partial Y}\right)}_X \ dR & {\left({\partial R}/{\partial Y}\right)}_X \end{array} \right|}{J\left(\dfrac{Q,R}{X,Y}\right)}=\dfrac{{\left(\dfrac{\partial R}{\partial Y}\right)}_XdQ-{\left(\dfrac{\partial Q}{\partial Y}\right)}_XdR}{J\left(\dfrac{Q,R}{X,Y}\right)} \nonumber$
and
$dY=\dfrac{\left| \begin{array}{cc} {\left({\partial Q}/{\partial X}\right)}_Y & dQ \ {\left({\partial R}/{\partial X}\right)}_Y & dR \end{array} \right|}{J\left(\dfrac{Q,R}{X,Y}\right)}=\dfrac{{-\left(\dfrac{\partial R}{\partial X}\right)}_YdQ+{\left(\dfrac{\partial Q}{\partial X}\right)}_YdR}{J\left(\dfrac{Q,R}{X,Y}\right)} \nonumber$
where $J\left({\left(Q,R\right)}/{\left(X,Y\right)}\right)$ is the Jacobian of the transformation of variables $X$ and $Y$ to variables$\ Q$ and $R$:
\begin{align*} J\left(\dfrac{Q,R}{X,Y}\right) &= \left| \begin{array}{cc} {\left({\partial Q}/{\partial X}\right)}_Y & {\left({\partial Q}/{\partial Y}\right)}_X \ {\left({\partial R}/{\partial X}\right)}_Y & {\left({\partial R}/{\partial Y}\right)}_X \end{array} \right| \[4pt] &= {\left(\dfrac{\partial Q}{\partial X}\right)}_Y{\left(\dfrac{\partial R}{\partial Y}\right)}_X + {\left(\dfrac{\partial Q}{\partial Y}\right)}_X{\left(\dfrac{\partial R}{\partial X}\right)}_Y \end{align*}
To find
$dM={\left(\dfrac{\partial M}{\partial Q}\right)}_RdQ+{\left(\dfrac{\partial M}{\partial R}\right)}_QdR \nonumber$
We substitute these results for $dX$ and $dY$ into the total differential of $M=M\left(X,Y\right):$
\begin{aligned} dM &={\left(\dfrac{\partial M}{\partial X}\right)}_YdX+{\left(\dfrac{\partial M}{\partial Y}\right)}_XdY \[4pt] &=\dfrac{{\left(\dfrac{\partial M}{\partial X}\right)}_Y\left[{\left(\dfrac{\partial R}{\partial Y}\right)}_XdQ-{\left(\dfrac{\partial Q}{\partial Y}\right)}_XdR\right]}{J\left(\dfrac{Q,R}{X,Y}\right)}+\dfrac{{\left(\dfrac{\partial M}{\partial Y}\right)}_X\left[{-\left(\dfrac{\partial R}{\partial X}\right)}_YdQ+{\left(\dfrac{\partial Q}{\partial X}\right)}_YdR\right]}{J\left(\dfrac{Q,R}{X,Y}\right)} \[4pt] &=\left[\dfrac{{\left(\dfrac{\partial M}{\partial X}\right)}_Y{\left(\dfrac{\partial R}{\partial Y}\right)}_X-{\left(\dfrac{\partial M}{\partial Y}\right)}_X{\left(\dfrac{\partial R}{\partial X}\right)}_Y}{J\left(\dfrac{Q,R}{X,Y}\right)}\right]dQ +\left[\dfrac{-{\left(\dfrac{\partial M}{\partial X}\right)}_Y{\left(\dfrac{\partial Q}{\partial Y}\right)}_X+{\left(\dfrac{\partial M}{\partial Y}\right)}_X{\left(\dfrac{\partial Q}{\partial X}\right)}_Y}{J\left(\dfrac{Q,R}{X,Y}\right)}\right]dR \end{aligned} \nonumber
where the coefficients of $dQ$ and $dR$ are ${\left({\partial M}/{\partial Q}\right)}_R$ and ${\left({\partial M}/{\partial R}\right)}_Q$, respectively. In §5, we find the other total differentials in terms of $dP$ and $dT$. If we set $X=T$ and $Y=P$, we can use these relationships to find the total differential for any state function expressed in terms of any two other state functions.
To illustrate this point, let us use these relationships to find the total differential of $S$ expressed as a function of $P$ and $V$, $S=S\left(P,V\right)$. In this case, we are transforming from the variables $\left(P,T\right)$ to the variables $\left(P,V\right)$. This is a one-variable transformation. To effect it, we make the additional substitutions $M=S$, $Q=V$, and $R=P$. Since we have $Y=R=P$, the transformation equations simplify substantially. We have
$\left(\dfrac{\partial R}{\partial Y}\right)_X=\left(\dfrac{\partial P}{\partial P}\right)_X=1 \nonumber$
$\left(\dfrac{\partial R}{\partial X}\right)_Y=\left(\dfrac{\partial P}{\partial T}\right)_P=0 \nonumber$
$\left(\dfrac{\partial Q}{\partial Y}\right)_X=\left(\dfrac{\partial V}{\partial P}\right)_T \nonumber$
$\left(\dfrac{\partial Q}{\partial X}\right)_Y=\left(\dfrac{\partial V}{\partial T}\right)_P \nonumber$
The Jacobian becomes
$J\left(\dfrac{Q,R}{X,Y}\right) = \left(\dfrac{\partial V}{\partial T}\right)_P \nonumber$
and the partial derivatives of $S$ become
${\left(\dfrac{\partial S}{\partial V}\right)}_P={{\left(\dfrac{\partial S}{\partial T}\right)}_P}/{{\left(\dfrac{\partial V}{\partial T}\right)}_P={\dfrac{C_P}{T}\left(\dfrac{\partial T}{\partial V}\right)}_P} \nonumber$
and
\begin{align} \left(\dfrac{\partial S}{\partial P}\right)_V &= \dfrac{-{\left(\dfrac{\partial S}{\partial T}\right)}_P{\left(\dfrac{\partial V}{\partial P}\right)}_T+{\left(\dfrac{\partial S}{\partial P}\right)}_T{\left(\dfrac{\partial V}{\partial T}\right)}_P}{{\left(\dfrac{\partial V}{\partial T}\right)}_P} \[4pt] &={\dfrac{C_P}{T}\left(\dfrac{\partial T}{\partial P}\right)}_V+{\left(\dfrac{\partial S}{\partial P}\right)}_T \end{align} \nonumber
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We have found total differentials for the principal thermodynamic functions with $V$ and $T$ and with $P$ and $T$ as the independent variables. In §6, we see how to find such differentials, $dM = M_xdX + M_YdY$, for any pair of independent variables, $X\mathrm{\ }$and $Y$. These equations express our physical theory as a mathematical structure. Because $E$, $T$, $S$, $P$, $V$, $H$, $A$, and $G$ are state functions, the mathematical properties of state functions enable us to obtain the relationships $M_X = {\left({\partial M}/{\partial X}\right)}_Y$, $M_Y = {\left({\partial M}/{\partial Y}\right)}_X$, and $\left({\partial M_Y}/{\partial X}\right) = \left({\partial M_X}/{\partial Y}\right)$ that we find in §1. These equations apply to reversible processes in closed systems when only pressure–volume work is possible. Using these equations to describe reversible processes involves a number of important ideas. Let us consider four cases: [GrindEQ__1_] None of $M$, $X$, and $Y$ are constant; [GrindEQ__2_] $M$ is constant, but $X$ and Y are not; [GrindEQ__3_] $X$ is constant, but $M$ and $Y$ are not; [GrindEQ__4_] $X$ and $Y$ are constant, but $M\mathrm{\ }i$s not.
None of $\boldsymbol{M}$, $\boldsymbol{X}$, and $\boldsymbol{Y}$ is constant.– The properties of state functions and the existence of the exact differential , $dM = M_XdX + M_YdY$ imply that $M$ is a function of $X$ and $Y$, $M = M\left(X,Y\right)$. If $M_X$ and $M_Y$ are single-valued and continuous along some path in the $XY$-plane, we can evaluate the line integral of $dM$ along this path. Given $M$ at a first point, $M\left(X_{\mathrm{1}},Y_{\mathrm{1}}\right)$, we can find $M$ at a second point, $M\left(X_{\mathrm{2}},Y_{\mathrm{2}}\right)$, by evaluating this line integral along some path in the $XY$-plane between the points $\left(X_{\mathrm{1}},Y_{\mathrm{1}}\right)$ and $\left(X_{\mathrm{2}},Y_{\mathrm{2}}\right)$. Given the values of $X$ and $Y$, the value of$\mathrm{\ }M$ is uniquely determined; $M\left(X,Y\right)$ is a two-dimensional surface in the three-dimensional space whose dimensions are $M$, $X$, and $Y$.
The $M\left(X,Y\right)$ surface is just the set of points that are accessible to the system in reversible processes in which $X$ and $Y$ change. Conversely, the only values of $M$ that are accessible to the system in reversible processes in which $X$ and $Y$ change correspond to those points that lie on the surface. In §5, we find the total differentials for $dE$, $dS$, $dH$, $dA$, and $dG$ using $P$ and $T$ as the independent variables. Evidently, for a reversible process in a closed system, there is a surface representing each of $E$, $S$, $H$, $A$, and $G$ over the $P--T$-plane. Since the system is also characterized by an equation of state that relates the values of $P$, V, and $T$, there is also a surface representing $V$ over the $P--T$-plane.
In general, a given system can also undergo spontaneous changes. Suppose that a system is originally at equilibrium at temperature $T_{\mathrm{1}}$ and pressure $P_{\mathrm{1}}$. If we contact this system with surroundings at some arbitrary $\hat{T}$ and we arrange for the pressure applied to the system to have some arbitrary value, $P_{applied}$, the system will respond, eventually reaching equilibrium with the system temperature equal to $\hat{T}$ and the system pressure equal to $P_{applied}$. (The change-enabled state in which $P_{system}\mathrm{\ }{ = P}_{\mathrm{1}}$ and ${T_{system} = T}_{\mathrm{1}}$, while the applied pressure is $P_{applied}$ and the surroundings temperature is $\hat{T}$, is a hypothetical state. It is not an equilibrium state, because $P_{system}\mathrm{\neq }P_{applied}$ and $T_{system}\mathrm{\neq }\hat{T}$. The change-enabled state can undergo spontaneous change; however, its thermodynamic functions have the same values as they have in the original equilibrium state, in which $P_{system}\mathrm{\ }{ = P}_{\mathrm{1}}$ and ${T_{system} = T}_{\mathrm{1}}$.) Since this change is spontaneous, it may not be possible to trace the path of the system in the $P--T$-plane as the change occurs. If we can trace the path in the $P--T$-plane, the energy of the system can be described as a line in $E--P--T$-space, but this line will not lie on the reversible-process surface specified by the function $E = E\left(P,T\right)$. Nevertheless, we can select paths in the $P--T$-plane that connect the initial point $\left(P_{\mathrm{1}},T_{\mathrm{1}}\right)$ to the final point $\left(P_{applied},\hat{T}\right)$. There are reversible processes that correspond to these paths. By evaluating the line integral for any state function along any of these paths, we can find the change that occurs in the state function during the spontaneous process.
Cases arise in which $M$ is not single-valued or continuous along some or all of the paths that connect points $\left(X_{\mathrm{1}},Y_{\mathrm{1}}\right)$ and $\left(X_{\mathrm{2}},Y_{\mathrm{2}}\right)$. Then $M_X$ or $M_Y$ may not exist for some points $\left(X,Y\right)$. In this case, it may not be possible to evaluate the line integral to find the change $M\left(X_{\mathrm{2}},Y_{\mathrm{2}}\right) + M\left(X_{\mathrm{1}},Y_{\mathrm{1}}\right)$. This can occur when there is a phase change. If $\left(P_{vp},T_{bp}\right)$ specifies a state of liquid–vapor equilibrium, the enthalpy of the system is not single-valued. Below, we consider the thermodynamic surfaces of water when a phase change occurs. In §8, we see that $M\mathrm{\ }$may not be single-valued when $V$ and $P$ or when $T$ and $S$ are the independent variables.
$\boldsymbol{M}$ is constant, $\boldsymbol{X}$ and $\boldsymbol{Y}$ are not.– If $M$ is constant, we have $dM\mathrm{=0}$. If $X$ and $Y$ are not constant, if $M_X$ and $M_Y$ are defined, and if $M_Y\mathrm{\neq }\mathrm{0}$, we can apply the divide-through rule to obtain
${\left(\frac{\partial Y}{\partial X}\right)}_M\mathrm{=-}\frac{M_X}{M_Y} \nonumber$
Such relationships are useful. In Chapter 12, we discuss the Clapeyron and Clausius-Clapeyron equations, which we obtain from $dG\mathrm{=-}SdT + VdP$ using this argument.
$\boldsymbol{X}$ is constant, $\boldsymbol{M}$ and $\boldsymbol{Y}$ are not.– If $X$ is constant, we have $dX\mathrm{=0}$. Instead of , $dM = M_XdX + M_YdY$, we have , $dM = M_XdX$, and we must ask whether $M$ is indeed expressible as a function of $Y$ only. If $M$ can be expressed as a function of $Y$ only, so that $M_Y$ is single-valued and continuous, we can integrate to find
$M_{\mathrm{2}} + M_{\mathrm{1}} = \int^{Y_{\mathrm{2}}}_{Y_{\mathrm{1}}}{M_Y}dY \nonumber$
$\boldsymbol{X}$ and $\boldsymbol{Y}$ are constant, $\boldsymbol{M}$ is not constant.– An interesting and important case arises when $X$ and $Y$ are constant, but $M$ is not constant. When $X$ and $Y$ are constant, $dX=dY=0$, and from $dM=M_XdX+M_YdY$, it follows that $dM=0$. Nevertheless, we can readily identify processes in which some state function, $M$, changes while two (or more) others, $X$ and $Y$, remain constant. In this case, it is clear that $X$ and $Y$ are not sufficient to model the change in $M$. Recall from our discussion of Duhem’s Theorem that, while two independent variables are sufficient to describe a reversible process in which pressure-volume work is the only work, which pair of variables is adequate depends on the system.
Below we discuss the reversible vaporization of water at constant $P$ and $T$. For this process we have $dG=0$. However, we know that $dS>0$. For independent variables $P$ and $T$, our differential expressions for $dG$ and $dS$ are
$dG=VdP-SdT \nonumber$ and $dS=\frac{1}{T}{\left(\frac{\partial H}{\partial T}\right)}_PdT-{\left(\frac{\partial V}{\partial T}\right)}_TdP \nonumber$
Setting $dP=dT=0$ in these equations correctly gives $dG=0$; however, $dS=0$ is false. Evidently, variables $P$ and $T$ are not sufficient to model the entropy change in this process.
However, at constant $P$ and $T$, variables $V$ and $T$ are adequate. We have
$dS=\frac{1}{T}{\left(\frac{\partial E}{\partial T}\right)}_VdT+{\left(\frac{\partial P}{\partial T}\right)}_VdV \nonumber$
In Section 12.10 we develop the Clausius-Clapeyron equation for this vaporization process; we find
$\left(\frac{\partial P}{\partial T}\right)_V=\frac{P{\Delta }_{vap}\overline{H}}{RT^2} \nonumber$
(${\Delta }_{vap}\overline{H}$ is the enthalpy of vaporization per mole.) Since the volume of the system is essentially the volume of the gas phase, we have, assuming the vapor behaves as an ideal gas, $V={n_gRT}/{P}$, and
$dV=\frac{RT}{P}dn_g \nonumber$
The entropy equation then becomes
\begin{aligned} dS & = \left(\frac{\partial P}{\partial T}\right)_VdV \ ~ & =\frac{P{\Delta }_{vap}\overline{H}}{RT^2} \cdot \frac{RT}{P}dn_g \ ~ & =\frac{\Delta _{vap}\overline{H}}{T} dn_g \end{aligned} \nonumber
so that the entropy change for this reversible process is directly proportional to the number of moles of vapor produced.
We see that we must introduce an extensive variable to model the entropy change in the vaporization process. The system volume serves this purpose, although we wind up expressing this volume in terms of the number of moles of vapor in the system.
From another perspective, we can write the entropy as a function of $P$, $T$, and $n_g$: $S=S\left(P,T,n_g\right)$. Then
$dS=\frac{1}{T} \left(\frac{\partial H}{\partial T}\right)_PdT- \left(\frac{\partial V}{\partial T}\right)_TdP+ \left(\frac{\partial S}{\partial n_g}\right)_{PT}dn_g \nonumber$
which becomes
$\left(dS\right)_{PT}= \left(\frac{\partial S}{\partial n_g}\right)_{PT}dn_g \nonumber$
with
$\left(\frac{\partial S}{\partial n_g}\right)_{PT}=\frac{\Delta _{vap}\overline{H}}{T} \nonumber$
In Chapter 14, we extend all of our thermodynamic models to include variables that specify the system composition—the number of moles of the substances present in the system.
Thermodynamic surfaces in the reversible vaporization of water
To illustrate the fact that $dT = dP\mathrm{=0}$ has different implications for $dS$ than it does for $dG$, let us consider the reversible vaporization of one mole of water at constant $P$ and $T$. $\mathrm{\Delta }G$ for this process is zero, but $\mathrm{\Delta }S$ and $\mathrm{\Delta }H$ are not. We can describe a system comprised of one mole of liquid water using pressure and temperature as the independent variables. There is a wide range of pressure and temperature values that is consistent with the system remaining entirely liquid. Every combination of pressure and temperature at which the system remains entirely liquid can be reached by a reversible process from any other such combination of pressure and temperature. For every combination of pressure and temperature within this range, there is one and only one value for every other thermodynamic function. Choosing the enthalpy function to be specific, we can say that the set of enthalpy–pressure–temperature-points for which the system remains entirely liquid is a thermodynamic surface on which reversible change is possible.
We can say all of these same things about a system that consists of one mole of gaseous water. Of course, the enthalpy surface for gaseous water is a different surface from the enthalpy surface for liquid water. At any given temperature, there is a pressure at which liquid and gaseous water are at equilibrium. Above this pressure, the system is entirely liquid; below it, the system is entirely gas. The enthalpy surface of the liquid lies over a different part of the pressure–temperature-plane than does that of the gas. (If the liquid can be superheated or the gas can be supercooled, a given pressure and temperature may be represented by a point on the enthalpy surfaces for both the pure liquid and the pure gas.) The enthalpy surface for the gaseous system lies at higher energies than that for the liquid system; the two enthalpy surfaces do not intersect.
To reversibly transform pure liquid water to pure gaseous water, we must move on the enthalpy surface of the liquid to a pressure and temperature at which the liquid and the gas are at equilibrium. At this pressure and temperature, we can reversibly increase the volume of the system, causing the reversible vaporization of liquid water, and we can continue this process until all of the liquid has been vaporized. When all of the liquid has been vaporized, the system is on the enthalpy surface of the gas. Thereafter, we can change the pressure and temperature of the system to reversibly change the state of the pure gas. While we can describe this process in terms of the successive changes that we impose on the state functions of a system that consists of one mole of water, we are considering three different systems when we describe the overall process from the perspective afforded by Gibbs’ phase rule.
1. The first system is one mole of pure liquid. This system has one phase. There are two degrees of freedom, which we take to be pressure and temperature.
2. The second system is one mole of water, of which $x_{\ell }$ mole is liquid and $\mathrm{1-}x_{\ell }$ mole is gas, at equilibrium, at a fixed pressure and temperature. In the vaporization process, the pressure and temperature are constant while the volume of the system increases ($x_{\ell }$ decreases) reversibly. The one mole of water is described by this system from the time the first bubble of gas appears to the time the last drop of liquid vaporizes. There are two phases and one degree of freedom. When we reversibly vaporize water at a fixed pressure and temperature, one variable must describe the composition: We can take it to be the volume of the system or the liquid mole fraction, $x_{\ell }$. (Of course, we can reversibly vaporize water in a process in which the pressure, temperature, and composition all change; however, because there is only one degree of freedom, specifying a temperature change uniquely determines the pressure change, and conversely. In $H--P--T$-space, a reversibly vaporizing system traces a path on a vertical plane between the enthalpy surfaces of the liquid and the gas. If pressure and temperature are constant, this path is a vertical line. If reversibility is achieved through synchronous variation of pressure and temperature, the path is not vertical, but it remains in a vertical plane.)
3. The third system describes the mole of water after all of the water has been converted to the gas. This system has one phase and two degrees of freedom, which we again take to be pressure and temperature.
We can say that this description of the reversible conversion of liquid water to gaseous water involves three Gibbsian $H-P-T$-manifolds. Two of these are the enthalpy surfaces for the gas and the liquid. The third is a line of enthalpies at constant pressure and temperature; successive points on this line represent different mole fractions of liquid water.
We can track the reversible conversion of one mole of pure liquid water to pure gas on other thermodynamic surfaces. For example, if we consider enthalpy as a function of volume and temperature, the entire process can be traced on a single $H\left(V,T\right)$ surface; that is, every volume–temperature point, $\left(V,T\right)$, specifies a unique state of the system, and conversely.
When we use pressure and temperature as the independent variables, the Gibbs free energy provides the criterion for reversibility. Unlike the corresponding enthalpy surfaces, which never meet, the Gibbs free energy surfaces for the pure liquid and the pure gas intersect along a line of pressure and temperature values. At an equilibrium pressure and temperature, the Gibbs free energy change for the reversible vaporization of water is zero, which means that the Gibbs free energy for a mole of liquid water is the same as the Gibbs free energy for a mole of gaseous water at that pressure and temperature.
When we trace the reversible conversion of a mole of liquid water to a mole of gaseous water on the Gibbs free energy surfaces, the point representing the state of the mole of water moves on the Gibbs free energy surface of the liquid from the initial pressure and temperature to the pressure–temperature equilibrium line. The pressure–temperature equilibrium line is formed by the intersection of the Gibbs free energy surface of the liquid with the Gibbs free energy surface of the gas. (The projection of this line of intersection onto the $P--T$-plane is a line of points in the $P--T$-plane that satisfies the differential relationship $\mathrm{0=-}{\mathrm{\Delta }}_{vap}SdT + {\mathrm{\Delta }}_{vap}VdP$. The point $P\mathrm{\ =\ 1\ }\mathrm{atm}$, $T = \mathrm{373.15\ K}$, lies on this line.)
The conversion of liquid to gas can occur while the mole of water remains at the same point in pressure–temperature–Gibbs free energy space, and the mole fractions, $x_{\ell }$ and $\mathrm{1-}x_{\ell }$, vary continuously over the range $\mathrm{0<}x_{\ell }\mathrm{<1}$. During this reversible vaporization process, $dG = dT = dP\mathrm{=0}$, while ${\mathrm{\Delta }}_{vap}V\mathrm{>0}$, ${\mathrm{\Delta }}_{vap}H\mathrm{>0}$, and ${\mathrm{\Delta }}_{vap}S\mathrm{>0}$. (We find ${\mathrm{\Delta }}_{vap}H$ and ${\mathrm{\Delta }}_{vap}S$ by measuring the heat required to vaporize a mole of water at $P$ and $T$. Then $q_P = {\mathrm{\Delta }}_{vap}H$, and ${q_P}/{T} = {\mathrm{\Delta }}_{vap}S$. Since the process is reversible, we have ${\mathrm{\Delta }}_{vap}S\mathrm{=-}{\mathrm{\Delta }}_{vap}\hat{S}$, and ${\mathrm{\Delta }}_{vap}S + {\mathrm{\Delta }}_{vap}\hat{S}\mathrm{=0}$.) When the conversion of liquid to gas is complete, reversible changes to the one mole of gaseous water correspond to motion of a point on the Gibbs free energy surface of the gas.
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When only pressure–volume work is possible, various pairs of state functions can specify the state of any closed equilibrium system. A given pair may be sufficient to specify the state on a given Gibbsian manifold but not to specify the state of the same system on a different Gibbsian manifold. (In Section 10.7, we see that the set $\left\{T,P\right\}$ is sufficient to specify the state of a closed system containing only liquid water or only gaseous water. However, specifying T and P is not sufficient to establish the amount of water vapor in a closed system in which the liquid is vaporizing reversibly.)
It can also happen that a given pair of state functions specifies the state of a closed system over part but not all of a given Gibbsian manifold. Specifying the values of such a pair is not sufficient to describe the entire manifold. In particular, we can show that neither the set $\left\{P,V\right\}$ nor the set $\left\{T,S\right\}$ is always a sufficient pair in this sense.
When we say that specifying $P$ and $V$ is sufficient to specify the state of a system on a particular Gibbsian manifold, we mean that any state function, $M$, must be uniquely specified when $P$ and $V$ are specified; a single-valued function, $M\left(P,V\right)$, must exist. Conversely, if for any choice of $M$ in any system, $M\left(P,V\right)$ is not single-valued,$\ P$ and $V$ are not always a sufficient set. In §6, we see how to find the total differential, $dM={\left(\frac{\partial M}{\partial P}\right)}_VdP+{\left(\frac{\partial M}{\partial V}\right)}_PdV \nonumber$
It might seem that this is sufficient to ensure that specifying P and V always enables us to find $M\left(P,V\right)$ relative to its value in an initial reference state $M\left(P_1,V_1\right)$. To do so, we need only evaluate $dM$ as a line integral along some reversible path that leads from $\left(P_1,V_1\right)$ to $\left(P_2,V_2\right)$. However, we can evaluate this line integral only if both of the partial derivatives can be integrated. If one of the partial derivatives is undefined along any path that connects $\left(P_1,V_1\right)$ to $\left(P_2,V_2\right)$, we cannot find $M\left(P,V\right)$ by this method.
Let us consider a closed reversible system that consists of one mole of liquid water. At ordinary pressures, the density of liquid water is not a monotonic function of temperature. At one atmosphere, the density of liquid water reaches a maximum at 4 C. Therefore, at a pressure of one atmosphere, the molar volume of water is a minimum at 4 C, as indicated in Figure 2. This means that, at one atmosphere and a range of volumes, liquid water can be at either of two temperatures for specified values of $P$ and $\overline{V}$. Therefore, specifying $P$ and $\overline{V}$ does not specify $T$; temperature is not a single-valued function of pressure and volume; we cannot uniquely express the temperature as the required function $T=T\left(P,\overline{V}\right)$. Moreover, because the density has a maximum, we have
${\left(\frac{\partial \overline{V}}{\partial T}\right)}_P=0 \nonumber$
at this maximum, and it follows that
${\left(\frac{\partial T}{\partial \overline{V}}\right)}_P \nonumber$
is not defined at this temperature and pressure. In Section 10.6, we find
${\left(\frac{\partial \overline{S}}{\partial \overline{V}}\right)}_P={\frac{C_P}{T}\left(\frac{\partial T}{\partial \overline{V}}\right)}_P \nonumber$ so that ${\left({\partial \overline{S}}/{\partial \overline{V}}\right)}_P$ is also undefined. Hence, we cannot evaluate $\Delta \overline{S}$ by evaluating the line integral of $d\overline{S}\left(P,\overline{V}\right)$ along any path that includes a point of maximum density. These examples show that pressure and volume are not sufficient to describe the entire Gibbsian manifold for liquid water.
Temperature and entropy are likewise not sufficient. Since we have
$d\overline{S} = \frac{C_P}{T}dT + {\left(\frac{\partial \overline{V}}{\partial T}\right)}_PdP \nonumber$
the total differential for pressure as a function of entropy and temperature is
$dP = \left[\frac{C_P}{T}dT + d\overline{S}\right]{\left(\frac{\partial T}{\partial \overline{V}}\right)}_P \nonumber$
so that $dP$ is not defined at pressures and temperatures of maximum density. Consequently, we cannot express the pressure as $P=P\left(\overline{S},T\right)$ over the entire liquid region.
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In Chapter 7, we derive the relationship between $C_P$ and $C_V$ for an ideal gas. It is useful to have a relationship between these quantities that is valid for any substance. We can derive this relationship from the equations for $dS$ that we develop in Sections 10.4 and 10.5. If we apply the divide-through rule to $dS$ expressed as a function of $dT$ and dV, at constant pressure, we have
${\left(\frac{\partial S}{\partial T}\right)}_P = \frac{C_V}{T} + {\left(\frac{\partial P}{\partial T}\right)}_V{\left(\frac{\partial V}{\partial T}\right)}_P \nonumber$
From $dS$ expressed as a function of $T$ and $P$,
$dS = \frac{C_P}{T}dT + {\left(\frac{\partial V}{\partial T}\right)}_PdP \nonumber$
we have
${\left(\frac{\partial S}{\partial T}\right)}_P = \frac{C_P}{T} \nonumber$ so that $\frac{C_P}{T} = \frac{C_V}{T} + {\left(\frac{\partial P}{\partial T}\right)}_V{\left(\frac{\partial V}{\partial T}\right)}_P \nonumber$
and
$C_P + C_V = T{\left(\frac{\partial P}{\partial T}\right)}_V{\left(\frac{\partial V}{\partial T}\right)}_P \label{eq10}$
For an ideal gas, the right side of Equation \ref{eq10} reduces to $R$, in agreement with our previous result. Note also that, for any substance, $C_P$ and $C_V$ become equal when the temperature goes to zero.
The partial derivatives on the right hand side can be related to the coefficients of thermal expansion, $\alpha$, and isothermal compressibility, $\beta$. Using
$\frac{\alpha }{\beta } = {\left(\frac{\partial P}{\partial T}\right)}_V \nonumber$
we can write the relationship between $C_P$ and $C_V$ as
$C_P + C_V = \frac{VT{\alpha }^{\mathrm{2}}}{\beta } \nonumber$
10.10: The Dependence of Cv on Volume and of Cp on Pressu
The heat capacities of a substance increase with temperature. The rate of increase decreases as the temperature increases. To achieve adequate accuracy in calculations, we often need to know how heat capacities depend on temperature. In contrast, the dependence of heat capacities on pressure and volume is usually negligible; that is, the dependence of $C_V$ on $V$ and the dependence of $C_P$ on $P$ can usually be ignored. Nevertheless, we need to know how to find them.
An exact equation for the dependence of $C_V$ on $V$ follows readily from $dS$ expressed as a function of $dT$ and $dV$
$dS=\frac{C_V}{T}dT+{\left(\frac{\partial P}{\partial T}\right)}_VdV \nonumber$
Since the mixed second-partial derivatives must be equal, we have
${\left[\frac{\partial }{\partial V}\left(\frac{C_V}{T}\right)\right]}_T = {\left[\frac{\partial }{\partial T}{\left(\frac{\partial P}{\partial T}\right)}_V\right]}_V \nonumber$
and thus
$\left(\frac{\partial C_V}{\partial V}\right)_T = T \left(\frac{\partial^{\mathrm{2}}P}{\partial T^{\mathrm{2}}}\right)_V \nonumber$
Similarly, the dependence of $C_P$ on $P$ follows from $dS$ expressed as a function of $dT$ and $dP$,
$dS = \frac{C_P}{T}dT + {\left(\frac{\partial V}{\partial T}\right)}_PdP \nonumber$ Equating the mixed second-partial derivatives, we have
$\left[\frac{\partial }{\partial P}\left(\frac{C_P}{T}\right)\right]_T = \left[ + \frac{\partial }{\partial T}{\left(\frac{\partial V}{\partial T}\right)}_P\right]_P \nonumber$
and thus
$\left(\frac{\partial C_P}{\partial P}\right)_T = -T\left(\frac{\partial ^{\mathrm{2}}V}{\partial T^{\mathrm{2}}}\right)_P \nonumber$
For an ideal gas, it follows that $C_V$ is independent of $V$, and $C_P$ is independent of $P$.
When we use the coefficient of thermal expansion to describe the variation of volume with temperature, we have ${\left(\frac{\partial V}{\partial T}\right)}_P=\alpha V \nonumber$
When it is adequate to approximate $\alpha$ as a constant, another partial differentiation with respect to temperature gives
${\left(\frac{\partial C_P}{\partial P}\right)}_T = -T{\left(\frac{\partial \left(\alpha V\right)}{\partial T}\right)}_P = -{\alpha }^{\mathrm{2}}TV \nonumber$
Since $\mathrm{\alpha }$ is normally small, this result predicts weak dependence of $C_P$ on $P$. If $\mathrm{\alpha }$ and $\mathrm{\beta }$ are both adequately approximated as constants, we have from
${\left(\frac{\partial P}{\partial T}\right)}_V=\frac{\alpha }{\beta } \nonumber$ that ${\left(\frac{\partial C_V}{\partial V}\right)}_T = T{\left(\frac{\partial \left({\alpha }/{\beta }\right)}{\partial T}\right)}_V\mathrm{=0} \nonumber$
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When temperature and pressure are the independent variables, the Gibbs free energy is the change criterion that takes the most simple form:
$dG =-SdT + VdP. \nonumber$
In chemical applications, temperature and pressure are often the most convenient choice of independent variables, making the Gibbs free energy a particularly useful function. Constant Gibbs free energy is the criterion for equilibrium at constant pressure and temperature. The Gibbs free energy of the system does not change when the system goes from one equilibrium state to another at the same temperature and pressure, ${\left(dG\right)}_{PT}\mathrm{=0}$. An equilibrium system of ice and water is one example; we can melt a portion of the ice, changing the composition of the system, while maintaining equilibrium at constant pressure and temperature. Similarly, we may be able to change the equilibrium composition of an equilibrium system that consists of reacting gases by changing the volume of the system while maintaining constant pressure and temperature.
Consider a system that undergoes some arbitrary change from a state A, in which its Gibbs free energy is $G_A$, to a second state B, in which its Gibbs free energy is $G_B$. In general, $G_A\mathrm{\neq }G_B$; in the most general case, the pressures and temperatures of states A and B are different. (For example, state A might be a mole of ice at $-\mathrm{10}$ C and $\mathrm{0.5\ bar}$, while state B is a mole of water at $+ \mathrm{10\ C}$ and $\mathrm{2.0\ bar}$. Either of these states can be converted to the other; however, they are not at equilibrium with one another, and their Gibbs free energies are not equal.) Representing the pressures and temperatures as $P_{\mathrm{1}}$, $T_{\mathrm{1}}$, $P_{\mathrm{2}}$, $T_{\mathrm{2}}$, we can express the Gibbs free energies of these two state as $G_A = G_A\left(P_{\mathrm{1}},T_{\mathrm{1}}\right)$ and $G_B = G_B\left(P_{\mathrm{2}},T_{\mathrm{2}}\right)$, respectively. The difference is the change in the Gibbs free energy when the system passes from state A to state B:
$\mathrm{\Delta }_{AB}G = G_B\left(P_{\mathrm{2}},T_{\mathrm{2}}\right) - G_A\left(P_{\mathrm{1}},T_{\mathrm{1}}\right) \nonumber$
Often, we are interested in Gibbs free energy differences between states that are at the same pressure and temperature, say $P_{\mathrm{1}}$ and $T_{\mathrm{1}}$. Then the Gibbs free energy difference is
$\mathrm{\Delta }_{AB}G\left(P_1,T_{\mathrm{2}}\right) = G_B\left(P_1,T_{\mathrm{2}}\right) - G_A\left(P_1,T_1\right) \nonumber$
(For example, state A might be a mole of ice at $-\mathrm{10\ C}$ and$\mathrm{\ 0.5\ bar}$, while state B is a mole of water, also at $-\mathrm{10\ C}$ and $\mathrm{0.5\ bar}$. This would be a super-cooled state of liquid water. These states are not at equilibrium with one another, and their Gibbs free energies are not equal. The difference between the Gibbs free energies of these states is the change in the Gibbs free energy when ice goes to super-cooled water at $-\mathrm{10\ C}$ and $\mathrm{0.5\ bar}$.)
Similar considerations apply to expressing differences between the enthalpies and the entropies of two states that are available to a system. The Gibbs free energy is defined by
$G = H - TS. \nonumber$
When we are interested in a process that converts some state A to a second state B at constant pressure and temperature, we usually write
$\mathrm{\Delta }G = \mathrm{\Delta }H - T\mathrm{\Delta }S \nonumber$
which relies on the context for the information about the pressure and temperature and the initial and final states. To explicitly denote that the change is one that occurs at a constant temperature, $T_0$, we can write
$\mathrm{\Delta }G\left(T_0\right) = \mathrm{\Delta }H\left(T_0\right) - T_0\mathrm{\Delta }S\left(T_0\right). \nonumber$
Frequently we are interested in the way that $\mathrm{\Delta }G$, $\mathrm{\Delta }H$, and $\mathrm{\Delta }S$ vary with temperature at constant pressure. If we know how $G$, $H$, and $S$ vary with temperature for each of the two states of interest, we can find the temperature dependence of $\mathrm{\Delta }G$, $\mathrm{\Delta }H$, and $\mathrm{\Delta }S$. The Gibbs-Helmholtz equation is a frequently useful expression of the temperature dependence of $G$ or $\mathrm{\Delta }G$. Since it is a mathematical consequence of our thermodynamic relationships, we derive it here.
At constant pressure, the temperature derivative of the Gibbs free energy is $+ S$; that is,
$\left(\frac{\partial G}{\partial T}\right)_P\mathrm{=} -S \nonumber$
Using this result and the definition, $G = H - TS$, we obtain the temperature dependence of $G$ as
$\left(\frac{\partial G}{\partial T}\right)_P = \frac{G - H}{T} \nonumber$
However, the Gibbs-Helmholtz equation can be expressed most compactly as the temperature derivative of ${G}/{T}$. As a matter of calculus, we have
$\left(\frac{\partial \left(G/T\right)}{ \partial T} \right)_P =- \frac{G}{T^2} + \frac{1}{T} \left( \frac{\partial G}{\partial T}\right)_P \nonumber$
Using the relationships above, this becomes
$\left(\frac{\partial \left(G/T\right)}{\partial T}\right)_P\mathrm{=-}\frac{H + TS}{T^{\mathrm{2}}} + \frac{S}{T}\mathrm{=-}\frac{H}{T^{\mathrm{2}}} \nonumber$
Since $\mathrm{\Delta }_{AB}G = G_B + G_A$, we have
$\left( \partial \mathrm{\Delta }_{AB}G/ \partial T \right)_P =- \left(S_B + S_A\right) =- \mathrm{\Delta }_{AB}S \nonumber$
and
\begin{aligned} \left(\frac{\partial \left( \mathrm{\Delta }_{AB}G/T\right)}{ \partial T} \right)_P & = \left(\frac{\partial \left(G_B/T\right)}{\partial T}\right)_P + \left(\frac{\partial \left(G_A/T\right)}{ \partial T} \right)_P \ ~ & =- \left(\frac{H_B}{T^{\mathrm{2}}} + \frac{H_A}{T^{\mathrm{2}}}\right) \ ~ & =- \frac{\mathrm{\Delta }_{AB}H}{T^2} \end{aligned} \tag{The Gibbs-Helmholtz equation}
If we know the temperature dependence of $\mathrm{\Delta }S$ or $\mathrm{\Delta }H$, we can find the temperature dependence of $\mathrm{\Delta }G$ by integrating the relationships above. That is, given $\mathrm{\Delta }G$ at $T_{\mathrm{1}}$, we can find $\mathrm{\Delta }G$ at $T_{\mathrm{2}}$. Thus, from ${\left({\partial G}/{\partial T}\right)}_P\mathrm{=-}S$, we have
$\int^{\mathrm{\Delta }G\left(T_2\right)}_{\mathrm{\Delta }G\left(T_1 \right)} \left(\frac{\partial \mathrm{\Delta }G}{\partial T}\right)_PdT = \mathrm{\Delta }G\left(T_2 \right) - \mathrm{\Delta }G\left(T_1 \right) =- \int^{T_2}_{T_1} \mathrm{\Delta }S dT \nonumber$
and from $\left( \partial \left( \mathrm{\Delta }G/T\right)/\partial T\right)_P =- \mathrm{\Delta }H/T^2$, we have
$\int^{\mathrm{\Delta }G\left(T_2 \right)/T_2}_{\mathrm{\Delta }G\left(T_1 \right)/T_1} \left(\frac{\partial \left(\mathrm{\Delta }G/T\right)}{ \partial T} \right)_P dT = \frac{\mathrm{\Delta }G\left(T_2 \right)}{T_2} - \frac{\mathrm{\Delta }G\left(T_1 \right)}{T_1} =- \int^{T_2}_{T_1}{\frac{\mathrm{\Delta }H}{T^2}}dT \nonumber$
For small temperature differences, $\mathrm{\Delta }H$ is often approximately constant. Then, we can evaluate the change in $\mathrm{\Delta }G$ from
$\frac{\mathrm{\Delta }G\left(T_{\mathrm{2}}\right)}{T_{\mathrm{2}}} - \frac{\mathrm{\Delta }G\left(T_{\mathrm{1}}\right)}{T_{\mathrm{1}}} = \mathrm{\Delta }H\left(\frac{\mathrm{1}}{T_{\mathrm{2}}} + \frac{\mathrm{1}}{T_{\mathrm{1}}}\right) \nonumber$
Another common application arises when we know $\mathrm{\Delta }G$ at several temperatures. A plot of ${\mathrm{\Delta }G}/{T}$ versus ${\mathrm{1}}/{T}$ is then approximately linear with a slope that approximates the average value of $\mathrm{\Delta }H$ in the temperature interval.
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We make extensive use of the principle that the energy of an ideal gas depends only on temperature when only pressure–volume work is possible. In Chapter 2, we consider the Joule experiment, which provides weak evidence that this principle is correct. In the Joule experiment, no temperature change is observed during the adiabatic free expansion of a gas whose behavior is approximately ideal at the initial temperature and pressure. While this observation supports the principle, the accuracy attainable in the Joule experiment is poor. Otherwise, the most compelling evidence for this principle that we have developed is the theoretical relationship between the pressure–volume product of an ideal gas model and the mean-square velocity of its molecules. We derive this relationship from the Maxwell-Boltzmann distribution law for gas velocities and use the ideal gas equation to find that the mean squared velocity depends only on temperature.
To appreciate the importance of this principle, let us review some of the important steps in our development of the second law. In Chapter 7, we observe that, since its energy depends only on temperature, $C_V$ for an ideal gas must also depend only on temperature; this follows immediately from the definition,$\ C_V={\left({\partial E}/{\partial T}\right)}_V$. In Chapter 9, we use the conclusion that $C_V$ depends only on temperature in our development of the relationships among the heat, work, volume, and temperature changes for an ideal gas traversing a Carnot cycle. In considering these relationships, we observe that the values of the terms ${q^{rev}}/{T}$ for the steps in this cycle sum to zero, as required for a state function. This leads us to define entropy by the differential expression $dS=dq^{rev}/T$ and to infer that the entropy so defined is a state function. Reasoning from the machine-based statement of the second law, we conclude that this inference is correct.
That the energy of an ideal gas depends only on temperature is therefore of central importance to the internal consistency of the thermodynamic theory we have developed. It is easy to demonstrate this internal consistency. From the ideal gas equation and the relationships developed earlier in this chapter, we can show that the quantities ${\left({\partial E}/{\partial V}\right)}_T$, ${\left({\partial E}/{\partial P}\right)}_T$, ${\left({\partial H}/{\partial V}\right)}_T$, ${\left({\partial H}/{\partial P}\right)}_T$, ${\left({\partial C_V}/{\partial V}\right)}_T$, and ${\left({\partial C_P}/{\partial P}\right)}_T$ are all identically zero.
The fact that our theory passes this test of internal consistency is independent of the properties of real gases. However, since we want to make predictions about the behavior of real gases, we need to be able to measure these quantities for real gases. Moreover, because we want to understand the properties of real gases in terms of their molecular characteristics, we want to be able to interpret these quantities for real gases using real-gas models that explain the differences between real gas molecules and ideal gas molecules. The van der Waals equation of state provides a simple model for the effects of attractive and repulsive molecular interactions. In the next section, we first consider simple qualitative arguments about the effects of intermolecular interactions on the energy of a real gas. We then investigate these effects for a van der Waals gas. We see that the van der Waals model and our qualitative arguments are consistent.
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Let us consider the effects that intermolecular forces of attraction and repulsion must have in the adiabatic free expansion $\left(P_{applied}=0\right)$ of a real gas. In such an expansion, no energy can be exchanged between the gas and its surroundings.
Suppose that the molecules of the gas are attracted to one another. Then energy must be expended to separate the molecules as the expansion takes place. (To achieve the expansion, work must be done against the intermolecular attractive forces.) Since the system cannot obtain this energy from its surroundings, it must be obtained by decreasing the translational kinetic energy (and the rotational and vibrational energy) of the gas molecules themselves. This means that the temperature of the gas must decrease during the expansion.
Conversely, if the molecules repel one another, energy is released as the expansion takes place, and the temperature of the gas increases during the expansion. The temperature can remain unchanged after the adiabatic free expansion only if the effects of the intermolecular forces of attraction and repulsion offset one another exactly.
We can express these conclusions more precisely by saying that we expect ${\left({\partial T}/{\partial P}\right)}_E\mathrm{>0}$ if forces of attraction dominate the intermolecular interactions. We expect ${\left({\partial T}/{\partial P}\right)}_E\mathrm{<0}$ if forces of repulsion dominate. Now, as a matter of mathematics, we have
${{{\left(\frac{\partial E}{\partial P}\right)}_T\mathrm{=-}\left(\frac{\partial E}{\partial T}\right)}_P\left(\frac{\partial T}{\partial P}\right)}_E \nonumber$
As a matter of experience, increasing the temperature of any gas at constant pressure always increases the energy of the gas; that is, we observe ${\left({\partial E}/{\partial T}\right)}_P\mathrm{>0}$. It follows that we can expect
${\left(\frac{\partial E}{\partial P}\right)}_T<0 \nonumber$
(attraction dominates)
when intermolecular forces of attraction dominate and
${\left(\frac{\partial E}{\partial P}\right)}_T\mathrm{>0} \nonumber$
(repulsion dominates)
when forces of repulsion dominate.
In the Joule experiment, a gas is allowed to expand into an initially evacuated container. The Joule experiment is a direct test of these ideas; however, as we have noted, it is difficult to carry out accurately. Fortunately, a simple modification of the Joule experiment produces an experiment that is much more sensitive. Instead of allowing the gas to expand freely into a fixed volume, we allow it to expand adiabatically against a constant applied pressure. This is the Joule-Thomson experiment. In the next section, we show that the enthalpy of the gas does not change in such a process. We measure the temperature change as the gas expands adiabatically from an initial, constant, higher pressure to a final, constant, lower pressure. Since this process occurs at constant enthalpy, the Joule-Thomson experiment measures
${\left(\frac{\partial T}{\partial P}\right)}_H \nonumber$
from which we can obtain
${\left(\frac{\partial H}{\partial P}\right)}_T \nonumber$
To interpret the Joule-Thomson experiment in terms of intermolecular forces, we need to show that
${\left(\frac{\partial H}{\partial P}\right)}_T<0 \nonumber$
(attraction dominates)
at pressures and temperatures where intermolecular forces of attraction dominate and
${\left(\frac{\partial H}{\partial P}\right)}_T>0 \nonumber$
(repulsion dominates)
where forces of repulsion dominate. To do this using an explicit mathematical model for a real gas, let us find
${\left(\frac{\partial E}{\partial P}\right)}_T \nonumber$ and ${\left(\frac{\partial H}{\partial P}\right)}_T \nonumber$
for a van der Waals gas. Writing van der Waals equation in terms of the molar volume, $\left(P+{a}/{{\overline{V}}^2}\right)\left(\overline{V}-b\right)=RT$, and introducing
$\gamma \left(P,\overline{V}\right)=P-\frac{a}{{\overline{V}}^2}+\frac{2ab}{{\overline{V}}^3} \nonumber$
so that we can express the results more compactly, we find
${\left(\frac{\partial P}{\partial \overline{V}}\right)}_T\mathrm{=-}\frac{\gamma \left(P,\overline{V}\right)}{\overline{V} + b} \nonumber$ so that ${\left(\frac{\partial \overline{V}}{\partial P}\right)}_T\mathrm{=-}\frac{\overline{V} + b}{\gamma \left(P,\overline{V}\right)} \nonumber$ and ${\left(\frac{\partial \overline{V}}{\partial T}\right)}_P = \frac{R}{\gamma \left(P,\overline{V}\right)} \nonumber$
Substituting into results we develop in Section 10.5, we have
$\left(\frac{\partial \overline{E}}{\partial P}\right)_T =-P\left(\frac{\partial \overline{V}}{\partial P}\right)_T - T \left(\frac{\partial \overline{V}}{\partial T}\right)_P =- \frac{a\left(\overline{V} - b\right)}{\overline{V}^2 \gamma \left(P,\overline{V}\right)} \nonumber$
and
$\left(\frac{\partial \overline{H}}{\partial P}\right)_T = \overline{V} + T \left(\frac{\partial \overline{V}}{\partial T}\right)_P = \overline{V} + \frac{RT}{\gamma \left(P,\overline{V}\right)} \nonumber$
We introduce van der Waals equation in Section 2.12. By the argument we make there, the $\left(P+a/ \overline{V}^2 \right)$ term models the effects of attractive intermolecular interactions when $a>0$. By a parallel argument, we can see that it models the effects of repulsive interactions when $a<0$. Parameter $b$ models the effects of intermolecular repulsive interactions that come into play when the molecules come into contact with one another. For present purposes, we can consider molecules for which $b=0$; this simplifies our equations without affecting the description they give of the phenomena that are of current interest.
This gives us a model in which the effects of intermolecular interactions are described by the values of a single parameter that has a straightforward physical interpretation. Thus, we can write
$\left(P + a/ \overline{V}^2 \right)\overline{V} = RT \nonumber$
to describe a gas of point-mass molecules that experience intermolecular forces. When $a>0$, these forces are attractive; when $a<0$ they are repulsive. (For any given real gas, our equation can only be an approximation that is valid over a limited range of conditions. In some ranges, $a>0$; in others, $a<0$.) With $b=0$ we must have
$\gamma \left(P,\overline{V}\right)=P-\frac{a}{\overline{V}^2}>0 \nonumber$
(If $\gamma \left(P,\overline{V}\right)\le 0$, we have $\left( \partial P/ \partial \overline{V}\right)_T\ge 0$. As a matter of experience, the pressure of a gas always decreases with increasing volume at constant temperature. It follows that van der Waals equation with $b=0$ and $\left(P-a/ \overline{V}^2\right)\le 0$ cannot describe any gas.) With $b=0$ we have
$\left(\frac{\partial \overline{E}}{\partial P}\right)_T=\frac{-a}{\overline{V}\left(P-a/\overline{V}^2 \right)} \nonumber$
and
$\left(\frac{\partial \overline{H}}{\partial P}\right)_T=RT\left[\frac{1}{P+a/\overline{V}^2}- \frac{1}{P-a/\overline{V}^2} \right] \nonumber$
For a gas at conditions in which forces of attraction dominate, we have $a>0$, so that
$\left(\partial \overline{E}/ \partial P\right)_T<0$ and $\left( \partial \overline{H}/ \partial P\right)_T<0$
(attraction dominates)
Conversely, at conditions in which forces of repulsion dominate, we have $a<0$, and
$\left(\partial \overline{E}/\partial P\right)_T>0$ and $\left( \partial \overline{H}/\partial P\right)_T>0$
(repulsion dominates)
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In practice, the Joule-Thomson experiment is done by allowing gas from a pressure vessel to pass through an insulated tube. The tube contains a throttling valve or a porous plug through which gas flows slowly enough so that the gas upstream from the plug is at a uniform pressure $P_{\mathrm{1}}$, and the gas downstream is at a uniform pressure $P_{\mathrm{2}}$. In general, the temperature of the downstream gas is different from that of the upstream gas. Depending on the initial temperature and pressure, the pressure drop, and the gas, the temperature of the gas can either decrease or increase as it passes through the plug. (We see below that it must be constant if the gas is ideal.)
The temperature change is called the Joule-Thomson effect. The enthalpy of the gas remains constant. If the measured temperature and pressure changes are $\mathrm{\Delta }T$ and $\mathrm{\Delta }P$, their ratio is called the Joule-Thomson coefficient, ${\mu }_{JT}$. We define
$\mathrm{\ }{\mu }_{JT} = {\left(\frac{\partial T}{\partial P}\right)}_H\mathrm{\approx }\frac{\mathrm{\Delta }T}{\mathrm{\Delta }P} \nonumber$
To see that the enthalpy of the gas is the same on both sides of the plug, we consider an idealized version of the experiment, in which the flow of gas through the plug is controlled by the coordinated movement of two pistons. (See Figure 3.) We suppose that the gas is pushed through the plug in such a way that the upstream pressure remains constant at $P_1$ and the downstream pressure remains constant at $P_2$. Let us consider the changes that result when one mole of gas passes through the plug under these conditions. Initially, there are $n_1+1$ moles of gas on the upstream side at a pressure $P_1$, occupying a volume $\left(n_1+1\right){\overline{V}}_1$, at a temperature $T_1$, and having an energy per mole of ${\overline{E}}_1$. On the downstream side, there are $n_2$ moles of gas at a pressure $P_2$, occupying a volume $n_2{\overline{V}}_2$, but having a temperature $T_2$ and an energy per mole of ${\overline{E}}_2$.
When the process is complete, there are $n_1$ moles of gas on the upstream side, still at a pressure $P_1$ and temperature $T_1$, but occupying a volume $n_1{\overline{V}}_1$. On the downstream side, there are $n_2+1$ moles of gas at pressure $P_2$, occupying volume $\left(n_2+1\right){\overline{V}}_2$ at a temperature $T_2$ and with an energy per mole of ${\overline{E}}_2$. On the upstream side, $\Delta E_1=-\overline{E}_1$ and $w_1=-P_1\left[n_1 \overline{V}_1-\left(n_1+1\right) \overline{V}_1 \right]=P_1 \overline{V}_1$
On the downstream side, $\Delta E_2={\overline{E}}_2$, and
$w_2=-P_2\left[\left(n_2+1\right){\overline{V}}_1-n_2{\overline{V}}_1\right]=-P_2{\overline{V}}_2 \nonumber$
Since the process is adiabatic, any heat taken up by the upstream gas must be surrendered by the downstream gas, so that $q_1+q_2=0$. For the process of moving the mole of gas across the plug,
$\Delta E=\Delta E_1+\Delta E_2=-\overline{E}_1+\overline{E}_2=q_1+q_2+w_1+w_2=P_1\overline{V}_1-P_2\overline{V}_2 \nonumber$
from which
${\overline{E}}_1+P_1{\overline{V}}_1={\overline{E}}_2+P_2{\overline{V}}_2 \nonumber$ or ${\overline{H}}_1={\overline{H}}_2 \nonumber$
so that we have $\Delta H=0$ for the expansion.
In practice, it is convenient to measure downstream pressures and temperatures, $P_2$ and $T_2$, in a series of experiments in which the upstream pressure and temperature, $P_1$ and $T_1$, are constant. The enthalpy of the gas is the same at each of these pressure-temperature points. A graph of these points is an isenthalpic (constant enthalpy) curve. At any given pressure and temperature, the Joule-Thomson coefficient, ${\mu }_{JT}$, is the slope of this curve.
We can also express $\mathrm{\ }{\mu }_{JT}$ as a function of the heat capacity, $C_P$, and the coefficient of thermal expansion, $\alpha$, where $\alpha =V^{-1}{\left({\partial V}/{\partial T}\right)}_P$. We begin by expressing $d\overline{H}$ as a function of temperature and pressure:
$d\overline{H}={\left(\frac{\partial \overline{H}}{\partial T}\right)}_PdT+{\left(\frac{\partial \overline{H}}{\partial P}\right)}_TdP \nonumber$
If we divide through by $dP$ and hold $\overline{H}$ constant, we obtain
$0={\left(\frac{\partial \overline{H}}{\partial T}\right)}_P{\left(\frac{\partial T}{\partial P}\right)}_{\overline{H}}+{\left(\frac{\partial \overline{H}}{\partial P}\right)}_T \nonumber$
so that
${\mu }_{JT}={\left(\frac{\partial T}{\partial P}\right)}_{\overline{H}}=-{{\left(\frac{\partial \overline{H}}{\partial P}\right)}_T}/{{\left(\frac{\partial \overline{H}}{\partial T}\right)}_P}=-\frac{1}{C_P}{\left(\frac{\partial \overline{H}}{\partial P}\right)}_T \nonumber$
If we substitute the coefficient of thermal expansion into the expression for ${\left({\partial \overline{H}}/{\partial P}\right)}_T$ that we develop in Section 10.5, we have
${\left(\frac{\partial \overline{H}}{\partial P}\right)}_T=\overline{V}-T{\left(\frac{\partial \overline{V}}{\partial T}\right)}_P=\overline{V}-\alpha \overline{V}T=\overline{V}\left(1-\alpha T\right) \nonumber$
For an ideal gas, ${\left({\partial \overline{V}}/{\partial T}\right)}_P={\overline{V}}/{T}$, so that both ${\left({\partial \overline{H}}/{\partial P}\right)}_T$ and ${\mu }_{JT}$ are zero. For real gases, we substitute into the expression for ${\mu }_{JT}$ to find
${\mu }_{JT}={\left(\frac{\partial T}{\partial P}\right)}_{\overline{H}}=-\frac{1}{C_P}{\left(\frac{\partial \overline{H}}{\partial P}\right)}_T=-\frac{\overline{V}}{C_P}\left(1-\alpha T\right) \nonumber$
Given $\overline{V}$ and any two of ${\mu }_{JT}$, $C_P$, or $\alpha$, we can find the third from this relationship.
Making the same substitutions using the partial derivatives we found above for a van der Waals gas, we find
${\mu }_{JT}=-\frac{1}{C_P}\left(\overline{V}-\frac{RT}{\gamma \left(P,\overline{V}\right)}\right) \nonumber$
Given that the van der Waals equation oversimplifies the effects of intermolecular forces, we can anticipate that calculation of the Joule-Thomson coefficient from the van der Waals parameters is likely to be qualitatively correct, but in poor quantitative agreement with experimental results. Figure 4 compares calculated and experimental curves for the Joule-Thomson coefficient of nitrogen gas at 0 C from 1 to 200 bar. (Calculated values take $a=0.137\mathrm{\ Pa}\ {\mathrm{m}}^{\mathrm{6}}\mathrm{\ }{\mathrm{mol}}^{--\mathrm{2}}$ and $b=3.81\times {10}^{-5}\ {\mathrm{m}}^{\mathrm{3}}\mathrm{\ }{\mathrm{mol}}^{--\mathrm{1}}$. The experimental data are from reference 1.)
We anticipate that the Joule-Thomson coefficient becomes zero at pressures and temperatures where the effects of intermolecular attractions and repulsions exactly offset one another. For interactions between molecules, attractive forces have the dominant effect at long distances, while repulsive forces dominate at short distances. The lower the pressure, the greater the average distance between gas molecules. Therefore, at any given temperature and a sufficiently low pressure, the effects of intermolecular attractive forces are more important than those of intermolecular repulsive forces. At low pressures, the Joule-Thomson coefficient should be positive. As the pressure increases, the effects of both attractive and repulsive forces must both increase, but at a sufficiently high pressure, the average intermolecular distance becomes so small that the effects of intermolecular repulsive forces become dominant. Therefore, we anticipate that the Joule-Thomson coefficient decreases as the pressure increases, eventually becoming negative.
Experiments confirm these expectations. A temperature and pressure at which the Joule-Thomson coefficient becomes zero is called a Joule-Thomson inversion point. The experimentally determined curve for nitrogen gas${}^{1}$ is graphed in Figure 5. The van der Waals model also exhibits this effect. The inversion curve can be found from the expression for ${\mu }_{JT}$ developed above for a van der Waals gas. The inversion curve for nitrogen that is found in this way is also graphed in Figure 5. Qualitatively, the agreement is a satisfying confirmation of the basic interpretation that we have given for the role of intermolecular forces. Quantitatively, the agreement is poor, as we expect given the overly simple character of the van der Waals model.
The Joule-Thomson coefficient for an ideal gas is zero, and we normally expect the properties of real gases to approach those of an ideal gas as the pressure falls to zero. However, both experiment and the van der Waals model indicate that the Joule-Thomson coefficient converges to a finite value as the pressure decreases to zero at a fixed temperature. A statistical thermodynamic model${}^{2}$ also predicts this outcome. This model calculates the coefficients in the virial equation of state. In it, the second virial coefficient reflects the net effect of attractive and repulsive forces between a pair of molecules, and it is the second virial coefficient and its temperature derivative determine that the value of ${\left({\partial \overline{H}}/{\partial P}\right)}_T$. (Higher-order virial coefficients reflect interactions among larger numbers of molecules.)
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1. Show that ${\left(\frac{\partial \left({A}/{T}\right)}{\partial T}\right)}_V=-\frac{E}{T^2} \nonumber$
2. At 60 C, the vapor pressure of water is 19,932 Pa, and the enthalpy of vaporization is 42.482 kJ mol${}^{-1}$.
(a) Is the vaporization of water at these conditions impossible, spontaneous, or reversible? What is $\Delta G$ for this process?
(b) Estimate $\Delta G$ for the vaporization of liquid water at 19,932 Pa and 70 C. Is this process impossible, spontaneous, or reversible?
(c) Estimate $\Delta G$ for the vaporization of liquid water at 19,932 Pa and 50 C. Is this process impossible, spontaneous, or reversible?
3. At 298.15 K and 1 bar, the Gibbs free energy of one mole of $N_2O_4$ is 4.729 kJ less than the Gibbs free energy of two moles of $NO_2$. The enthalpy of one mole of $N_2O_4$ is 57.111 kJ less than the enthalpy of two moles of $NO_2$. We customarily express these facts by saying that the Gibbs free energy and the enthalpy changes for the reaction $2\ NO_2\to N_2O_4$ are ${\Delta }_rG^o\left(298.15\ \mathrm{K}\right)=-4.729\ \mathrm{kJ}$ and ${\Delta }_rH^o\left(298.15\ \mathrm{K}\right)=-57.111\ \mathrm{kJ}$. Assume that the enthalpy change for this process is independent of temperature. Estimate the Gibbs free energy change for this reaction at 500 K and 1 bar, ${\Delta }_rG^o\left(500\ \mathrm{K}\right)$.
4. Over the temperature range $300\ \mathrm{K}, the Gibbs free energy change for the formation of ammonia from the elements, \({\frac{1}{2}}N_2+{\frac{3}{2\ }} H_2\to NH_3$, is well approximated by ${\Delta }_fG^o\left(NH_3\right)=a+b\left(T-600\right)+c{\left(T-600\right)}^2+d{\left(T-600\right)}^3 \nonumber$ where $a=15.824\ \mathrm{kJ}$, $b=0.1120\ \mathrm{kJ\ }{\mathrm{K}}^{--\mathrm{1}}$, $c=1.316\times {10}^{-5}\ \mathrm{kJ\ }{\mathrm{K}}^{--\mathrm{2}}$, and $d=-1.324\times {10}^{-8}\ \mathrm{kJ\ }{\mathrm{K}}^{--\mathrm{3}}$. Estimate the enthalpy change for this process, ${\Delta }_fH^o\left(NH_3\right)$, at 600 K.
5. Consider the total differentials for $S=S\left(P,T\right)$, $E=E\left(P,T\right)$, $H=H\left(P,T\right)$, $A=A\left(P,T\right)$, and $G=G\left(P,T\right)$. Can we ever encounter an undefined integrand when we evaluate the line integral of one of these total differentials between any two points $\left(P_1T_1\right)$ and $\left(P_2T_2\right)$? (In the next chapter, we find that, because of the third law of thermodynamics, no real system can ever reach the absolute zero of temperature.)
6. Consider the total differentials for $S=S\left(V,T\right)$, $E=E\left(V,T\right)$, $H=H\left(V,T\right)$, $A=A\left(V,T\right)$, and $G=G\left(V,T\right)$. Can we ever encounter an undefined integrand when we evaluate the line integral of one of these total differentials between any two points $\left(V_1T_1\right)$ and $\left(V_2T_2\right)$?
7. The normal boiling point of methanol is 337.8 K at 1 atm. The enthalpy of vaporization at the normal boiling point is ${\Delta }_{vap}H=35.21\ \mathrm{kJ}\mathrm{\ }{\mathrm{mol}}^{--\mathrm{1}}$. Is the process impossible, spontaneous, or reversible? Find $q$, $w$, ${\Delta }_{vap}E$, ${\Delta }_{vap}S$, ${\Delta }_{vap}A$, ${\Delta }_{vap}G$ for the vaporization of one mole of methanol at the normal boiling point. Assume that methanol vapor behaves as an ideal gas.
8. For $S=S\left(P,V\right)$, we obtain ${\left(\frac{\partial S}{\partial V}\right)}_P=\frac{C_P}{T}{\left(\frac{\partial T}{\partial V}\right)}_P \nonumber$ For $S=S\left(P,T\right)$, we obtain ${\left(\frac{\partial S}{\partial T}\right)}_P=\frac{C_P}{T} \nonumber$ For temperatures near 4 C and at a pressure of 1 atm, the molar volume of water is given by $\overline{V}={\overline{V}}_4+a{\left(T-277.15\right)}^2 \nonumber$ where ${\overline{V}}_4=1.801575\times {10}^{-6}\ {\mathrm{m}}^{\mathrm{3}}\mathrm{\ }{\mathrm{mol}}^{--\mathrm{1}}$ and $a=1.45\times {10}^{-11}\ {\mathrm{m}}^{\mathrm{3}}\mathrm{\ }{\mathrm{K}}^{-\mathrm{1}}$. The heat capacity of liquid water is 75.49 J mol${}^{-1}$.
(a) Using ${\left({\partial S}/{\partial T}\right)}_P$, calculate the entropy change when one mole of water is warmed from 2 C to 6 C while the pressure is constant at 1 atm.
(b) Repeat the calculation in (a), for warming the water from 4 C to 6 C.
(c) Can we calculate the entropy change when one mole of water is warmed from 2 C to 6 C using $\left(\partial S/\partial V\right)_P$? Why, or why not? The required integral can be transformed to $\int \frac{A\ du}{u^{1/2}+Bu}=\left(\frac{2A}{B}\right) \ln \left(1+\beta u^{1/2}\right) +C \nonumber$
where $C$ is an arbitrary constant.
(d) Using ${\left({\partial S}/{\partial V}\right)}_P$, calculate the entropy change when one mole of water is warmed from 4 C to 6 C. Compare this result to the value obtained in (b).
9. For an ideal gas, show that ${\left({\partial E}/{\partial V}\right)}_T$, ${\left({\partial E}/{\partial P}\right)}_T$, ${\left({\partial H}/{\partial V}\right)}_T$, ${\left({\partial H}/{\partial P}\right)}_T$, ${\left({\partial C_V}/{\partial V}\right)}_T$, and ${\left({\partial C_P}/{\partial P}\right)}_T$
are all zero.
10. Find ${\left({\partial E}/{\partial P}\right)}_T$ for a gas that obeys the virial equation of state $P\left[\overline{V}-B\left(T\right)\right]=RT$, in which $B(T$) is a function of temperature.
11. Derive the following relationships for an ideal gas:
(a) $dE=C_VdT$
(b) $dS=\left({C_V}/{T}\right)dT+\left({R}/{V}\right)dV$
(c) $dS=\left({C_P}/{T}\right)dT-\left({R}/{P}\right)dP$
12. Derive the following relationships for a gas that obeys the virial equation, $P\left[\overline{V}-B\left(T\right)\right]=RT$, where $B\left(T\right)$ is a function of temperature:
(a) $d\overline{E}=C_VdT-\left[\frac{RT}{\overline{V}-B}+\frac{RT}{\overline{V}-B}\left(\frac{dB}{dT}\right)-\frac{RT^2}{{\left(\overline{V}-B\right)}^2}\left(\frac{dB}{dT}\right)\right]d\overline{V} \nonumber$ (b) $d\overline{E}=\left[C_P-R-P\left(\frac{dB}{dT}\right)\right]dT-T\left(\frac{dB}{dT}\right)dP \nonumber$ (c) $d\overline{S}=\frac{C_V}{T}dT+\left[\frac{R}{\overline{V}-B}+\frac{RT}{{\left(\overline{V}-B\right)}^2}\left(\frac{dB}{dT}\right)\right]d\overline{V} \nonumber$ (d) $\ d\overline{S}=\frac{C_P}{T}dT-\left[\frac{R}{P}+\left(\frac{dB}{dT}\right)\right]dP \nonumber$
13. One mole of a monatomic ideal gas ($C_V={3R}/{2}$), originally at 10 bar and 300 K (state A), undergoes an adiabatic free expansion against a constant applied pressure of 1 bar to reach state B. Thereafter the gas is warmed reversibly at constant volume back to 300 K, reaching state C. Finally, the warmed gas is compressed reversibly and isothermally to the original pressure. What is are the temperature and volume in state B, following the original adiabatic free expansion? Find $q$, $w$, $\Delta E$, $\Delta H$, and $\Delta S$ for each of the steps and for the cycle A$\mathrm{\to }$B$\mathrm{\to }$C$\mathrm{\to }$A.
14. As in problem 13, one mole of a monatomic ideal gas ($C_V={3R}/{2}$), originally at 10 bar and 300 K (state A), undergoes an adiabatic free expansion against a constant applied pressure of 1 bar to reach state B. The gas is then returned to its original state in a different two-step process. From state A a reversible constant-pressure warming takes the gas to state D at the original temperature of 300 K. The gas is then returned to state A by an isothermal compression to the original volume. What are the temperature and volume after the constant-pressure warming step? Find $q$, $w$, $\Delta E$, $\Delta H$, and $\Delta S$ for each of the steps and for the cycle A$\mathrm{\to }$B$\mathrm{\to }$D$\mathrm{\to }$A.
15. As in problem 13, one mole of a monatomic ideal gas ($C_V={3R}/{2}$), originally at 10 bar and 300 K (state A), undergoes an adiabatic free expansion against a constant applied pressure of 1 bar to reach state B. Now consider a reversible adiabatic expansion from the same initial state, A, that reaches the same temperature as the gas in state B. Call this state F. Find $q$, $w$, $\Delta E$, $\Delta H$, and $\Delta S$ for the step A$\mathrm{\to }$F. Find $q$, $w$, $\Delta E$, $\Delta H$, and $\Delta S$ for reversible isothermal expansion from state F to state B. What are $q$, $w$, $\Delta E$, $\Delta H$, $\Delta S$, and $\Delta \hat{S}$ for the cycle A$\mathrm{\to }$F$\mathrm{\to }$B$\to$A. Does this cycle violate the machine-based statement of the second law?
16. One mole of carbon dioxide, originally at 10 bar and 300 K, is taken around the cycle in problem 13. Find the energy and entropy changes for the steps in this cycle using the ideal gas equation and the temperature-dependent heat capacity. The constant-volume heat capacity is $C_V=14.7+0.046\times T$. Find $q$, $w$, $\Delta E$, $\Delta H$, and $\Delta S$ for each of the steps and for the cycle when $CO_2$ is taken around the cycle A$\mathrm{\to }$B$\mathrm{\to }$C$\mathrm{\to }$A.
17. Ten moles of a monatomic ideal gas, initially occupying a volume of 30 L at 25 C, is expanded against a constant applied pressure of 2 bar. The final temperature is 25 C.
(a) What is the initial pressure? The final volume?
(b) Is this process impossible, spontaneous, or reversible?
(c) Find $q$, $w$, $\Delta E$, $\Delta H$, $\Delta A$, $\Delta S$, and $\Delta G$ for this process.
18. One mole of $CO_2$, originally at 1.00 bar and 300 K, expands adiabatically against a constant applied pressure of 0.200 bar. Assume that $CO_2$ behaves as an ideal gas with constant heat capacity, $C_V=28.5\ \ \mathrm{J\ }{\mathrm{mol}}^{--\mathrm{1}}\mathrm{\ }{\mathrm{K}}^{--\mathrm{1}}$.
(a) For the spontaneous expansion, we have $dE=C_VdT-P_{applied}dV$. Find the final temperature and volume for this spontaneous expansion. What is $\Delta E$ for this process?
(b) Find the volume and pressure after the gas is compressed adiabatically and reversibly to the original temperature of 300 K. What are $\Delta S$ and $\Delta E$ for this step?
(c) Find $\Delta E$ when the gas in the final state of part (b) is compressed isothermally to the original volume. What is $\Delta S$ for this step?
(d) What are $\Delta E$ and $\Delta S$ for the cycle comprised of the spontaneous expansion of part (a), the adiabatic compression of part (b), and the isothermal compression of part (c)?
(e) What are $\Delta S$, $\Delta \hat{S}$, and $\Delta S_{universe}$ for the spontaneous expansion?
19. Consider the energy surface depicted in Figure 1. As sketched, $E$ increases monotonically as $S$ increases. $E$ decreases monotonically as $V$ increases. Could the energy surface decrease as $S$ increases or increase as $V$ increases?
20. At 298.15 K, the vapor pressure of water is $3.169\times {10}^{-3}\ \mathrm{Pa}$. Some thermodynamic properties for liquid and gaseous water at this temperature and pressure are given in the table below.
liquid gas
$\overline{G},\ \mathrm{kJ\ }{\mathrm{mol}}^{--\mathrm{1}}$ – 237.1 – 237.1
$\overline{S},\ \mathrm{J\ }{\mathrm{mol}}^{--\mathrm{1}}\mathrm{\ }{\mathrm{K}}^{--\mathrm{1}}$ 70.0 217.5
$\overline{E},\ \mathrm{kJ\ }{\mathrm{mol}}^{--\mathrm{1}}$ – 285.5 – 245.1
$C_P,\ \mathrm{J\ }{\mathrm{mol}}^{--\mathrm{1}}\mathrm{\ }{\mathrm{K}}^{--\mathrm{1}}$ 75.3 33.6
$C_V,\ \mathrm{J\ }{\mathrm{mol}}^{--\mathrm{1}}\mathrm{\ }{\mathrm{K}}^{--\mathrm{1}}$ 67.0 25.3
1. Find ${\Delta }_{vap}\overline{G}$, ${\Delta }_{vap}\overline{S}$, ${\Delta }_{vap}\overline{E}$ for water at this temperature and pressure. Is this process reversible, spontaneous, or impossible?
(b) Sketch $\overline{G}\left(\ell \right)$ and $\overline{G}\left(g\right)$ vs. $T$ for $288.15. What path is followed when one mole of water at 288.15 K and \(3.169\times {10}^3\ \mathrm{Pa}$ goes reversibly to 308.15 K at the same pressure?
(c) On the graph of part (b), indicate the transition in which superheated liquid water at 300 K and $3.169\times {10}^3\ \mathrm{Pa}$ goes to gaseous water at 300 K and the same pressure. Is this process spontaneous, reversible, or impossible? Is $\Delta \overline{G}$ for this process positive, zero, or negative?
(d) Sketch $\overline{E}\left(\ell \right)$ and $\overline{E}\left(g\right)$ vs. $T$ for $288.15. What path is followed when one mole of water at 288.15 K and \(3.169\times {10}^3\ \mathrm{Pa}$ goes reversibly to 308.15 K at the same pressure?
(e) On the graph of part (b), indicate the transition in which superheated liquid water at 300 K and $3.169\times {10}^3\ \mathrm{Pa}$ goes to gaseous water at 300 K and the same pressure. Is $\Delta \overline{E}$ for this process positive, zero, or negative?
21. At 273.15 K and 1 bar, the enthalpy of fusion of ice is $6010\ \mathrm{J\ }{\mathrm{mol}}^{--\mathrm{1}}$. Estimate the Gibbs free energy change for the fusion of ice at 283.15 K and 1 bar.
Notes
${}^{1}$ J. R. Roebuck and H. Osterberg, The Joule-Thomson Effect in Nitrogen, Phys. Rev., Vol. 48, pp 450-457 (1935).
${}^{2}$ See T. L. Hill, An Introduction to Statistical Thermodynamics, Addison-Wesley Publishing Co., Reeding, MA, 1960, pp 266-268.
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• 11.1: Heat Capacity as a Function of Temperature
The heat capacity of the solid substance decreases to zero as the absolute temperature decreases to zero; the curve meets the abscissa at the zero of temperature and does so asymptotically. That this is true for all substances seems like an odd sort of coincidence. Why should all solid substances exhibit essentially the same heat capacity (zero) at one temperature (absolute zero)?
• 11.2: Enthalpy as a function of Temperature
The fact that Cₚ goes to zero asymptotically as the temperature goes to zero has no practical ramifications for the measurement or use of enthalpy. We can only measure changes in energy and enthalpy; no particular state of any system is a uniquely useful reference state for the enthalpy function. Experimental convenience is the only consideration that makes one reference state a better choice than another.
• 11.3: The Third Law
The idea that the entropy change for a pure substance goes to zero as the temperature goes to zero finds expression as the third law of thermodynamics: If the entropy of each element in some crystalline state be taken as zero at the absolute zero of temperature, every substance has a positive finite entropy; but at the absolute zero of temperature the entropy may become zero, and does so become in the case of perfect crystalline substances.
• 11.4: Genesis of the Third Law - the Nernst Heat Theorem
The third law arises in a natural way in the development of statistical thermodynamics. It is probably fair to say that the classical thermodynamic treatment of the third law was shaped to a significant degree by the statistical thermodynamic treatment that developed about the same time. Nevertheless, we can view the third law as an inference from thermochemical observations.
• 11.5: Absolute Entropy
At any given temperature, the entropy value that is obtained in this way is called the substance’s absolute entropy or its third-law entropy. When the entropy value is calculated for one mole of the substance in its standard state, the resulting absolute entropy is called the standard entropy. The standard entropy is usually given the symbol So . It is usually included in compilations of thermodynamic data for chemical substances.
• 11.6: The Standard State for Third-law Entropies
The standard state for entropies is essentially the same as the standard state for enthalpies. For liquids and solids, the standard state for entropies is identical to that for enthalpies: At any given temperature, the standard state is the most stable form of the substance at that temperature and a pressure of 1 bar.
• 11.7: The Fugacity of a Gas
For an ideal gas, the Gibbs free energy is a simple function of its pressure. It turns out to be useful to view the integral as a contribution to a “corrected pressure.” The “correction” is an adjustment to the pressure that, in our calculations, makes the real gas behave as an ideal gas. The idea is that we can express the Gibbs free energy as a function of this corrected pressure, which we call the fugacity. Fugacity is therefore a function of pressure.
• 11.8: A General Strategy for Expressing the Thermodynamic Properties of a Substance
Our goal is to create a scheme in which the enthalpy, the entropy, or the Gibbs free energy of any substance in any arbitrary state is equal to the change in that thermodynamic property when the substance is produced, in that state, from its pure, separate, constituent elements, in their standard states at the same temperature.
• 11.9: The Standard Entropy and the Gibbs Free Energy of Formation
• 11.10: The Nature of Hypothetical States
The hypothetical ideal gas standard state is a wholly theoretical construct. We create this “substance” only because it is convenient to have a name for the “unreal” state of substance A.
• 11.11: The Fugacity and Gibbs Free Energy of A Substance in Any System
When we define the chemical activity of a substance in a particular system, we also introduce a new standard state. The primary criterion for our choice of this activity standard state is that we be able to measure how much the Gibbs energy of the substance differs between the activity standard state and other states of the system. A principal object of the next chapters is to introduce ideas for measuring the difference between the Gibbs energy of a substance in two states of a given system.
• 11.12: Evaluating Entropy Changes Using Thermochemical Cycles
As for the standard enthalpy of reaction, we can obtain the standard entropy of reaction at a new temperature by evaluating entropy changes around a suitable thermochemical cycle. To do so, we need the standard entropy change at one temperature. We also need heat capacity data for all of the reactants and products.
• 11.13: Absolute Zero is Unattainable
The third law postulates that the entropy of a substance is always finite and that it approaches a constant as the temperature approaches zero. The value of this constant is independent of the values of any other state functions that characterize the substance. For any given substance, we are free to assign an arbitrarily selected value to the zero-temperature limiting value. However, we cannot assign arbitrary zero-temperature entropies to all substances.
• 11.14: Problems
11: The Third Law Absolute Entropy and the Gibbs Free Energy of Formation
It is relatively easy to measure heat capacities as a function of temperature. If we measure the constant-pressure heat capacity of a pure substance over a wide temperature range, we typically observe a curve like that in Figure 1. The heat capacity is a smooth, continuous function of temperature except for a small number of discontinuities. These occur at temperatures where the substance undergoes phase changes. These can be changes from one solid phase to another, melting to convert a solid phase to the liquid, or vaporization to convert the liquid to the gas. The details of the curve are pressure dependent; for example, at a low pressure, we might observe sublimation of the material from a solid phase directly into its gas phase.
Another general feature of these curves is that the heat capacity of the solid substance decreases to zero as the absolute temperature decreases to zero; the curve meets the abscissa at the zero of temperature and does so asymptotically. That this is true for all substances seems like an odd sort of coincidence. Why should all solid substances exhibit essentially the same heat capacity (zero) at one temperature (absolute zero)?
As it turns out, this result has a straightforward molecular interpretation in the theory of statistical thermodynamics. In Section 22.6, we consider a theory of low-temperature heat capacity developed by Einstein. Einstein’s theory explains all of the qualitative features that are observed when we measure heat capacities at low temperatures, but its predictions are not quantitatively exact. Debye extended the Einstein model and developed a theory that gives generally excellent quantitative predictions. The Debye theory predicts that, at temperatures near absolute zero, the heat capacity varies as the cube of temperature: $C_P=AT^3$, where $A$ is a constant. If we have heat capacity data down to a temperature near absolute zero, we can estimate the value of $A$ from the value of $C_P$ at the lowest available temperature.
Anticipating results that we develop in Chapter 22, we can characterize the statistical interpretation as follows: When a system of molecules gives up heat to its surroundings, some of the molecules move from higher energy levels to lower ones. Statistical thermodynamics posits that the fraction of the molecules that are in the lowest energy level approaches one as the temperature goes to zero. If nearly all of the molecules are already in the lowest energy level, decreasing the temperature still further has a negligible effect on the energy and enthalpy of the system.
Given such heat capacity data, we can find the enthalpy or entropy change that occurs as we change the temperature of a quantity of the substance from some reference temperature to any other value. When we use pressure and temperature as the independent variables, we have
$dH=C_PdT+\left[V-T{\left(\frac{\partial V}{\partial T}\right)}_P\right]dP \nonumber$
and
$dS=\frac{C_P}{T}dT-{\left(\frac{\partial V}{\partial T}\right)}_PdP \nonumber$
At constant pressure, we have
$\left(dH\right)_P=C_PdT \nonumber$
so that
$H\left(T\right)-H\left(T_{ref}\right)=\int^T_{T_{ref}}{C_PdT} \nonumber$
and
${\left(dS\right)}_P=\frac{C_P}{T}dT \nonumber$ so that $S\left(T\right)-S\left(T_{ref}\right)=\int^T_{T_{ref}}{\frac{C_P}{T}dT} \nonumber$
If phase transitions occur as the temperature goes from the reference temperature to the temperature of interest, these integrations must be carried out in steps. Also, we must include the enthalpy and entropy changes that occur during these phase changes.
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The fact that $C_P$ goes to zero asymptotically as the temperature goes to zero has no practical ramifications for the measurement or use of enthalpy. We can only measure changes in energy and enthalpy; no particular state of any system is a uniquely useful reference state for the enthalpy function. Experimental convenience is the only consideration that makes one reference state a better choice than another. In Chapter 8, we define standard enthalpies of formation for elements and compounds. For this purpose, we choose to define the standard enthalpy of formation of each element to be zero at every temperature. For standard enthalpies of formation, the reference state is different at every temperature.
Compilations of thermodynamic data often choose 298.15 K and one bar as the zero of enthalpy for pure substances. (In citing data from such compilations, 298.15 K is frequently abbreviated to “298 K.”) Because it is near the ambient temperature of most laboratories, much thermochemical data has been collected at or near this temperature. Choosing a reference temperature near ambient laboratory temperatures helps minimize the errors introduced when we extrapolate experimental thermodynamic data to the reference temperature. Expressed relative to a reference temperature, the substance’s enthalpy at any other temperature is the change in enthalpy that occurs when the substance is taken from the reference temperature to that temperature. Such enthalpy changes are often called absolute enthalpies. Enthalpy-data tables frequently include values for “$H^o_T-H^o_{\mathrm{298K}}$” at a series of temperatures. These data should not be confused with enthalpies of formation.
11.03: The Third Law
For entropy on the other hand, the fact that the heat capacity goes to zero as the temperature decreases has important consequences. Consider the change in the entropy of a pure substance whose heat capacity approaches some finite limiting value as its temperature decreases to absolute zero. For such a substance, ${C_P}/{T}$ becomes arbitrarily large as the temperature decreases, and the entropy integral $\int^0_T{\frac{C_P}{T}}dT \nonumber$
approaches minus infinity as the temperature approaches zero. For real substances, this does not occur. In the neighborhood of absolute zero, heat capacities decrease more rapidly than temperature. The entropy change approaches zero as the temperature approaches zero.
The idea that the entropy change for a pure substance goes to zero as the temperature goes to zero finds expression as the third law of thermodynamics. In 1923, Lewis and Randall${}^{1}$ gave a statement of the third law that is particularly convenient in chemical applications:
If the entropy of each element in some crystalline state be taken as zero at the absolute zero of temperature, every substance has a positive finite entropy; but at the absolute zero of temperature the entropy may become zero, and does so become in the case of perfect crystalline substances.
Implicitly, the Lewis and Randall statement defines the entropy of any substance, at any temperature, $T$, to be the difference between the entropy of the constituent elements, at absolute zero, and the entropy of the substance at temperature $T$. Equivalently, we can say that it is the entropy change when the substance is formed at temperature $T$ from its constituent elements at absolute zero. Arbitrarily, but very conveniently, the statement sets the entropy of an element to zero at absolute zero.
The distinction between perfect crystalline substances and less-than-perfect crystalline substances lies in the regularity of the arrangement of the molecules within the crystal lattice. In any lattice, each molecule of the substance is localized at a specific site in the lattice. In a perfect crystal, all of the molecules are in oriented the same way with respect to the lattice. Some substances form crystals in which the molecules are not all oriented the same way. This can happen when the molecule can fit into a lattice site of the same shape in more than one way. For example, in solid carbon monoxide, the individual molecules occupy well-defined lattice sites. If the carbon monoxide crystal were perfect, all of the molecules would point in the same direction, as diagrammed in Figure 2. Instead, they point randomly in either of two possible directions.
11.04: Genesis of the Third Law - the Nernst H
The third law arises in a natural way in the development of statistical thermodynamics. It is probably fair to say that the classical thermodynamic treatment of the third law was shaped to a significant degree by the statistical thermodynamic treatment that developed about the same time. Nevertheless, we can view the third law as an inference from thermochemical observations.
Walther Nernst was the first to recognize the principle that underlies the third law. From published experimental results, Nernst inferred a postulate known as the Nernst heat theorem. The experimental results that inspired Nernst were measurements of enthalpy and Gibbs free energy differences, \({\Delta }_rH\) and \({\Delta }_rG\), for particular reactions at a series of temperatures. (We define \({\Delta }_rH^o\) in Section 8.6. We define \({\Delta }_rH\) the same way, except that the reactants and products are not all in their standard states. Likewise, \({\Delta }_rG\) and \({\Delta }_rS\) are differences between Gibbs free energies and entropies of reactants and products. We give a more precise definition for \({\Delta }_rG^o\) in Section 13.2.) As the temperature decreased to a low value, the values of \({\Delta }_rH\) and \({\Delta }_rG\) converged. Since \(T{\Delta }_rS = {\Delta }_rH-{\Delta }_rG\), this observation was consistent with the fact that the temperature was going to zero. However, Nernst concluded that the temperature factor in \({T\Delta }_rS\) was not, by itself, adequate to explain the observed dependence of \({\Delta }_rH-{\Delta }_rG\) on temperature. He inferred that the entropy change for these reactions decreased to zero as the temperature decreased to absolute zero and postulated that this observation would prove to be generally valid. The Nernst heat theorem asserts that the entropy change for any reaction of pure crystalline substances goes to zero as the temperature goes to zero.
Subsequently, Max Planck suggested that the entropy of reaction goes to zero because of a still more basic phenomenon: the entropy of every crystalline substance goes to zero as the temperature goes to zero. Further investigation then showed that Planck’s formulation fails for substances like carbon monoxide, in which the crystalline solid does not become perfectly ordered at the temperature goes to zero. The Lewis and Randall statement refines the Planck formulation by recognizing that non-zero entropies will be observed at absolute zero for solids that are not crystalline and for crystalline solids that are not perfectly ordered. The Lewis and Randall statement also makes a choice (implicit also in the Planck formulation) of the zero point for the entropies of chemical substances—namely, “some crystalline state” of each element at absolute zero. This choice ensures that, at any temperature greater than zero, the entropy of every substance will be greater than zero.
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By the Lewis and Randall statement of the third law, the entropy of a substance that forms a perfect crystal is identically equal to zero at absolute zero. Much as the ideal gas temperature scale has a natural zero at the temperature at which the volume extrapolates to zero, a perfect crystalline substance has a natural zero of entropy at this same temperature. We can choose a non-zero value for the absolute zero of temperature. The Centigrade scale is based on such a choice. However, for thermodynamic purposes, any such choice is much less convenient. Similarly, we could choose arbitrary values for the entropies of the elements at the absolute zero of temperature. The entropy of a perfect crystalline substance at absolute zero would then be the sum of the entropies of its constituent elements. (See problem 5.) However, choosing non-zero values proves to be much less convenient.
Given the entropy of a substance at absolute zero, its entropy at any higher temperature can be calculated from the entropy changes that occur as the substance is warmed to the new temperature. At the very lowest temperatures, this entropy change is calculated by integrating ${C_P}/{T}$, using Debye’s theoretical relationship, $C_P=AT^3$; $A$ is obtained from the value of $C_P$ at the lowest temperature for which an experimental value of $C_P$ is available. In temperature ranges where experimental heat capacity data are available, the entropy change is obtained by integration using these data.
Phase changes are isothermal and reversible. Where the substance undergoes phase changes, the contribution that the phase change makes to the entropy of the substance is equal to the enthalpy change for the phase change divided by the temperature at which it occurs.
At any given temperature, the entropy value that is obtained in this way is called the substance’s absolute entropy or its third-law entropy. When the entropy value is calculated for one mole of the substance in its standard state, the resulting absolute entropy is called the standard entropy. The standard entropy is usually given the symbol $S^o$. It is usually included in compilations of thermodynamic data for chemical substances.
We write $S^o_A\left(T\right)$ to indicate the absolute entropy of substance $A$ in its standard state at temperature $T$. $S^o_A\left(T\right)$ is the entropy of the substance in its standard state at absolute zero plus the entropy increase that occurs as the substance changes reversibly to its standard state at $T$. So long as substance $A$ forms a perfect crystal at absolute zero, $S^o_A\left(T\right)$ is the difference between its molar entropy at $T$ and its molar entropy at absolute zero—as calculated from heat capacity and phase-change enthalpy data.
If substance $A$ does not form a perfect crystal at absolute zero, the true value of $S^o_A\left(T\right)$ exceeds the calculated value. The excess is the molar entropy of the imperfect crystal at absolute zero. We observe the discrepancy when measured values, at $T$, of entropies of reactions that involve $A$ fail to agree with those calculated using the incorrect value of $S^o_A\left(T\right)$.
In Section 11.2 we note that many tables of thermochemical properties present “absolute enthalpy” data for chemical substances. An absolute enthalpy is the difference between the enthalpies of a substance at two different temperatures, but the reference temperature is not absolute zero. In Section 6.4 and Section 6.5, we introduce enthalpy standard states and the standard enthalpy of formation of substance $A$ at temperature $T$, which we designate as ${\Delta }_fH^o_A\left(T\right)$. We define the standard enthalpy of formation of any element at any temperature to be zero. In Section 8.6, we find that the enthalpy difference between reactants in their standard states and products in their standard states is readily calculated from the standard enthalpies of formation of the participating substances. As illustrated in Figure 8-2, this calculation is successful because it utilizes an isothermal cycle which begins and ends in a common set of elements, all of which are at the same temperature.
We can also define the standard entropy of formation of any substance to be the difference between its standard entropy, $S^o_A\left(T\right)$, and those of its pure constituent elements in their standard states at the same temperature. This definition is embedded in the Lewis and Randall statement of the third law. For example, the standard entropy of formation of water at 400 K is the difference
$\Delta_fS^o\left(H_2O,\ \mathrm{400}\mathrm{\ }\mathrm{K}\right)=\Delta S^o\left(H_2O,\ \mathrm{400}\mathrm{K}\right) \nonumber$
$-\Delta S^o\left(H_2,\ \mathrm{400}\mathrm{\ }\mathrm{K}\right)-{\frac{1}{2}}\Delta S^o\left(O_2,\mathrm{400}\mathrm{K}\right) \nonumber$
Because of this definition, the standard entropy of formation of an element in its standard state is zero. We can calculate the standard entropy change for any reaction, ${\Delta }_rS^o\left(T\right)$, either as the difference between the standard entropies of formation (the ${\Delta }_fS^o\left(T\right)$ values) of the reactants and products or as the difference between their standard entropies (the $S^o_A\left(T\right)$ values). Either calculation is successful because it begins and ends with a common set of elements, all of which are at the same temperature. When we compute ${\Delta }_rS^o\left(T\right)$ using values of $S^o_A\left(T\right)$ for the reactants and products the reference temperature for the elements is absolute zero. When we compute ${\Delta }_rS^o\left(T\right)$ using values of ${\Delta }_fS^o\left(T\right)$ for the reactants and products, the reference temperature is $T$.
Given ${\Delta }_fH^o$$\left(T\right)$ and ${\Delta }_fS^o\left(T\right)$, the standard Gibbs free energy of formation is immediately obtained from ${\Delta }_fG^o\left(T\right)={\Delta }_fH^o\left(T\right)-T{\Delta }_fS^o\left(T\right)$. For any element at any temperature, we have ${\Delta }_fH^o=0$ and ${\Delta }_fS^o=0$; it follows that the standard Gibbs free energy of formation of an element in its standard state is zero. Tables of thermodynamic data usually give values for${\Delta }_fH^o$, ${\Delta }_fG^o$, and $S^o$. (A set of standard entropies contains the same information as the corresponding set of entropies of formation. Entropies of formation are seldom tabulated. If ${\Delta }_fS^o$ is needed, it can be calculated either from ${\Delta }_fH^o$ and ${\Delta }_fG^o$ or from the absolute entropies of the substance and the elements from which it is formed.)
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The standard state for entropies is essentially the same as the standard state for enthalpies. For liquids and solids, the standard state for entropies is identical to that for enthalpies: At any given temperature, the standard state is the most stable form of the substance at that temperature and a pressure of 1 bar.
For gases, the ideal gas standard state for entropy tabulations is the hypothetical ideal gas state at a pressure of 1 bar. A substance in a hypothetical ideal gas state is a creature of theory, and we obtain its thermodynamic properties from calculations that use the experimentally determined properties of the real gas. The idea behind these calculations is that we can—by calculation—remove the effects of intermolecular interactions from the measured properties. When we want the properties of the real gas, we put these effects back. From this perspective, we can say that a substance in its hypothetical ideal gas standard state has the entropy that the real gas would have if it behaved as an ideal gas at pressures of 1 bar and below.
In principle, this differs from the standard state chosen for enthalpy, because the enthalpy standard state is defined to be an arbitrary low pressure at which the substance behaves as an ideal gas. However, because the enthalpy of an ideal gas is independent of pressure, we can consider the standard state for either enthalpy or entropy to be the hypothetical ideal gas at a pressure of 1 bar. (Below, we write “$HIG^o$” to designate this state.) Since the Gibbs free energy is defined by $G=H-TS$, we can also describe the standard state for the Gibbs free energy of gases as the hypothetical ideal gas at 1 bar.
To see precisely what we mean by the hypothetical ideal gas standard state, let us consider the conversion of one mole of a real gas, initially at pressure $P$ and temperature $T$, to its hypothetical ideal gas state at 1 bar and the same temperature. We accomplish this in a three-step process. In the first step, we reversibly and isothermally expand the real gas to some arbitrary low pressure at which the real gas behaves ideally. The second step is purely conceptual: We suppose that the real gas changes into a hypothetical gas that behaves ideally. The third step is the reversible isothermal compression of the hypothetical ideal gas to a pressure of 1 bar. Letting $A$ denote the gas, these steps are
1. $A\left(\mathrm{real\ gas},\ P\right)\to A\left(\mathrm{real\ gas},\ P^*\approx 0\right)$
2. $A\left(\mathrm{real\ gas},\ P^*\approx 0\right)\to A\left(\mathrm{id}\mathrm{eal\ gas},\ P^*\approx 0\right)$
3. $A\left(\mathrm{id}\mathrm{eal\ gas},\ P^*\approx 0\right)\to A\left({HIG}^o,\ P^o=1\ \mathrm{bar}\right)$
We use the symbol “$P^o$” to designate the standard-state pressure. $P^o$ is a constant, whose value is 1 bar.
Let the enthalpy, entropy, and Gibbs free energy changes for the i${}^{th}$ step be ${\Delta }_i\overline{H}$, ${\Delta }_i\overline{S}$, and ${\Delta }_i\overline{G}$. (We use the overbar to emphasize that the system consists of one mole of a pure substance. Since the superscript zero implies that $H^o_A\left(HIG^o\right)$, for example, is a property of one mole of $A$ in its standard state, we omit the overbar from standard state properties.) From our expressions for $dH$ and $dS$ as functions of pressure and temperature, we have, at constant temperature,
${\left(d\overline{H}\right)}_T=\left[\overline{V}-T{\left(\frac{\partial \overline{V}}{\partial T}\right)}_P\right]dP \nonumber$ and ${\left(d\overline{S}\right)}_T=-{\left(\frac{\partial \overline{V}}{\partial T}\right)}_PdP \nonumber$
For an ideal gas, ${\left(d\overline{H}\right)}_T=0 \nonumber$ and ${\left(d\overline{S}\right)}_T=-{R}/{P} \nonumber$ For step [GrindEQ__1_], the enthalpy change is
${\Delta }_1\overline{H}=\int^{P^*}_P{\left[\overline{V}-T{\left(\frac{\partial \overline{V}}{\partial T}\right)}_P\right]dP} \nonumber$
the entropy change is ${\Delta }_1\overline{S}=\int^{P^*}_P{-{\left(\frac{\partial \overline{V}}{\partial T}\right)}_PdP} \nonumber$
We must evaluate the partial derivative using an equation of state that describes the real gas.
Step [GrindEQ__2_] is merely a change in our perspective; nothing actually happens to the gas. There is no enthalpy or entropy change: ${\Delta }_2\overline{S}=0$ and ${\Delta }_2\overline{H}=0$.
For step [GrindEQ__3_], the enthalpy change is zero because the gas is ideal, ${\Delta }_3\overline{H}=0$. We evaluate the entropy change using the ideal gas equation; then
${\Delta }_3\overline{S}=\int^{P^o}_{P^*}{-\frac{R}{P}dP} \nonumber$
Let ${\overline{H}}_A\left(P\right)$, ${\overline{S}}_A\left(P\right)$, and ${\overline{G}}_A\left(P\right)$ be the molar enthalpy, entropy, and Gibbs free energy of real gas $A$ at pressure $P$. Let us express the enthalpy, entropy, and Gibbs free energy of the standard state substance relative to the corresponding properties of the constituent elements in their standard states at temperature $T$. Then the molar enthalpy, entropy, and Gibbs free energy of $A$ in its hypothetical ideal gas standard state are ${\Delta }_fH^o_A\left({HIG}^o\right)$, ${\Delta }_fS^o_A\left({HIG}^o\right)$, and ${\Delta }_fG^o_A\left({HIG}^o\right)$. Since these are state functions, we have
${\Delta }_fH^o_A\left({HIG}^o\right)={\overline{H}}_A\left(P\right)+{\Delta }_1\overline{H}+{\Delta }_2\overline{H}+{\Delta }_3\overline{H}={\overline{H}}_A\left(P\right)+\int^{P^*}_P{\left[\overline{V}-T{\left(\frac{\partial \overline{V}}{\partial T}\right)}_P\right]dP} \nonumber$ and
${\Delta }_fS^o_A\left({HIG}^o\right)={\overline{S}}_A\left(P\right)+{\Delta }_1\overline{S}+{\Delta }_2\overline{S}+{\Delta }_3\overline{S}={\overline{S}}_A\left(P\right)-\int^{P^*}_P{{\left(\frac{\partial \overline{V}}{\partial T}\right)}_PdP}-\int^{P^o}_{P^*}{\frac{R}{P}dP} \nonumber$
These equations relate the enthalpy and entropy of the hypothetical ideal gas standard state to the enthalpy and entropy of the real gas at pressure $P$, at the same temperature.
To evaluate the enthalpy change, we can use the virial equation for the volume of a real gas, set $P^*=0$, and evaluate the resulting integral. (See problem 1a.) For the entropy, there is a complication. If we set $P^*=0$, neither integral is finite. We can overcome this difficulty by choosing a small positive value for $P^*.$ Then both integrals are finite, and the value of their sum remains finite in the limit as $P^*\to 0$. This occurs because the molar volume of any gas approaches the molar volume of an ideal gas at a sufficiently low pressure. (See problem 1.)
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As it turns out, the pressure-dependence of the Gibbs free energy is useful more often than that of the enthalpy or entropy. From the defining relationship, $G=H-TS$, and the results above, we find
\begin{align*} {\Delta }_fG^o_A\left({HIG}^o\right) &={\Delta }_fH^o_A\left({HIG}^o\right)-T{\Delta }_fS^o_A\left({HIG}^o\right) \[4pt] &={\overline{H}}_A\left(P\right)+\int^{P^*}_P{\left[\overline{V}-T{\left(\frac{\partial \overline{V}}{\partial T}\right)}_P\right]dP} -T{\overline{S}}_A\left(P\right)+\int^{P^*}_P{T{\left(\frac{\partial \overline{V}}{\partial T}\right)}_PdP}+\int^{P^o}_{P^*}{\frac{RT}{P}dP} \[4pt] &={\overline{G}}_A\left(P\right)-\int^P_{P^*}{\overline{V}dP}-\int^{P^*}_{P^o}{\frac{RT}{P}dP} \end{align*}
where we have reversed the direction of integration in the remaining integrals. Again, we cannot evaluate the integrals with $P^* = 0$. To overcome this difficulty, we introduce a clever expedient: we add and subtract the same integral, obtaining
${\Delta }_fG^o_A\left({HIG}^o\right)={\overline{G}}_A\left(P\right)-\int^P_{P^*}{\overline{V}dP}-\int^{P^*}_{P^o}{\frac{RT}{P}dP} \nonumber$ $-\int^P_{P^*}{\frac{RT}{P}dP}+\int^P_{P^*}{\frac{RT}{P}dP} \nonumber$
$={\overline{G}}_A\left(P\right)-\int^P_{P^*}{\left[\overline{V}-\frac{RT}{P}\right]dP}-\int^P_{P^o}{\frac{RT}{P}dP} \nonumber$
This result is pivotal. We will derive it again later by a slightly different argument. For now, we evaluate the last integral and rearrange the equation to the form in which we write it most often:
${\overline{G}}_A\left(P\right)={\Delta }_fG^o_A\left({HIG}^o\right)+RT{ \ln \left(\frac{P}{P^0}\right)\ }+RT\int^P_0{\left[\frac{\overline{V}}{RT}-\frac{1}{P}\right]dP} \nonumber$
In doing so, we use the fact that
$\frac{\overline{V}}{RT}-\frac{1}{P} \nonumber$
vanishes for any gas when the pressure is sufficiently low, which means that the integral from $P^*$to $P$ remains finite when we let $P^*\to 0$.
Recapitulating: The last equation relates the Gibbs free energy of one mole of a real gas at pressure $P$ and temperature $T$, ${\overline{G}}_A\left(P\right)$, to the Gibbs free energy of formation of the gas in its standard state, ${\Delta }_f$$G^o_A\left({HIG}^o\right)$, where the standard state is the hypothetical ideal gas at $P=P^o=1\ \mathrm{bar}$ and the same temperature. (We write “$RT{ \ln \left({P}/{P^o}\right)\ }$” rather than “$RT{ \ln P\ }$” to emphasize that the argument of the logarithm is a dimensionless quantity.)
If the gas is ideal, the integrand is zero at any pressure. For an ideal gas, the Gibbs free energy, at pressure $P$ and temperature $T$, is related to the Gibbs free energy in the standard state by
${\overline{G}}_A\left(P\right)=G^o_A\left(\mathrm{1\ bar}\right)+RT{ \ln \left(\frac{P}{P^o}\right)\ } \nonumber$ (ideal gas)
This observation enables us to give a physical interpretation to the integral
$\int^P_0{\left[\overline{V}-\frac{RT}{P}\right]dP}=RT\int^P_0{\left[\frac{\overline{V}}{RT}-\frac{1}{P}\right]dP} \nonumber$
Evidently, this integral is the difference between the Gibbs free energy of one mole of the real gas and the Gibbs free energy that it would have if it behaved as an ideal gas. When we add it to the Gibbs free energy of the hypothetical ideal gas, we get the Gibbs free energy of the real gas.
For an ideal gas, the Gibbs free energy is a simple function of its pressure. It turns out to be useful to view the integral as a contribution to a “corrected pressure.” The “correction” is an adjustment to the pressure that, in our calculations, makes the real gas behave as an ideal gas. The idea is that we can express the Gibbs free energy as a function of this corrected pressure, which we call the fugacity, and to which we give the symbol “$f$ ”. Fugacity is therefore a function of pressure. Fugacity also has the units of pressure, which we always take to be bars. We define the fugacity of $A$ at pressure $P$, $f_A\left(P\right)$, by
${\overline{G}}_A\left(P\right)={\Delta }_fG^o_A\left({HIG}^o\right)+RT{ \ln \left(\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)}\right)\ } \nonumber$ (real gas)
The fugacity, $f_A\left({HIG}^o\right)$, and the standard Gibbs free energy of formation, ${\Delta }_fG^o_A\left({HIG}^o\right)$, describe the same state. We define the fugacity in this standard state to be one bar. That is, the fugacity of a substance in its hypothetical ideal-gas standard state, $f_A\left({HIG}^o\right)$, is a constant whose value is 1 bar. (If we want to express fugacity in units of, say, pascals, then $P^o=f_A\left({HIG}^o\right)={10}^5\ \mathrm{Pa}$.)
We can calculate the fugacity of a real gas, at pressure $P$ and temperature $T$, from
$RT{ \ln \left(\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)}\right)\ }=\ RT{ \ln \left(\frac{P}{P_0}\right)\ }+RT\int^P_0{\left[\frac{\overline{V}}{RT}-\frac{1}{P}\right]dP} \nonumber$
We find it useful to introduce a function of pressure that we call the fugacity coefficient, ${\gamma }_A\left(P\right)$. We define it as
${\gamma }_A\left(P\right)=\frac{f_A\left(P\right)}{P} \nonumber$
Since $f_A\left({HIG}^o\right)=P^o=1\ \mathrm{bar}$, we have
${ \ln {\gamma }_A\left(P\right)\ }=\int^P_0{\left[\frac{\overline{V}}{RT}-\frac{1}{P}\right]dP} \nonumber$
Figure 3 exhibits these quantities graphically. The shaded area is ${ \ln {\gamma }_A\left(P\right)\ }$.
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The equations we develop in Sections 11.7 and 11.8 express the differences ${\Delta }_fH^o_A\left({HIG}^o\right)-{\overline{H}}_A\left(P\right)$, ${\Delta }_fS^o_A\left({HIG}^o\right)-{\overline{S}}_A\left(P\right)$, and ${\Delta }_fG^o_A\left({HIG}^o\right)-{\overline{G}}_A\left(P\right)$ between thermodynamic functions for one mole of a gas at two different pressures and the same temperature. They follow from the properties of gases and the relationships that result when we express the enthalpy, entropy, and Gibbs free energy as functions of temperature and pressure. Because ${\Delta }_fH^o_A\left({HIG}^o\right)$, ${\Delta }_fS^o_A\left({HIG}^o\right)$, and ${\Delta }_fG^o_A\left({HIG}^o\right)$ are measured relative to the constituent elements of substance $A$ at the same temperature, ${\overline{H}}_A\left(P\right)$, ${\overline{S}}_A\left(P\right)$, and ${\overline{G}}_A\left(P\right)$ are the differences in these properties between the real substance at pressure $P$ and temperature $T$ and the real constituent elements in their standard states at the same temperature.
We have just found a way to express the thermodynamic functions of a pure real gas at any pressure and temperature. This development shows us the way toward a broader goal. Ultimately, we want to be able to express—and to find the values of—the thermodynamic functions of any substance in any system at any temperature and pressure.
Our goal is to create a scheme in which the enthalpy, the entropy, or the Gibbs free energy of any substance in any arbitrary state is equal to the change in that thermodynamic property when the substance is produced, in that state, from its pure, separate, constituent elements, in their standard states at the same temperature.
The scheme we create uses two steps to convert the constituent elements into the substance in the arbitrary state. The elements are first converted into the pure substance in its standard state at the same temperature. The substance is then taken from its standard state to the state it occupies in the arbitrary system, at the same temperature. While straightforward in principle, finding changes in thermodynamic properties for this last step is often difficult in practice.
The value of the Gibbs free energy of substance $A$ in any arbitrary system, $\overline{G}_A$, becomes equal to the sum of its Gibbs free energy in the standard state,$\ \Delta_fG^o_A$, and the Gibbs free energy change when the substance passes from its standard state into its state in the arbitrary system, $\Delta \overline{G}_A$. That is,
${\overline{G}}_A={\Delta }_fG^o_A+\Delta {\overline{G}}_A \nonumber$
and $\overline{G}_A$ is the same thing as the Gibbs free energy change when the substance is formed, as it is found in the arbitrary system, from its constituent elements. The same relationships apply to the enthalpies and entropies of these states:
${\overline{H}}_A={\Delta }_fH^o_A+\Delta {\overline{H}}_A \nonumber$
and
${\overline{S}}_A={\Delta }_fS^o_A+\Delta \overline{S}_A \nonumber$
Measuring the differences $\Delta \overline{H}_A$, $\Delta \overline{S}_A$, and $\Delta \overline{G}_A$ become important objectives—and major challenges—in the study of the thermodynamics of chemical systems.
In Section 11.6, we find the changes that occur in the enthalpy, entropy, and Gibbs free energy when one mole of a pure gas is taken from its real gas state at any pressure to its hypothetical ideal-gas standard state, at the same temperature. In Chapter 13, we extend this development to express the thermodynamic properties of any mixture of ideal gases in terms of the properties of the individual pure gases. As a result, we can find the equilibrium position for any reaction of ideal gases from the thermodynamic properties of the individual pure gases. This application is successful because we can find both ${\Delta }_fG^o_A$ and $\Delta {\overline{G}}_A$ for an ideal gas at any pressure. Beginning in Chapter 14, we extend this success by finding ways to measure $\Delta {\overline{G}}_A$ for the process of taking $A$ from its standard state as a pure substance to any arbitrary state in which $A$ may be only one component of a solution or mixture.
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Given ${\Delta }_fG^o$ for all of the species involved in a reaction, we can calculate the difference between the Gibbs free energies of formation of the pure separate products and those of the pure separate reactants. We call this difference the standard Gibbs free energy change for the reaction, ${\Delta }_rG^o$. A standard Gibbs free energy of formation is the standard Gibbs free energy change for the reaction that forms a substance from its elements. Likewise, from the absolute entropies, $S^o$, of the reactants and products, we can calculate the standard entropy change for the reaction, ${\Delta }_rS^o$. In doing so, we utilize the thermochemical cycle that we introduced to calculate ${\Delta }_rH^o$ from the values of ${\Delta }_fH^o$ for the reacting species. For
$a\ A+b\ B\to c\ C+d\ D \nonumber$
we have
${\Delta }_rG^o=c\ {\Delta }_fG^o\left(C\right)+d\ {\Delta }_fG^o\left(D\right)-a\ {\Delta }_fG^o\left(A\right)-b\ {\Delta }_fG^o\left(B\right) \nonumber$
and
${\Delta }_rS^o=c\ S^o\left(C\right)+d\ S^o\left(D\right)-a\ S^o\left(A\right)-b\ S^o\left(B\right) \nonumber$
We use ${\Delta }_fG^o_A\left({HIG}^o\right)$ to denote the Gibbs free energy of one mole of a gas in its hypothetical ideal-gas standard state. Because the fugacity of the ideal gas standard state is 1 bar, $f_A\left({HIG}^o\right)=P^o$, the Gibbs free energy of a gas at unit fugacity becomes the Gibbs free energy change for the formation of the substance in its hypothetical ideal gas standard state. For an ideal gas, unit fugacity occurs at a pressure of one bar. For real gases, the standard state of unit fugacity occurs at a real-gas pressure that is, in general, different from one bar.
The Gibbs free energy of the gas at any other pressure, ${\overline{G}}_A\left(P\right)$, becomes identical to the difference between the Gibbs free energy of the gas in that state and the Gibbs free energy of its constituent elements in their standard states at the same temperature. This convention makes the Gibbs free energy of the elements the “zero point” for the Gibbs free energy of the gas. As indicated in Figure 4, the Gibbs free energy of the gas at any pressure, P, becomes
\begin{align*} \overline{G}_A\left(P\right) &={\Delta }_fG^o_A\left({HIG}^o\right)+RT{ \ln \left(\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)}\right)\ } \[4pt] &={\Delta }_fG^o_A\left({HIG}^o\right)+RT{ \ln \left(\frac{f_A\left(P\right)}{P^o}\right)\ } \end{align*}
11.10: The Nature of Hypothetical States
It is worthwhile to call attention to some important aspects of this development. The hypothetical ideal gas standard state is a wholly theoretical construct. We create this “substance” only because it is convenient to have a name for the “unreal” state of substance A, whose Gibbs free energy we have denoted as $G^o_A\left({HIG}^o\right)={\Delta }_fG^o_A\left({HIG}^o\right)$. We have developed procedures for calculating ${\Delta }_fG^o_A\left({HIG}^o\right)$ from the properties of the corresponding real gas. Given the properties of real gas A, these procedures determine ${\Delta }_fG^o_A\left({HIG}^o\right)$ uniquely. ${\Delta }_fG^o_A\left({HIG}^o\right)$ is a useful quantity; the calculation of ${\Delta }_fG^o_A\left({HIG}^o\right)$ is “real” even though the substance it putatively describes is not.
Other hypothetical states are frequently useful. Problem 8 in this chapter considers a hypothetical liquid state of methanol at 500 K and 1 atm—conditions at which the real substance is a gas. Alternative approximations enable us to calculate the Gibbs free energy of this hypothetical state in different ways. The results have predictive value. Not surprisingly, however, the alternative approximations produce Gibbs free energy values whose quantitative agreement is poor. Were it useful to do so, we could select one particular approximation and define the Gibbs free energy of the hypothetical superheated liquid methanol to be the value produced using that approximation. This would not make the superheated liquid methanol any more real, but it would uniquely define the Gibbs free energy of the hypothetical substance.
Later in our development, we create other hypothetical reference states. As for the hypothetical ideal gas standard state, we specify unique ways to calculate the properties of these hypothetical states from measurements that we can make on real systems.
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We can find the Gibbs free energy of formation for substances whose standard states are condensed phases. As indicated in Section 11.9, we adopt the same rule for any substance; we set $G^o_A={\Delta }_fG^o\left(A\right)$ for any substance, whether its standard state is a gas, liquid, or solid. The Gibbs free energy of the elements becomes the “zero point” for the Gibbs free energy of any substance.
In Chapters 14 and 15, we see that we can also define the fugacity of any substance in any system; that is, we can define the fugacity for a pure liquid, a pure solid, or for one component in any mixture. When we do so, the Gibbs free energy of one mole of the substance in the system, ${\overline{G}}_A\left(\mathrm{system},P\right)$, is given by the same relationship we developed for the molar Gibbs free energy of a pure gas. We find
${\overline{G}}_A\left(\mathrm{system},P\right)={\Delta }_fG^o_A\left({HIG}^o\right)+RT{ \ln \left(\frac{f_A\left(\mathrm{system},P\right)}{f_A\left({HIG}^o\right)}\right)\ } \nonumber$
To obtain this result and to see how to find the fugacity of $A$ in any system, $f_A\left(\mathrm{system},P\right)$, we must introduce a number of additional ideas. For now, let us note some of the consequences.
The essential consequence is that the difference between the Gibbs free energy of one mole of a substance in two different systems, say system X and system Y, can be expressed using the ratio of the corresponding fugacities. That is,
${\overline{G}}_A\left(\mathrm{system}\mathrm{\ \ X},P_X\right)-{\overline{G}}_A\left(\mathrm{system}\mathrm{\ \ Y},P_Y\right)=RT{ \ln \left(\frac{f_A\left(\mathrm{system},P_X\right)}{f_A\left(\mathrm{system}\mathrm{\ \ Y},P_Y\right)}\right)\ } \nonumber$
where $P_X$ and $P_Y$ are the pressures of systems X and Y, respectively, and both systems are at the same temperature.
For liquids and solids, the standard state is the pure substance in its most stable form at one bar and the temperature of interest. The fugacity in the standard state must be determined experimentally. If the liquid or solid has negligible vapor pressure, this may not be possible. Since we intend “any system” to include all manner of mixtures and solutions, it can be very difficult to find the Gibbs free energy change for taking the substance from its standard state to the arbitrary system in which its fugacity is $f_A\left(\mathrm{system},P\right)$. In Chapter 14, we introduce the chemical activity of the substance to cope with such cases.
When we define the chemical activity of a substance in a particular system, we also introduce a new standard state. The primary criterion for our choice of this activity standard state is that we be able to measure how much the Gibbs free energy of the substance differs between the activity standard state and other states of the system. A principal object of the next seven chapters is to introduce ideas for measuring the difference between the Gibbs free energy of a substance in two states of a given system. Even so, our treatment of the issues involved in this step is quite incomplete.
11.12: Evaluating Entropy Changes Using Thermo
As for the standard enthalpy of reaction, we can obtain the standard entropy of reaction at a new temperature by evaluating entropy changes around a suitable thermochemical cycle. To do so, we need the standard entropy change at one temperature. We also need heat capacity data for all of the reactants and products. For the reaction
$a\ A+b\ B\to c\ C+d\ D \nonumber$
we can evaluate the entropy change at a second temperature by summing the individual contributions to the change in entropy around the cycle in Figure 5. For this cycle, we have
${\Delta }_rS^o\left(T_2\right)={\Delta }_rS^o\left(T_1\right)+c\int^{T_2}_{T_1}{\frac{C_P\left(C\right)}{T}dT} +d\int^{T_2}_{T_1}{\frac{C_P\left(D\right)}{T}dT}-a\int^{T_2}_{T_1}{\frac{C_P\left(A\right)}{T}dT}-b\int^{T_2}_{T_1}{\frac{C_P\left(B\right)}{T}dT} \nonumber$
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The third law postulates that the entropy of a substance is always finite and that it approaches a constant as the temperature approaches zero. The value of this constant is independent of the values of any other state functions that characterize the substance. For any given substance, we are free to assign an arbitrarily selected value to the zero-temperature limiting value. However, we cannot assign arbitrary zero-temperature entropies to all substances. The set of assignments we make must be consistent with the experimentally observed zero-temperature limiting values of the entropy changes of reactions among different substances. For perfectly crystalline substances, these reaction entropies are all zero. We can satisfy this condition by assigning an arbitrary value to the zero-temperature molar entropy of each element and stipulating that the zero-temperature entropy of any compound is the sum of the zero-temperature entropies of its constituent elements. This calculation is greatly simplified if we let the zero-temperature entropy of every element be zero. This is the essential content of the third law.
The Lewis and Randall statement incorporates this selection of the zero-entropy reference state for entropies, specifying it as “a crystalline state” of each element at zero degrees. As a result, the entropy of any substance at zero degrees is greater than or equal to zero. That is, the Lewis and Randall statement includes a convention that fixes the zero-temperature limiting value of the entropy of any substance. In this respect, the Lewis and Randall statement makes an essentially arbitrary choice that is not an intrinsic property of nature. We see, however, that it is an overwhelmingly convenient choice.
We have discussed alternative statements of the first and second laws. A number of alternative statements of the third law are also possible. We consider the following:
It is impossible to achieve a temperature of absolute zero.
This statement is more general than the Lewis and Randall statement. If we consider the application of this statement to the temperatures attainable in processes involving a single substance, we can show that it implies, and is implied by, the Lewis and Randall statement.
The properties of the heat capacity, $C_P$, play a central role in these arguments. We have seen that $C_P$ is a function of temperature. While it is not useful to do so, we can apply the defining relationship for $C_P$ to a substance undergoing a phase transition and find $C_P=\infty$. If we think about a substance whose heat capacity is less than zero, we encounter a contradiction of our basic ideas about heat and temperature: If $q>0$ and ${q}/{\Delta T}<0$, we must have $\Delta T<0$; that is, heating the substance causes its temperature to decrease. In short, the theory we have developed embeds premises that require $C_P>0$ for any system on which we can make measurements.
Let us characterize a pure-substance system by its pressure and temperature and consider reversible constant-pressure processes in which only pressure–volume work is possible. Then ${\left({\partial S}/{\partial T}\right)}_P={C_P}/{T}$ and $dS={C_PdT}/{T}$. We now want to show: the Lewis and Randall stipulation that the entropy is always finite requires that the heat capacity go to zero when the temperature goes to zero. (Since we are going to show that the third law prohibits measurements at absolute zero, this conclusion is consistent with our conclusion in the previous paragraph.) That the heat capacity goes to zero when the temperature goes to zero is evident from $S={C_PdT}/{T}.$ If $C_P$ does not go to zero when the temperature goes to zero, $dS$ becomes arbitrarily large as the temperature goes to zero, which contradicts the Lewis and Randall statement.
To develop this result more explicitly, we let the heat capacities at temperatures $T$ and zero be $C_P\left(T\right)$ and $C_P\left(0\right)$, respectively. Since $C_P\left(T\right)>0$ for any $T\ >\ 0$, we have $S\left(T\right)-S\left(T^*\right)>0$ for any $T>T^*>0$. Since the entropy is always finite, $\infty >S\left(T\right)-S\left(T^*\right)>0$, so that
$\infty >{\mathop{\mathrm{lim}}_{T^*\to 0} \left[S\left(T\right)-S\left(T^*\right)\right]\ }>0 \nonumber$
and
$\infty >{\mathop{\mathrm{lim}}_{T^*\to 0} \int^T_{T^*}{\frac{C_P}{T}}\ }dT>0 \nonumber$
For temperatures in the neighborhood of zero, we can expand the heat capacity, to arbitrary accuracy, as a Taylor series polynomial in $T$:
$C_P\left(T\right)=C_P\left(0\right)+\left(\frac{\partial C_P\left(0\right)}{\partial T}\right)_PT +\frac{1}{2} \left(\frac{{\partial }^2C_P\left(0\right)}{\partial T^2}\right)_PT^2+\dots \nonumber$
The inequalities become
$\infty >{\mathop{\mathrm{lim}}_{T^*\to 0} \left\{C_P\left(0\right){ \ln \frac{T}{T^*}\ }+{\left(\frac{\partial C_P\left(0\right)}{\partial T}\right)}_P\left(T-T^*\right)+\frac{1}{4}{\left(\frac{{\partial }^2C_P\left(0\right)}{\partial T^2}\right)}_P{\left(T-T^*\right)}^2+\dots \right\}\ }>0 \nonumber$
The condition on the left requires $C_P\left(0\right)=0$.
We could view the third law as a statement about the heat capacities of pure substances. We infer not only that $C_P>0$ for all $T>0$, but also that
${\mathop{\mathrm{lim}}_{T\to 0} \left(\frac{C_P}{T}\right)=0\ } \nonumber$
More generally, we can infer corresponding assertions for closed reversible systems that are not pure substances: ${\left({\partial H}/{\partial T}\right)}_P>0$ for all $T>0$, and ${\mathop{\mathrm{lim}}_{T\to 0} T^{-1}{\left({\partial H}/{\partial T}\right)}_P=0\ }$. (The zero-temperature entropies of such systems are not zero, however.) In the discussion below, we describe the system as a pure substance. We can make essentially the same arguments for any system; we need only replace $C_P$ by ${\left({\partial H}/{\partial T}\right)}_P$. The Lewis and Randall statement asserts that the entropy goes to a constant at absolute zero, irrespective of the values of any other thermodynamic functions. It follows that the entropy at zero degrees is independent of the value of the pressure. For any two pressures, $P_1$ and $P_2$, we have $S\left(P_2,0\right)-S\left(P_1,0\right)=0$. Letting ${P=P}_1$ and $P_2=P+\Delta P$ and, we have
$\frac{S\left(P+\Delta P,0\right)-S\left(P,0\right)}{\Delta P}=0 \nonumber$
for any $\Delta P$. Hence, we have
${\left(\frac{\partial S}{\partial P}\right)}_{T=0}=0 \nonumber$
In Chapter 10, we find ${\left({\partial S}/{\partial }P\right)}_T=-{\left({\partial V}/{\partial T}\right)}_P$, so both the entropy and the volume approach their zero-temperature values asymptotically.
When we say that absolute zero is unattainable, we mean that no system can undergo any change in which its final temperature is zero. To see why absolute zero must be unattainable, let us consider processes that can decrease the temperature of a system. In general, we have heat reservoirs available at various temperatures. We can select the available reservoir whose temperature is lowest, and bring the system to this temperature by simple thermal contact. This much is trivial; clearly, the challenge is to decrease the temperature further. To do so, we must effect some other change. Whatever this change may be, it cannot be aided by an exchange of heat with the surroundings. Once we have brought the system to the temperature of the coldest available portion of the surroundings, any further exchange of heat with the surroundings can only be counter-productive. We conclude that any process suited to our purpose must be adiabatic. Since an adiabatic process exchanges no heat with the surroundings, $\Delta \hat{S}=0$.
The process must also be a possible process, so that $\Delta S+\Delta \hat{S}\ge 0$, and since it is adiabatic, $\Delta S\ge 0$. Let us consider a reversible process and an irreversible process in which the same system${}^{2}$ goes from the state specified by $P_1$ and $T_1$ to a second state in which the pressure is $P_2$. The final temperatures and the entropy changes of these processes are different. For the reversible process, $\Delta S=0$; we designate the final temperature as $T_2$. For the irreversible process, $\Delta S>0$; we designate the final temperature as $T^*_2$. As it turns out, the temperature change is less for the irreversible process than for the reversible process; that is, $T_2-T_1<t^*_2-t_1$>. Equivalently, the reversible process reaches a lower temperature: ${T_2<t}^*_2$>. From
$dS=\frac{C_P}{T}dT-{\left(\frac{\partial V}{\partial T}\right)}_PdP \nonumber$
we can calculate the entropy changes for these processes. For the reversible process, we calculate $\Delta S^{rev}=S\left(P_2,T_2\right)-S\left(P_1,T_1\right) \nonumber$
To do so, we first calculate
${\left(\Delta S\right)}_T=S\left(P_2,T_1\right)-S\left(P_1,T_1\right) \nonumber$
for the isothermal reversible transformation from state $P_1$, $T_1$ to the state specified by $P_2$ and $T_1$. For this step, $dT$ is zero, and so
${\left(\Delta S\right)}_T=\int^{P_2}_{P_1}{{\left(\frac{\partial V}{\partial T}\right)}_PdP} \nonumber$
We then calculate
${\left(\Delta S\right)}_P=S\left(P_2,T_2\right)-S\left(P_2,T_1\right) \nonumber$
for the isobaric reversible transformation from state $P_2$, $T_1$ to state $P_2$, $T_2$. For this transformation, $dP$ is zero, and
${\left(\Delta S\right)}_P=-\int^{T_2}_{T_1}{\frac{C_P}{T}dT} \nonumber$
Then,
$\Delta S^{rev}=S\left(P_2,T_2\right)-S\left(P_1,T_1\right)=\int^{T_2}_{T_1}{\frac{C_P}{T}dT}-\int^{P_2}_{P_1}{{\left(\frac{\partial V}{\partial T}\right)}_PdP}=0 \nonumber$
Because $\Delta S^{rev}=0$, the reversible process is unique; that is, given $P_1$, $T_1$, and $P_2$, the final temperature of the system is determined. We find $T_2$ from
$\int^{T_2}_{T_1}{\frac{C_P}{T}dT}=\int^{P_2}_{P_1}{{\left(\frac{\partial V}{\partial T}\right)}_PdP} \nonumber$
To understand the entropy change for the irreversible process, we note first that there are an infinite number of such processes. There is nothing unique about the final temperature. Given $P_1$, $T_1$, and $P_2$, the final temperature, $T^*_2$, can have any value consistent with the properties of the substance. To specify a particular irreversible process, we must specify all four of the quantities $P_1$, $T_1$, $P_2$, and $T^*_2$. Having done so, however, we can calculate the entropy change for the irreversible process,
$\Delta S^{irrev}=S\left(P_2,T^*_2\right)-S\left(P_1,T_1\right)>0 \nonumber$
by computing the entropy changes as we reversibly carry the system along the isothermal two-step path from $P_1$, $T_1$ to $P_2$, $T_1$ and then along the isobaric path from $P_2$, $T_1$ to $P_2$, $T^*_2$. The calculation of $\Delta S^{irrev}$ for this reversible path from $P_1$, $T_1$ to $P_2$, $T^*_2$ employs the same logic as the calculation, in the previous paragraph, of $\Delta S$ for the reversible path from $P_1$, $T_1$ to $P_2$, $T_2$. The difference is that $T^*_2$ replaces $T_2$ as the upper limit in the temperature integral. The pressure integral is the same. We have
$\Delta S^{irrev}=S\left(P_2,T^*_2\right)-S\left(P_1,T_1\right)=\int^{T^*_2}_{T_1}{\frac{C_P}{T}dT}-\int^{P_2}_{P_1}{{\left(\frac{\partial V}{\partial T}\right)}_PdP} >0 \nonumber$
From $\Delta S^{irrev}>\Delta S^{rev}$, we have
$\int^{T^*_2}_{T_1}{\frac{C_P}{T}dT}>\int^{T_2}_{T_1}{\frac{C_P}{T}dT} \nonumber$
Since the integrands are the same and positive, it follows that $T^*_2>T_2$, as asserted above.
Figure 6 shows the relationships among the various quantities discussed in this argument. In the first instance, Figure 6 shows a plot of two of the system’s isobars in temperature—entropy space. That is, the line labeled $P=P_1$ depicts the set of temperature—entropy points at which the equilibrated system has pressure $P_1$; the line labeled $P=P_2$, depicts the equilibrium positions at pressure $P_2$. Other lines in this sketch represent paths along which the system can undergo reversible changes at constant entropy or constant temperature. The dotted line represents the irreversible process in which the system goes from the state specified by $P_1$, $T_1$ to the state specified by $P_2$, $T^*_2$. This line is dotted to represent the fact that the system’s temperature may not be well defined during the irreversible process.
Effective cooling can be achieved using pressure changes if the system is a gas. However, for liquids and solids, ${\left({\partial V}/{\partial T}\right)}_P$ is small; consequently, the temperature change for a reversible pressure change is also small. At temperatures near absolute zero, nearly all substances are solid; to achieve effective cooling we must change a thermodynamic variable for which a solid’s temperature coefficient is as large as possible. To consider the general problem of decreasing the temperature of a system by varying something other than pressure, we must consider a system in which some form of non-pressure–volume work is possible. Such a system is subject to an additional force, and its energy changes as this force changes.
Adiabatic Demagnetization
The practical method by which extremely low temperatures are achieved is called adiabatic demagnetization. This method exploits the properties of paramagnetic solids. In such solids, unpaired electrons localized on individual atoms give rise to a magnetic moment. Quantum mechanics leads to important conclusions about the interaction between such magnetic moments and an applied magnetic field: In an applied magnetic field, the magnetic moment of an individual atom is quantized. In the simplest case, it can be aligned in only two directions; it must be either parallel or anti-parallel to the applied magnetic field. When an atom’s magnetic moment is parallel to the magnetic field the energy of the system is less than when the alignment is anti-parallel. The applied magnetic field exerts a force on the magnetic moments associated with individual atoms. The energy of the system depends on the magnitude of the applied magnetic field.
Rather than focus on the particular case of adiabatic demagnetization, let us consider the energy and entropy changes associated with changes in a generalized potential, ${\mathit{\Phi}}_{\theta }$, and its generalized displacement, $\theta$. (For adiabatic demagnetization, $\theta$ would be the applied magnetic field.) Three variables are required to describe reversible changes in this system. We can express the energy and entropy as functions of temperature, pressure, and $\theta$:
$E=E\left(T,P,\theta \right)$ and $S=S\left(T,P,\theta \right)$. The total differential of the entropy includes a term that specifies the dependence of entropy on $\theta$. We have
$dS={\left(\frac{\partial S}{\partial T}\right)}_{P,\theta }dT+{\left(\frac{\partial S}{\partial P}\right)}_{T,\theta }dP+{\left(\frac{\partial S}{\partial \theta }\right)}_{T,P}d\theta =\frac{C\left(T,P,\theta \right)}{T}dT-{\left(\frac{\partial V}{\partial T}\right)}_{P,\theta }dP+{\left(\frac{\partial S}{\partial \theta }\right)}_{T,P}d\theta \nonumber$
where we write $C\left(T,P,\theta \right)$ to emphasize that our present purposes now require that we measure the heat capacity at constant pressure and constant $\theta$.
For constant pressure, P, and constant displacement, $\theta$, the entropy depends on temperature as
$S\left(T,P,\theta \right)=S\left(0,P,\theta \right)+\int^T_0 \left(\frac{\partial S}{\partial T}\right)_{P,\theta }dT=S\left(0,P,\theta \right)+\int^T_0 \frac{C\left(T,P,\theta \right)}{T}dT \nonumber$
The postulate that the entropy be finite at any temperature implies that the pressure- and $\theta$-dependent heat capacity becomes zero at absolute zero. That is, at absolute zero, the heat capacity vanishes whatever the values of P and $\theta$. The argument is exactly the same as before. Earlier, we wrote $C_P\left(0\right)=0$; for the present generalized case, we write $C\left(0,P,\theta \right)=0$.
Similarly, from the postulate that the entropy goes to a constant at absolute zero for all values of the other thermodynamics variables, it follows that, for any two pressures $P_1$ and $P_2$, and for any two values of the generalized displacement, ${\theta }_1$ and ${\theta }_2$,
$S\left(0,P_1,{\theta }_1\right)=S\left(0,P_2,{\theta }_1\right)=S\left(0,P_1,{\theta }_2\right)=S\left(0,P_2,{\theta }_2\right)=0 \nonumber$
and hence that
${\left(\frac{\partial S}{\partial P}\right)}_{T=0,\theta }={\left(\frac{\partial S\left(0,P,\theta \right)}{\partial P}\right)}_{T,\theta }=0 \nonumber$ and ${\left(\frac{\partial S}{\partial \theta }\right)}_{T=0,P}={\left(\frac{\partial S\left(0,P,\theta \right)}{\partial \theta }\right)}_{T,P}=0 \nonumber$
We want to consider a process in which a system goes from the lowest temperature available in the surroundings to a still lower temperature. To minimize the final temperature, this process must be carried out adiabatically. It must also be a possible process, so that $dS\ge 0$. For simplicity, let us now assume that we carry out this process at a constant pressure, $P$, and that the system goes from the state specified by $P$, $T_1$, ${\theta }_1$ to the state specified by $P$, $T_2$, ${\theta }_2$ where $T_1>T_2$. The entropies of these two states are
$S\left(T_1,P,{\theta }_1\right)=S\left(0,P,{\theta }_1\right)+\int^{T_1}_0{\frac{C\left(T,P,{\theta }_1\right)}{T}}dT \nonumber$ and $S\left(T_2,P,{\theta }_2\right)=S\left(0,P,{\theta }_2\right)+\int^{T_2}_0{\frac{C\left(T,P,{\theta }_2\right)}{T}}dT \nonumber$
The entropy change for this process is
$S\left(T_2,P,{\theta }_2\right)-S\left(T_1,P,{\theta }_1\right)=S\left(0,P,{\theta }_2\right)-S\left(0,P,{\theta }_1\right) \nonumber$ $+\int^{T_2}_0{\frac{C\left(T,P,{\theta }_2\right)}{T}}dT-\int^{T_1}_0{\frac{C\left(T,P,{\theta }_1\right)}{T}}dT\ge 0 \nonumber$
Now, let us suppose that the final temperature is zero; that is, $T_2=0$, so that
$\int^{T_2}_0{\frac{C\left(T,P,{\theta }_2\right)}{T}}dT=0 \nonumber$ It follows that $S\left(0,P,{\theta }_2\right)-S\left(0,P,{\theta }_1\right)\ge \int^{T_1}_0{\frac{C\left(T,P,{\theta }_1\right)}{T}}dT>0 \nonumber$
where the inequality on the right follows from the fact that that $C\left(T,P,{\theta }_1\right)>0$. Then, it follows that
$S\left(0,P,{\theta }_2\right)-S\left(0,P,{\theta }_1\right)>0 \nonumber$
which contradicts the Lewis and Randall statement of the third law. The assumption that the system can reach absolute zero leads to a contradiction of the Lewis and Randall statement of the third law. Therefore, if the Lewis and Randall statement is true, absolute zero is unattainable.
The converse applies also; that is, from the proposition that absolute zero is unattainable, we can show that the Lewis and Randall statement is true. To do so, we rearrange the above equation for $\Delta S$,
$\int^{T_2}_0{\frac{C\left(T,P,{\theta }_2\right)}{T}}dT\ge \nonumber$ $\int^{T_1}_0{\frac{C\left(T,P,{\theta }_1\right)}{T}}dT-S\left(0,P,{\theta }_2\right)+S\left(0,P,{\theta }_1\right) \nonumber$
If we now assume that the Lewis and Randall statement is false, the expression on the right can be less than or equal to zero. The integral on the left can then be zero, in which case the system can reach absolute zero. If the Lewis and Randall statement is false, it is true that the system can reach absolute zero. Therefore: If the system cannot reach absolute zero, the Lewis and Randall statement is true.
Figures 7 and 8 depict these ideas using contour plots in temperature–entropy space. Each figure shows two contour lines. One of these contour lines is a set of temperature and entropy values along which the pressure is constant at $P$ and $\theta$ is constant at ${\theta }_1$. The other contour line is a set of temperature and entropy values along which the pressure is constant at $P$ and $\theta$ is constant at ${\theta }_2$. The slope of a contour line is
${\left(\frac{\partial T}{\partial S}\right)}_{P,\theta }=\frac{T}{C\left(T,P,\theta \right)} \nonumber$
Because the heat capacity is always positive, this slope is always positive.
In Figure 7, the Lewis and Randall statement is satisfied. When the temperature goes to zero, the contour lines meet at the same value of the entropy; these contours satisfy the relationship
$S\left(0,P,{\theta }_1\right)=S\left(0,P,{\theta }_2\right) \nonumber$
An adiabatic (vertical) path from the contour for $P$ and ${\theta }_1$ meets the contour for $P$ and ${\theta }_2$ at a positive temperature, $T_2>0$. Since this is evidently true for any $P$ and any ${\theta }_2$, the final state for any adiabatic process will have $T_2>0$. Because the Lewis and Randall statement is satisfied, the system cannot reach absolute zero, and vice versa.
In Figure 8, the Lewis and Randall statement is violated, because we have $S\left(0,P,{\theta }_1\right)$. In this case, an adiabatic process initiated from a low enough initial temperature, $T_1$, will reach absolute zero without intersecting the contour for constant $P$ and ${\theta }_2$. Because the Lewis and Randal statement is violated, the system can reach absolute zero, and vice versa.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/11%3A_The_Third_Law_Absolute_Entropy_and_the_Gibbs_Free_Energy_of_Formation/11.13%3A_Absolute_Zero_is_Unattainable.txt
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1. The relationships between $H^o$, $S^o$, and $G^o$for the standard state of a gas and the molar enthalpy, $\overline{H}\left(P\right)$, entropy, $\overline{S}\left(P\right)$, and Gibbs free energy, $\overline{G}\left(P\right)$, of the real gas at pressure $P$ and temperature $T$ involve several integrals. Given the virial equation for a real gas,
$Z=\frac{P\overline{V}}{RT}=1+B^*\left(T\right)P+C^*\left(T\right)P^2+D^*\left(T\right)P^3+\dots \nonumber$
evaluate the following:
(a)$\int^P_0{\left[\overline{V}-T{\left(\frac{\partial \overline{V}}{\partial T}\right)}_P\right]dP} \nonumber$
(b) $\int^P_0{\left[\frac{\overline{V}}{RT}-\frac{1}{P}\right]dP} \nonumber$
(c) $\int^P_{P^*} \left(\frac{\partial \overline{V}}{\partial T}\right)_PdP-\int^{\mathrm{1\ bar}}_{P^*} \frac{R}{P} dP \nonumber$
2. Why does ${\left({\partial S}/{\partial P}\right)}_{T=0}={-\left({\partial V}/{\partial T}\right)}_{P,\ T=0}$ imply that both the entropy and the volume approach their zero-temperature values asymptotically? Is this consistent with defining absolute zero to be the temperature at which the volume of an ideal gas extrapolates to zero—at constant pressure?
3. Prove that $\overline{S}_A\left(P\right)-S^0_A=\int^{P^*}_P \left(\frac{\partial \overline{V}}{\partial T}\right)_PdP+\int^{\mathrm{1\ bar}}_{P^*} \frac{R}{P} dP \nonumber$
remains finite for any gas in the limit as $P^*\to 0$. Hint: Express the integral from $P$ to $P^*$ as the sum of integrals from $P^*$ to 1 bar and from 1 bar to $P$.
4. Let $A$, $B$, $C$, and $D$ be elements, whose absolute entropies at 1 bar and temperature $T$ are $S^o\left(A,T\right)$, $S^o\left(B,T\right)$, $S^o\left(C,T\right)$, and $S^o\left(D,T\right)$, respectively. Let $AB$, $CD$, $AC$, and $BD$ be binary compounds of these elements, and represent their absolute entropies at these conditions by $S^o\left(AB,T\right)$, $S^o\left(CD,T\right)$, $S^o\left(AC,T\right)$, and $S^o\left(BD,T\right)$.
(a) What is the entropy change, ${\Delta }_rS^o\left(T\right)$, for the reaction $AB+CD\to AC+BD$?
(b) What are ${\Delta }_fS^o\left(AB,T\right)$, ${\Delta }_fS^o\left(CD,T\right)$, ${\Delta }_fS^o\left(AC,T\right)$, and ${\Delta }_fS^o\left(BD,T\right)$?
(c) Show that
${\Delta }_rS^o\left(T\right)={\Delta }_fS^o\left(AC,T\right)+{\Delta }_fS^o\left(BD,T\right) -{\Delta }_fS^o\left(AB,T\right)-{\Delta }_fS^o\left(CD,T\right) \nonumber$
5. Let $A$ and $B$ be elements; let $A_aB_b$ be a binary compound of these elements. At temperature $T$ and 1 bar, let the entropy of these substances be $S^o\left(A,T\right)$, $S^o\left(B,T\right)$, and $S^o\left(A_aB_b,T\right)$, respectively. At absolute zero and 1 bar, let these entropies be $S^o\left(A,0\right)$, $S^o\left(B,0\right)$, and $S^o\left(A_aB_b,0\right)$. Represent the change in entropy when these substances are warmed from 0 K to $T$, at a constant pressure of 1 bar, as ${\Delta }_{0\to T}S^o\left(A\right)$, ${\Delta }_{0\to T}S^o\left(B\right)$, and ${\Delta }_{0\to T}S^o\left(A_aB_b\right)$. These quantities are related by the following equations:
$S^o\left(A,T\right)=S^o\left(A,0\right)+ {\Delta }_{0\to T}S^o\left(A\right) \nonumber$
$S^o\left(B,T\right)=S^o\left(B,0\right)+{\Delta }_{0\to T}S^o\left(B\right) \nonumber$
$S^o\left(A_aB_b,T\right)=S^o\left(A_aB_b,0\right)+{\Delta }_{0\to T}S^o\left(A_aB_b\right) \nonumber$
By the Nernst Heat Theorem, the entropy change for formation of $A_aB_b$ at absolute zero must be zero:
${\Delta }_fS^o\left(A_aB_b,0\right)=S^o\left(A_aB_b,0\right)-a\ S^o\left(A,0\right) -b\ S^o\left(B,0\right) \nonumber$
When, following Planck and Lewis and Randall, we choose to let the entropies of the elements be zero at absolute zero, the entropy of $A_aB_b$ and its entropy of formation also become zero at absolute zero:
${\Delta }_fS^o\left(A_aB_b,0\right)=S^o\left(A_aB_b,0\right)=0 \nonumber$
Suppose that we decide to create an alternative set of absolute entropies by assigning non-zero values to the entropies of the elements at absolute zero. Let us distinguish entropy values in this new scheme with a tilde. Then, the non-zero values that we assign to the elements at absolute zero are $\tilde{S}^o\left(A,0\right)\neq 0$ and $\tilde{S}^o\left(B,0\right)\neq 0$. By the Nernst Heat Theorem, we have
${\Delta }_f \tilde{S}^o\left(A_aB_b,0\right)= \tilde{S}^o\left(A_aB_b,0\right)-a\ \tilde{S}^o\left(A,0\right) -b \tilde{S}^o\left(B,0\right) \nonumber$
Evidently, we have
${\Delta }_f \tilde{S}^o\left(A_aB_b,0\right)={\Delta }_fS^o\left(A_aB_b,0\right)=0 \nonumber$
and, since the values of $\tilde{S}^o\left(A,0\right)$ and $\tilde{S}^o\left(B,0\right)$ are arbitrary, we can choose them so that $\tilde{S}^o\left(A_aB_b,0\right)$ is non-zero also
$\tilde{S}^o\left(A_aB_b,0\right)=a \tilde{S}^o\left(A,0\right) + b\ \tilde{S}^o\left(B,0\right)\mathrm{\neq }\mathrm{0\ } \nonumber$
(a) What are $\tilde{S}^o\left(A,T\right)$, $\tilde{S}^o\left(B,T\right)$, and $\tilde{S}^o\left(A_aB_b,T\right)$?
(b) What is ${\Delta }_f \tilde{S}^o\left(A_aB_b,T\right)$?
(c) Show that ${\Delta }_f \tilde{S}^o\left(A_aB_b,T\right)={\Delta }_fS^o\left(A_aB_b,T\right)$.
(d) Consider a reaction in which $A_aB_b$ is a reactant or a product. How will the alternative choice of values for the entropies of $A$ and $B$ at absolute zero affect the values we calculate for ${\Delta }_rS^o$, ${\Delta }_rH^o$, ${\Delta }_rG^o$?
(e) Can you think of any circumstance in which there would be an advantage to choosing $\tilde{S}^o\left(A,0\right)\neq 0$ and $\tilde{S}^o\left(B,0\right)\neq 0$?
6. Find $\ln \gamma$ for a gas that obeys the equation of state $P\left(\overline{V}-b\right)=RT$. For $CO_2$ at 300 K, the value of $b$ (the second virial coefficient) is $-1.26\times {10}^{-4}\ \mathrm{m}^3\ \mathrm{mol}^{-1}$.
Calculate the fugacity coefficient and the fugacity of $CO_2$ at 300 K and pressures of 1, 10, and 100 bar.
7. Consider the following sequence of steps that convert a van der Waals gas, vdwg, at an arbitrary pressure to the corresponding hypothetical ideal gas in its standard state.
(I) $A\left(\mathrm{vdwg},P,T\right) \begin{array}{c} {\Delta }_IG \ \to \ \end{array} A\left(\mathrm{vdwg},P^*\approx 0,T\right)$
(II) $A\left(\mathrm{vdwg},P^*\approx 0,T\right) \begin{array}{c} {\Delta }_{II}G \ \to \ \end{array} A\left(\mathrm{ideal\ gas},P^*\approx 0,T\right)$ (III) $A\left(\mathrm{ideal\ gas},P^*\approx 0,T\right) \begin{array}{c} {\Delta }_{III}G \ \to \ \end{array} A\left({HIG}^o,T\right)$
Show that the fugacity of the van der Waals gas is given by $\ln f_{vdw}=\frac{b}{\overline{V}-b} -\frac{2a}{RT\overline{V}}+ \ln \frac{RT}{\overline{V}-b} \nonumber$
Hint: Find $\Delta G={\Delta }_IG+{\Delta }_{II}G+{\Delta }_{III}G$. To calculate ${\Delta }_IG$, use integration by parts:
${\Delta }_IG=\int^{P^*}_P \overline{V}_{vdw}dP=\left[P\overline{V}\right]^{P^*}_P-\int^{\overline{V}^*}_{\overline{V}} Pd\overline{V} \nonumber$
When $P^*\to 0$, $\overline{V}^*\to \infty$.
8. The normal boiling point of methanol is 337.8 K at 1 atm. The enthalpy of vaporization at the normal boiling point is ${\Delta }_{vap}H=35.21\ \mathrm{kJ}\ \mathrm{mol}^{-1}$.
(a) What are ${\Delta }_{vap}G$ and ${\Delta }_{vap}S$ for methanol at its normal boiling point?
(b) At 1 bar, the heat capacity of gaseous methanol depends on temperature as
$C_P\left(CH_3OH,\mathrm{g},1\ \mathrm{bar}\right)=21.737+0.07494\ T\ \left[\mathrm{J}\ \mathrm{K}^{-1}\ \mathrm{mol}^{-1}\right] \nonumber$
Assume that the heat capacity at 1 atm is the same as it is at 1 bar. Calculate the enthalpy change and the entropy change when one mole of gaseous methanol is heated from the normal boiling point to 500 K at 1 atm.
(c) At 1 bar, the absolute entropy of gaseous methanol depends on temperature as
$S^o\left(CH_3OH,\mathrm{g},1\ \mathrm{bar}\right)=192.8+0.1738\ T -\left(5.367\times {10}^{-5}\right)T^2 \left[\mathrm{J}\ {\mathrm{K}}^{-1}\ {\mathrm{mol}}^{-1}\right] \nonumber$
Assume that the absolute entropy at 1 atm is the same as it is at 1 bar. Calculate the Gibbs Free Energy change, $\Delta G$, when one mole of gaseous methanol is heated from the normal boiling point to 500 K at a constant pressure of 1 atm.
(d) The heat capacity of liquid methanol, $C_P\left(CH_3OH,\mathrm{liq},298.15\ \mathrm{K},1\ \mathrm{bar}\right) \nonumber$ is $1.1\ \mathrm{J}\ {\mathrm{K}}^{-1}\ {\mathrm{mol}}^{-1}$. Assume that this heat capacity remains constant for superheated liquid methanol. Calculate the enthalpy and entropy changes when liquid methanol is heated from the normal boiling point to 500 K.
(e) The molar entropy of liquid methanol at 1 bar and 298.15 K is $126.8\ \mathrm{J}\ {\mathrm{K}}^{-1}\ {\mathrm{mol}}^{-1}$. We can estimate the molar entropy of superheated liquid methanol by using the heat capacity at 298.15 K to estimate the entropy at higher temperatures:
$S^o\left(T\right)\approx S^o\left(298.15\ \mathrm{K}\right)+\int^T_{298.15}{\frac{C_P}{T}}dT \nonumber$
Find $S^o\left(T\right)$ and use this equation to calculate $\Delta G$ for heating liquid methanol from the normal boiling point to 500 K.
(f) Devise a cycle that enables you to use the results you obtain in parts (a)–(e) to calculate ${\Delta }_{vap}H\left(500\ \mathrm{K}\right)$, ${\Delta }_{vap}S\left(500\ \mathrm{K}\right)$, and ${\Delta }_{vap}G\left(500\ \mathrm{K}\right)$ when one mole of methanol vaporizes at 1 atm and 500 K.
(g) Use the values you obtain for ${\Delta }_{vap}H\left(500\ \mathrm{K}\right)$ and ${\Delta }_{vap}S\left(500\ \mathrm{K}\right)$ in part (f) to calculate ${\Delta }_{vap}G\left(500\ \mathrm{K}\right)$.
(h) Use the Gibbs-Helmholtz equation to calculate the Gibbs free energy change, ${\Delta }_{vap}G\left(500\ \mathrm{K}\right)$, when one mole of methanol vaporizes at 1 atm and 500 K. Compare this result to those you obtained in parts (f) and (g). Is this process impossible, spontaneous, or reversible?
(i) How much heat is taken up by the system when one mole of methanol is vaporized reversibly at its normal boiling point and the resulting vapor is heated reversibly at 1 atm to 500 K? How much work is done on the system in this process? What are $\Delta E$ and $\Delta S$ for this process?
(j) Suppose that the change of state in part (i) is effected irreversibly by contacting the liquid methanol with surroundings at 500 K, while maintaining the applied pressure constant at 1 atm. How much work is done on the system in this spontaneous process? How much heat is taken up by the system? What are $\Delta \hat{S}$ and $\Delta S_{universe}$ for this spontaneous process?
9. To maximize the temperature change for a given change in system pressure, the value of $\left| \left( \partial T / \partial P \right) _S\right|$ should be as large as possible. If only pressure–volume work is possible, we have
$dS=\frac{C_P}{T}dT- \left(\frac{\partial V}{\partial T}\right)_PdP \nonumber$
Show that
$\left(\frac{\partial T}{\partial P}\right)_S=\frac{T}{C_P} \left(\frac{\partial V}{\partial T}\right)_P \nonumber$
What happens to $T/C_P$ as the temperature approaches absolute zero? For condensed phases, we find $\left( \partial V/ \partial T \right)_P\ll V$. Consider the behavior of the molecules in a lattice as the temperature approaches absolute zero. Is it reasonable to expect
$\mathop{\mathrm{lim}_{T\to 0}} \left(\frac{\partial V}{\partial T}\right)_P=0 \nonumber$
Why?
Notes
${}^{1\ }$Lewis, G.N., Randall, M., K. S. Pitzer, and L. Brewer, Thermodynamics, 2${}^{nd}$ Edition, McGraw-Hill, New York, 1981, p 130.
${}^{2}$ More precisely, we consider a reversible process and a spontaneous process whose initial state is a “change-enabled” modification of the reversible-process initial state. The state functions are the same in both initial states.
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The equations we derive in Chapters 9 and 10 are the core of chemical thermodynamics. However, we have yet to deal with the effects of changing the concentrations of the substances present in the system. To apply our theory to chemical changes, we must extend our theory so that it can model these effects. In this chapter, we consider some basic applications that do not involve chemical reactions and in which both intermolecular interactions and the effects of mixing can be ignored. In Chapters 13-16, we develop the application of thermodynamic concepts to processes in which a chemical reaction occurs. We do so in two steps. In Chapter 13, we consider an approximation in which the properties of a multi-component system are determined by the effects of mixing pure substances whose molecules neither attract nor repel one another. In Chapter 14, we begin to consider the general case in which intermolecular interactions can be important.
12: Applications of the Thermodynamic Criteria for Change
When we talk about a purely mechanical process, we have in mind a system in which one or more unchanging objects can move relative to some reference frame. Their movements are described completely by Newton’s laws of motion. The objects are characterized by their masses, locations, velocities, and accelerations. They may be subject to the effects of force fields, whose magnitudes can vary with location and time. We stipulate that the volume, pressure, temperature, entropy, composition, and the internal energy, $U$, of an individual object remain constant. Since entropy and volume are extensive state functions, we can obtain the entropy and volume of the objects in aggregate by summing up those for the individual objects. Moreover, the total entropy and the total volume are constants.
The energy of a purely mechanical system is the sum of its internal energy, $U$, its kinetic energy, $\tau$, and its mechanical potential energy, $\upsilon$; that is,
$E=U+\tau +\upsilon. \nonumber$
$U$ is the sum of the internal energies of the unchanging constituent objects. Since it is constant, the internal energy can be given an arbitrary value that we usually choose to be zero. When we do so, the energy of the system becomes the sum of its kinetic and potential energies. Noting explicitly that the entropy and volume are constant, we have
${\left(\Delta E\right)}_{SV}=\tau +\upsilon. \nonumber$
The essential distinction between a purely mechanical system and a thermodynamic system is that our models for mechanical systems focus on the motions of unchanging objects; our models for thermodynamic systems focus on the internal changes of stationary objects.
An important aspect of this distinction is that our definitions of equilibrium, reversibility, and spontaneous change in mechanical systems are not wholly congruent with the definitions we use in developing the principles of thermodynamics. Thus, an equilibrium state of a mechanical system is one in which the objects comprising the system are stationary with respect to some reference frame. For a mechanical system at equilibrium, the kinetic energy is constant and can usually be taken to be zero, $\tau =0$. The motion of a system generally has no bearing on whether the substances that comprise the system are at equilibrium in a thermodynamic sense.
We normally consider that a sufficient condition for a mechanical system to be reversible is that—following some excursion to other states—the initial conditions of both the system and the surroundings can be restored exactly${}^{1}$. In our thermodynamic view of reversibility, this condition is necessary but not sufficient: We say that a thermodynamic system is reversible only if the direction in which it is changing can be reversed at any time by an arbitrarily small change in its interaction with its surroundings. The initial conditions can be restored after any excursion.
Our treatments of the frictionless harmonic oscillator illustrate this imperfect congruence. Viewed as a mechanical system, a frictionless harmonic oscillator is a reversible system. If we adopt the view that it is continuously undergoing spontaneous change, our thermodynamic principles mean that its entropy is continuously increasing. However, since its state and that of its surroundings reproduce themselves exactly after every period of oscillation, our thermodynamic principles mean also that there is zero net change in the entropy over any complete oscillation. Clearly, this is a contradiction. If we attempt to salvage the situation by supposing that the entropy of an isolated freely moving harmonic oscillator is constant, our thermodynamic principles require us to say that it is at equilibrium. This contradicts the view that a mechanical system is at equilibrium only if its kinetic energy is zero. Since neither formulation is satisfactory, we recognize that we cannot expect to describe every mechanical system in purely thermodynamic terms.
Nor can we expect to describe every thermodynamic system in mechanical terms. This becomes obvious when we observe that the second law of thermodynamics is essential to our description of thermodynamic systems, but it is not among the principles of mechanics. Beginning in Chapter 20, we find that we can model the thermodynamic properties of a system that is itself a collection of a large number of subsystems by focusing on the average values of the properties of the subsystems. The laws of motion model the movements of the individual particles of a system. The laws of thermodynamics model the average properties of the particles in a system that contains a very large number of particles. While we cannot usefully describe an individual harmonic oscillator as a thermodynamic system, we see in §22-6 that the thermodynamic properties of a system composed of many identical harmonic oscillators can be modeled very successfully.
In short, mechanics and thermodynamics model different kinds of systems from fundamentally different perspectives. Nevertheless, when we limit our consideration of mechanical systems to the prediction of spontaneous change from one equilibrium state to another, we can recognize that the criteria we are accustomed to apply to mechanical systems are analogous to our thermodynamic criteria as they apply to such processes.
From our thermodynamic perspective, a purely mechanical process involves no change in the entropy or volume of the system, and the criterion for irreversible change is ${\left(\Delta E\right)}_{SV}<w_{npv}$>. From our mechanical perspective, an irreversible transition between equilibrium states is one in which various objects interact with one another or with various force fields. We consider that the process can occur if the change in the potential energy of the system is less than the work done on it; that is $\Delta \upsilon <w_{npv}$>. In the mechanical system, some of the work done on the system is dissipated by frictional forces as heat that appears in the surroundings. (If the process involves no exchange of work with the surroundings, the criterion becomes $\Delta \upsilon <0$.) If the mechanical system begins and ends at rest, we have ${\left(\Delta E\right)}_{SV}<\Delta \upsilon$, so that again ${\left(\Delta E\right)}_{SV}<w_{npv}$.
If no work is exchanged with the surroundings, potential energy minimization, $\Delta \upsilon <0$, is a sufficient condition for spontaneous change to be possible in mechanical systems under the circumscribed conditions we have outlined. In Section 14-2, we find that minimization of the chemical potential energy,
$\sum^{\omega }_{j=1}{{\mu }_j}dn_j<0 \nonumber$
is a necessary and sufficient condition for spontaneous change to be possible in a thermodynamic system. These conditions are parallel, but they are not equivalent to one another.
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The idea that thermal energy can be transferred from a warmer body to a colder one, but not in the opposite direction, is a fundamental assumption in our development of the thermodynamic criteria for change. Therefore, if our theory is to be internally consistent, we must be able to deduce this principle from the criteria we have developed. Let us consider one way in which this can be done: We consider an isolated system comprised of two subsystems, \(A\) and \(B\), that are in thermal contact with one another. We suppose that the temperatures are \(T_A\) and \(T_B\) and that \(T_A\neq T_B\). If the energy of subsystem \(A\) increases, heat transfers from subsystem \(B\) to subsystem \(A\). In this case, we know that \(q_A>0\) and \(T_B>T_A\).
When we seek to analyze this process using our thermodynamic theory, we encounter a problem that arises for any spontaneous process: Since the process is not reversible, we must introduce approximating assumptions. For the present analysis, we want to estimate the entropy change that occurs in each subsystem. To do so, we suppose that an increment of heat, \(dq\), can pass from one subsystem to the other without significantly changing the temperature of either one. It is evident that we could—by some other process—effect this change in either subsystem as nearly reversibly as we wish. (In §5, we consider such a process.) Even though the present process is not reversible, we have good reason to assume that the entropy changes in the subsystems are well approximated as \({dq_A}/{T_A}\) and \({dq_B}/{T_B}=-{dq_A}/{T_B}\). Since the system is isolated, the process can be spontaneous only if its entropy change is positive; that is, the relevant thermodynamic criterion is \({dq_A}/{T_A}-{dq_A}/{T_B}>0\). With \({dq}_A>0\), we find \({1}/{T_A}-{1}/{T_B}>0\), or \(T_B>T_A\). When heat is spontaneously transferred from \(B\) to \(A\), our thermodynamic criterion also requires that subsystem \(B\) be warmer than subsystem \(A\).
12.03: Phase Changes - the Fusion of Ice
Let us consider processes in which transfer of heat from the surroundings melts one mole of ice. We suppose that the ice is initially at 0 ºC and one bar. At these conditions, the enthalpy change for melting a mole of ice is 6010 J. If the ice melts reversibly at these conditions, the temperature of the surroundings is also 0 ºC. As it melts, the ice takes up 6010 J of heat, which is given up by the surroundings. For this process, we have $q^{rev}_P={\Delta }_{fus}H=6010\ \mathrm{J}$. The temperature is constant, and the entropy change for the system is
$\Delta S=\frac{q^{rev}_P}{T}=\frac{{\Delta }_{fus}H}{T} \nonumber$
Since ${\hat{q}=-q}^{rev}_P$, we have
$\Delta \hat{S}=\frac{-q^{rev}_P}{\hat{T}}=\frac{-{\Delta }_{fus}H}{\hat{T}} \nonumber$
so that
$\Delta S={6010\ \mathrm{J}}/{273.15\ \mathrm{K}}=22.00\ \mathrm{J}\ {\mathrm{K}}^{-1} \nonumber$ and $\Delta \hat{S}=-{6010\ \mathrm{J}}/{273.15\ \mathrm{K}}=-22.00\ \mathrm{J}\ {\mathrm{K}}^{-1} \nonumber$ As required for a reversible process, we have $\Delta S+\Delta \hat{S}=0$. The Gibbs free energy change is
${\left(\Delta G\right)}_{PT}={\left(\Delta H\right)}_{PT}-T{\left(\Delta S\right)}_{PT}=q^{rev}_P-T\left({q^{rev}_P}/{T}\right)=0 \nonumber$
which is also as required for a reversible process.
Now let us consider a spontaneous process, in which the ice melts while in thermal contact with surroundings at 10 ºC. To reach equilibrium, the system must reach the temperature of the surroundings, which we assume to be constant. In this process, the ice melts and the melt water warms to 10 ºC. To find the entropy change, we must find a reversible process that effects the same change. A two-step process effects this conveniently. The first step is the one we have just considered: Surroundings at 0 ºC transfer $6010$ J of heat to the system, reversibly melting the ice to water at 0 ºC. We have $q_1=6010\ \mathrm{J}$ and $\Delta S_1=22.00\ \mathrm{J}\ {\mathrm{K}}^{-1} \nonumber$
In the second reversible step, surroundings that are always at the same temperature as the system transfer heat to the system as the temperature increases from 273.15 K to 283.15 K. The heat capacity of liquid water is 75.3$\ \mathrm{J}\ {\mathrm{mol}}^{-1}$. For this step,
$q_2=\int^{283.15}_{273.15}{C_PdT}=\left(75.3\ \mathrm{J}\ {\mathrm{K}}^{-1}\right)\left(10\ \mathrm{K}\right)=753\ \mathrm{J} \nonumber$
and ${\Delta S}_2=\int^{283.15}_{273.15}{\frac{C_P}{T}dT}=\left(75.3\ \mathrm{J}\ {\mathrm{K}}^{-1}\right){ \ln \frac{283.15}{273.15}\ }=2.71\ \mathrm{J}\mathrm{\ }{\mathrm{K}}^{-1} \nonumber$
For these reversible changes in the system, we have $\Delta S=\Delta S_1+\Delta S_2=24.71\mathrm{\ J}\ {\mathrm{K}}^{-1}$. This is also the value of $\Delta S$ for the spontaneous process. The heat taken up by the system in the two-step reversible process is $q=q_1+q_2=6763\ \mathrm{J}$. This heat is surrendered by the surroundings, and we could effect identically the same change in the surroundings by exchanging this quantity of heat reversibly. For the spontaneous process, therefore, we have $\hat{q}=-6763\ \mathrm{J}$ and
$\Delta \hat{S}={-6763\ \mathrm{J}}/{283.15\ \mathrm{K}}=-23.88\ \mathrm{J}\ {\mathrm{K}}^{-1} \nonumber$
For the universe, we have
$\Delta S_{universe}=\Delta S+\Delta \hat{S}=+0.83\ \mathrm{J}\ {\mathrm{K}}^{-1} \nonumber$
which is greater than zero, as required for a spontaneous process.
Because this reversible two-step process does not occur at a constant temperature, its Gibbs free energy change is not zero. However, we can use the Gibbs-Helmholtz equation to estimate the Gibbs free energy change for the related process in which ice at 10 ºC and 1 bar (a hypothetical substance) melts to form liquid water at the same temperature and pressure. For this process, we estimate $\Delta G=-220\ \mathrm{J}\ {\mathrm{mo}\mathrm{l}}^{-1}\ {\mathrm{K}}^{-1} \nonumber$
(See problem 10-21.) Since we have $\Delta G<0$ for the process, our change criterion asserts that, in agreement with our experience, superheated ice melts spontaneously at 10 C.
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We define entropy in terms of its differential as
$dS=dq^{rev}/T. \nonumber$
To measure an entropy change using this definition requires that the process be reversible, which means that the system and surroundings must be at the same temperature as the heat transfer occurs. We understand reversible heat transfer to be the limiting case in which the temperature difference between system and surroundings becomes arbitrarily small. Here we want to consider the conceptual problems associated with keeping the temperature of the surroundings arbitrarily close to the temperature of the system while the system undergoes an arbitrary reversible change, which may include a temperature change.
We can illuminate one necessary aspect by posing a trivial dilemma: Since system and surroundings jointly comprise the universe, the requirement that system and surroundings be at the same temperature might seem to require that the entire universe be at a single temperature. Plainly, this condition is not met; the temperature of the universe varies from place to place. In fact, the requirement is only that the system and that portion of the surroundings with which the system exchanges heat be at the same temperature. We can satisfy this requirement by permitting the exchange of heat between the system and a portion of the surroundings, under conditions in which the combination of the two is thermally isolated from the rest of the universe.
The non-trivial aspect of this problem arises from the requirement that the temperature of the surroundings remain arbitrarily close to the temperature of the system, while both temperatures change and heat is exchanged between the system and the surroundings. A clumsy solution to this problem is to suppose that we exchange one set of surroundings (at temperature $T$) to a new set (at temperature $T+\Delta T$) whenever the temperature of the system changes by $\Delta T$. A more elegant solution is to use a machine that can measure the entropy change associated with an arbitrary reversible change in any closed system. This is a conceptual device, not a practical machine. We can use it in gedanken experiments to make arbitrarily small changes in the temperature and pressure of the system along any reversible path. At every step along this path, the entropy change is
$dS=\frac{1}{T}{\left(\frac{\partial H}{\partial T}\right)}_PdT-{\left(\frac{\partial V}{\partial T}\right)}_PdP \nonumber$
The entropy-measuring machine is sketched in Figure 1. In this device, the portion of the surroundings with which the system can exchange heat is a quantity of ideal gas, which functions as a heat reservoir. This heat reservoir is in thermal contact with the system. The combination of system and ideal-gas heat reservoir is thermally isolated from the rest of the universe. We consider the case in which only pressure–volume work can be done on either the system or the ideal-gas heat reservoir. In this device, all changes are driven by changes in the pressures applied to the surroundings (the ideal-gas heat reservoir) and the system. The pressure applied to the system and the pressure applied to the ideal-gas heat reservoir can be varied independently. We suppose that the system is initially at equilibrium and that changes in the applied pressures are effected in such a manner that all changes in the system and in the ideal-gas heat reservoir occur reversibly. For any change effected in the entropy-measuring machine, the heat and entropy changes in the heat reservoir are $\hat{q}$ and $\Delta \hat{S}$.
Isothermal Process
Let us consider first a process in which a quantity of heat must be transferred from the surroundings to the system while both are at the constant temperature $T_P$. To be specific, let this be a process in which a mole of pure liquid vaporizes at constant pressure, taking up a quantity of heat equal to the molar enthalpy of vaporization. We can supply this heat by reversibly and isothermally compressing the ideal-gas heat reservoir. To keep the temperature of the ideal-gas heat reservoir constant, we reversibly withdraw the piston that controls the pressure of the system, causing the vaporization of liquid in the system and absorption by the system of the heat given up by the ideal-gas heat reservoir. Overall, we apply forces to the two pistons to achieve reversible isothermal compression of the ideal gas in the heat reservoir and reversible isothermal vaporization of a quantity of liquid in the system. Because $q=-\hat{q}$, the magnitude of the entropy change for the surroundings is equal to that for the ideal-gas heat reservoir. The entire process is reversible, the entropy change for the system and the entropy change for the surroundings sum to zero: $\Delta S+\Delta \hat{S}=0$.
We can calculate the entropy change for the ideal-gas heat reservoir. Overall, the $\hat{n}$ moles of ideal gas in the heat reservoir go from the state specified by ${\hat{P}}_1$ and $T_P$ to the state specified by ${\hat{P}}_2$ and $T_P$. From the general relationship, $dS=T^{-1}{\left({\partial H}/{\partial T}\right)}_PdT-{\left({\partial V}/{\partial T}\right)}_PdP$, with $dT=0$, we have $d\hat{S}=-\left(\hat{n}R/P\right)dP$ and
$\Delta \hat{S}=-\hat{n}R \ln \frac{\hat{P}_2}{\hat{P}_1} =\hat{n}R \ln \frac{\hat{V}_2}{\hat{V}_1} \nonumber$
The entropy change for the system is $\Delta S=-\Delta \hat{S}$. So long as we carry out the process isothermally and reversibly, we can determine the entropy change for the system simply by measuring the initial and final pressures (or volumes) of the ideal-gas heat reservoir.
Any Reversible Process
If the system undergoes a reversible change in which the temperature of the system is not constant, we can operate the entropy-measuring machine in essentially the same manner as before. The only difference is that we must adjust the pressure applied to the system so that the system temperature changes in the manner required to keep the process reversible—that is, to maintain the system at equilibrium. Then the change to the system and the change to the ideal-gas heat reservoir both take place reversibly. Even though the temperatures change, appropriate control of the applied pressures assures that the system is always in an equilibrium state and that the temperature of the system is always arbitrarily close to the temperature of the ideal-gas heat reservoir.
We can calculate the entropy change for the ideal-gas heat reservoir. Overall, the $\hat{n}$ moles of ideal gas in the heat reservoir go from the state specified by ${\hat{P}}_1$ and ${\hat{T}}_1$ to the state specified by ${\hat{P}}_2$ and ${\hat{T}}_2$. We can evaluate the entropy change for taking the ideal gas from state 1 to state 2 by a two-step path. We first compress the gas isothermally at ${\hat{T}}_1$ from ${\hat{P}}_1$ to ${\hat{P}}_2$. We then warm the gas at constant pressure ${\hat{P}}_2$ from ${\hat{T}}_1$ to ${\hat{T}}_2$. For the first step, $d\hat{T}=0$, and, as before, we find
$\Delta {\hat{S}}_{isothermal}=-\hat{n}R{ \ln \frac{{\hat{P}}_2}{{\hat{P}}_1}\ }=\hat{n}R{ \ln \frac{{\hat{V}}_2}{{\hat{V}}_1}\ } \nonumber$
For the second step, $d\hat{P}=0$, and
$\Delta \hat{S}_{isobaric}=\int^{\hat{T}_2}_{\hat{T}_1} \frac{\hat{n}C_P}{T}dT= \hat{n}C_P \ln \frac{\hat{T}_2}{\hat{T}_1} \nonumber$
The entropy change for the ideal-gas heat reservoir is thus
$\Delta \hat{S}=\Delta \hat{S}_{isothermal}+\Delta \hat{S}_{isobaric}=-\hat{n}R \ln \frac{\hat{P}_2}{ \hat{P}_1}+ \hat{n}C_P \ln \frac{\hat{T}_2}{\hat{T}_1} \nonumber$
and we have $\Delta S=-\Delta \hat{S}$.
The essential point of the entropy-measuring machine is that we can determine the entropy change for any process without knowing anything about the process except how to control the system pressure and temperature so that the process occurs reversibly. Of course, this one reason that the entropy-measuring machine is not a practical device. To control the machine in the required manner, we must know how the thermodynamic properties of the system are related to one another on the Gibbsian manifold that defines the system’s equilibrium states. If we know this, then we know ${\left({\partial H}/{\partial T}\right)}_P$ and ${\left({\partial V}/{\partial T}\right)}_P$ for the system along any reversible path, and we can calculate the entropy change for the system in the same way that we calculate the entropy change for the ideal-gas heat reservoir. If we have the information needed to perform the measurement, we can calculate the entropy change without using the machine.
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When we apply Le Chatelier’s principle, we imagine an equilibrium system on which we impose some step-wise change. Immediately following the imposition of this change, we isolate the system from all further interactions with its surroundings. This isolated system is a hypothetical construct, which can be only approximated in any real experiment. It has peculiar features: While the changed and isolated system has the properties of the original system, it is also free to undergo a further change that the original system could not. The hypothetical isolated system is no longer at equilibrium; it can undergo a spontaneous process of further change until it reaches a new position of equilibrium. The principle asserts that this further change opposes the imposed change.
The principle is inherently qualitative. This contributes to its utility in that we do not have to have quantitative data in order to use it. However, a qualitative prediction is less useful than a quantitative one. Let us now attempt to apply our second-law based quantitative models to the sequence of changes envisioned by Le Chatelier’s principle. We begin by restating the principle in more mathematical language. We then illustrate these ideas for the specific case of vapor–liquid equilibrium with temperature and pressure as the independent variables.
Let us suppose that $W$, $X$, $Y$, and $Z$ are a set of thermodynamic variables that is adequate to specify the state of the system. In any equilibrium state, the entropy of the system is then a function of these variables; we have $S=S\left(W,X,Y,Z\right)$. For present purposes, we assume that we know the function $S\left(W,X,Y,Z\right)$. Given small changes, $dW$, $dX$, $dY$, and $dZ$, in the independent variables, we can find the change in $dS$ for a reversible transition from $\left(W,X,Y,Z\right)$ to $\left(W+dW,X+dX,Y+dY,Z+dZ\right)$:
$dS={\left(\frac{\partial S}{\partial W}\right)}_{XYZ}dW+{\left(\frac{\partial S}{\partial X}\right)}_{WYZ}dX+{\left(\frac{\partial S}{\partial Y}\right)}_{WXZ}dY+{\left(\frac{\partial S}{\partial Z}\right)}_{WXY}dZ \nonumber$
When we impose the change creating the hypothetical isolated system, we imagine that some characteristic of the system changes instantaneously, and that it does so without changing the other properties of the system. Since we suppose that nothing about the system changes in the perturbation and isolation step, the entropy of the perturbed, isolated, hypothetical system remains the same as that of the original equilibrium system.
Figure 2 shows the entropies for three states in the cycle that comprises the Le Chatelier model for change. The entropy of the original equilibrium system is $S\left(W,X,Y,Z\right)$ and that of the final equilibrium system is $S\left(W+dW,X+dX,Y+dY,Z+dZ\right)$. The same final equilibrium state is reached by both the irreversible transition from the change-enabled hypothetical state and by a reversible transition from the initial equilibrium state. Since entropy is a state function, its change around this cycle must be zero. Hence, the incremental changes $dW$, $dX$, $dY$, and $dZ$ that occur in the thermodynamic variables must satisfy the inequality
${\left(\frac{\partial S}{\partial W}\right)}_{XYZ}dW+{\left(\frac{\partial S}{\partial X}\right)}_{WYZ}dX+{\left(\frac{\partial S}{\partial Y}\right)}_{WXZ}dY +{\left(\frac{\partial S}{\partial Z}\right)}_{WXY}dZ>0 \nonumber$
We can view the application of this inequality to the hypothetical, change-enabled, isolated system as a mathematical expression of Le Chatelier’s principle. To see this more clearly, let us suppose that we are able to keep $W$ and $Z$ constant. We suppose that the imposed change requires that the final value of $X$ be $X+dX$. For the system to remain at equilibrium, the remaining variable, $Y$, must change by an amount, $dY$, that satisfies this inequality. That is, to reach the new equilibrium state, the change in $Y$ must satisfy
${\left(\frac{\partial S}{\partial X}\right)}_{WYZ}dX+{\left(\frac{\partial S}{\partial Y}\right)}_{WXZ}dY>0 \nonumber$
In this model, variables $X$ and $Y$ drive the entropy change as the hypothetical system moves toward its new equilibrium position. The imposed change in $X$ changes the entropy of the system by
$dS_{imposed}={\left(\frac{\partial S}{\partial X}\right)}_{WYZ}dX \nonumber$
Since the effect of the imposed change is to drive the system away from its original equilibrium position, we have $dS_{imposed}<0$. The system’s response changes the entropy of the system by
$dS_{response}={\left(\frac{\partial S}{\partial Y}\right)}_{WXZ}dY \nonumber$
We have $dS_{response}>-dS_{imposed}>0$, so that we can reasonably describe the response, $dY$, that makes $dS_{response}>0$, as a change that opposes the imposed change, $dX$, that makes $dS_{imposed}<0$.
Applying Le Chatelier’s principle is something of an art. Central to this art is an ability to devise a hypothetical, change-enabled, isolated, non-equilibrium state that is a good model for the initial state of the spontaneous process. In Section 6.6, we use qualitative arguments to apply Le Chatelier’s principle to vapor–liquid equilibrium. To relate these qualitative arguments to the mathematical model we have developed, let us consider the gedanken experiment depicted in Figure 3. We suppose that the initial equilibrium system contains the liquid and vapor of a pure substance at pressure, $P$, and temperature, $T$. We imagine that we can create the hypothetical isolated system by imposing a step change to the applied pressure without changing the pressure of the system.
To model the pressure perturbation, let us view the liquid–vapor mixture as a subsystem, which is enclosed in a vertical cylinder that is sealed by a frictionless piston. A mass, $M$, rests on top of the piston. For present purposes, we consider this mass to be a portion of a second subsystem. The gravitational force on this mass creates the pressure applied to the liquid–vapor mixture. Since this is an equilibrium state of the substance, this applied pressure is equal to the pressure, $P$, of the liquid–vapor subsystem. A small mass, $dM$, is also a part of the second subsystem. In this original equilibrium state of the system, this smaller mass is supported in some manner, so that it does not contribute to the applied pressure. We assume that the piston is a perfect thermal insulator, so that no heat can be exchanged between the two subsystems.
As sketched in Figure 3, we create the hypothetical change-enabled system by moving the smaller mass so that it too rests on top of the piston. Immediately thereafter, we completely isolate the system from the rest of the universe. We suppose that the applied pressure instantaneously increases to $P+dP$. However, since the liquid–vapor subsystem is unchanged, we suppose that the pressure, entropy, and all other thermodynamic properties of the liquid–vapor subsystem remain unchanged in this hypothetical state. The system is not at equilibrium in this hypothetical state, because the applied pressure is not equal to that of the liquid–vapor subsystem. Spontaneous change to a new equilibrium state can occur. Because the system is isolated, $\Delta \hat{S}=0$. Therefore, we have $\Delta S>0$. The final equilibrium temperature is $T+dT$.
With pressure and temperature as the independent variables, this model for Le Chatelier’s principle gives rise to the following mathematical requirement:
$dS=T^{-1}{\left({\partial H}/{\partial T}\right)}_PdT-{\left({\partial V}/{\partial T}\right)}_PdP>0. \nonumber$
We know that $T$, ${\left({\partial H}/{\partial T}\right)}_P$, and ${\left({\partial V}/{\partial T}\right)}_P$ are positive. Therefore we can rearrange the inequality to find
$dT>T\left[{{\left(\frac{\partial V}{\partial T}\right)}_P}/{{\left(\frac{\partial H}{\partial T}\right)}_P}\right]dP \nonumber$
If we have $dP>0$, it follows that $dT>0$; that is, the liquid–vapor equilibrium temperature increases with pressure.
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In Sections 12.4 and 12.6, we explore two approaches to using the entropy-based criterion for spontaneous change. In discussing the melting of ice at $\mathrm{+10}$ C, we calculate the entropy changes for both the system and the surroundings to show that $\Delta S_{universe}>0$, as the second law requires for a spontaneous process. In discussing the pressure-dependence of a liquid’s boiling point in Section 12.6, we relate the second law criterion for spontaneous change to Le Chatelier’s principle. We turn now to specifying the pressures and temperatures at which two phases of a pure substance are in equilibrium. When we choose pressure and temperature as the independent variables, the Gibbs free energy criteria specify the equilibrium state and the direction of spontaneous change.
For a reversible process in which all the work is pressure–volume work and in which the pressure and temperature change by $dP$ and $dT$, the change in the Gibbs free energy is $dG=VdP-SdT$. Let us apply this relationship to the liquid–vapor equilibrium problem that we discuss in Section 12.6. To do so, we view the process from a slightly different perspective. We suppose that we have two systems. These are identical to the initial and final states of the system in our discussion above. One of these systems is at liquid–vapor equilibrium at a particular pressure, $P$, and temperature,$\ T$. The other is at liquid–vapor equilibrium at $P+dP$ and $T+dT$. We consider the change in the Gibbs free energy of a mole of the substance as it reversibly traverses the cycle sketched in Figure 4.
The pressure and temperature are constant in each of the two equilibrium states. In either of these equilibrium states, the Gibbs free energy does not change when a mole of liquid is converted to its gas, ${\Delta }_{vap}G\left(P,T\right)=0$ and ${\Delta }_{vap}G\left(P+dP,T+dP\right)=0$. When the pressure and temperature of one mole of liquid change from $P$ and $T$ to $P+dP$ and $T+dT$, the Gibbs free energy change is $d\overline{G}\left(\ell \right)=\overline{V}_{\ell }dP- \overline{S}_{\ell }dT$. For a mole of gas, this change in the pressure and temperature change the Gibbs free energy by $d\overline{G}\left(g\right)= \overline{V}_gdP- \overline{S}_gdT$. ($\overline{V}_{\ell }$, $\overline{V}_g$, $\overline{S}_{\ell }$, and $\overline{S}_g$ are evaluated at $P$ and $T$. However, since $dP$ and $dT$ are small, these quantities are essentially constant over the pressure and temperature ranges involved.) For the individual steps in this cycle, we have
$\left(\ell ,\ P,\ T\right)\to \left(\ell ,\ P+dP,\ T+dT\right) d\overline{G}\left(\ell \right)=\overline{V}_{\ell }dP- \overline{S}_{\ell }dT \nonumber$
$\left(\ell ,\ P+dP,\ T+dT\right)\to \left(g,\ P+dP,\ T+dT\right) {\Delta }_{vap}G\left(P+dP,T+dT\right)=0 \nonumber$
$\left(g,\ P+dP,\ T+dT\right)\to \left(g,\ P,\ T\right) -d\overline{G}\left(g\right)=-\left(\overline{V}_gdP- \overline{S}_gdT\right) \nonumber$
$\left(g,\ P,\ T\right)\to \left(\ell ,\ P,\ T\right) -{\Delta }_{vap}G\left(P,T\right)=0 \nonumber$
Since the Gibbs free energy is a state function, the sum of these terms is zero. We have
${\Delta }_{vap}G\left(P+dP,T+dT\right)-{\Delta }_{vap}G\left(P,T\right) +d\overline{G}\left(\ell \right)-d\overline{G}\left(g\right) =d\overline{G}\left(\ell \right) -d\overline{G}\left(g\right) =0 \nonumber$
so that $d\overline{G}\left(\ell \right)=d\overline{G}\left(g\right)$. That is, the Gibbs free energy of the liquid changes by the same amount as the Gibbs free energy of the gas when a mole of either is taken from one liquid–vapor equilibrium state to another. Substituting, we find a condition that the pressure and temperature changes must satisfy when the system goes from the liquid–vapor equilibrium state at $\left(P,T\right)$ to the liquid–vapor equilibrium state at $\left(P+dP,T+dT\right)$:
$\left( \overline{V}_g- \overline{V}_{\ell }\right)dP-\left(\overline{S}_g-\overline{S}_{\ell }\right)dT=0 \nonumber$
We let ${\Delta }_{vap}\overline{V}= \overline{V}_g- \overline{V}_{\ell }$ and ${\Delta }_{vap}\overline{S}=\overline{S}_g-\overline{S}_{\ell }$, where ${\Delta }_{vap}\overline{V}$ and ${\Delta }_{vap}\overline{S}$ are the volume and entropy changes that accompany the vaporization of one mole of the liquid at $P$ and $T$. ${\Delta }_{vap}\overline{V}$ and ${\Delta }_{vap}\overline{S}$ are essentially constant over the small pressure and temperature ranges involved. Substituting, we have ${\Delta }_{vap}\overline{V}dP-{\Delta }_{vap}\overline{S}dT=0$, which we can rearrange to give
$\frac{dP}{dT}=\frac{\Delta _{vap}\overline{S}}{\Delta _{vap}\overline{V}} \nonumber$
As one mole of liquid vaporizes reversibly at $P$ and $T$, the system accepts heat $q^{rev}_P=\Delta _{vap}\overline{H}$. Hence, the entropy of vaporization at $P$ and $T$ is ${\Delta }_{vap}\overline{S}={\Delta }_{vap}\overline{H}/T$, and the relationship between $dP$ and $dT$ becomes
$\frac{dP}{dT}=\frac{\Delta _{vap}\overline{H}}{T \Delta _{vap}\overline{V}} \nonumber$
Below we see that such a relationship holds for any equilibrium between two pure phases. The general relationship is called the Clapeyron equation.
This analysis is successful because the constituents are pure phases; the properties of the liquid are independent of how much vapor is present and vice versa. When we analyze the equilibrium between a liquid solution and a gas of the solution’s components, the problem is more complex, because the properties of the phases depend on their compositions.
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We can also represent reversible changes by paths on contour maps. In Figure 5, a Gibbs free energy surface is represented as a contour map.
For small changes in $T$ and $P$, we can evaluate
$dG=G\left(P_0+dP,T_0+dT\right)-G\left(P_0,T_0\right) \nonumber$
from
$dG=-S\left(P_0,T_0\right)dT+V\left(P_0,T_0\right)dP \nonumber$
For larger changes, we can integrate along the paths $P=P_0$ and $T=T_0+\Delta T$ to find
$\Delta G=\int^{T_0+\Delta T}_{T_0}{-S\left(P_0,T\right)dT+}\int^{P_0+\Delta P}_{P_0}{V\left(P,T_0+\Delta T\right)dP} \nonumber$
The calculation of $\Delta S$ in Section 12.5 could be similarly represented as a path in the temperature–pressure plane that connects two constant-entropy contours.
Analysis of solid–liquid equilibrium parallels that of liquid–vapor equilibrium. Let us again consider the equilibrium between ice and water. Given that ice and water are at equilibrium at a particular temperature and pressure, and supposing that we increase the pressure from this equilibrium value, how must the temperature change in order that the system remain at equilibrium? In Section 6.6, we use Le Chatelier’s principle to answer this question qualitatively. Now, we find a quantitative answer by an argument that closely parallels that in Section 12.7.
Figure 6 depicts the line of pressures and temperatures along which ice and water are in equilibrium. We can view this as a contour map. In this case, the contours are sets of pressures and temperatures for which ${\Delta }_{fus}\overline{G}$ is constant. Only the contour for ${\Delta }_{fus}\overline{G}=0$ is shown. The figure also depicts paths along which ice and water can individually be taken from their equilibrium state at $P$ and $T$ to their equilibrium state at $P+dP$ and $T+dT$. The Gibbs free energy change for the ice must equal that for water. Letting ${\overline{G}}_{\ell }$, ${\overline{S}}_{\ell }$, and ${\overline{V}}_{\ell }$ be the Gibbs free energy, the entropy, and the volume of one mole of water at temperature $T$ and pressure $P$, the equation
$d{\overline{G}}_{\ell }={\overline{V}}_{\ell }dP-{\overline{S}}_{\ell }dT \nonumber$
specifies the change in the Gibbs free energy of one mole of water when the pressure changes $P$ to $P+dP$ and the temperature changes from $T$ to $T+dT$. Similarly, using the subscript “s” to denote ice, we have
$d{\overline{G}}_s={\overline{V}}_sdP-{\overline{S}}_sdT \nonumber$
Since these Gibbs free energy changes connect states of ice–water equilibrium, they must be equal, and we have
$d{\overline{G}}_{\ell }-d{\overline{G}}_s=\left({\overline{V}}_{\ell }-{\overline{V}}_s\right)dP-\left({\overline{S}}_{\ell }-{\overline{S}}_s\right)dT={\Delta }_{fus}\overline{V}dP-{\Delta }_{fus}\overline{S}dT=0 \nonumber$
where we introduce ${\Delta }_{fus}\overline{S}$ and ${\Delta }_{fus}\overline{V}$ to represent the entropy and volume changes that occur when one mole of ice melts reversibly at $P$ and $T$. Rearranging gives
$\frac{dP}{dT}=\frac{\Delta_{fus}\overline{S}}{\Delta_{fus}\overline{V}} \nonumber$
Since
$\Delta_{fus}\overline{S}={\Delta_{fus}\overline{H}}/{T}, \nonumber$
the Clapeyron equation becomes
$\frac{dP}{dT}=\frac{\Delta_{fus}\overline{H}}{T\ {\Delta }_{fus}\overline{V}} \nonumber$
At a pressure of one bar and a temperature of 273.15 K, the enthalpy of fusion is $\mathrm{6010}\ \mathrm{J}\ {\mathrm{mol}}^{\mathrm{-1}}\ {\mathrm{K}}^{-1}$. The enthalpy value changes only slowly as the equilibrium temperature changes. The volumes of one mole of ice and one mole of water are 19.651 and 18.019 ${\mathrm{cm}}^3$, respectively. At 273.15 K, we obtain
$\frac{dP}{dT}=-143.7\ \mathrm{bar}\ {\mathrm{K}}^{-1} \nonumber$
If the pressure increases to 1000 bar, the change in the melting point is about –6.96 K, so that $T_{mp}\left(1000\ \mathrm{bar}\right)=266.2\ \mathrm{K}$.
Again, this analysis is successful because the constituents are pure phases; the properties of the ice are independent of how much water is present and vice versa. When we analyze the equilibrium between ice and salt water, the properties of the salt water depend on the kind of salt present and on its concentration.
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The analysis in the two previous sections can be repeated for any phase change of a pure substance. Let $\alpha$ and $\beta$ denote the two phases that are at equilibrium.
$\alpha \rightleftharpoons \beta \nonumber$
Let $\overline{G}_{\alpha }$, $\overline{S}_{\alpha }$, and ${\overline{V}}_{\alpha }$ represent the Gibbs free energy, the entropy, and the volume of one mole of pure phase $\alpha$ at pressure $P$ and temperature $T$. Let $\overline{G}_{\beta }$, $\overline{S}_{\beta }$, and $\overline{V}_{\beta }$ represent the corresponding properties of one mole of pure phase $\beta$. The equations
$d\overline{G}\left(\alpha \right)=\overline{V}_\alpha dP-\overline{S}_{\alpha }dT \nonumber$
and
$d\overline{G}\left(\beta \right)= \overline{V}_{\beta } dP- \overline{S}_{\beta }dT \nonumber$
describe the changes in the Gibbs free energy of a mole of $\alpha$ and a mole of $\beta$ when they go from one $\alpha-\beta$-equilibrium state at $P$ and $T$ to a second $\alpha-\beta$-equilibrium state at $P+dP$ and $T+dT$. Since these Gibbs free energy changes must be equal, we have
\begin{align*} d\overline{G}\left(\beta \right)-d\overline{G}\left(\alpha \right) &=\left({\overline{V}}_{\beta }-{\overline{V}}_{\alpha }\right)dP-\left({\overline{S}}_{\beta }-\overline{S}_{\alpha }\right)dT \[4pt] &=\Delta \overline{V}dP-\Delta \overline{S}dT \[4pt] &=0 \end{align*}
and
$\frac{dP}{dT}=\frac{\Delta \overline{S}}{\Delta \overline{V}} \nonumber$
where $\Delta \overline{S}$ and $\Delta \overline{V}$ are the entropy and volume changes that occur when one mole of the substance goes from phase $\alpha$ to phase $\beta$. Since $\Delta \overline{S}=\Delta \overline{H}/T$, the condition for equilibrium between phases $\alpha$ and $\beta$ becomes
$\frac{dP}{dT}=\frac{\Delta \overline{H}}{T\ \Delta \overline{V}} \label{Clap1}$
Equation \ref{Clap1} is known as the Clapeyron equation.
12.09: The Clausius-Clapeyron Equation
To use the Clapeyron equation we must know the enthalpy and volume differences at one equilibrium temperature and pressure. In general, these properties are readily measured. If we fix the pressure, we can measure the corresponding equilibrium temperature. We can obtain the enthalpy change at this pressure by measuring the heat required to convert a mole of the substance from one phase to the other. We can obtain the volume change from the molar volumes, which we can obtain by measuring the density of each phase. The enthalpy of the phase change varies only weakly as the equilibrium pressure and temperature vary. Similarly, for condensed phases, the densities are weak functions of temperature. This means that, for transitions between condensed phases, ${\Delta \overline{H}}/{\Delta \overline{V}}$ is approximately constant over a modest temperature range.
For a sublimation or vaporization process, the product is a gas. Then the molar volume of the product is a sensitive function of both pressure and temperature. However, the molar volume of the product phase is much greater than the molar volume of the initial solid or liquid phase. To a good approximation, the volume change for the process equals the volume of the gas produced. If we have an equation of state for the gas, the volume calculated from the equation of state is a good approximation to $\Delta V$ for the phase change. The ideal gas equation is usually adequate for this purpose. Then, $\Delta \overline{V}\approx {RT}/{P}$, and
$\frac{dP}{dT}=\frac{P\Delta \overline{H}}{{RT}^2} \tag{Clausius-Clapeyron equation}$
This equation for the pressure–temperature relationship for a phase equilibrium involving a gas is called the Clausius-Clapeyron equation. Dividing both sides by the pressure, we can put the Clausius-Clapeyron equation into an alternative and often-useful form:
$\frac{d{ \ln P\ }}{dT}=\frac{\Delta \overline{H}}{{RT}^2} \nonumber$
If we can assume that $\Delta \overline{H}$ is independent of pressure, we can separate variables and integrate to obtain the Clausius-Clapeyron equation in integrated form. If we can assume further that $\Delta \overline{H}$ is constant, the integration yields
$\int^P_{P_0}{\frac{dP}{P}}=\frac{\Delta \overline{H}}{R}\int^T_{T_0}{\frac{dT}{T^2}} \nonumber$ and ${ \ln \frac{P}{P_0}\ }=-\frac{\Delta \overline{H}}{R}\left(\frac{1}{T}-\frac{1}{T_0}\right) \nonumber$
where $P_0$ and $T_0$ are the initial equilibrium position.
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1. For any change in a reversible system, we have $dG=-SdT+VdP$. Consider two systems, $\alpha$ and $\beta$, where system $\alpha$ can be converted to system $\beta$. (Below, we will let $\alpha$ and $\beta$ be the solid and liquid phases of the same pure substance, but this is not a necessary restriction.) For incremental changes in temperature and pressure, represented by $dT$ and $dP$, we have
$dG_{\alpha }=V_{\alpha }dP-S_{\alpha }dT \nonumber$ and $dG_{\beta }=V_{\beta }dP-S_{\beta }dT$
We can subtract to find
$d\left(G_{\beta }-G_{\alpha }\right)=\left(V_{\beta }-V_{\alpha }\right)dP-\left(S_{\beta }-S_{\alpha }\right)dT \nonumber$ or $d\left({\Delta }_{\alpha \to \beta }G\right)=\left({\Delta }_{\alpha \to \beta }V\right)dP-\left({\Delta }_{\alpha \to \beta }S\right)dT \nonumber$
which we usually write as
$d\left(\Delta G\right)=\Delta VdP-\Delta SdT \nonumber$
Here ${\Delta }_{\alpha \to \beta }X$ (or $\Delta X$) is the change in the state function $X$ that occurs when system $\alpha$ is converted to system $\beta$. For many inter-convertible systems, it is a good approximation to say that ${\Delta }_{\alpha \to \beta }S$ (or $\Delta S$) and ${\Delta }_{\alpha \to \beta }V$ (or $\Delta V$) are constant for modest changes in temperature or pressure. Then, representing the pressure and temperature in the initial and final states as $\left(P_1,T_1\right)$ and $\left(P_2,T_2\right)$, respectively, the change in ${\Delta }_{\alpha \to \beta }G$ (or $\Delta G$) can be obtained by integration:
$\int^{P_2,T_2}_{P_1,T_1}{d\left(\Delta G\right)=-\Delta S\int^{T_2}_{T_1}{dT}+\Delta V\int^{P_2}_{P_1}{dP}} \nonumber$ or $\Delta \left(\Delta G\right)=\Delta G\left(P_2,T_2\right)-\Delta G\left(P_1,T_1\right)=-\Delta S\left(T_2-T_1\right)+\Delta V\left(P_2-P_1\right) \nonumber$
Note that $\alpha$ and $\beta$ need not be in equilibrium with one another at either the condition specified by $\left(P_1,T_1\right)$ or that specified by $\left(P_2,T_2\right)$.
However, in the important special case that $\boldsymbol{\alpha }$ and $\boldsymbol{\beta }$ are in equilibrium at $\left({\boldsymbol{P}}_{\boldsymbol{1}},{\boldsymbol{T}}_{\boldsymbol{1}}\right)$, we have ${\boldsymbol{\Delta }}_{\boldsymbol{\alpha }\boldsymbol{\to }\boldsymbol{\beta }}\boldsymbol{G}\left({\boldsymbol{P}}_{\boldsymbol{1}},{\boldsymbol{T}}_{\boldsymbol{1}}\right)\boldsymbol{=}\boldsymbol{\Delta }\boldsymbol{G}\left({\boldsymbol{P}}_{\boldsymbol{1}},{\boldsymbol{T}}_{\boldsymbol{1}}\right)\boldsymbol{=}\boldsymbol{0}$. Then
$\Delta \left(\Delta G\right)=\Delta G\left(P_2,T_2\right)=-\Delta S\left(T_2-T_1\right)+\Delta V\left(P_2-P_1\right) \nonumber$
Consider the application of these observations to the case where $\alpha$ and $\beta$ are solid and liquid aluminum metal, respectively. At one bar, aluminum melts at 933.47 K. At the melting point, the enthalpy of fusion is ${\Delta }_{fus}\overline{H}=+10.71\ \mathrm{kJ}\ {\mathrm{mol}}^{-1}$. The atomic weight of aluminum is $26.9815\ \mathrm{g}\ {\mathrm{mol}}^{-1}$. The density of liquid aluminum at the melting point is $2.375\ \mathrm{g}\ {\mathrm{cm}}^{-1}$; the molar volume of the liquid is therefore $1.1361\times {10}^{-5}\ {\mathrm{m}}^3\ {\mathrm{mol}}^{-1}$. At 20 C, the density of solid aluminum is $2.70\ \mathrm{g}\ {\mathrm{cm}}^{-1}$, and the molar volume is $0.9993\times {10}^{-5}\ {\mathrm{m}}^3\ {\mathrm{mol}}^{-1}$. Assuming the molar volume of the solid to be independent of temperature, the change in the molar volume that occurs when aluminum melts is ${\Delta }_{fus}\overline{V}=1.368\times {10}^{-6}\ {\mathrm{m}}^3\ {\mathrm{mol}}^{-1}$.
(a) What is ${\Delta }_{fus}\overline{V}$ at the melting point at one bar (point A on the diagram)?
(b) Is the conversion of solid aluminum to liquid aluminum a reversible, spontaneous, or impossible process at (933.47 K, 1 bar)—that is, at point A?
(c) What is ${\Delta }_{fus}\overline{S}$ at the melting point at one bar—that is, at point A?
(d) What is $\Delta \left({\Delta }_{fus}\overline{G}\right)$ when the pressure and temperature go from (933.47 K, 1 bar) to (934.47 K, 1 bar)? (That is from point A to point B on the diagram.) What is ${\Delta }_{fus}\overline{G}$ at (934.47 K, 1 bar)?
(e) Is the conversion of solid aluminum to liquid aluminum a reversible, spontaneous, or impossible process at (934.47 K, 1 bar)—that is, at point B?
(f) What is $\Delta \left({\Delta }_{fus}\overline{G}\right)$ when the pressure and temperature go from (933.47 K, 1 bar) to (932.47 K, 1 bar)? (That is from point A to point C on the diagram.) What is ${\Delta }_{fus}\overline{G}$ at (932.47 K, 1 bar)?
(g) Is the conversion of solid aluminum to liquid aluminum a reversible, spontaneous, or impossible process at (932.47 K, 1 bar)—that is, at point C?
(h) What is $\Delta \left({\Delta }_{fus}\overline{G}\right)$ when the pressure and temperature go from (933.47 K, 1 bar) to (933.47 K, 101 bar)? (That is from point A to point D on the diagram.) What is ${\Delta }_{fus}\overline{G}$ at (933.47 K, 101 bar)?
(i) Is the conversion of solid aluminum to liquid aluminum a reversible, spontaneous, or impossible process at (933.47 K, 101 bar)—that is, at point D?
(j) If we maintain the pressure constant at 101 bar, how much would we have to change the temperature to just offset the change in $\Delta \left({\Delta }_{fus}\overline{G}\right)$ that occurred in part h? Note that this change will reach the conditions represented by point E on the diagram
(k) Is the conversion of solid aluminum to liquid aluminum a reversible, spontaneous, or impossible process at point E?
(l) What is $\Delta \left({\Delta }_{fus}\overline{G}\right)$ in going from point D (933.47 K, 101 bar) to point B (934.47 K, 1 bar)? Is this value equal to the difference between the Gibbs free energy of a mole of liquid aluminum at point B and a mole of solid aluminum at point D?
2. In the temperature interval $0.01>t>-10\ \mathrm{C}$, the vapor pressure of water (in Pa) above pure ice is aproximated by ${ \ln P\ }=a+bt+ct^2$, where $a=6.41532$, $b=8.229\times {10}^{-2}$, $c=-3.2\times {10}^{-4}$, $t=T-273.15$, and $T$ is the temperature in degrees kelvin. Estimate the enthalpy of sublimation of ice at $273.15\ \mathrm{K}$.
3. In the temperature interval $373.15>T>273.15\ \mathrm{K}$, the vapor pressure of water (in Pa) is approximated by ${ \ln P\ }=a+bt+ct^2+dt^3+et^4$, where $a=9.42095$, $b=4.960\times {10}^{-2}$, $c=-1.7536\times {10}^{-4}$, $d=6.02\times {10}^{-7}$, $e=-2.0\times {10}^{-9}$, $t=T-323.15$, and $T$ is the temperature in degrees kelvin. Estimate the enthalpy of vaporization of water at 323.15 K and 373.15 K.
4. The normal (1 atm) boiling point of acetone is 56.05 C. The enthalpy of vaporization at the normal boiling point is $29.10\ \mathrm{kJ}\ {\mathrm{mol}}^{-1}$. What is the entropy of vaporization of acetone at the normal boiling point? Estimate the vapor pressure of acetone at 25.0 C.
5. Two allotropic forms of tin, gray tin and white tin, are at equilibrium at 13.2 C and 1 atm. The density of gray tin is $5.769\ \mathrm{g}\ {\mathrm{cm}}^{-1}$; the density of white tin is 7.265$\ \mathrm{g}\ {\mathrm{cm}}^{-1}$. Assume that the densities are independent of temperature. At 298.15 K and 1 bar, $S^o$ and $C_P$ for gray tin are $44.1\mathrm{\ J}\ {\mathrm{mol}}^{-1}\ {\mathrm{K}}^{-1}$ and $25.8\mathrm{\ J}\ {\mathrm{mol}}^{-1}\ {\mathrm{K}}^{-1}$, respectively. For white tin, $S^o$ and $C_P$ are $51.2\mathrm{\ J}\ {\mathrm{mol}}^{-1}\ {\mathrm{K}}^{-1}$ and $27.0\mathrm{\ J}\ {\mathrm{mol}}^{-1}\ {\mathrm{K}}^{-1}$, respectively. Estimate $\Delta \overline{S}$ and $\Delta \overline{V}$ for the conversion of gray tin to white tin at 13.2 C. At what temperature are gray tin and white tin at equilibrium at a pressure of 100 atm?
6. At 1000K, the standard Gibbs free energies of formation of graphite and diamond are 0.000 and $+5.905\ \mathrm{kJ}\ {\mathrm{mol}}^{-1}$, respectively. At 298.15 K and 1.000 bar, the molar volumes of graphite and diamond are $5.46\times {10}^{-6\ }{\mathrm{m}}^3\ {\mathrm{mol}}^{-1}$ and $3.42\times {10}^{-6\ }{\mathrm{m}}^3\ {\mathrm{mol}}^{-1}$, respectively. Let $P^{\#}$ be the pressure at which graphite and diamond are at equilibrium at 1000 K.
(a) What is the value of $\Delta \overline{G}$ for
$C\left(\mathrm{graphite}, 1\ \mathrm{bar},\ 1000\mathrm{K}\right)\to C\left(\mathrm{diamond},\ 1\ \mathrm{bar},\ 1000\mathrm{K}\right)? \nonumber$
(b) Express $\Delta \overline{G}$ for
$C\left(\mathrm{graphite},\ 1\ \mathrm{bar},\ 1000\mathrm{K}\right)\to C\left(\mathrm{graphite},\ P^{\#},\ 1000\mathrm{K}\right) \nonumber$
as a function of $P^{\#}$.
(c) Express $\Delta \overline{G}$ for
$C\left(\mathrm{diamond},\ 1\ \mathrm{bar},\ 1000\mathrm{K}\right)\to C\left(\mathrm{diamond},\ P^{\#},\ 1000\mathrm{K}\right) \nonumber$
as a function of $P^{\#}$.
(d) What is the value of $\Delta \overline{G}$ for
$C\left(\mathrm{graphite},\ P^{\#},\ 1000\mathrm{K}\right)\to C\left(\mathrm{diamond},\ P^{\#},\ 1000\mathrm{K}\right)? \nonumber$
(e) Assume that the molar volumes are independent of pressure and temperature. Estimate the value of the equilibrium pressure, $P^{\#}$.
Notes
See Robert Bruce Lindsay and Henry Margenau, Foundations of Physics, Dover Publications, Inc., New York, 1963, p 195.
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In Chapter 11, we find a general equation for the molar Gibbs free energy of a pure gas. We adopt the Gibbs free energy of formation of the hypothetical ideal gas, in its standard state at 1 bar, $P^o$, as the reference state for the Gibbs free energy of the gas at other pressures and the same temperature. Then, the molar Gibbs free energy of pure gas $A$, at pressure $P$, is
${\overline{G}}_A\left(P\right)={\Delta }_fG^o\left(A,{HIG}^o\right)+RT{ \ln \left(\frac{P}{P^o}\right)\ }+RT\int^P_0{\left(\frac{\overline{V}}{RT}-\frac{1}{P}\right)dP} \nonumber$ (any pure gas)
${\overline{G}}_A\left(P\right)$ is the difference between the Gibbs free energy of the gas at pressure $P$ and that of its constituent elements at 1 bar and the same temperature. If gas $A$ is an ideal gas, the integral is zero, and the standard-state Gibbs free energy of formation is that of an “actual” ideal gas, not a “hypothetical state” of a real gas. To recognize this distinction, let us write ${\Delta }_fG^o\left(A,P^o\right)$, rather than ${\Delta }_fG^o\left(A,{HIG}^o\right)$, when the gas behaves ideally. In a mixture of ideal gases, the partial pressure of gas $A$ is given by $P_A=x_AP$, where $x_A$ is the mole fraction of $A$ and $P$ is the pressure of the mixture. In §3, we find that the Gibbs free energy of one mole of pure ideal gas $A$ at pressure $P_A$ has the same Gibbs free energy as one mole of gas $A$ in a gaseous mixture in which the partial pressure of $A$ is $P_A=x_AP$. Recognizing these properties of an ideal gas, we can express the molar Gibbs free energy of an ideal gas—pure or in a mixture—as
$\overline{G}_A\left(P_A\right)=\Delta_fG^o\left(A,P^o\right)+RT \ln \left(\frac{P_A}{P^o}\right)\ \nonumber$
(ideal gas)
Note that we can obtain this result for pure gas $A$ directly from ${\left({\partial \overline{G}}/{\partial P}\right)}_T=\overline{V}={RT}/{P}$ by evaluating the definite integrals
$\int^{\overline{G}_A\left(P_A\right)}_{\Delta_fG^o\left(A,P^o\right)}{d\overline{G}}=\int^{P_A}_{P^o}{\frac{RT}{P}dP} \nonumber$
Including the constant, $P^o$, in these relationships is a useful reminder that $RT{ \ln \left({P_A}/{P^o}\right)\ }$ represents a Gibbs free energy difference. Including $P^o$ makes the argument of the natural-log function dimensionless; if we express $P$ in bars, including${\ P}^o=1\ \mathrm{bar}$ leaves the numerical value of the argument unchanged. If we express $P$ in other units, $P^o$ becomes the conversion factor for converting those units to bars; if we express $P$ in atmospheres, we have ${\ P}^o=1\ \mathrm{bar}=0.986923\ \mathrm{atm}$.
However, including the “$P^o$” is frequently a typographical nuisance. Therefore, let us introduce another bit of notation; we use a lower-case “$p$” to denote the ratio “${P}/{P^0}$”. That is, $p_A$ is a dimensionless quantity whose numerical value is that of the partial pressure of $A$, expressed in bars. The molar Gibbs free energy becomes
${\overline{G}}_A\left(P_A\right)={\Delta }_fG^o\left(A,P^o\right)+RT{ \ln p_A\ } \nonumber$ (ideal gas)
13.02: The Gibbs Free Energy Change for A Reaction of Ideal Gases
Let us consider a system that consists of a mixture of ideal gases $A$, $B$, $C$, and $D$, at a particular fixed temperature. We suppose that reaction occurs according to
$a\ A+b\ B\ \rightleftharpoons \ c\ C+d\ D \nonumber$
We want to think about what happens when $a$ moles of $A$ (at pressure $P_A$) and $b$ moles of $B$ (at pressure $P_B$) react to form $c$ moles of $C$ (at pressure $P_C$) and $d$ moles of $D$ (at pressure $P_D$) under conditions in which the partial pressures in the mixture remain constant at $P_A$, $P_B$, $P_C$, and $P_D$. At first encounter, these conditions may appear to be impossible; if $A$ reacts, its partial pressure must change. However, on reflection, we recognize that by making the system very large, $a$ moles of $A$ and $b$ moles of$\ B$ can disappear without changing $P_A$ or $P_B$ very much. In fact, we can make the change in $P_A$ and $P_B$ as small as we like just by making the original system large enough. The same considerations apply to $P_C$ and $P_D$. We let ${\Delta }_rG$ be the Gibbs free energy change under these conditions. We want to know how the Gibbs free energy of this system, $G$, and the Gibbs free energy change, ${\Delta }_rG$, depend on the partial pressures of the gases involved in the reaction.
We have observed repeatedly that the temperature and pressure of a system undergoing spontaneous change may not be well defined, and the concentration of a component may vary from point to point within a given phase. If any of these inhomogeneities are substantial, the reaction conditions in the previous paragraph are not met. On the other hand, if any pressure and temperature variations are too small to have observable effects, and there are no point-to-point concentration variations, it is entirely reasonable to suppose that the Gibbs free energies of the system, and of its individual components, are described by the thermodynamic models we have developed for reversible systems.
To see that there can be no objection in principle to measuring Gibbs free energies in a non-equilibrium system, we need only find a hypothetical equilibrium system whose state functions must have the same values. Thus, if the reaction $a\ A+b\ B\ \rightleftharpoons \ c\ C+d\ D$ does not occur, mixtures of gases $A$, $B$, $C$, and $D$ in any proportions can be at equilibrium, and the thermodynamic properties of any such mixture are well defined. In concept, we can produce a hypothetical equilibrium state equivalent to any intermediate state in the spontaneous process if we suppose that the reaction occurs only in the presence of a solid catalyst. By introducing and then removing the catalyst from the reaction mixture, we can produce a quasi-equilibrium state whose composition is identical to that of the spontaneous reaction at any particular extent of reaction. In this quasi-equilibrium state, the pressure is equal to the applied pressure, and the temperature is equal to that of the surroundings. If the spontaneous uncatalyzed reaction is slow compared to the rate at which the pressure and temperature of the system can equilibrate with the applied pressure and the surroundings temperature, the state functions of a spontaneously reacting system and a static, quasi-equilibrium system with the same composition must be essentially identical.
It turns out that we can find the dependence of ${\Delta }_rG$ on concentrations by considering a fundamentally different system—one that is composed of exactly the same amount of each of these gases, but in which the gases are not mixed. Each gas occupies its own container. For the partial pressures $P_A$, $P_B$, $P_C$, and $P_D$, the Gibbs free energies per mole are
${\overline{G}}_A\left(P_A\right)={\Delta }_fG^o\left(A,P^o\right)+RT{ \ln p_A\ } \nonumber$
${\overline{G}}_B\left(P_B\right)={\Delta }_fG^o\left(B,P^o\right)+RT{ \ln p_B\ } \nonumber$
${\overline{G}}_A\left(P_C\right)={\Delta }_fG^o\left(C,P^o\right)+RT{ \ln p_C\ } \nonumber$
${\overline{G}}_D\left(P_D\right)={\Delta }_fG^o\left(D,P^o\right)+RT{ \ln p_D\ } \nonumber$
The Gibbs free energy of this system, which is just a composite of the separated gases, is
$G={n_A\overline{G}}_A\left(P_A\right)+{n_B\overline{G}}_B\left(P_B\right)+{n_C\overline{G}}_C\left(P_C\right)+{n_D\overline{G}}_D\left(P_D\right) \nonumber$
If we subtract the Gibbs free energies of $a$ moles of reactant $A$ and $b$ moles of reactant $B$ from the Gibbs free energies of $c$ moles of product $C$ and d moles of product $C$, we find ${c\ \overline{G}}_C\left(P_C\right)+{d\ \overline{G}}_D\left(P_D\right)-{a\ \overline{G}}_A\left(P_A\right)+{b\ \overline{G}}_B\left(P_B\right) \nonumber$ $=c\ {\Delta }_fG^o\left(C,P^o\right)+d\ {\Delta }_fG^o\left(D,P^o\right)-{a\ \Delta }_fG^o\left(A,P^o\right)-b\ {\Delta }_fG^o\left(B,P^o\right)+RT{ \ln \frac{p_Cp_D}{p_Ap_B}\ } \nonumber$
To represent these free energy differences when the gases are in separate containers, we introduce the abbreviations
${\Delta }_{sep}G={c\ \overline{G}}_C\left(P_C\right)+{d\ \overline{G}}_D\left(P_D\right)-{a\ \overline{G}}_A\left(P_A\right)-{b\ \overline{G}}_B\left(P_B\right) \nonumber$
and
${\Delta }_rG^o=c\ {\Delta }_fG^o\left(C,P^o\right)+d\ {\Delta }_fG^o\left(D,P^o\right)-{a\ \Delta }_fG^o\left(A,P^o\right)-b\ {\Delta }_fG^o\left(B,P^o\right) \nonumber$
so that the difference between the Gibbs free energies of the separated reactants and products can be written more compactly as
${\Delta }_{sep}G={\Delta }_rG^o+RT{ \ln \frac{p_Cp_D}{p_Ap_B}\ } \nonumber$
${\Delta }_{sep}G$ is the difference in the Gibbs free energies when the pressures of the separated gases are fixed at $P_A$, $P_B$, $P_C$, and $P_D$. Note that, if any of the pressures changes, ${\Delta }_{sep}G$ changes. When we introduce ${\Delta }_rG^o$ in Section 11.10, we emphasize that this quantity is the difference between the standard Gibbs free energies of the separated products
$c \text{ moles of } C \text{ at 1 bar and } d \text{ moles of } D \text{ at 1 bar.} \nonumber$
and the separated reactants
$a \text{ moles of } A \text{ at 1 bar and } b \text{ moles of } B \text{ at 1 bar.} \nonumber$
We call ${\Delta }_rG^o$ the standard Gibbs free energy change for the reaction. At a given temperature, ${\Delta }_rG^o$ is a constant. Our choice of standard state for the Gibbs free energy of a compound means that we can calculate the standard free energy change for a reaction from the standard free energies of formation of the products and reactants:
${\Delta }_rG^o=c\ {\Delta }_fG^o\left(C,P^o\right)+d\ {\Delta }_fG^o\left(D,P^o\right)-{a\ \Delta }_fG^o\left(A,P^o\right)-b\ {\Delta }_fG^o\left(B,P^o\right) \nonumber$
Let us recapitulate: In the first system, we are interested in the Gibbs free energy change, ${\Delta }_rG$, for a process in which $a$ moles of $A$ and $b$ moles of $B$ are converted to $c$ moles of $C$ and $d$ moles of $D$ in a (large) mixed system where the partial pressures are constant at the values $P_A$, $P_B$, $P_C$, and $P_D$. In the second system, there is actually no process at all. The Gibbs free energy change, ${\Delta }_{sep}G$, is merely a computed difference between the Gibbs free energies of the specified quantities of product and reactant gases, where each gas is in its own container at its specified pressure. When the gases are ideal, the Gibbs free energy differences (changes) for these two systems turn out to be the same. That is, ${{\Delta }_{sep}G=\Delta }_rG$.
This relationship is valid for systems in which the properties of one substance are not affected by the concentrations of other substances present. It is not true in general. On the other hand, ${\Delta }_rG^o$ is always the computed difference between the standard Gibbs free energies of the pure separated products and reactants. For ideal gases,
${\Delta }_rG={\Delta }_rG^o+RT{ \ln \frac{p_Cp_D}{p_Ap_B}\ } \nonumber$ (reaction of ideal gases)
We have asserted that we can equate ${\mathrm{\Delta }}_{sep}G$ and ${\mathrm{\Delta }}_rG$ for systems composed of ideal gases. Now we need to show that this is true. This is easy if we first understand a related problem—the thermodynamics of mixing ideal gases.
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When we talk about the thermodynamics of mixing, we have a very particular process in mind. By convention, the process of mixing two gases, call them $A$ and $B$, is the process in which the two gases initially occupy separate containers, but are both at a common pressure and temperature. (We denote the common initial pressure by “$P_0$”. $P_0$ is not to be confused with the constant $P^o$.) The final state after the mixing process is one in which there is a homogeneous mixture of $A$ and $B$ at the same temperature as characterized the initial state. The final volume is the sum of the initial volumes. If the gases are ideal, the final pressure is the same as the initial pressure, and the partial pressures are $P_A={n_ART}/{\left(V_A+V_B\right)}$ and $P_B={n_BRT}/{\left(V_A+V_B\right)}$.
The mixing process is represented by the change on the right side of Figure 1. The gases are always in thermal contact with constant-temperature surroundings. We imagine that we bring the initially separate containers together and then remove the overlapping walls. (Or we can imagine connecting the two containers by a tube—whose volume is negligibly small—that allows the molecules to move from one container to the other.) Molecular diffusion eventually causes the concentration of either gas to be the same in any macroscopic portion of the combined volumes. This diffusive mixing begins as soon as we provide a path for the molecules to move between their containers. Isothermal mixing is a spontaneous process. The reverse process does not occur. Isothermal mixing is irreversible.
Since the temperature is constant and the gases are ideal, the energy of the $A$ molecules is constant; likewise, the energy of the $B$ molecules constant. It follows that the energy of a system containing molecules of $A$ and $B$ is independent of their concentrations and that the energy of the mixture is the sum of the energies of the separated components. That is, we have ${\Delta }_{mix}E=0$. Since the volume and the pressure of the two-gas system are constant, it follows that ${\Delta }_{mix}\left(PV\right)=0$ and ${\Delta }_{mix}H=0$. Constant volume also means that $w_{mix}=0$. It follows that $q_{mix}=0$, and that ${\Delta }_{mix}\hat{S}=0$. Then, because the mixing process is spontaneous, we have ${\Delta }_{mix}S>0$. Finally, since ${\Delta }_{mix}H=0$, we have ${\Delta }_{mix}G=-T{\Delta }_{mix}S<0$. However, in order to calculate the values of ${\Delta }_{mix}S=0$ and ${\Delta }_{mix}G<0$, we must find a reversible path for the state change that occurs during mixing.
The two-step path represented by the changes along the left side and the bottom of Figure 1 is such a reversible path. The first step is simply the reversible isothermal expansion of the separate gases from their initial volumes ($V_A$ and $V_B$, respectively) to the common final volume, $V_A+V_B$. At the end of this step, the two gases are still in separate containers. Reversible isothermal expansion of an ideal gas is a familiar process; for this step, we know:
$\Delta E_A=\Delta H_A=0 \nonumber$ $\Delta E_B=\Delta H_B=0 \nonumber$
$\Delta S_A=-n_AR \ln \left(P_A/P_0\right)>0 \nonumber$ $\Delta S_B=-n_BR \ln \left(P_B/P_0\right) > 0 \nonumber$
$\Delta G_A=-T\Delta S_A<0 \nonumber$ $\Delta G_B=-T\Delta S_B<0 \nonumber$
We call the change along the bottom of the diagram the merging process. The merging process is the isothermal, reversible blending of the separate gas samples, each initially occupying a volume $V_A+V_B$, in such a way that the final state has all of the molecules of the two gases in the same container, whose volume is also $V_A+V_B$. While the final state of the merging process is identical to the final state of the mixing process, the initial state is distinctly different.
For ideal gases, it turns out that all of the thermodynamic functions are unchanged during the merging process. Consequently, the thermodynamic functions for mixing are just the sums of the thermodynamic functions for the reversible expansions of $A$ and $B$ separately. Equating thermodynamic functions for the two paths to the mixed state, we have
${\Delta }_{mix}E=\Delta E_A+\Delta E_B+{\Delta }_{merge}E=0 \nonumber$ ${\Delta }_{mix}H=\Delta H_A+\Delta H_B+{\Delta }_{merge}H=0 \nonumber$
${\Delta }_{mix}S=\Delta S_A+\Delta S_B+{\Delta }_{merge}S=-n_AR \ln \left(P_A/P_0\right)-n_BRT \ln \left(P_B/P_0\right) >0 \nonumber$
${\Delta }_{mix}G=-T{\Delta }_{mix}S<0 \nonumber$
The pressure ratios equal the mole fractions of the compounds in the mixture. Therefore, the entropy of mixing is also given by
${\Delta }_{mix}S=-n_AR{ \ln x_A\ }-n_BR{ \ln x_B\ }>0 \nonumber$
If we calculate the entropy of mixing per mole of $A$–$B$-mixture, ${\Delta }_{mix}\overline{S}$, we find
${\Delta }_{mix}\overline{S}=\frac{{\Delta }_{mix}S}{n_A+n_B}=-\left(\frac{n_A}{n_A+n_B}\right)R \ln x_A-\left(\frac{n_B}{n_A+n_B}\right)R \ln x_B=-x_AR \ln x_A-x_BR \ln x_B \nonumber$
It remains to prove our assertion that the thermodynamic functions are unchanged during the merging process. That the energy is unchanged follows from the fact that the energy of an ideal gas depends only on temperature; it is therefore independent of pressure and of the presence of any other substance. We can give a qualitative argument for the idea that other thermodynamic quantities are also unchanged. In this argument, the enthalpy, entropy, and free energy functions are unchanged because ideal gas molecules do not interact with one another. If $A$ molecules do not interact with $B$ molecules, it follows that the properties of the $A$ molecules are independent of whether the $B$ molecules share the same container or are present in a separate container of identical volume. At the same temperature, the $A$ molecules generate a pressure $P_A$, and the $B$ molecules independently generate a pressure $P_B$. Since these pressures are generated independently, we conclude that the total pressure is the sum of the two partial pressures—which is, of course, just Dalton’s law of partial pressures. The same argument applies to the enthalpy, entropy, and free energy functions, so these should also be unchanged during the merging process.
We can also create a device in which we can—in concept—carry out the reversible isothermal merging process and calculate thermodynamic-function changes. (We rely on the argument above to establish Dalton’s law of partial pressures and the conclusion that ${\Delta }_{merge}E=0$. We use the device to show from these results that there is no change in the other thermodynamic functions.) The device is sketched in Figure 2, at an intermediate stage of the process. It consists of three cylinders, each closed by a frictionless piston. The first contains unmerged gas $A$, the second contains unmerged gas $B$, and the third contains the mixture that results from merging them. The head of the $A$ cylinder is shared with part of the head of the merging cylinder. Likewise, the head of the $B$ cylinder is shared with another part of the head of the merging cylinder. We suppose that the $A$ and $B$ cylinder heads are comprised in part of molecule-selective gas-permeable membranes. The membrane in the head of the $A$ cylinder allows the diffusion of $A$ molecules in either direction, but does not permit $B$ molecules to pass. The membrane in the head of the $B$ cylinder allows the diffusion of $B$ molecules in either direction, but does not permit $A$ molecules to pass.
In the merging cylinder, we reversibly accumulate $A$ molecules from the $A$ cylinder and $B$ molecules from the $B$ cylinder. We accomplish this by controlling the pressures in the three cylinders. The pressure in the $A$ cylinder is always infinitesimally greater than $P_A$. The pressure in the $B$ cylinder is always infinitesimally greater than $P_B$. The partial pressure of $A$ in the merging cylinder is infinitesimally less than $P_A$; the partial pressure of $B$ in the merging cylinder is infinitesimally less than $P_B$; and the total pressure in the merging cylinder is infinitesimally less than $P_A+P_B$. (There is always a difference in the total pressure across each membrane.) The system consists of the contents of the cylinders. Work is done on the system by the forces acting on the $A$ and $B$ cylinders. Work is done by the system on the surroundings as the merging cylinder fills with the mixture.
As far as the $A$ cylinder is concerned, the process is reversible because the pressure due to the $A$ molecules is just infinitesimally greater in the $A$ cylinder than in the merging cylinder. Therefore, a very small decrease in the pressure of the $A$ cylinder would cause the net flow of $A$ molecules to change direction. The same is true for the $B$ cylinder.
The initial and final states of the apparatus when we use it to merge $n_A$ moles of $A$ at $P_A$ with $n_B$ moles of $B$ at $P_A$ are shown in Figure 3. Let $V_{mix}=V_A+V_B$ be the sum of the initial volumes of cylinders $A$ and $B$. This is also the final volume of the merging cylinder. The reversible work done in the $A$ cylinder is
$w^{rev}_A=-\int^0_{V_A+V_B}{P_AdV}=-P_A\left(0-V_A V_B\right)=P_A\left(V_A+V_B\right)=n_ART \nonumber$
Similarly, the work in the $B$ cylinder is
$w^{rev}_B=P_B\left(V_A+V_B\right)=n_BRT \nonumber$
In the merging cylinder it is
$w^{rev}_{A+B}=-\left(P_A+P_B\right)\left(V_A+V_B\right)=-\left(n_B+n_B\right)RT \nonumber$
For the merging process the net work is
$w^{rev}_{merge}=w^{rev}_A+w^{rev}_B+w^{rev}_{A+B}=0 \nonumber$
and the net change in the pressure-volume product is
${\Delta }_{merge}\left(PV\right)=\left(P_A+P_B\right)\left(V_A+V_B\right) \nonumber$ $-P_A\left(V_A+V_B\right)-P_B\left(V_A+V_B\right) \nonumber$ $=0 \nonumber$
Since constant temperature ensures that ${\Delta }_{merge}E=0$, it follows from ${\Delta }_{merge}\left(PV\right)=0$ that ${\Delta }_{merge}H=0$, and from $w^{rev}_{merge}=0$ that $q^{rev}_{merge}=0$. Since the merging process is reversible, we have also that ${\Delta }_{merge}S=0$ and ${\Delta }_{merge}G=0$.
These arguments can be extended to merging any number of ideal gases. In the initial state for this merging process each gas is at the same temperature, but occupies a separate container; all of these containers have the same volume. Each gas can be at a different pressure. In the final state, all of the gases occupy a common container, whose volume is the same as the common volume of their initial containers. The temperature of the mixture in the final state is equal to the common initial temperature of the separate gases. In the final state, the partial pressure of each gas is equal to its pressure in the initial state. For any number of gases, we have
${\Delta }_{merge}E={\Delta }_{merge}H={\Delta }_{merge}S={\Delta }_{merge}G=0 \nonumber$
Likewise, these arguments can be extended to the mixing of multiple ideal gases, all at the same original pressure and temperature, into a final volume that is equal to the sum of the initial volumes—and is at the original temperature. If there are $\omega$ such gases,
${\Delta }_{mix}E={\Delta }_{mix}H=0 \nonumber$
${\Delta }_{mix}S=\sum^{\omega }_{i=1}{-n_iR \ln x_i}>0 \nonumber$
${\Delta }_{mix}G=-T{\Delta }_{mix}S<0 \nonumber$
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Now we can compare the difference between the Gibbs free energies of reactants and products in the reaction
$a\ A+b\ B\ \rightleftharpoons \ c\ C+d\ D \nonumber$
when all of the gases are present in the same system to the same difference when each gas is in its own container.
In the first case, the gases are present in a mixture, and their partial pressures remain constant at $P_A$, $P_B$, $P_C$, and $P_D$ respectively. We called the Gibbs free energy change under these conditions ${\Delta }_rG$. These conditions can be satisfied if we suppose that the system is very large. That is, ${\Delta }_rG$ is the limiting Gibbs free energy change for the conversion of an initial mixture into a final mixture. The initial mixture contains $\left(n_A+a\right)$ moles of $A$, $\left(n_B+b\right)$ moles of $B$, $n_C$ moles of $C$, and $n_D$ moles of D. The final mixture contains $n_A$ moles of $A$, $n_B$ moles of $B$, $\left(n_C+c\right)$ moles of $C$, and $\left(n_D+d\right)$ moles of $D$. ${\Delta }_rG$ is the Gibbs free energy change for this conversion in the limit as $n_A$, $n_B$, $n_C$, and $n_D$ become arbitrarily large. In this limit, we have $n_A\gg a$, $n_B\gg b$, $n_C\gg c$, , and $n_D\gg d$. Since the partial pressures remain (essentially) constant, the $n_i$ must also satisfy
$P_A={n_A}/{\left(n_A+n_B+n_C+n_D\right)} \nonumber$ $P_B={n_B}/{\left(n_A+n_B+n_C+n_D\right)} \nonumber$ $P_C={n_C}/{\left(n_A+n_B+n_C+n_D\right)} \nonumber$ $P_D={n_D}/{\left(n_A+n_B+n_C+n_D\right)} \nonumber$
When the ideal gases are separated from one another, the Gibbs free energy difference is the Gibbs free energy of $c$ moles of gas $C$ (at pressure $P_C$) plus the Gibbs free energy of $d$ moles of gas $D$ (at pressure $P_D$) minus the Gibbs free energy of $a$ moles of gas $A$ (at pressure $P_A$) and minus the Gibbs free energy of $b$ moles of gas $B$ (at pressure $P_B$). In §2, we call the Gibbs free energy change under these conditions ${\Delta }_{sep}G$.
In Section 13.2, we assert that ${\Delta }_{sep}G={\Delta }_rG$. Now we can see why this is so. The cycle shown in Figure 4 relates these two Gibbs free energy differences.
We have
${\Delta }_{sep}G\left(P_A,P_B,P_C,P_D\right)+{\Delta }_{merge}G_{final} \nonumber$ $={\Delta }_{merge}G_{initial}+{\Delta }_rG\left(P_A,P_B,P_C,P_D\right) \nonumber$
and since
${\Delta }_{merge}G_{final}={\Delta }_{merge}G_{initial} \nonumber$ we have
${\Delta }_{sep}G\left(P_A,P_B,P_C,P_D\right)={\Delta }_rG\left(P_A,P_B,P_C,P_D\right) \nonumber$
Letting ${\overline{G}}_A\left(P_A\right)$ be the Gibbs free energy of one mole of pure ideal gas $A$ at pressure $P_A$, ${\overline{G}}_B\left(P_B\right)$ be the Gibbs free energy of one mole of pure ideal gas $B$ at pressure $P_B$, etc., we have
\begin{aligned} {\Delta }_{sep}G\left(P_A,P_B,P_C,P_D\right) & = n_A \overline{G}_A\left(P_A\right)+n_B\ \overline{G}_B\left(P_B\right) \ ~ & +\left(n_C+c\right)\ \overline{G}_C\left(P_C\right)+\left(n_D+d\right)\overline{G}_D\left(P_D\right) \ ~ & -\left(n_A+a\right) \overline{G}_A\left(P_A\right)-\left(n_B+b\right) \overline{G}_B\left(P_B\right) \ ~ & -n_C\ \overline{G}_C\left(P_C\right)-n_D \overline{G}_D\left(P_D\right) \ ~ & =c \overline{G}_C\left(P_C\right)+d \overline{G}_D\left(P_D\right)-a\ \overline{G}_A\left(P_A\right)-b\ \overline{G}_B\left(P_B\right) \end{aligned} \nonumber
Since ${\Delta }_{sep}G={\Delta }_rG$, we have
${\Delta }_rG=c\ \overline{G}_C\left(P_C\right)+d\ \overline{G}_D\left(P_D\right)-a\ \overline{G}_A\left(P_A\right)-b\ \overline{G}_B\left(P_B\right) \nonumber$
13.05: rG is the rate at which the Gibbs Free Energy Changes with The Extent
For the reaction $a\ A+b\ B\ \rightleftharpoons \ c\ C+d\ D$, let us call the consumption of $a$ moles of $A$ one “unit of reaction.” ${\Delta }_rG$ corresponds to the actual Gibbs free energy change for one unit of reaction only in the limiting case where the reaction occurs in an arbitrarily large system. For a closed system of specified initial composition, $n^o_A$, $n^o_B$, $n^o_C$, and $n^o_D$, whose composition at any time is specified by $n_A$, $n_B$, $n_C$, and $n_D$, the extent of reaction, $\mathrm{\xi }$, is
$\xi =-\left(\frac{n_A-n^o_A}{a}\right)=-\left(\frac{n_B-n^o_B}{b}\right)=\ \ \ \ \ \ \ \frac{n_C-n^o_C}{c}=\ \ \ \ \ \ \ \frac{n_D-n^o_D}{d} \nonumber$
At constant pressure and temperature, every possible state of this system is represented by a point on a plot of ${\Delta }_rG$ versus $\mathrm{\xi }$. Every such state is also represented by a point on a plot of $G_{system}$ versus $\mathrm{\xi }$.
From the general result that ${\left({dG}_{system}\right)}_{PT}\mathrm{=0}$ if and only if the system is at equilibrium, it follows that ${\mathrm{\Delta }}_rG\left({\xi }_{eq}\right)\mathrm{=0}$ if and only if ${\xi }_{eq}$ specifies the equilibrium state. (We can arrive at the same conclusion by considering the heat exchanged for one unit of reaction in an infinitely large system at equilibrium. This process is reversible, and it occurs at constant pressure and temperature, so we have ${\mathrm{\Delta }}_rH = q^{rev}_P$, ${\mathrm{\Delta }}_rS = {q^{rev}_P}/{T}$, and ${\mathrm{\Delta }}_rG = q^{rev}_P + T\left({q^{rev}_P}/{T}\right)\mathrm{=0}$.)
Below, we show that
${\mathrm{\Delta }}_rG\left(\xi \right) = {\left(\frac{\partial G_{system}}{\partial \xi }\right)}_{PT} \nonumber$
for any value of $\mathrm{\xi }$. (In Section 15.9, we use essentially the same argument to show that this conclusion is valid for any reaction among any substances.) Given this result, we see that the equilibrium composition corresponds to the extent of reaction, ${\xi }_{eq}$, for which the Gibbs free energy change for one unit of the reaction is zero
${\mathrm{\Delta }}_rG\left({\xi }_{eq}\right)\mathrm{=0} \nonumber$
and
${\left(\frac{\partial G_{system}}{\partial \xi }\right)}_{PT}=0 \nonumber$
So that the Gibbs free energy of the system is a minimum.
In the next section, we show that the condition ${\mathrm{\Delta }}_rG\left({\xi }_{eq}\right)\mathrm{=0}$ makes it easy to calculate the equilibrium extent of reaction, ${\xi }_{eq}$. Given the stoichiometry and initial composition, the equation for ${\mathrm{\Delta }}_rG\left({\xi }_{eq}\right)$ specifies the equilibrium composition and the partial pressures $P_A$, $P_B$, $P_C$, and $P_D$. This is the usual application of these results. Setting ${\Delta }_rG=0$ enables us to answer the question: If we initiate reaction at a given composition, what will be the equilibrium composition of the system? Usually this is what we want to know. The amount by which the Gibbs free energy changes as the reaction goes to equilibrium is seldom of interest.
To show that ${\mathrm{\Delta }}_rG={\left({\partial G}/{\partial \xi }\right)}_{PT}$ for any reaction, it is helpful to introduce modified stoichiometric coefficients, ${\nu }_j$, defined such that ${\nu }_j>0$ if the j-th species is a product and ${\nu }_j<0$ if the j-th species is a reactant. That is, for the reaction $a\ A+b\ B\ \rightleftharpoons \ c\ C+d\ D$, we define ${\nu }_A=-a$, ${\nu }_B=-b$, ${\nu }_C=c$, and ${\nu }_D=d$. Associating successive integers with the reactants and products, we represent the j-th chemical species as $X_j$ and an arbitrary reaction as
$\left|{\nu }_1\right|X_1+\left|{\nu }_2\right|X_2+\dots +\left|{\nu }_i\right|X_i\rightleftharpoons {\nu }_jX_j+\dots +{\nu }_{\omega }X_{\omega } \nonumber$
Let the initial number of moles of ideal gas $X_j$ be $n^o_j$; then $n_j=n^o_j+{\nu }_j\xi$. (For species that are present but do not participate in the reaction, we have ${\nu }_j=0$.)
We have shown that the Gibbs free energy of a mixture of ideal gases is equal to the sum of the Gibbs free energies of the components. In calculating ${\Delta }_rG$, we assume that this is as true for a mixture undergoing a spontaneous reaction as it is for a mixture at equilibrium. In doing so, we assume that the reacting system is homogeneous and that its temperature and pressure are well defined. In short, we assume that the Gibbs free energy of the system is the same continuous function of temperature, pressure, and composition, $G=G\left(T,P,n_1,n_2,\dots \right)$, whether the system is at equilibrium or undergoing a spontaneous reaction. For the equilibrium system, we have ${\left({\partial G}/{\partial T}\right)}_{P{,n}_j}=-S$ and ${\left({\partial G}/{\partial P}\right)}_{T{,n}_j}=V$. When we assume that these functions are the same for a spontaneously changing system as they are for a reversible system, it follows that
$dG={\left(\frac{\partial G}{\partial T}\right)}_{P,n_j}dT+{\left(\frac{\partial G}{\partial P}\right)}_{T,n_j}dP+\sum_j{{\left(\frac{\partial G}{\partial n_j}\right)}_{P,T,n_{i\neq j}}dn_j}=-SdT+VdP+\sum_j{{\left(\frac{\partial G}{\partial n_j}\right)}_{P,T,n_{i\neq j}}dn_j} \nonumber$
whether the system is at equilibrium or undergoing spontaneous change. At constant temperature and pressure, when pressure–volume work is the only work, the thermodynamic criteria for change, ${dG}_{TP}\le 0$ become
$\sum_j{{\left(\frac{\partial G}{\partial n_j}\right)}_{P,T,n_{i\neq j}}dn_j}\le 0 \nonumber$
When a reaction occurs in the system, the composition is a continuous function of the extent of reaction. We have $G=G\left(T,P,n^o_1+{\nu }_1\xi ,n^o_2+{\nu }_2\xi ,\dots \right)$. At constant temperature and pressure, the dependence of the Gibbs free energy on the extent of reaction is
$\left(\frac{\partial G}{\partial \xi }\right)_{P,T,n_m} =\sum_j \left(\frac{\partial G}{\partial \left(n^o_J+{\nu }_j\xi \right)}\right)_{P,T,n_{m\neq j}} \left(\frac{\partial \left(n^o_J+{\nu }_j\xi \right)}{ \partial \xi} \right)_{P,T,n_{m\neq j}} \nonumber$
Since
${\left(\frac{\partial G}{\partial \left(n^o_J+{\nu }_j\xi \right)}\right)}_{P,T,n_{m\neq j}}={\left(\frac{\partial G}{\partial n_j}\right)}_{P,T,n_{m\neq j}}={\overline{G}}_j \nonumber$
and
${\left(\frac{\partial \left(n^o_J+{\nu }_j\xi \right)}{\partial \xi }\right)}_{P,T,n_{m\neq j}}={\nu }_j \nonumber$
it follows that
${\left(\frac{\partial G}{\partial \xi }\right)}_{P,T,n_m}=\sum_j{{\nu }_j{\overline{G}}_j}={\Delta }_rG \nonumber$
Moreover, we have
${{\left(dG\right)}_{PT}=\left(\frac{\partial G}{\partial \xi }\right)}_{P,T,n_m}d\xi \nonumber$
The criteria for change, ${\left(dG\right)}_{PT}\le 0$, become
${\left(\frac{\partial G}{\partial \xi }\right)}_{P,T,n_m}d\xi \le 0 \nonumber$
From our definition of $\xi$, we have $d\xi >0$ for a process that proceeds spontaneously from left to right, so the criteria become
${\left(\frac{\partial G}{\partial \xi }\right)}_{P,T,n_m}\le 0 \nonumber$
or, equivalently,
$\sum_j{{\nu }_j{\overline{G}}_j}={\Delta }_rG\le 0 \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/13%3A_Equilibria_in_Reactions_of_Ideal_Gases/13.04%3A_The_Gibbs_Free_Energy_Change_for_Reaction_at_Constant_Partial_Pressure.txt
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The relationship between ${\Delta }_rG$ and ${\Delta }_rG^o$ is evident from the cycle in Figure 5. Since we have shown that ${\Delta }_{sep}G\left(P_A,P_B,P_C,P_D\right)={\Delta }_rG\left(P_A,P_B,P_C,P_D\right)$, we can consider the bottom equation in this cycle to represent the reaction occurring in a mixture while calculating its free energy change as the free energy difference between pure products and pure reactants. Since ${\Delta }_{cycle}G=0$,
${\Delta }_{cycle}G={\Delta }_rG^o-{\Delta }_rG+RT\left({ \ln p^c_C\ }+{ \ln p^d_D\ }-{ \ln p^a_A\ }-{ \ln p^B_B\ }\right)=0 \nonumber$
which can be rearranged to the result obtained in §2:
${\Delta }_rG={\Delta }_rG^o+RT{ \ln \frac{p^c_Cp^d_D}{p^a_Ap^b_B}\ } \nonumber$
${\Delta }_rG$ is the Gibbs free energy change for one unit of the reaction occurring in a system whose composition is specified by $P_A$, $P_B$, $P_C$, and $P_D$. In this spontaneously reacting system, the molar Gibbs free energy of ideal gas $A$ is
${\overline{G}}_A\left(P_A\right)={\Delta }_fG^o\left(A,P^o\right)+RT{ \ln p_A\ } \nonumber$
If the system is at equilibrium, $P_A$, $P_B$, $P_C$, and $P_D$ are equilibrium pressures; these values characterize an equilibrium state. Then ${\Delta }_rG$ is the free energy change for a reaction occurring at equilibrium at constant pressure and temperature, and ${\Delta }_rG$ is zero. The equation
$0={\Delta }_rG^o+RT{ \ln \frac{p^c_Cp^d_D}{p^a_Ap^b_B}\ } \nonumber$
is exact. We have, when the partial pressures are those for a system at equilibrium,
${\Delta }_rG^o=-RT{ \ln \frac{p^c_Cp^d_D}{p^a_Ap^b_B}\ } \nonumber$
Since ${\Delta }_rG^o$ is a constant, it follows that
$\frac{p^c_Cp^d_D}{p^a_Ap^b_B} \nonumber$
is a constant. It is, of course, the equilibrium constant. We have
$K_P=\frac{p^c_Cp^d_D}{p^a_Ap^b_B} \nonumber$
and ${\Delta }_rG^o=-RT{ \ln K_P\ } \nonumber$
or, solving for $K_P$
$K_P = \mathrm{exp}\left( + \frac{{\mathrm{\Delta }}_rG^o}{RT}\right) \nonumber$
Note that the value of the equilibrium constant is calculated from the Gibbs free energy change at standard conditions, not the Gibbs free energy change at equilibrium, which is zero.
13.07: The Gibbs Free Energy of Formation and Equilibrium in Ideal Gas Reacti
In Section 13.6, we develop a relationship between the standard Gibbs free energy change for a reaction and the equilibrium constant for that reaction. The standard Gibbs free energy change for a reaction is a specific quantity of energy, which depends only on the temperature. In Chapter 11, we develop the Gibbs free energy of formation for a substance in its standard state. We assert that it is convenient to set the Gibbs free energy of every substance in its standard state equal to its Gibbs free energy of formation. Now we can see the reason: When we express the molar Gibbs free energy of an ideal gas as
${\overline{G}}_A\left(P_A\right)={\Delta }_fG^o\left(A,P^o\right)+RT{ \ln \frac{P_A}{P^o}\ } \nonumber$
${\overline{G}}_A\left(P_A\right)$ is the Gibbs free energy change for producing one mole of ideal gas $A$, at pressure $P_A$, from the elements at the same temperature. Because the Gibbs free energies of the reactants and products are measured from a common starting point, we can use them to calculate the Gibbs free energy change for their reaction.
To demonstrate the role of our choice of the elements as the reference state for the Gibbs free energies of chemical substances, we need only expand the Gibbs free energy cycle in Figure 4. In Figure 6, we introduce the Gibbs free energy change for producing the separate components of the very large system.
Computing the Gibbs free energy change in a clockwise direction around the cycle in which the elements are converted first to isolated reactants, then to components of a large equilibrium system, then to separated products, and finally back to the elements, we find
$\mathrm{0=}{\mathrm{\Delta }}_fG\left(initial\right) + {\mathrm{\Delta }}_{merge}G_{initial} \nonumber$ $+ {\mathrm{\Delta }}_rG\left(P_A\mathrm{,\ }P_B\mathrm{,\ }P_C\mathrm{,\ }P_D\right) + {\mathrm{\Delta }}_{merge}G_{final} + {\mathrm{\Delta }}_fG\left(final\right) \nonumber$
Since the partial pressures $P_A$, $P_B$, $P_C$, and $P_D$ characterize the system at equilibrium, ${\mathrm{\Delta }}_rG\left(P_A\mathrm{,\ }P_B\mathrm{,\ }P_C\mathrm{,\ }P_D\right)\mathrm{=0}$. Also, ${\mathrm{\Delta }}_{merge}G_{initial} = {\mathrm{\Delta }}_{merge}G_{final}$= 0, so the equation for the Gibbs free energy change around the cycle simplifies to
\begin{aligned} 0 & ={\Delta }_fG\left(initial\right)-{\Delta }_fG\left(final\right) \ ~ & =a \overline{G}_A\left(p_A\right)+b \overline{G}_B\left(p_B\right)-c \overline{G}_C\left(p_C\right)-d \overline{G}_D\left(p_D\right) \ ~ & =a\ {\Delta }_fG^o\left(A,P^o\right)+b\ {\Delta }_fG^o\left(B,P^o\right)-c {\Delta }_fG^o\left(C,P^o\right) -d {\Delta }_fG^o\left(D,P^o\right) +RT \ln p^a_A\ +RT \ln p^b_B -RT \ln p^c_C-RT \ln p^d_D \ ~ & =-{\Delta }_rG^o-RT \ln \frac{p^c_Cp^d_D}{p^a_Ap^b_B} \end{aligned} \nonumber
So that we have
${\Delta }_rG^o=-RT{ \ln \frac{p^c_Cp^d_D}{p^a_Ap^b_B}\ } \nonumber$
To illustrate the application of these ideas, let us consider equilibrium in the oxidation of nitric oxide to nitrogen dioxide
$2\ NO+O_2\rightleftharpoons 2\ NO_2 \nonumber$
At 800 K, the standard Gibbs free energy of formation of nitric oxide is $+81.298\ \mathrm{kJ}\ {\mathrm{mol}}^{-1}$, and that of nitrogen dioxide is $+83.893\ \mathrm{kJ}\ {\mathrm{mol}}^{-1}$. Hence, the standard Gibbs free energy change for the oxidation is $+5,190\ \mathrm{J}\ {\mathrm{mol}}^{-1}$, and the equilibrium constant is
$K_P=\mathrm{exp}\left(-\frac{{\Delta }_rG^o}{RT}\right)=0.458 \nonumber$
Suppose that one mole of $NO_2$ and one mole of $O_2$ are mixed, that the mixture is thermostated at 800 K, and that the applied pressure remains constant at 10 bar while the reaction goes to equilibrium.
Let the number of moles of $NO$ present at equilibrium be $n_{NO}$. Then the moles of $NO_2$ and $O_2$ present at equilibrium are $n_{NO_2}\mathrm{=1-}n_{NO}$ and $n_{O_2}=1+{n_{NO}}/{2}$. The partial pressures are $P_{NO}={n_{NO}RT}/{V}$, $P_{NO_2}={n_{NO_2}RT}/{V}$, and $P_{O_2}={n_{O_2}RT}/{V}$, so that
$K_P=\frac{{\left(\mathrm{1-}n_{NO}\mathrm{\ }\right)}^2}{n^2_{NO}\left(1+{n_{NO}}/{2}\right)\left({RT}/{V}\right)} \nonumber$ The value of $RT/V$ depends on the total pressure; we have
$P=P_{NO}+P_{NO_2}+P_{O_2}=\left(2+{n_{NO}}/{2}\right)\left({RT}/{V}\right)=10\ \mathrm{bar} \nonumber$ so that
$\frac{RT}{V}=\frac{20}{4+n_{NO}} \nonumber$
and
$K_P=\frac{{\left(\mathrm{1-}n_{NO}\mathrm{\ }\right)}^2\left(4+n_{NO}\right)}{10n^2_{NO}\left(2+n_{NO}\right)}=0.458 \nonumber$
Solving, we find $n_{NO}=0.388$.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/13%3A_Equilibria_in_Reactions_of_Ideal_Gases/13.06%3A_The_Standard_Gibbs_Free_Energy_Change_and_Equilibrium_in_Ideal_Gas_Rea.txt
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Suppose that the very large equilibrium system with ideal gas components at pressures $P_A$, $P_B$, $P_C$, and $P_D$, also contains a quantity of liquid $A$. (For the present, we assume that this is pure liquid $A$; components $B$, $C$, and $D$ are insoluble in liquid $A$.) If this augmented system is at equilibrium, we know that liquid $A$ is in phase equilibrium with ideal gas $A$ at pressure $P_A$. That is, $P_A$ is the vapor pressure of liquid $A$ at the fixed temperature that we are considering, and the Gibbs free energy change for converting liquid $A$ to its ideal gas at $P_A$ is zero. As long as the liquid is in phase equilibrium with its ideal gas, the relationship between the ideal gas partial pressures and the standard Gibbs free energy change for the reaction is not affected by the presence of the liquid. These same considerations apply when the very large equilibrium system contains both ideal gas $A$ and solid $A$, so long as the equilibrium sublimation pressure is equal to the partial pressure of ideal gas A, $P_A$.
Now, let us suppose that substance $A$ is a non-volatile liquid or solid. In this case, we may not be able to measure the standard Gibbs free energy of formation of ideal gas $A$. From a practical standpoint, this is an important consideration; if we cannot find the standard Gibbs free energy of formation of the ideal gas, we cannot use it to calculate equilibrium constants. From a theoretical standpoint, it is less crucial; we can reasonably imagine that any substance has a finite vapor pressure at any temperature, even if the value is much too small to measure experimentally. We can reason about the relationship of the equilibrium vapor pressure to other quantities whether we can measure it or not.
If substance $A$ is a non-volatile liquid (or solid) at the temperature of interest, it is useful to modify the cycle introduced in the preceding section. We can find the standard Gibbs free energy of formation for the liquid from thermal measurements, and we can use liquid $A$ rather than ideal gas $A$ as the standard state. If $A$ is non-volatile, it is present in the equilibrium system only as the liquid. For present purposes, we again assume that the other components are insoluble in liquid $A$; then, the only difference between $A$ in its standard state and $A$ in the equilibrium system is that the pressure on $A$ in its standard state is one bar and the pressure on liquid $A$ in the equilibrium system is $P=P_B+P_C+P_D$. Therefore, when $a$ moles of liquid $A$ go from their standard state to the equilibrium system, the Gibbs free energy change is
${\Delta }_{press}G_A=a\int^P_{P^o}{{\overline{V}}^{\textrm{⦁}}_A}\left(\ell \right)dP \nonumber$
where ${\overline{V}}^{\textrm{⦁}}_A\left(\ell \right)$ is the molar volume of pure liquid $A$. When $a$ moles of liquid $A$ are produced at the equilibrium pressure from the elements, the Gibbs free energy change is
$a{\overline{G}}_A\left(\ell ,P\right)=a\ {\Delta }_fG^o\left(A,\ell \right)+a\int^P_{P^o}{{\overline{V}}^{\textrm{⦁}}_A}\left(\ell \right)dP \nonumber$
In Section 15.3, we give further attention to the value of this integral; for now, let us simply note that it is negligible in essentially all circumstances. These considerations mean that we can modify the left side of the cycle in Figure 5 as indicated in Figure 7.
When we sum around the modified cycle in the same manner as before, we find
$0=a\ {\Delta }_fG^o\left(A,\ell \right)+b\ {\Delta }_fG^o\left(B,P^o\right)-c\ {\Delta }_fG^o\left(C,P^o\right) \nonumber$ $-d\ {\Delta }_fG^o\left(D,P^o\right)+a\int^P_{P^o}{{\overline{V}}^{\textrm{⦁}}_A}\left(\ell \right)dP+RT{ \ln p^b_B\ } \nonumber$ $-RT{ \ln p^c_C-RT{ \ln p^d_D\ }\ } \nonumber$
where we have supplemented our notation to emphasize that $A$ is a non-volatile liquid while $B$, $C$, and$\ D$ are ideal gases. We continue to use ${\Delta }_rG^o$ to represent the difference between the standard Gibbs free energies of the products and those of the reactants. In the present circumstances, we have
${\Delta }_rG^o=\ c\ {\Delta }_fG^o\left(C,P^o\right)+d\ {\Delta }_fG^o\left(D,P^o\right)-a\ {\Delta }_fG^o\left(A,\ell \right)-b\ {\Delta }_fG^o\left(B,P^o\right) \nonumber$
Taking the value of the integral as zero, our result simplifies to
${\Delta }_rG^o=-RT{ \ln \frac{p^c_Cp^d_D}{p^b_B}\ } \nonumber$
We have again arrived at the conclusion that the “concentration” of a pure solid or liquid can be set equal to unity in the equilibrium constant expression for a reaction in which it participates. When we do so, we must use the Gibbs free energy of formation of the condensed phase in the calculation of ${\Delta }_rG^o$.
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Finally, let us consider a very large equilibrium system that contains ideal gas components $A$, $B$, $C$, and $D$, at pressures $P_A$, $P_B$, $P_C$, and $P_D$. We suppose that this system also contains a very large quantity of an inert solvent. This solvent is a liquid in which components $A$, $B$, $C$, and $D$ are soluble. Let the concentrations of components $A$, $B$, $C$, and $D$ in this solvent be $\left[A\right]$, $\left[B\right]$, $\left[C\right]$, and $\left[D\right]$. Since this solvent-containing system is at equilibrium, we know that $A$ dissolved in the solvent at concentration $\left[A\right]$ is in equilibrium with ideal gas $A$ at pressure $P_A$. Since we assume that both the gas and the solution phases are very large, the transport of $a$ moles of $A$ from one to the other does not significantly change any concentration in either phase. The Gibbs free energy change is zero for this phase-transfer process. The Gibbs free energy change for the ideal-gas reaction and its equilibrium position are unaffected by the presence of the solution.
In Chapter 6, we observe that a distribution equilibrium is characterized by an equilibrium constant; the ratio of the concentration of a given chemical species in one phase to its concentration in a second phase is (approximately) a constant. In the present instance, the partial pressure of component $A$ is a measure of its gas-phase concentration and $\left[A\right]$ is a measure of its solution-phase concentration. Letting
${\kappa }_A= \left(\frac{P_A/P^o}{\left[A\right]}\right)=\frac{p_A}{\left[A\right]} \nonumber$
be the distribution constant for component $A$, we have $p_A={\kappa }_A\left[A\right]$, and parallel relationships for components$\ B$, $C$, and $D$. Substituting the distribution equilibrium relationships into the equilibrium constant equation, we have
${\Delta }_rG^o=-RT{ \ln \frac{p^c_Cp^d_D}{p^a_Ap^b_B}\ }=-RT{ \ln \frac{{\kappa }^c_C{\kappa }^d_D}{{\kappa }^a_A{\kappa }^b_B}\ }-RT{ \ln \frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b}\ } \nonumber$
Evidently, we can characterize the position of equilibrium in this system using either the pressure-based constant,
$K_P=\frac{p^c_Cp^d_D}{p^a_Ap^b_B} \nonumber$
or the concentration based constant,
$K_C=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b} \nonumber$
The relationship between the pressure-based and the concentration-based constants is
$K_C=\left(\frac{{\kappa }^c_C{\kappa }^d_D}{{\kappa }^a_A{\kappa }^b_B}\right)K_P \nonumber$
In our present discussion, the concentrations can be expressed in any convenient units. The numerical value of the concentration-based constant depends on the units of concentration and the values of the distribution-equilibrium constants as well as the standard Gibbs free energy change for the reaction of the ideal gases.
When all of the reacting species are non-volatile, all of the reacting substances are present in the solution. The partial pressures of these species in any gas phase above the solution is immeasurably small, and the Gibbs free energies of formation of the ideal gases are not accessible by thermal measurements. Since ${\Delta }_rG^o$ for the ideal-gas reaction is not available, it cannot be used to find $K_P$. Nevertheless, our thermodynamic model presumes that these parameters have finite—albeit immeasurable—values. The concentration-based constant
$K_C=\frac{{\left[C\right]}^c{\left[D\right]}^d}{{\left[A\right]}^a{\left[B\right]}^b} \nonumber$
characterizes the position of equilibrium in solution even when the data to characterize the gas-phase equilibrium process are immeasurable.
We have again arrived at the same function of concentrations to characterize the position of equilibrium. It is the same whether the equilibrating species are present in the gas phase or in an inert-liquid solvent. To obtain this result, we have used our general thermodynamic model for equilibrium, but we have made special assumptions about the properties of the reacting species. We have assumed that the gases behave ideally and that the distribution-equilibrium constants can be expressed using the species’ concentrations. In Chapter 15, we return to this subject and develop a rigorous model for chemical equilibrium that does not require these special assumptions.
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1. Ethylene is the most important feedstock for the organic chemicals industry. The ethylene-production process with the lowest processing cost is the “thermal cracking” of ethane to produce hydrogen and ethylene. The table below gives ${\Delta }_fG^o$ for ethane and ethylene at 1000 K, 1100 K, and 1200 K.
T, K $C_6H_6$ $C_6H_4$
1000 110,750 119,067
1100 132,385 127,198
1200 154,096 135,402
(a) Calculate ${\Delta }_rG^o$ and $K_P$ at each temperature.
(b) At each temperature, calculate the extent of reaction when one mole of pure ethane reacts to reach equilibrium, while the pressure of the system is maintained constant at one bar.
(c) At each temperature, calculate the extent of reaction when one mole of pure ethane reacts to reach equilibrium, while the pressure of the system is maintained constant at 0.100 bar.
(d) To minimize side reactions, it is desirable to operate a cracking reactor at the lowest possible temperature. The feed to a cracking reactor contains many moles of water (steam) for every mole of ethane. Steam is simply an inexpensive inert gas in this system. Why is steam fed to cracking reactors?
2. A system initially contains one mole of ethane. The cracking reaction occurs while the system is maintained at a constant pressure of 0.1 bar and a constant temperature of 1000 K.
(a) Write the equations for the molar Gibbs free energies of ethane, ethylene, and hydrogen as a function of the extent of reaction, $\xi$.
(b) Write the equation for ${\Delta }_rG^o$ as a function of the extent of reaction. For a constant system pressure of 0.1 bar, calculate ${\Delta }_rG^o$ for $\xi$ = 0, 0.4, 0.7, 0.8, 0.9, and 1.0. Roughly, what extent of reaction corresponds to ${\Delta }_rG^o=0$?
(c) Write the equation for the Gibbs free energy of the system, $G_{system}$, as a function of the extent of reaction, $\xi$. For a constant system pressure of 0.1 bar, calculate $G_{system}$ for $\xi$ = 0, 0.4, 0.7, 0.8, 0.9, and 1.0. Roughly, at what extent of reaction is $G_{system}$ a minimum?
(d) From the equation for $G_{system}$ in part (c), find ${\left({\partial G_{system}}/{\partial \xi }\right)}_{T,P,n_j}$. (When the $n_j$ are constant, the composition of the system is fixed: ${\overline{G}}_{C_6H_6}$, ${\overline{G}}_{C_6H_4}$, and ${\overline{G}}_{H_2}$ are constants.)
3. At –56.6 C, the vapor pressure of solid $CO_2$ is 5.18 bar. (This is the triple point.) At –78.5 C, the vapor pressure of solid $CO_2$ is 1.01 bar.
(a) A cold bath is prepared by mixing solid $CO_2$ with methanol in a Dewar flask that is open to the atmosphere. What is the temperature of this cold bath?
(b) Use the Clausius-Clapeyron equation to estimate the enthalpy of sublimation of $CO_2$.
(c) Assume that we know the Gibbs free energy of formation for both solid and ideal-gas $CO_2$ at a particular temperature, T. That is, we know ${\Delta }_fG^o\left(CO_2,\mathrm{g},T\right)$ and ${\Delta }_fG^o\left(CO_2,\mathrm{s},T\right)$. What is the Gibbs free energy of gaseous $CO_2$ as a function of pressure at constant temperature, T?
(d) What is the Gibbs free energy of solid $CO_2$ at a constant temperature, T?
(e) Assume that the variation of the Gibbs free energy of the solid with pressure is negligible. Using your answers to (c) and (d), write the Gibbs free energy change, $\Delta G$, for the process $CO_2\left(\mathrm{solid},\ P,T\right)\to CO_2\left(\mathrm{gas},\ P,T\right)$. If $\left(P,T\right)$ is a point of solid—gas equilibrium, what is the value of $\Delta G$? How is the equilibrium constant for this process related to $P_{CO_2}$?
(f) Use the Gibbs-Helmholtz equation to estimate the enthalpy of sublimation of $CO_2$.
(g) Compare the equation you used in part (b) to the equation that you used in part (f).
4. Ethanol is manufactured by the addition of water to ethylene over a solid-acid catalyst. The process is known as “the hydration of ethylene.” One mole of ethylene, 10 moles of water, and a quantity of solid catalyst are charged to a reactor. When equilibrium is reached at 400 K, the system pressure is 10.00 bar and the conversion to ethanol is 64.17%. Assume that ethylene, water, and ethanol behave as ideal gases. What is ${\Delta }_rG^o$ for the hydration of ethylene at 400 K?
5. At 298.15 K, the Gibbs free energies for formation of $NO_2$ and $N_2O_4$, in their hypothetical ideal gas standard states, are 51.3 and 99.8 kJ mol${}^{--1}$, respectively.
(a) At this temperature, what is the value of the equilibrium constant for the reaction $N_2O_4\rightleftharpoons 2\ NO_2$?
(b) A 10 L sample of gas initially contains one mole of $N_2O_4$. The balance of the sample is $N_2$. The pressure and temperature of the system are maintained constant at 3 bar and 298.15 K. How many moles of $NO_2$ are present at equilibrium? What fraction of the $N_2O_4$ is converted to $NO_2$?
(c) A 10 L sample of gas initially contains only one mole of $N_2O_4$ at 298.15 K. The volume and temperature of the system are maintained constant. What fraction of the $N_2O_4$ is converted to $NO_2$?
(d) A 10 L sample of gas initially contains only ${10}^{-1}$ mole of $N_2O_4$ at 298.15 K. The pressure and temperature of the system are maintained constant. What fraction of the $N_2O_4$ is converted to $NO_2$?
(e) A 10 L sample of gas initially contains only ${10}^{-1}$ mole of $N_2O_4$ at 298.15 K. The volume and temperature of the system are maintained constant. What fraction of the $N_2O_4$ is converted to $NO_2$?
6. At 800 K, the Gibbs free energies for formation of methanol and formaldehyde, in their hypothetical ideal gas standard states, are –88.063 and –87.893 $\mathrm{kJ}\ {\mathrm{mol}}^{-1}$, respectively.
(a) At 800 K, what is the value of the equilibrium constant for the reaction $CH_3OH\rightleftharpoons CH_2O+H_2$?
(b) Initially, one mole of pure methanol occupies 1 ${\mathrm{m}}^3$ at 800 K. The volume and temperature of this system are maintained constant. Assume that the presence of a catalyst allows this reaction to occur selectively. How many moles of formaldehyde are present at equilibrium? What fraction of the methanol is converted to formaldehyde?
(c) The initial system in (b) is allowed to equilibrate at constant pressure and temperature. What fraction of the methanol is converted to formaldehyde?
(d) Initially, one mole of hydrogen and one mole of methanol occupy 1 ${\mathrm{m}}^3$ at 800 K. The volume and temperature of this system are maintained constant. What fraction of the methanol is converted to formaldehyde at equilibrium?
7. At 298.15 K, the Gibbs free energies of formation of $Br_2$ and $BrCl$, in their hypothetical ideal gas standard states, are +3.1 and –1.0 $\mathrm{kJ}\ {\mathrm{mol}}^{-1}$, respectively. Why is ${\Delta }_fG^o\left(Br_2,{HIG}^o,298.15\ \mathrm{K}\right)$ not zero? A 1 ${\mathrm{m}}^3$ vessel initially contains 0.100 mole of $Br_2$ and 0.200 mole of ${Cl}_2$ at 298.15 K. The volume and temperature are maintained constant. What are the partial pressures of $Br_2$, ${Cl}_2$, and $BrCl$ at equilibrium?
8. Devise a non-isothermal process for the reversible mixing of two gases. (Suppose that one of the substances can be condensed to a liquid, whose vapor pressure is negligible, at a temperature at which the other substance remains a gas.)
9. The following thermodynamic data are available: ${\Delta }_fG^o\left({SO}_2,\mathrm{g},298.15\ \mathrm{K}\right)=-300.1\ \mathrm{kJ}\ {\mathrm{mo}l}^{-1} \nonumber$ ${\Delta }_fG^o\left({SO}_3,\mathrm{g},298.15\ \mathrm{K}\right)=-371.1\ \mathrm{kJ}\ {\mathrm{mo}l}^{-1} \nonumber$ ${\Delta }_fH^o\left({SO}_2,\mathrm{g},298.15\ \mathrm{K}\right)=-296.8\ \mathrm{kJ}\ {\mathrm{mo}l}^{-1} \nonumber$ ${\Delta }_fH^o\left({SO}_3,\mathrm{g},298.15\ \mathrm{K}\right)=-395.7\ \mathrm{kJ}\ {\mathrm{mo}l}^{-1} \nonumber$ (a) What are ${\Delta }_rG^o$ and ${\Delta }_rH^o$ for the reaction $2\ SO_2+O_2\rightleftharpoons 2\ {SO}_3$ at 298.15 K?
(b) Estimate the temperature, $T_0$, at which the standard Gibbs free energy change for this reaction is zero.
(c) A mixture is prepared to contain 1.00 mole of ${SO}_2$ and 1.00 mole of $O_2$. The mixture is allowed to equilibrate at $T_0$ and a pressure of 10.0 bar. What fraction of the ${SO}_2$ is converted to ${SO}_3$?
(d) A mixture is prepared to contain 1.00 mole of ${SO}_2$ and 10.0 mole of $O_2$. The mixture is allowed to equilibrate at $T_0$ and a pressure of 10.0 bar. What fraction of the ${SO}_2$ is converted to ${SO}_3$?
10. At 400 K, the Gibbs free energies of formation for carbon monoxide and methanol are $-\mathrm{146.341\ kJ}\ {\mathrm{mol}}^{-1}$ and $-\mathrm{148.509}\ \mathrm{kJ}\ {\mathrm{mol}}^{-1}$, respectively. What is ${\Delta }_rG^o$ for the reaction $CO+2\ H_2\rightleftharpoons CH_3OH$? A mixture of carbon monoxide and hydrogen in the stoichiometric proportions is prepared. Assume that the presence of a catalyst allows this reaction to occur selectively. When this mixture reaches equilibrium at 400 K and 1.00 bar, what fraction of the carbon monoxide is converted to methanol?
11. At 500 K, the Gibbs free energies of formation for carbon monoxide and methanol are $-155.412\ \mathrm{kJ}\ {\mathrm{mol}}^{-1}$ and $-134.109\ \mathrm{kJ}\ {\mathrm{mol}}^{-1}$, respectively. What is ${\Delta }_rG^o$ for the reaction $CO+2\ H_2\rightleftharpoons CH_3OH$? Assume that the presence of a catalyst allows this reaction to occur selectively. What ratio of hydrogen to carbon monoxide must be charged to a reactor in order for 50% of the carbon monoxide to be converted to methanol when the system reaches equilibrium at a pressure of 100.0 bar?
12. The reaction of carbon monoxide with water to produce carbon dioxide and hydrogen, $CO+H_2O\rightleftharpoons CO_2+H_2$, is known as “the water-gas shift reaction.” It is used in the commercial manufacture of hydrogen. The Gibbs free energy change for this reaction becomes less favorable as the temperature increases. At 1100 K, the standard Gibbs free energy change for this reaction is ${\Delta }_rG^o=+0.152\ \mathrm{kJ}\ {\mathrm{mol}}^{-1}$. Equal numbers of moles of carbon monoxide and water are charged to a reactor. What fraction of the water is converted to hydrogen when the system reaches equilibrium at 1100 K and 10.0 bar?
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/13%3A_Equilibria_in_Reactions_of_Ideal_Gases/13.10%3A_Problems.txt
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We develop the thermodynamic criteria for change in a closed system without specifying the composition of the system. It is clear, therefore, that the validity of the results is not restricted in any way by the composition of the system. For the most part, our development proceeds as if the system is a single substance. However, the validity of the thermodynamic criteria for change is independent of whether the system comprises a single substance in a single phase or multiple substances in multiple phases. The criteria for reversible change are independent of whether the change involves interconversions among the substances comprising the system, so long as these interconversions occur reversibly.
While we develop these criteria without specifying the composition of the system undergoing change, we use them to make predictions about processes in which system compositions do change. In Chapter 12, we apply the Gibbs free energy criterion for reversible change to equilibria between two phases of a pure substance. This application is successful because the thermodynamic properties of one pure phase are independent of how much of any other pure phase is present.
In Chapter 13, we find a relationship between the standard Gibbs free energy change and the equilibrium constant for a reaction of ideal gases. We also use the equation for the Gibbs free energy of an ideal gas to find the value of the Gibbs free energy change for a spontaneous reaction of ideal gases at a constant temperature. This is a noteworthy result. Because we can find this value, we can predict the spontaneous transformation of one non-equilibrium state to a second non-equilibrium state. These applications are successful because ideal gas molecules do not interact with one another; the laws describing the behavior of an ideal gas do not depend on the properties of other substances that may be present.
In general, the behavior of a system depends on the molecular characteristics of all of the components and on their concentrations. For example, the pressure at which pure ice is in equilibrium with a salt solution depends not only on the temperature but also on the choice of salt and its concentration. When it is expressed as a function of reagent concentrations, the equilibrium constant for a reaction of real gases varies to some extent with the composition and with the total pressure of the system. Such observations are consistent with the idea that the thermodynamic properties of a system depend on the attractive and repulsive forces among its molecules. Such forces act between any pair of real molecules, whether they are molecules of the same substance or molecules of different substances.
Our treatment of ideal-gas reactions shows that, if we can model the effects of compositional changes on the values of a system’s thermodynamic functions, we can use our thermodynamic theory to predict spontaneous changes in chemical composition. To extend to real substances the treatment that we develop for ideal gases in Chapter 13, we must find general relationships between a system’s thermodynamic properties and its chemical composition. These relationships must reflect the dependence of the thermodynamic properties of one substance on the molecular characteristics of the other substances present. To find them, we must introduce additional inferences and assumptions. Since we want to describe spontaneous processes, these ideas must apply to non-equilibrium systems.
To introduce these ideas, let us consider an open system that can undergo a spontaneous change. We begin by considering its energy. We want to model any change in the energy, $dE$, of this system as a function of a sufficient set of independent variables. Thus far, we have developed equalities that relate thermodynamic functions only for closed systems undergoing reversible change. Consequently, when we now seek an equation for $dE$ that is valid for an open, spontaneously changing system, our task is one of scientific inference, not mathematical deduction from the reversible-process equations. Nevertheless, the reversible-process equations are our primary resource.
For a closed, reversible system, we interpret the fundamental equation to mean that $dS$ determines the effect of thermal processes and that $dV$ determines the effect of pressure–volume work on the energy of the system. We infer that entropy and volume changes will play these same roles in the description of open, spontaneously changing systems. If $\lambda$ different kinds of non-pressure–volume work are possible, the energy of the system is also a function of $\lambda$ generalized extensive thermodynamic variables, ${\theta }_1$, ${\theta }_2$,, ${\theta }_k$,, ${\theta }_{\lambda }$. If the system contains $\omega$ substances, adding or removing some amount of any of these substances changes the energy of the system. Evidently, the energy of the system is a function of the number of moles of each substance present. Let $n_1$, $n_2$,,$\ n_{\omega }$, represent the number of moles of each component in the system. We can recognize these dependencies formally by writing
$E=E\left(S,V,{\theta }_1\mathrm{,\ }{\theta }_2\mathrm{,\dots ,}{\theta }_{\lambda },n_1\mathrm{,\ }n_2\mathrm{,\dots ,}\ n_{\omega }\right) \nonumber$
This formalism applies to both open and closed systems. If the system is open, all of these variables are independent; any one of them can be changed independently of the others. If the system is closed, changes in the $n_j$ are not independent, because the mass of the system is constant. (Letting $\overline{M}_j$ be the molar mass of the j-th substance, we have $0=\sum^{\omega }_1{\overline{M}_j}dn_j$.) If the energy is a continuous and differentiable function of each of these variables, the total differential becomes
\begin{aligned} dE= \left(\frac{\partial E}{\partial S}\right)_{V,{\theta }_m,n_j} & dS+\left(\frac{\partial E}{\partial V}\right)_{S,{\theta }_m,n_j}dV \ ~ & +\sum^{\lambda }_{k=1} \left(\frac{\partial E}{\partial {\theta }_k}\right)_{S,V,{\theta }_{m\neq k},n_j}d{\theta }_k \ ~ & +\sum^{\omega }_{j=1} \left(\frac{\partial E}{\partial n_j}\right)_{S,V,{\theta }_m,n_{p\neq j}}dn_j \end{aligned} \nonumber
(Inclusion of “${\theta }_{m\neq k}$” and “$n_{p\neq j}$” in the subscripted variable lists indicates that the partial taken with respect to any ${\theta }_k$ is taken with the other $\theta$-values held constant, and the partial taken with respect to any of the $n_j$ is taken with the other $n$-values held constant.)
We hypothesize that this total differential describes energy changes in any system whose energy, entropy, volume, work variables, ${\theta }_k$, and composition variables, $n_j$, vary in a continuous manner. In particular, we hypothesize that it describes the energy change that results from any change in composition, specified by the ${dn}_j$, whether the system is open or closed. All of these partial derivatives are intensive variables. Each of them specifies how much the energy of the system changes when its associated extensive variable undergoes an incremental change. We find it convenient to refer to the partial derivatives as potentials. In this book, we restrict our attention to systems in which a given potential has the same value at every location within the system.
Since ${\left(\frac{\partial E}{\partial n_j}\right)}_{S,V,{\theta }_m,n_{p\neq j}} \nonumber$
is the energy change that occurs when the amount of the j-th chemical substance changes by ${dn}_j$, we call this partial derivative a chemical potential. We introduce the new symbol ${\mu }_j$ to denote the chemical potential of the j-th substance; that is,
${\mu }_j={\left(\frac{\partial E}{\partial n_j}\right)}_{S,V,{\theta }_m,n_{p\neq j}} \nonumber$
The value of ${\mu }_j$ depends exclusively on the properties of the system. When the amount of the j-th substance changes by ${dn}_j$, the contribution to the energy change is ${dE={\mu }_jdn}_j$. When the system is open, the amount of the j-th substance can change either because of a process that occurs entirely within the system or because some of the substance moves across the system’s boundary.
Reversible processes in closed systems
In Section 9.14, we find, for a reversible process in a closed system, that $dE=TdS-PdV+dw_{NPV}$, where
$dw_{NPV}=\sum^{\lambda }_{k=1}{{\mathit{\Phi}}_k}d{\theta }_k \nonumber$
In our hypothesized total differential, the contribution that composition changes make to $dE$ is
$\sum^{\omega }_{j=1} \left(\frac{\partial E}{\partial n_j}\right)_{S,V,{\theta }_m,n_{p\neq j}}dn_j=\sum^{\omega }_{j=1} {\mu }_j{dn}_j \nonumber$
From the empirical fact that the energy change for a reversible process in a closed system can be specified by specifying $dS$, $dV$, and the $d{\theta }_k$, it follows that
$\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}=0 \nonumber$
for any such process. Implicitly using this fact in Section 10.1, we find that ${\left({\partial E}/{\partial S}\right)}_V=T$ and ${\left({\partial E}/{\partial V}\right)}_S=-P$ for reversible processes in which all of the work is pressure–volume work. We can extend the argument to a reversible process involving any form of work in a closed system. It follows that
${\left(\frac{\partial E}{\partial S}\right)}_{V,{\theta }_m,n_j}=T \nonumber$ ${\left(\frac{\partial E}{\partial V}\right)}_{S,{\theta }_m,n_j}=-P \nonumber$ ${\left(\frac{\partial E}{\partial {\theta }_k}\right)}_{S,V,{\theta }_{m\neq k},n_j}={\mathit{\Phi}}_k \nonumber$ and $\sum^{\lambda }_{k=1}{{\left(\frac{\partial E}{\partial {\theta }_k}\right)}_{S,V,{\theta }_{m\neq k},n_j}d{\theta }_k}=dw_{NPV} \nonumber$
Substituting into
\begin{aligned} dE= \left(\frac{\partial E}{\partial S}\right)_{V,{\theta }_m,n_j} & dS+ \left(\frac{\partial E}{\partial V}\right)_{S,{\theta }_m,n_j}dV \ ~ & +\sum^{\lambda }_{k=1} \left(\frac{\partial E}{\partial {\theta }_k}\right)_{S,V,{\theta }_{m\neq k},n_j}d{\theta }_k \ ~& +\sum^{\omega }_{j=1} \left(\frac{\partial E}{\partial n_j}\right)_{S,V,{\theta }_m,n_{p\neq j}}dn_j \end{aligned} \nonumber
we have
$dE=TdS-PdV+dw_{NPV}+\sum^{\omega }_{j=1}{{\mu }_j{dn}_j} \nonumber$
with
$\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}=0 \nonumber$
for any reversible process in a closed system.
Reversible processes in open systems
Let us suppose that an open system is undergoing a reversible change and that we instantaneously stop any further exchange of matter between the system and its surroundings. Before this action, the system is open; thereafter, it is closed. The act of closure does not change the system’s state functions. If, after closure, the system continues to change reversibly, $dE=TdS-PdV+dw_{NPV}$ must be valid. Closure of the system introduces a constraint that reduces the number of independent variables. However, $S$, $V$, and the ${\theta }_k$ remain independent after closure, and the act of closure cannot alter the laws that describe the interaction of the system with its surroundings during a reversible process. The relationships ${\left({\partial E}/{\partial S}\right)}_V=T$, ${\left({\partial E}/{\partial V}\right)}_S=-P$, and $\sum^{\lambda }_{k=1}{{\left({\partial E}/{\partial {\theta }_k}\right)}_{S,V,{\theta }_{m\neq k},n_j}d{\theta }_k}=dw_{NPV}$ are valid for the closed system. We infer that they are also valid for the open system, so that the total differential for a system undergoing reversible change is
$dE=TdS-PdV+dw_{NPV}+\sum^{\omega }_{j=1}{{\mu }_j{dn}_j} \nonumber$
whether the system is open or closed. This comparison implies that $\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}=0 \nonumber$
for any reversible process, whether the system is open or closed. (As noted below, however, our model implicitly assumes that ${\mu }_j={\widehat{\mu }}_j$; that is, ${\mu }_j$ has the same value whether the change ${dn}_j$ is the result of internal transformation, movement of the j-th substance across the system boundary, or a combination of the two.)
Spontaneous processes
We hypothesize that the same mathematical model describes energy changes for both spontaneous and reversible processes in both open and closed systems, so long as the process occurs in such a way that all of the system’s state functions behave as continuous variables. For reversible systems, we conclude that this model is $dE=TdS-PdV+dw_{NPV}+\sum^{\omega }_{j=1}{{\mu }_j{dn}_j} \nonumber$
By hypothesis, the model also describes spontaneous processes, so long as the state functions behave as continuous variables. We base the development of our thermodynamic theory for spontaneous reactions on this hypothesis. It is, of course, the utility of the resulting theory, not the logical force of the reasoning by which we reach it, that justifies our acceptance of the hypothesis.
In Section 9.19, we find that ${\left(dE\right)}_{SV}<{dw}_{NPV}$ for a spontaneous process, and we reach this conclusion by an argument that is independent of the composition of the system. For a spontaneous process at constant entropy and volume, our model becomes
${\left(dE\right)}_{SV}=dw_{NPV}+\sum^{\omega }_{j=1}{{\mu }_j{dn}_j} \nonumber$
and since ${\left(dE\right)}_{SV}<{dw}_{NPV}$, it follows that
$\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}<0 \nonumber$
for any spontaneous process.
All of the variables that appear in our model for $dE$ are properties of the system. The surroundings can affect the behavior of the system by imposing particular values on one or more of the system’s thermodynamic functions. (For example, immersing a reacting system in a constant-temperature bath ensures that the system reaches equilibrium only when $T=\hat{T}$.) When the surroundings impose particular values on the system’s thermodynamic functions, the system’s state functions must approach the imposed values as the system approaches equilibrium.
Internal entropy and chemical potential
Substituting the internal and external entropies that we introduce in Section 9.16, our model for the incremental energy change during a spontaneous composition change becomes $dE=Td_iS+dq+{dw}_{PV}+dw_{NPV}+\sum^{\omega }_{j=1}{{\mu }_j{dn}_j} \nonumber$
Our first-law equation is $dE=dq+{dw}_{PV}+dw_{NPV}$. It follows that $Td_iS+\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}=0 \nonumber$
and we find $d_iS=-\frac{1}{T}\sum^{\omega }_{j=1}{{\mu }_j{dn}_j} \nonumber$ The change criteria $d_iS\ge 0 \nonumber$ and $\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}\le 0 \nonumber$ are equivalent.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/14%3A_Chemical_Potential_-_Extending_the_Scope_of_the_Fundamental_Equation/14.01%3A_Dependence_of_the_Internal_Energy_on_the.txt
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Using the appropriate sets of independent variables, we can obtain similar expressions for $dH$, $dA$, $dG$, and $dS$ under the conditions that we assume in Section 14.1. Since entropy and pressure are the natural variables for enthalpy in a closed system, we infer that a change in the enthalpy of any system can be expressed as a function of $dS$, $dP$, changes in the non-pressure–volume work variables, $d{\theta }_k$, and changes in the composition, ${dn}_j$. That is, from
$H=H\left(S,P,{\theta }_1\mathrm{,\ }{\theta }_2\mathrm{,\dots ,}{\theta }_{\lambda },n_1\mathrm{,\ }n_2\mathrm{,\dots ,}\ n_{\omega }\right) \nonumber$
we have
\begin{aligned} dH= \left(\frac{\partial H}{\partial S}\right)_{P,{\theta }_m,n_j}dS & +\left(\frac{\partial H}{\partial P}\right)_{S,{\theta }_m,n_j}dP \ ~ & +\sum^{\lambda }_{k=1} \left(\frac{\partial H}{\partial {\theta }_k}\right)_{S,P,{\theta }_{m\neq k},n_j}d{\theta }_k \ ~ & +\sum^{\omega }_{j=1} \left(\frac{\partial H}{\partial n_j}\right)_{S,P,{\theta }_m,n_{p\neq j}}dn_j \end{aligned} \nonumber
From $H=E+PV$, we have
$dH=dE+PdV+VdP \nonumber$
for any system. Substituting our result from Section 14.1.
$dE=TdS-PdV+dw_{NPV}+\sum^{\omega }_{j=1} {\mu }_j dn_j \nonumber$
where
${\mu }_j=\left(\frac{\partial E}{\partial n_j}\right)_{S,V,{\theta }_m,n_{p\neq j}} \nonumber$
we have
$dH=TdS+VdP+dw_{NPV}+\sum^{\omega }_{j=1} {\mu }_j dn_j \nonumber$
Since we assume that both of these equations for $dH$ describe open systems, all of the variables are independent, and the corresponding coefficients must be equal. For systems that satisfy our assumptions, we have
$T={\left(\frac{\partial H}{\partial S}\right)}_{P,{\theta }_m,n_j} \nonumber$ $V={\left(\frac{\partial H}{\partial P}\right)}_{S,{\theta }_m,n_j} \nonumber$ ${dw}_{NPV}=\sum^{\lambda }_{k=1}{{\left(\frac{\partial H}{\partial {\theta }_k}\right)}_{S,V,{\theta }_{m\neq k},n_j}d{\theta }_k} \nonumber$ and
${\mu }_j={\left(\frac{\partial H}{\partial n_j}\right)}_{S,P,{\theta }_m,n_{p\neq j}}={\left(\frac{\partial E}{\partial n_j}\right)}_{S,V,{\theta }_m,n_{p\neq j}} \nonumber$
By parallel arguments, we find
$dA=-SdT-PdV+dw_{NPV}+\sum^{\omega }_{j=1} {\mu }_j{dn}_j \nonumber$
with
${\mu }_j= \left(\frac{\partial A}{\partial n_j}\right)_{T,V,{\theta }_m,n_{p\neq j}} \nonumber$
and
$dG=-SdT+VdP+dw_{NPV}+\sum^{\omega }_{j=1} {\mu }_j{dn}_j \nonumber$
with
${\mu }_j= \left(\frac{\partial G}{\partial n_j}\right)_{T,P,{\theta }_m,n_{p\neq j}} \nonumber$
Rearranging the result for $dE$, we find
$dS=\frac{1}{T}dE+\frac{P}{T}dV-\frac{dw_{NPV}}{T}-\frac{1}{T}\sum^{\omega }_{j=1}{{\mu }_j{dn}_j} \nonumber$
so that ${\mu }_j=-T{\left(\frac{\partial S}{\partial n_j}\right)}_{E,V,{\theta }_m,n_{p\neq j}} \nonumber$
and from the result for $dH$, we have
$dS=\frac{1}{T}dH-\frac{V}{T}dP-\frac{dw_{NPV}}{T}-\frac{1}{T}\sum^{\omega }_{j=1}{{\mu }_j{dn}_j} \nonumber$
so that ${\mu }_j=-T{\left(\frac{\partial S}{\partial n_j}\right)}_{H,P,{\theta }_m,n_{p\neq j}} \nonumber$
Evidently, for any system whose condition corresponds to our assumptions, all of the relationships that we develop in Section 10.1 remain valid. The non-pressure–volume work can be expressed in terms of partial derivatives in several equivalent ways:
\begin{aligned} {dw}_{NPV} & =\sum^{\lambda }_{k=1}{{\left(\frac{\partial E}{\partial {\theta }_k}\right)}_{S,V,{\theta }_{m\neq k},n_j}d{\theta }_k} \ ~ & =\sum^{\lambda }_{k=1}{{\left(\frac{\partial H}{\partial {\theta }_k}\right)}_{S,P,{\theta }_{m\neq k},n_j}d{\theta }_k} \ ~ & =\sum^{\lambda }_{k=1}{{\left(\frac{\partial A}{\partial {\theta }_k}\right)}_{V,T,{\theta }_{m\neq k},n_j}d{\theta }_k} \ ~ & =\sum^{\lambda }_{k=1}{{\left(\frac{\partial G}{\partial {\theta }_k}\right)}_{P,T,{\theta }_{m\neq k},n_j}d{\theta }_k} \ ~ & =-T\sum^{\lambda }_{k=1}{{\left(\frac{\partial S}{\partial {\theta }_k}\right)}_{E,V,{\theta }_{m\neq k},n_j}d{\theta }_k} \ ~ & =-T\sum^{\lambda }_{k=1}{{\left(\frac{\partial S}{\partial {\theta }_k}\right)}_{H,P,{\theta }_{m\neq k},n_j}d{\theta }_k} \end{aligned} \nonumber
Most importantly, the chemical potential and the criteria for change can be expressed in several alternative ways. We have
\begin{aligned} {\mu }_j & ={\left(\frac{\partial E}{\partial n_j}\right)}_{S,V,{\theta }_m,n_{p\neq j}} & ={\left(\frac{\partial H}{\partial n_j}\right)}_{S,P,{\theta }_m,n_{p\neq j}} \ ~ & ={\left(\frac{\partial A}{\partial n_j}\right)}_{V,T,{\theta }_m,n_{p\neq j}} & ={\left(\frac{\partial G}{\partial n_j}\right)}_{P,T,{\theta }_m,n_{p\neq j}} \ ~ & =-T{\left(\frac{\partial S}{\partial n_j}\right)}_{E,V,{\theta }_m,n_{p\neq j}} & =-T{\left(\frac{\partial S}{\partial n_j}\right)}_{H,P,{\theta }_m,n_{p\neq j}} \end{aligned} \nonumber
For any system, open or closed, that satisfies the assumptions we introduce in §1, the criteria for change are
$\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}\le 0 \nonumber$
where the chemical potential is equivalently expressed as any of the partial derivatives above.
Let us review the scope and significance of these results. We develop the fundamental equation,$\ dE=TdS-PdV+dw_{NPV}$, by reasoning about the behavior of closed, reversible systems. For any process in a closed system, we develop the criteria for change
${\left(\Delta E\right)}_{SV}\le w_{NPV} \nonumber$ ${\left(\Delta H\right)}_{SP}\le w_{NPV} \nonumber$ ${\left(\Delta A\right)}_{VT}\le w_{NPV} \nonumber$ ${\left(\Delta G\right)}_{PT}\le w_{NPV} \nonumber$ ${\left(\Delta S\right)}_{EV}\ge {-w_{NPV}}/{\hat{T}} \nonumber$ ${\left(\Delta S\right)}_{HP}\ge {-w_{NPV}}/{\hat{T}} \nonumber$
and the corresponding relationships among differentials for incremental changes. Now we are extending these conclusions to produce equations for the changes in the various thermodynamic functions when a system undergoes a spontaneous composition change. To do so, we introduce the idea that composition variables must be included in a complete model for a thermodynamic function. When we write
$E=E\left(S,V,{\theta }_1\mathrm{,\ }{\theta }_2\mathrm{,\dots ,}{\theta }_{\lambda },n_1\mathrm{,\ }n_2\mathrm{,\dots ,}\ n_{\omega }\right) \nonumber$
we assert that every set of values, $\left\{S,V,{\theta }_1\mathrm{,\ }{\theta }_2\mathrm{,\dots ,}{\theta }_{\lambda },n_1\mathrm{,\ }n_2\mathrm{,\dots ,}\ n_{\omega }\right\}$, corresponds to a state of the system in which the system has a definite energy. (Of course, we are not asserting that every such set of values can actually be attained by the real system. Many such sets correspond to hypothetical states. The particular sets of values that do correspond to realizable states of the system lie on a manifold. Given one such set of values—one point on the manifold—our differential expressions specify all of the other states that lie on the same manifold.)
When we assume that the various partial derivatives, ${\left({\partial E}/{\partial S}\right)}_{V,{\theta }_k,n_j}$, ${\left({\partial E}/{\partial V}\right)}_{S,{\theta }_k,n_j}$, ${\left({\partial E}/{\partial {\theta }_k}\right)}_{S,V{\theta }_{m\mathrm{\neq }k},n_j}$, and ${\left({\partial E}/{\partial n_j}\right)}_{S,V,{\theta }_k,n_{p\mathrm{\neq }j}}$, exist, we are assuming that $E$ is a smooth, continuous function of each of these variables. Since some of these partial derivatives are synonymous with intensive variables, we are assuming that these intensive variables are well defined. Moreover, since these intensive variables characterize the system as a whole, we are assuming that each intensive variable has the same value in every part of the system${}^{1}$. When we assume that ${\mu }_j = {\left({\partial E}/{\partial n_j}\right)}_{S,V,{\theta }_m,n_{p\mathrm{\neq }j}}$ has a well-defined value, we are assuming that $E$ is a continuous function of $n_j$; we are assuming that ${\mu }_j$ exists for any arbitrary state of the system and not just for states undergoing reversible change.
When we assume that an arbitrary change is described by the total differential $dE=TdS-PdV+dw_{NPV}+\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}$, we are going beyond our conclusion that this total differential describes paths of reversible change. We are asserting that it describes any process in which a change in composition is the sole source of irreversibility. As a practical matter, we expect it to describe any process that occurs in a system whose potential functions are well defined. By well defined, we mean, of course, that that they can be measured and that the measurements are reproducible and consistent. We expect these conclusions to apply to multiple-phase, open systems, so long as each potential has the same value in every phase. In thus assuming that we can expand the scope of the fundamental equation, we are not modifying the change criteria that we develop in Chapter 9. Our conclusions that ${\left(dE\right)}_{SV}={dw}_{NPV}$ for a reversible process and ${\left(dE\right)}_{SV}<{dw}_{NPV}$ for a spontaneous process in a closed system are not affected.
Again, while our arguments for them are compelling, these results are not rigorously logical consequences of our earlier conclusions about reversible processes. As for any scientific principle, their validity depends on their predictive capability, not their provenance.
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Because they are easy to control in typical laboratory experiments, pressure, temperature, and the number of moles of each component are the independent variables that we find useful most often. Partial derivatives of thermodynamic quantities, taken with respect to the number of moles of a component, at constant pressure, temperature, and ${\boldsymbol{\theta }}_{\boldsymbol{k}}$, are given a special designation; they are called partial molar quantities. That is,
${\left(\frac{\partial E}{\partial n_A}\right)}_{P,T,{\theta }_m,n_{p\neq A}} \nonumber$
is the partial molar energy of component $A$,
${\left(\frac{\partial G}{\partial n_A}\right)}_{P,T,{\theta }_m,n_{p\neq A}} \nonumber$
is the partial molar Gibbs free energy, etc. All partial molar quantities are intensive variables.
Because partial molar quantities are particularly useful, it is helpful to have a distinctive symbol to represent them. We use a horizontal bar over a thermodynamic variable to represent a partial molar quantity. (We have been using the horizontal over-bar to mean simply per mole. When we use it to designate a partial molar quantity, it means per mole of a specific component.) Thus, we write
${\left(\frac{\partial E}{\partial n_A}\right)}_{P,T,{\theta }_m,n_{p\neq A}}={\overline{E}}_A \nonumber$ ${\left(\frac{\partial V}{\partial n_A}\right)}_{P,T,{\theta }_m,n_{p\neq A}}={\overline{V}}_A \nonumber$ ${\left(\frac{\partial G}{\partial n_A}\right)}_{P,T,{\theta }_m,n_{p\neq A}}={\overline{G}}_A \nonumber$ etc.
In Sectiona 14.1 and 14.2, we introduce the chemical potential for substance $A$, ${\mu }_A$, and find that the chemical potential of substance $A$ is equivalently expressed by several partial derivatives. In particular, we have
${\mu }_A={\left(\frac{\partial G}{\partial n_A}\right)}_{P,T,{\theta }_m,n_{p\neq A}}={\overline{G}}_A \nonumber$
that is, the chemical potential is also the partial molar Gibbs free energy.
It is important to recognize that the other partial derivatives that we can use to calculate the chemical potential are not partial molar quantities. Thus,
${\mu }_A={\left(\frac{\partial E}{\partial n_A}\right)}_{S,V,{\theta }_m,n_{p\neq A}}\neq {\left(\frac{\partial E}{\partial n_A}\right)}_{P,T,{\theta }_m,n_{p\neq A}} \nonumber$
That is, ${\mu }_A\neq {\overline{E}}_A$. Similarly, ${\mu }_A\neq {\overline{H}}_A$, ${\mu }_A\neq {\overline{A}}_A$, and ${\mu }_A\neq -T{\overline{S}}_A$.
We can think of a thermodynamic variable as a manifold—a “surface” in a multidimensional space. If there are two independent variables, the dependent thermodynamic variable is a surface in a three-dimensional space. Then we can visualize the partial derivative of the dependent thermodynamic variable with respect to an independent variable as the slope of a line tangent to the surface. This tangent lies in a plane in which the other independent variable is constant. If the independent variables are pressure, temperature, ${\theta }_k$-values, and compositions, the slope of the tangent line at $\left(P,T,{\theta }_1\mathrm{,\ }{\theta }_2\mathrm{,\dots ,}{\theta }_{\lambda },n_1\mathrm{,\ }n_2\mathrm{,\dots ,}\ n_{\omega }\right)$ is the value of a partial molar quantity at that point.
A more concrete way to think of a partial molar quantity for component $A$ is to view it as the change in that quantity when we add one mole of $A$ to a very large system having the specified pressure, temperature, ${\theta }_k$-values, and composition. When we add one mole of A to this system, the relative change in any of the system’s properties is very small; for example, the ratio of the final volume to the initial volume is essentially unity. Nevertheless, the volume of the system changes by a finite amount. This amount approximates the partial molar volume of substance $A$. This approximation becomes better as the size of the system becomes larger. We expect the change in the volume of the system to be approximately equal to the volume of one mole of pure $A$, but we know that in general it will be somewhat different because of the effects of attractive and repulsive forces between the additional $A$ molecules and the molecules comprising the original system.
Partial molar quantities can be expressed as functions of other thermodynamic variables. Because pressure and temperature are conveniently controlled variables, functions involving partial molar quantities are particularly useful for describing chemical change in systems that conform to the assumptions that we introduce in §1. Because the chemical potential is the same thing as the partial molar Gibbs free energy, it plays a prominent role in these equations.
To use these equations to describe a real system, we must develop empirical models that relate the partial molar quantities to the composition of the system. In general, these empirical models are non-linear functions of the system composition. However, simple approximations are sometimes adequate. The simplest approximation is a case we have already considered. If we can ignore the attractive and repulsive interactions among the molecules comprising the system, the effect of increasing $n_A$ by a small amount, ${dn}_A$, is simply the effect of adding $dn_A$ moles of pure component $A$ to the system. If we let ${\overline{E}}^{\textrm{⦁}}_A$ be the energy per mole of pure component $A$, the contribution to the energy of the system, at constant temperature and pressure, is
${\left(\frac{\partial E}{\partial n_A}\right)}_{P,T,{\theta }_m,n_{p\neq A}}dn_A={\overline{E}}^{\textrm{⦁}}_A\mathrm{\ }dn_A \nonumber$
In Chapter 12, we apply the thermodynamic criteria for change to the equilibria between phases of a pure substance. To do so, we use the Gibbs free energies of the pure phases. In Chapter 13, we apply these criteria to chemical reactions of ideal gases, using the Gibbs free energies of the pure gases. In these cases, the properties of a phase of a pure substance are independent of the amounts of any other substances that are present. That is, we use the approximation
${\left(\frac{\partial G}{\partial n_A}\right)}_{P,T,{\theta }_m,n_{p\neq A}}dn_A={\overline{G}}^{\textrm{⦁}}_A\mathrm{\ }dn_A \nonumber$
albeit without using the over-bar or the bullet superscript to indicate that we are using the partial molar Gibbs free energy of the pure substance. In Section 14.1, we develop the principle that $\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}\le 0$ are general criteria for change that are applicable not only to closed systems but also to open systems composed of homogeneous phases.
Thus far in this chapter, we have written each partial derivative with a complete list of the variables that are held constant. This is typographically awkward. Clarity seldom requires that we include the work-related variables, ${\theta }_k$, and composition variables, $n_j$, in this list. From here on, we usually omit them.
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Let us now apply our chemical-potential equilibrium criterion, $\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}=0$, to a simple, closed system in which phases $\alpha$ and $\beta$ of a single component are at equilibrium. The total number of moles, $n$, is fixed; we have $n=n_{\alpha }+n_{\beta }$. Reversible change is possible while the system remains at equilibrium. However, there is a stoichiometric constraint; growth in one phase is exactly matched by shrinkage in the other; d$n=dn_{\alpha }+{dn}_{\beta }=0$, so that $dn_{\alpha }=-{dn}_{\beta }$. The equilibrium criterion becomes$\left({\mu }_{\alpha }-{\mu }_{\beta }\right){dn}_{\alpha }=0$. If reversible change occurs, we have ${dn}_{\alpha }\neq 0$, and it follows that ${\mu }_{\alpha }={\mu }_{\beta }$.
We infer that this result must always be valid. That is, if a substance is present in two phases of an equilibrium system, the chemical potential of the substance is the same in each phase; no matter what other processes involving the substance may occur. This is so because all processes that occur in the system must be at equilibrium in order for the system as a whole to be at equilibrium. In our discussion of chemical kinetics, we saw this conclusion expressed as the principle of microscopic reversibility. In our development of Gibbs phase rule, we asserted that a distribution-equilibrium constant relates the concentrations of a species in any two phases. Our present conclusion that the chemical potential of a species must be the same in each phase is a more rigorous statement of the same principle.
For a process in which $dn$ moles of a substance transfer spontaneously from phase $\beta$ to phase $\alpha$, we have $dn=dn_{\alpha }=-{dn}_{\beta }$. The criterion for spontaneous change, $\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}<0$, becomes
${\mu }_{\alpha }dn_{\alpha }+{\mu }_{\beta }{dn}_{\beta }={\mu }_{\alpha }dn_{\alpha }-{\mu }_{\beta }{dn}_{\alpha }<0 \nonumber$
and it follows that ${\mu }_{\alpha }<{\mu }_{\beta }$. We infer that this result, too, must always be valid. That is, if a substance spontaneously transfers between any two phases, the chemical potential of the transferring substance must be greater in the donor phase than it is in the acceptor phase.
While $\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}\le 0$ summarizes our criteria for change, it does not specify the process that occurs in the system. In order to describe a particular chemical reaction, we must incorporate the constraints that the reaction stoichiometry imposes on the $dn_j$. For a description of the equilibrium system that does not require the size of the system to be specified, the position of equilibrium must be given as a function of intensive variables. While the ${\mu }_j$ are intensive variables, the $dn_j$ are not. When we include information about the stoichiometry of the equilibrium process, we obtain a criterion for equilibrium that is specific to that process.
Consider the reaction $a\ A+b\ B\rightleftharpoons c\ C+d\ D$ in a closed system. Let $d\xi$ be the incremental extent of reaction. Using the modified stoichiometric coefficients, ${\nu }_j$, that we introduce in Section 13.5, the incremental extent of reaction becomes
$d\xi =\frac{{dn}_A}{{\nu }_A}=\frac{{dn}_B}{{\nu }_B}=\frac{{dn}_C}{{\nu }_C}=\frac{{dn}_D}{{\nu }_D} \nonumber$ and we can use the stoichiometric coefficients to express the incremental composition changes, $dn_j$, as $dn_j={\nu }_jd\xi$. The criteria for change become
$\sum^{\omega }_{j=1}{{\mu }_j{\nu }_jd\xi }\le 0 \nonumber$
For the reaction to occur left to right, we must have $d\xi >0$, so that $\sum^{\omega }_{j=1}{{\mu }_j{\nu }_j}\le 0 \nonumber$
In this form, the criteria involve only intensive variables, and the stoichiometry of the equilibrium process is uniquely specified. In most circumstances, there are additional relationships among the ${\mu }_j$ for which we may have little or no specific information.
To see the nature of these additional relationships, let us consider a chemical reaction involving three species, $A+B\rightleftharpoons \ C$, at constant pressure and temperature. No other substances are present. There are two components; so there are three degrees of freedom. Pressure and temperature account for two of the degrees of freedom. The chemical potential of any one of the equilibrating species, say $A$, accounts for the third. Evidently, fixing pressure, temperature, and ${\mu }_A$ is sufficient to fix the values of the remaining chemical potentials, ${\mu }_B$ and ${\mu }_C$. ${\mu }_A$ must contain information about the other chemical potentials, ${\mu }_B$ and ${\mu }_C$.
If the same reaction occurs in the presence of a solvent, there are three components and four degrees of freedom. In this case, two of the chemical potentials can vary independently. Fixing pressure, temperature, and, say ${\mu }_A$ and ${\mu }_B$, fixes ${\mu }_C$ and the chemical potential of the solvent. We can expect the chemical potential of the solvent to be nearly constant, and we often omit it when we write $dG$ for this system. This omission notwithstanding, the chemical potentials of the equilibrating species include information about the chemical potential of the solvent. In general, the chemical potential of any component of an equilibrium system is a function of the chemical potentials of all of the other components.
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We conclude that $\sum^{\omega }_{j=1}{{\mu }_j{dn}_j}=0$ is a criterion for reversible change in any system. When the change involves equilibria among two or more phases or substances, it alters the number of moles of the components present. An extent of reaction, $\xi ={\left(n_j-n^o_j\right)}/{{\nu }_j}$, characterizes the displacement of every such equilibrium. The magnitude of each incremental equilibrium displacement is specified by composition changes, $dn_j={\nu }_jd\xi$, and conversely. The criterion for reversible change becomes $\sum^{\omega }_{j=1}{{\mu }_j{\nu }_j}d\xi =0$. When this criterion is satisfied because $\sum^{\omega }_{j=1}{{\mu }_j{\nu }_j}=0$, $d \xi$ is arbitrary, and the system can reversibly traverse a range of equilibrium states. In other words, $\sum^{\omega }_{j=1}{{\mu }_j{\nu }_j}=0$ defines a Gibbsian equilibrium manifold.
We can also have a reversible process for which $d\xi =0.$ If the process is reversible, the state of the system corresponds to a point on the Gibbsian manifold, but $d\xi =0$ stipulates that the system cannot change: it must remain at the specified point on the manifold. This corresponds to what we are calling a primitive equilibrium state. The system is constrained to remain in this state by the nature of its interactions with its surroundings: the system may be isolated, or the surroundings may act to maintain the system in a fixed state.
14.06: The Change Criteria in A System Composed
Let us now consider a closed, constant temperature, constant pressure system that is composed of open subsystems. Chemical substances can pass from one subsystem to another, but they cannot enter or leave the system. We assume that our model for $dG$ applies in every subsystem. Each subsystem is at the same temperature and pressure. For the r-th subsystem,
$G_r=G_r\left(P,T,{\theta }_{r,1} ~ {\theta }_{r,2}, ~ \cdots ,~ {\theta }_{r,\lambda },n_{r,1},\ n_{r,2}, \cdots ,\ n_{r,\omega }\right) \nonumber$
For a physical system in which all of these assumptions correspond closely to physical reality, we have, for the r-th subsystem,
$dG_r=-S_rdT+V_rdP+\left(dw_{NPV}\right)_r+\sum^{\omega }_{j=1}{\mu }_j dn_{r,j} \nonumber$
For the closed system, we have
$\sum_r{dG_r}=-dT\sum_r S_r +dP\sum_r V_r +\sum_r \left(dw_{NPV}\right)_r +\sum_r \sum^{\omega }_{j=1} {\mu }_j dn_{r,j} \nonumber$
Since Gibbs free energy, entropy, volume, and work are extensive variables, we have, for the closed system, $dG=\sum_r{dG_r}$, $dS=\sum_r{S_r}$, $V=\sum_r{V_r}$, and $dw_{NPV}=\sum_r \left(dw_{NPV}\right)_r$. Therefore,
$dG=-SdT+VdP+dw_{NPV}+\sum_r \sum^{\omega }_{j=1} {\mu }_j dn_{r,j} \nonumber$
For any process that occurs in this closed system at constant pressure and temperature, we have ${\left(dG\right)}_{PT}\le dw_{NPV}$, and
$\sum_r \sum^{\omega }_{j=1} {\mu }_j dn_{r,j} \le 0 \nonumber$
expresses the criteria for change in the closed system as a sum of conditions on the open subsystems.
Now let us consider the possibility that, for the $\rho$-th open subsystem, we have
$\sum^{\omega }_{j=1} {\mu }_j dn_{\rho ,j}>0 \nonumber$
If this were true, the sum over all of the subsystems could still be less than or equal to zero. In this case, the energy increase occurring in the $\rho$-th subsystem would have to be offset by energy decreases occurring in the other subsystems. This is at odds with the way that physical systems are observed to behave. To see this, let us suppose that the process is a chemical reaction. Then the composition changes are related to the extent of reaction as $dn_{\rho ,j}={\nu }_jd{\xi }_{\rho }$. For the open subsystem, we have
$\sum^{\omega }_{j=1} {\mu }_j{\nu }_jd{\xi }_{\rho }>0 \nonumber$
Now, we can alter the boundary of this subsystem to make it impermeable to matter, while keeping its state functions unchanged. This change converts the open subsystem to a closed system, for which we know that
$\sum^{\omega }_{j=1} {\mu }_j{\nu }_jd{\xi }_{\rho }<0 \nonumber$
If the criterion for spontaneous change switches from $\sum^{\omega }_{j=1} {\mu }_j{\nu }_jd{\xi }_{\rho }>0$ to $\sum^{\omega }_{j=1} {\mu }_j{\nu }_jd{\xi }_{\rho }<0$ the sign of $d{\xi }_{\rho }$ must change. The supposition that $\sum^{\omega }_{j=1} {\mu }_j{\nu }_jd{\xi }_{\rho }>0$ is possible in an open subsystem implies that the direction of a spontaneous change in a closed system can be opposite the direction of a spontaneous change in an otherwise identical open system. No such thing is ever observed. We conclude that the criterion for spontaneous change,
$\sum^{\omega }_{j=1} {\mu }_j dn_j<0 \nonumber$
must be satisfied in every part of any system in which the various potentials are the same throughout. Since
$d_iS=-\frac{1}{T}\sum^{\omega }_{j=1}{{\mu }_j{dn}_j} \nonumber$
it follows that $d_iS>0$ must also be satisfied in every part of the system.
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At constant temperature and pressure, the chemical potential of component $A$ is the contribution that one mole of $A$ makes to the Gibbs free energy of the system. The Gibbs free energy of the system is the sum of the contributions made by all of its components. The chemical potential change that occurs during a reaction,
$\mathrm{\ }{\Delta }_r\mu = \sum^{\omega }_{j=1}{{\mu }_j{\nu }_j} \nonumber$
is the same thing as the Gibbs free energy change. To establish these points, we introduce Euler’s theorem on homogenous functions. For simplicity, we consider systems in which only pressure–volume work occurs.
A function $f\left(x,y,z\right)$ is said to be homogeneous of order $n$ if
${\lambda }^nf\left(x,y,z\right)=f\left(\lambda x,\lambda y,\lambda z\right) \nonumber$ Then,
$\frac{d}{d\lambda }\left[{\lambda }^nf\left(x,y,z\right)\right]=n{\lambda }^{n-1}f\left(x,y,z\right)={\left(\frac{\partial f}{\partial \left(\lambda x\right)}\right)}_{yz}\frac{d\left(\lambda x\right)}{d\lambda }+{\left(\frac{\partial f}{\partial \left(\lambda y\right)}\right)}_{xz}\frac{d\left(\lambda y\right)}{d\lambda }+{\left(\frac{\partial f}{\partial \left(\lambda z\right)}\right)}_{xy}\frac{d\left(\lambda z\right)}{d\lambda } \nonumber$ $=x{\left(\frac{\partial f}{\partial \left(\lambda x\right)}\right)}_{yz}+y{\left(\frac{\partial f}{\partial \left(\lambda y\right)}\right)}_{xz}+z{\left(\frac{\partial f}{\partial \left(\lambda z\right)}\right)}_{xy} \nonumber$
Since this must be true for any $\lambda$, it must be true for $\lambda =1$. Making this substitution, we have Euler’s theorem for order $n$:
$nf\left(x,y,z\right)=x{\left(\frac{\partial f}{\partial x}\right)}_{yz}+y{\left(\frac{\partial f}{\partial y}\right)}_{xz}+z{\left(\frac{\partial f}{\partial z}\right)}_{xy} \nonumber$
An extensive state function is homogeneous of order one in its extensive variables. In particular, we have
$\lambda G\left(P,T,n_A,n_B,n_C,n_D\right)=G\left(P,T,{\lambda n}_A,{\lambda n}_B,{\lambda n}_C,{\lambda n}_D\right) \nonumber$
for any $\lambda$. (For, say, $\lambda ={1}/{2}$, this says only that, if we divide a homogeneous equilibrium system into two equal portions, all else remaining constant, the Gibbs free energy of each portion will be half of that of the original system. The pressure and temperature are independent of $\lambda$.) Taking the derivative with respect to $\lambda$, we find
$\frac{d}{d\lambda }\left[\lambda G\left(P,T,n_A,n_B,n_C,n_D\right)\right]=G\left(P,T,n_A,n_B,n_C,n_D\right)=n_A{\left(\frac{\partial G}{\partial \left(\lambda n_A\right)}\right)}_{PT}+n_B{\left(\frac{\partial G}{\partial \left(\lambda n_B\right)}\right)}_{PT}+n_C{\left(\frac{\partial G}{\partial \left(\lambda n_c\right)}\right)}_{PT}+n_D{\left(\frac{\partial G}{\partial \left(\lambda n_D\right)}\right)}_{PT} \nonumber$
Since this must be true for any $\lambda$, it must be true for $\lambda =1$. We find
$G\left(P,T,n_A,n_B,n_C,n_D\right)={\mu }_An_A+{\mu }_Bn_B+{\mu }_Cn_C+{\mu }_Dn_D \nonumber$
The same equation follows from applying Euler’s theorem to other state functions. For example, viewing the internal energy as a function of entropy, volume, and composition, we have
$\lambda E\left(S,V,n_A,n_B,n_C,n_D\right)=E\left(\lambda S,\lambda V,{\lambda n}_A,{\lambda n}_B,{\lambda n}_C,{\lambda n}_D\right) \nonumber$
so that
$\frac{d}{d\lambda }\left[\lambda E\left(S,V,n_A,n_B,n_C,n_D\right)\right]=E\left(S,V,n_A,n_B,n_C,n_D\right)=S{\left(\frac{\partial E}{\partial \left(\lambda S\right)}\right)}_V+V{\left(\frac{\partial E}{\partial \left(\lambda V\right)}\right)}_S+n_A{\left(\frac{\partial E}{\partial \left(\lambda n_A\right)}\right)}_{SV}+n_B{\left(\frac{\partial E}{\partial \left(\lambda n_B\right)}\right)}_{SV}+n_C{\left(\frac{\partial E}{\partial \left(\lambda n_c\right)}\right)}_{SV}+n_D{\left(\frac{\partial E}{\partial \left(\lambda n_D\right)}\right)}_{SV} \nonumber$
Setting $\lambda =1$, we have
$E\left(S,V,n_A,n_B,n_C,n_D\right) \nonumber$ $=TS-PV+{\mu }_An_A+{\mu }_Bn_B+{\mu }_Cn_C+{\mu }_Dn_D \nonumber$
which we can rearrange to $G=E-TS+PV={\mu }_An_A+{\mu }_Bn_B+{\mu }_Cn_C+{\mu }_Dn_D \nonumber$
This equation describes any system whose thermodynamic functions are continuous functions of one another. Evidently, we can model the thermodynamic functions of any such system by modeling the chemical potentials of its components. In the remainder of this chapter, we develop this idea.
By definition, ${\Delta }_rG$ is the Gibbs free energy change for converting $a$ moles of $A$ and $b$ moles of $B$ to $c$ moles of $C$ and $d$ moles of $D$ at constant pressure and temperature and while the composition remains constant at $n_A$, $n_B$, $n_C$, and $n_D$. That is,
${\Delta }_rG=G\left(n_A,n_B,n_C+c,n_D+d\right)-G\left(n_A+a,n_B+b,n_C,n_D\right) \nonumber$
where $n_A\gg a$, $n_B\gg b$, $n_C\gg c$, and $n_D\gg d$. Using the relationship we have just found between $G$ and the $n_i$, we have
${\Delta }_rG=\left[{\mu }_An_A+{\mu }_Bn_B+{\mu }_C\left(n_C+c\right)+{\mu }_D\left(n_D+d\right)\right] \nonumber$ $-\left[{\mu }_A\left(n_A+a\right)+{\mu }_B\left(n_B+b\right)+{\mu }_Cn_C+{\mu }_Dn_D\right] \nonumber$ $=c{\mu }_C+d{\mu }_D-a{\mu }_A-d{\mu }_d \nonumber$ $={\Delta }_r\mu \nonumber$
Since any other extensive thermodynamic function is also homogeneous of order one in the composition variables, a similar relationship will exist between the change in the function itself and its partial molar derivatives. For example, expressing the system volume as a function of pressure, temperature, and composition, we have
$\lambda V\left(P,T,n_A,n_B,n_C,n_D\right)=V\left(P,T,{\lambda n}_A,{\lambda n}_B,{\lambda n}_C,{\lambda n}_D\right) \nonumber$
and a completely parallel derivation shows that the volume of the system is related to the partial molar volumes of the components by
$V\left(P,T,n_A,n_B,n_C,n_D\right)={\overline{V}}_An_A+{\overline{V}}_Bn_B+{\overline{V}}_Cn_C+{\overline{V}}_Dn_D \nonumber$
The volume change for the reaction is given by
${\Delta }_rV=c{\overline{V}}_C+d{\overline{V}}_D-a{\overline{V}}_A-b{\overline{V}}_B \nonumber$
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An important relationship among the differentials of the chemical potentials for a system follows from the relationships we have just developed. From the fact that the Gibbs free energy, $G\left(P,T,n_A,n_B,n_C,n_D\right)$, is homogeneous of order one in the composition variables, we find that the Gibbs free energy of the system is related to its partial molar derivatives by
$G\left(P,T,n_A,n_B,n_C,n_D\right)={\mu }_An_A+{\mu }_Bn_B+{\mu }_Cn_C+{\mu }_Dn_D\nonumber$
The differential of the left hand side is
$dG=-SdT+VdP+{\mu }_Adn_A+{\mu }_Bdn_B+{\mu }_Cdn_C+{\mu }_Ddn_D\nonumber$
and the differential of the right hand side is
$dG={\mu }_Adn_A+n_Ad{\mu }_A+{\mu }_Bdn_B+n_Bd{\mu }_B+{\mu }_Cdn_C+n_Cd{\mu }_C+{\mu }_Ddn_D+n_Dd{\mu }_D\nonumber$
Since these differential expressions must be equal, we have
$-SdT+VdP=n_Ad{\mu }_A+n_Bd{\mu }_B+n_Cd{\mu }_C+n_Dd{\mu }_D\nonumber$
for any change in this system.
While we have considered the particular case of a system containing the species $A$, $B$, $C$, and $D$, it is clear that the same arguments apply to any system. For a system that contains $\omega$ species, we can write the result in general form as
$-SdT+VdP=\sum^{\omega }_{j=1}{n_jd{\mu }_j}\nonumber$
This relationship is known as the chemical-potential Gibbs-Duhem equation. It is a constraint on the $d{\mu }_j$ that must be satisfied when any change occurs in a system whose thermodynamic functions are continuous functions of its composition variables. If pressure and temperature are constant and this equation is satisfied, the system is at equilibrium—it is on a Gibbsian equilibrium manifold—and the chemical-potential Gibbs-Duhem equation becomes
$0=\sum^{\omega }_{j=1}{n_jd{\mu }_j}\nonumber$
In the next two sections, we develop a particularly useful expression for $d{\mu }_j$. We can obtain similar relationships for other partial molar quantities. (See problems 14.2 and 14.3.) These relationships are also called Gibbs-Duhem equations. Because the derivation requires only that the thermodynamic function be homogeneous of order one, the same relationships exist among the differentials of the partial molar derivatives of any extensive thermodynamic function. For partial molar volumes at constant pressure and temperature, we find
$0=\sum^{\omega }_{j=1}{n_jd{\overline{V}}_j}\nonumber$
14.09: The Dependence of Chemical Potential on
The chemical potential of a substance in a particular system is a function of all of the variables that affect the Gibbs free energy of the system. For component $A$, we can express this by writing
${\mu }_A={\mu }_A\left(P,T,n_1,n_2,\dots ,n_A,\dots {,n}_{\omega }\right) \nonumber$
for which the total differential is
$d{\mu }_A={\left(\frac{\partial {\mu }_A}{\partial T}\right)}_PdT+{\left(\frac{\partial {\mu }_A}{\partial P}\right)}_TdP+\sum^{\omega }_{j=1}{{\left(\frac{\partial {\mu }_A}{\partial n_j}\right)}_{PT}dn_j} \nonumber$
Recalling the definition of the chemical potential and the fact that the mixed second-partial derivatives of a state function are equal, we have
${\left(\frac{\partial {\mu }_A}{\partial T}\right)}_P={\left(\frac{\partial }{\partial T}\right)}_P{\left(\frac{\partial G}{\partial n_A}\right)}_{TP}={\left(\frac{\partial }{\partial n_A}\right)}_{TP}{\left(\frac{\partial G}{\partial T}\right)}_P=-{\left(\frac{\partial S}{\partial n_A}\right)}_{TP}=-{\overline{S}}_A \nonumber$ Similarly,
${\left(\frac{\partial {\mu }_A}{\partial P}\right)}_T={\left(\frac{\partial }{\partial P}\right)}_T{\left(\frac{\partial G}{\partial n_A}\right)}_{TP}={\left(\frac{\partial }{\partial n_A}\right)}_{TP}{\left(\frac{\partial G}{\partial P}\right)}_T={\left(\frac{\partial V}{\partial n_A}\right)}_{TP}={\overline{V}}_A \nonumber$ Thus, the total differential of the chemical potential for species $A$ can be written as
$d{\mu }_A=-{\overline{S}}_AdT+{\overline{V}}_AdP+\sum^{\omega }_{j=1}{{\left(\frac{\partial {\mu }_A}{\partial n_j}\right)}_{PT}dn_j} \nonumber$
To illustrate the utility of this result, we can use it to derive the Clapeyron equation for equilibrium between two phases of a pure substance. In Chapter 12, we derived the Clayeyron equation using a thermochemical cycle. We can now use the total differential of the chemical potential to present essentially the same derivation using a simpler argument. Letting the two phases be $\alpha$ and $\beta$, the total differentials for a system that contains both phases becomes
$d{\mu }_{\alpha }=-{\overline{S}}_{\alpha }dT+{\overline{V}}_{\alpha }dP+{\left(\frac{\partial {\mu }_{\alpha }}{\partial n_{\alpha }}\right)}_{PT}dn_{\alpha }+{\left(\frac{\partial {\mu }_{\alpha }}{\partial n_{\beta }}\right)}_{PT}dn_{\beta } \nonumber$ and $d{\mu }_{\beta }=-{\overline{S}}_{\beta }dT+{\overline{V}}_{\beta }dP+{\left(\frac{\partial {\mu }_{\beta }}{\partial n_{\alpha }}\right)}_{PT}dn_{\alpha }+{\left(\frac{\partial {\mu }_{\beta }}{\partial n_{\beta }}\right)}_{PT}dn_{\beta } \nonumber$
Since equilibrium between phases $\alpha$ and $\beta$ means that ${\mu }_{\alpha }={\mu }_{\beta }$, we have also that $d{\mu }_{\alpha }=d{\mu }_{\beta }$ for any process in which the phase equilibrium is maintained. Moreover, $\alpha$ and $\beta$ are pure phases, so that ${\mu }_{\alpha }$ and ${\mu }_{\beta }$ are independent of $n_{\alpha }$ and $n_{\beta }$. Then
${\left(\frac{\partial {\mu }_{\alpha }}{\partial n_{\alpha }}\right)}_{PT}={\left(\frac{\partial {\mu }_{\beta }}{\partial n_{\alpha }}\right)}_{PT}={\left(\frac{\partial {\mu }_{\alpha }}{\partial n_{\beta }}\right)}_{PT}={\left(\frac{\partial {\mu }_{\beta }}{\partial n_{\beta }}\right)}_{PT}=0 \nonumber$
Hence,
$-{\overline{S}}_{\alpha }dT+{\overline{V}}_{\alpha }dP=-{\overline{S}}_{\beta }dT+{\overline{V}}_{\beta }dP \nonumber$
and the rest of the derivation follows as before.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/14%3A_Chemical_Potential_-_Extending_the_Scope_of_the_Fundamental_Equation/14.08%3A_Gibbs-Duhem_Equation.txt
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In Section 5.17, we use mathematical models for the rates of chemical reactions to find the equilibrium constant equation, which specifies the equilibrium position of a reacting system. The variables in the reaction-rate models are the concentrations of reactants and products. Consequently, these concentrations are also the variables in the equilibrium constant equation. We develop our reaction-rate models from simple arguments about the dependence of collision frequencies on reagent concentrations. These arguments ignore the possibility that the effects of intermolecular forces on collision frequencies can change as the composition of the system changes. For a great many purposes, the concentration-based models describe experimental results with sufficient accuracy. However, as we note in our introductory discussion in Section 1.6, the concentration-based equilibrium constant equation proves to be only approximately constant: To obtain rigorously accurate results, we must introduce new quantities, having the character of “corrected concentrations”, that we call activities.
We are now in a position to define the chemical activity of a substance more precisely. We introduce the activity concept in the particular context of chemical equilibria, but its scope is broader. The activity of a component is a wholly new thermodynamic quantity. We view it as the property that determines the chemical potential, $\mu$, of a substance in a particular system at a given temperature. As such, the activity directly links the properties of the substance to the behavior of the system. Introducing the activity makes our equations for $\mu$ and $d\mu$ more compact. It also provides an avenue to relate qualitative ideas about the properties of the substance—ideas that we may not be able to express in mathematical models—to the effects that the substance has on the properties of the system.
The heuristic idea that an activity is a corrected concentration arises in the first instance because we use the pressure-dependence of the Gibbs free energy of an ideal gas as the model for our definition of activity. In Section 15.7, we find that the equilibrium constant for a chemical reaction can be specified using the activities of the reactants and products and that the form of the equilibrium constant is the same for activities as it is for concentrations. In Chapter 16 we see that assuming ideal behavior for solutions enables us to equate the activity and the mole fraction of a component in such solutions.
Our consideration of the chemical reactions of gases in Chapter 13 introduces the basic ideas that we use: The Gibbs free energy of reaction is the same thing as the change in the chemical potential of the system; a reaction reaches equilibrium when the Gibbs free energy of the system cannot decrease further; the Gibbs free energy of reaction is a function of the chemical potentials of the reactants and products; and the chemical potential of a gas is a function of its partial pressure. In ideal gas mixtures, the chemical potential of a component depends on the partial pressure of that component, but not on the partial pressures of other species that may be present. The model we develop for ideal gases becomes unsatisfactory for real gases under the kinds of conditions that cause the experimentally observed properties of gases to depart from ideal gas behavior—conditions in which intermolecular forces cannot be neglected.
In Section 14.9, we observe that the chemical potential of substance $A$ can be expressed as a function of temperature, pressure, and its molar composition: $\mu_A\left(P,T,n_1,n_2,\dots ,n_A,\dots {,n}_{\omega }\right) \nonumber$ The total differential becomes
$d\mu_A=-{\overline{S}}_AdT+{\overline{V}}_AdP+\sum^{\omega }_{j=1}{{\left(\frac{\partial \mu_A}{\partial n_j}\right)}_{PT}dn_j} \nonumber$ We can write this as
$d\mu_A=-{\overline{S}}_AdT+{\overline{V}}_AdP+{\left(d\mu_A\right)}_{PT} \nonumber$ where
${\left(d\mu_A\right)}_{PT}=\sum^{\omega }_{j=1}{{\left(\frac{\partial \mu_A}{\partial n_j}\right)}_{PT}dn_j} \nonumber$
In general, the change in the chemical potential of $A$ depends on partial derivatives with respect to the amount of every substance present. The dependence on the amount of substance $j$ is given by
${\left(\frac{\partial \mu_A}{\partial n_j}\right)}_{P,T,n_{m\neq j}} \nonumber$
and the dependence on the amount of $A$ is given by
${\left(\frac{\partial \mu_A}{\partial n_A}\right)}_{P,T,n_{m\neq A}} \nonumber$
In a multi-component system whose components do not interact, ${\left({\partial \mu_A}/{\partial n_A}\right)}_{P,T,n_{m\neq A}}$ is not zero, but all of the other partial derivatives ${\left({\partial \mu_A}/{\partial n_j}\right)}_{P,T,n_{m\neq j}}$ must, by definition, vanish. This is exemplified in an ideal-gas system, in which the chemical potential of component A is given by
$\mu_A\left(P_A\right)={\Delta }_fG^o\left(A,P^o\right)+RT{ \ln \left({P_A}/{P^o}\right)\ } \nonumber$
and the partial pressure of $A$ depends only on the mole fraction of $A$ in the system, $P_A=x_AP_{system}$. For an ideal gas at constant temperature, changes in the chemical potential do not depend on the amounts of other substances that may be present:
${\left({d\mu }_A\right)}_T=RT{\left(d{ \ln P_A\ }\right)}_T \nonumber$
In the special case that the system pressure is constant, this becomes:
${\left({d\mu }_A\right)}_T=RT{\left(d{ \ln x_A\ }\right)}_T \nonumber$
foreshadowing our eventual recognition that mole fraction is the most natural concentration unit for theoretical modeling of the chemical potential.
In real systems, the chemical potential of a component depends on its own concentration, but it also depends—more weakly—on the concentrations of the other species present. To treat real systems adequately, we need a general method to express the chemical potential of a component as a function of the component’s concentration, in a way that fully accounts for the effects of the other species. Our grand strategy is to develop the activity of a component as an abstract, dimensionless quantity. We do this by choosing a simple function to define the activity of a component in terms of its chemical potential. This makes it possible to express the thermodynamic properties of any system as relatively simple functions of the activities of its components. To apply these results to specific systems, we must then find empirical equations that express the component activities as functions of component concentrations. So, from one perspective, the activity is merely a convenient intermediary in our overall effort to express the chemical potential as a function of composition.
To begin expressing this strategy in mathematical notation, let us represent the activity and concentration of a component, $A$, as ${\tilde{a}}_A$ and $c_A$, respectively. For simplicity, we consider systems in which pressure–volume work is the only work. The chemical potential is a function of pressure, temperature, and composition; that is $\mu_A=\mu_A\left(P,T,c_A,c_B,c_C,\dots \right)$. We suppose that we have a large volume of such data available; that is, we have measured $\mu_A$ at many conditions represented by widely varying values of the variables $P,T,c_A,c_B,c_C,\dots$. Our strategy is to find two empirical functions. The first empirical function expresses the chemical potential of $A$ as a function of its activity and the temperature of the system, $\mu_A=\mu_A\left(T,{\tilde{a}}_A\right)$. The second expresses the activity of $A$ as a function of the concentrations of the species present and of the pressure and temperature of the system, ${\tilde{a}}_A={\tilde{a}}_A\left(P,T,c_A,c_B,c_C,\dots \right)$. The mathematical composition of these two functions expresses the chemical potential as a function of $P,T,c_A,c_B,c_C,\dots$; that is, the mathematical composition is the function $\mu_A=\mu_A\left(P,T,c_A,c_B,c_C,\dots \right)$.
If we focus only on finding a function, $\mu_A\left(P,T,c_A,c_B,c_C,\dots \right)$, that fits the data, we can let the dependence of chemical potential on activity be anything we please. All we have to do is fix up the activity function, ${\tilde{a}}_A={\tilde{a}}_A\left(P,T,c_A,c_B,c_C,\dots \right)$, so that the mathematical composition of the two accurately reproduces the experimental data. This is just another way of saying that, if all we want is an empirical correlation of the form $\mu_A=\mu_A\left(P,T,c_A,c_B,c_C,\dots \right)$, we do not need to introduce the activity at all.
However, our goal is more exacting. We want to construct the activity function so that it behaves as much like a concentration as possible. In particular, we want the activity of a component to become equal to its concentration in cases where intermolecular interactions vanish—or become independent of the component’s concentration for some other reason. To achieve these objectives, we must choose the form of the function $\mu_A=\mu_A\left(T,{\tilde{a}}_A\right)$ carefully. Here the behavior of ideal gases provides a valuable template. Our observation that the chemical potential is related to the partial pressure by the differential expression ${\left({d\mu }_A\right)}_T=RT{\left(d{ \ln P_A\ }\right)}_T$ suggests that it might be possible to develop the activity function using a relationship of the same form. In fact, this strategy proves to be successful.
We relate the chemical potential of a component to its activity by the differential expression
${\left({d\mu }_A\right)}_T=RT{\left(d{ \ln {\tilde{a}}_A\ }\right)}_T \nonumber$
The activity becomes a single function that captures the dependence of ${\left({d\mu }_A\right)}_T$ on the number of moles of each of the components of the system:
$RT{\left(d{ \ln {\tilde{a}}_A\ }\right)}_T=\sum^{\omega }_{j=1}{{\left(\frac{\partial \mu_A}{\partial n_j}\right)}_{PT}dn_j} \nonumber$
If we hold the temperature constant, we can integrate between some arbitrary base state (in which the activity and chemical potential are ${\tilde{a}}^{\#}_A$ and $\mu^{\#}_A$) and a second state (in which these quantities are ${\tilde{a}}_A$ and $\mu_A$). That is, at any given temperature, we can select a particular state that we define to be the base state of A for the purpose of creating an activity scale. We denote the activity in the base state as ${\tilde{a}}^{\#}_A$. We let the chemical potential of $A$ be $\mu^{\#}_A$ when $A$ is in its base state at this temperature. At a fixed temperature, $\mu^{\#}_A$ is fixed. However, even if we keep all the other variables the same, the properties of the base state change when the temperature changes. This means that the chemical potential in the base state is a function of temperature—and only of temperature.
At constant temperature, the relationship between chemical potential and activity in any other state becomes
$\int^{\mu_A}_{\mu^{\#}_A} d\mu_A = RT\int^{a_A}_{a^{\#}_A} d \ln \tilde{a}_A \nonumber$
or
$\mu_A=\mu^{\#}_A+RT{ \ln \left(\frac{{\tilde{a}}_A}{\mathrm{\ }{\tilde{a}}^{\#}_A}\right)\ } \nonumber$
where it is implicit in the last equation that the activity is a function of pressure, temperature, and the chemical composition of the system. The temperature and chemical potential are state functions, so this expression requires that the activity be a state function also.
It is convenient to let the activity be unity in the base state. When we do this, we give the base state another name; we call it the activity standard state. When we stipulate that the activity in the base state is unity, we use a superscript zero to designate the activity and chemical potential. That is, in the standard state, ${\tilde{a}}^o_A=1$, and the value of the chemical potential of $A$ is ${\widetilde\mu}^o_A$ when $A$ is in this state. (In many cases, it turns out to be convenient to choose a standard state that does not correspond to any physical condition that can actually be achieved. In such cases, the standard state is a hypothetical system, whose properties we establish by mathematical extrapolation of measurements made on a real system in non-standard states.) With this convention, the chemical potential and activity are related by the equation
$\mu_A={\widetilde\mu}^o_A+RT{ \ln {\tilde{a}}_A\ } \nonumber$
at the specified temperature. We adopt this equation as our formal definition of the activity of component $A$. The activity and the chemical potential depend on the same variables and contain equivalent information.
Note that we have done nothing to restrict the state we choose to designate as the activity standard state. This creates opportunity for confusion, because it allows us to choose an activity standard state for $A$ in a particular system that is different from the standard state for pure $A$ that we define for the tabulation of thermodynamic data for pure substances. This means that we choose to let the meaning of the words “standard state” be context dependent. The practical effect of this ambiguity is that, whenever we are dealing with a chemical activity, we must be careful to understand what state of the substance is being designated as the standard state—and thereby being assigned chemical potential ${\widetilde\mu}^o_A$ and unit activity. Because the standard state is a fixed composition, ${\widetilde\mu}^o_A$ is a function of temperature, but not of pressure or composition. At a given temperature, changes in the system pressure or the concentration of $A$ affect the chemical potential, $\mu_A$, only by their effect on the activity, ${\tilde{a}}_A$.
To complete our program, we define a function that we call the activity coefficient, ${\gamma }_A\left(P,T,c_A,c_B,c_C,\dots \right)$, of component $A$, in a system characterized by variables $P,T,c_A,c_B,c_C,\dots$, by the equation
${\tilde{a}}_A\left(P,T,c_A,c_B,c_C,\dots \right)=c_A{\gamma }_A\left(P,T,c_A,c_B,c_C,\dots \right) \nonumber$
This definition places only one constraint on the form of ${\gamma }_A$; the function ${\gamma }_A\left(P,T,c_A,c_B,c_C,\dots \right)$ can be anything that adequately accounts for the experimental data, so long as ${\gamma }_A\to 1$ when the effects of intermolecular interactions are negligible.
Introducing the activity coefficient does not simplify the job of finding a suitable, empirical, activity function; it just imposes a condition on its form. In Chapters 15 and 16, we find that standard states are best described using mole fraction or molality as the concentration unit. (Note that mole fraction is dimensionless and molality is proportional to mole fraction for dilute solutes.)
In Section 16.6 we consider the use of molality when A is a solute in a solution whose other components are in fixed proportions to one another. At very low concentrations of $A$, the environment around every $A$ molecule is essentially the same. The chemical potential of $A$ is observed to be a linear function of ${ \ln {\underline{m}}_A\ }$, because a small increase in the concentration of $A$ does not significantly change the intermolecular interactions experienced by $A$ molecules. As the concentration of $A$ increases further, intermolecular interactions among $A$ molecules become increasingly important, and the chemical potential ceases to be a simple linear function of ${ \ln {\underline{m}}_A\ }$. The activity coefficient is no longer equal to one.
In the remainder of this chapter, we focus on some of the general properties of the activity function. In the next two chapters, we develop a few basic applications of these ideas. First, however, we digress to relate the ideas that we have just developed about activity to the ideas that we developed in Chapter 11 about fugacity.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/14%3A_Chemical_Potential_-_Extending_the_Scope_of_the_Fundamental_Equation/14.10%3A_Chemical_Activity.txt
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In Chapter 11, we introduce the fugacity as an alternative measure of the difference between the Gibbs free energy of one mole of a pure gas in its hypothetical ideal gas standard state and its Gibbs free energy in any other state at the same temperature. This definition makes the fugacity of a gas an intensive function of pressure and temperature. At a fixed temperature, the state of one mole of a pure gas is specified by its pressure, and the fugacity is a function of pressure only. Fugacity has the units of pressure. Giving effect to our decision to let the fugacity of the gas be unity when the gas is in its hypothetical ideal gas standard state $({HIG}^o)$ and using the Gibbs free energy of formation for the gas in this state as the standard Gibbs free energy for fugacity, we define the fugacity of a pure gas, $A$, by the equation
${\overline{G}}_A\left(P\right)={\Delta }_fG^o\left(A,{HIG}^o\right)+RT{ \ln \left[\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$
For substance $A$ in any system, the chemical potential is the partial molar free energy; that is, ${\mu }_A\left(P,T\right)={\overline{G}}_A\left(P,T\right)$. Since the Gibbs free energy of formation is defined for one mole of pure substance at a specified pressure and temperature, it is a partial molar quantity. When we elect to use the hypothetical ideal gas at a pressure of one bar as the standard state for the Gibbs free energy of formation of the gas, we also establish the Gibbs free energy of formation of the hypothetical ideal gas in its standard state as the standard-state chemical potential; that is, ${\mu }^o_A={\Delta }_fG^o\left(A,{HIG}^o\right)$. Hence, we can also express the fugacity of a gas by the equation
${\mu }_A\left(P\right)={\Delta }_fG^o\left(A,{HIG}^o\right)+RT{ \ln \left[\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$ or ${\mu }_A\left(P\right)={\mu }^o_A+RT{ \ln \left[\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$ (pure real gas)
For a mixture of real gases, we can extend the definition of fugacity in a natural way. We want the fugacity of a component gas to measure the difference between its chemical potential in the mixture, $\mu$, and its chemical potential in its standard state, ${\mu }^o$, where its standard state is the pure hypothetical ideal gas at one bar pressure. If gas $A$ is a component of a constant-temperature mixed-gas system, we have ${d\mu }_A={\overline{V}}_AdP$, where ${\overline{V}}_A$ is the partial molar volume of $A$ in the system, and $P$ is the pressure of the system. Let us find ${\mu }_A$ in a binary mixture that contains one mole of $A$ and $n_B$ moles of a second component, $B$. Let the partial molar volume of $B$ be ${\overline{V}}_B$. The system volume is $V={\overline{V}}_A+n_B{\overline{V}}_B$. The mole fractions of $A$ and $B$ are $x_A={1}/{\left(n_B+1\right)}$ and $x_B={n_B}/{\left(n_B+1\right)}$.
To find the change in ${\mu }_A$, we need a reversible process that takes one mole of $A$ in its standard state to a mixture of one mole $A$ with $n_B$ moles of $B$, in which the pressure of the mixture is $P$. The four-step process described in Figure 1 answers our requirements: One mole of $A$ and $n_B$ moles of $B$ are separately expanded from their hypothetical ideal gas standard states, at $P^o$, to the arbitrary low pressures $x_AP^*$ and $x_BP^*$, respectively. For this expansion, the change in the Gibbs free energy of one mole of $A$, which remains in its hypothetical ideal gas state, is
${\Delta }_{exps} \overline{G}_A=\int^{x_AP^*}_{P^o} \overline{V}^{\textrm{⦁}}_A dP= \int^{x_AP^*}_{P^o} \frac{RT}{P}dP=RT \ln \left(\frac{x_AP^*}{P^o}\right) \nonumber$
Next, these low-pressure ideal gases are merged to form a mixture of one mole of $A$ with $n_B$ moles of $B$ at the total pressure $P^*$. For this merging process, ${\Delta }_{merge}{\overline{G}}_A=0$. Then, we suppose that the ideal gases become real gases in a mixture whose pressure is $P^*$. Since this is merely a conceptual change, we have ${\Delta }_{concept}{\overline{G}}_A=0$. Finally, we compress the mixture of real gases from $P^*$ to an arbitrary pressure,$\ P$. Since the volume of the mixture is $V={\overline{V}}_A+n_B{\overline{V}}_B$, the Gibbs free energy change for this compression of the mixture is
${\Delta }_{comp}G_{mixture}=\int^P_{P^*}{V}dP=\int^P_{P^*}{{\overline{V}}_A}dP+n_B\int^P_{P^*}{{\overline{V}}_B}dP \nonumber$
We see that the Gibbs free energy change for the real-gas system is the sum of the Gibbs free energy changes for the components; we have
${\Delta }_{comp}{\overline{G}}_A=\int^P_{P^*}{{\overline{V}}_A}dP \nonumber$
For this process, we have
\begin{aligned} {\mu }_A-{\mu }^o_A & =RT \ln \left[\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)}\right] \ ~ & ={\Delta }_{exps} \overline{G}_A+{\Delta }_{merge} \overline{G}_A+{\Delta }_{concept} \overline{G}_A+{\Delta }_{comp} \overline{G}_A \ ~ & =RT \ln \left[\frac{x_AP^*}{P^o}\right] +\int^P_{P^*} \overline{V}_AdP-\int^P_{P^*} \frac{RT}{P}dP+\int^P_{P^*} \frac{RT}{P}dP \end{aligned} \nonumber
where we have added and subtracted the quantity
$\int^P_{P^*}{\frac{RT}{P}}dP \nonumber$
Dividing by $RT$ and evaluating the last integral, we find
\begin{aligned} \frac{{\mu }_A-{\mu }^o_A}{RT} & = \ln \left[\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)}\right] \ ~ & = \ln x_A\ + \ln P^*\ - \ln P^o +\int^P_{P^*} \left(\frac{\overline{V}_A}{RT}-\frac{1}{P}\right) dP+ \ln P - \ln P^* \end{aligned} \nonumber
$P^*$ is a finite pressure arbitrarily near zero. At very low pressures, real gas $A$ behaves as an ideal gas; hence, at very low pressures, the partial molar volume of the real gas is well approximated by the partial molar volume of pure gas $A$. That is, we have ${\overline{V}}_A\approx \overline{V^{\textrm{⦁}}_A}$, and
\begin{aligned} \int^{P^*}_0 \left(\frac{\overline{V}_A}{RT}-\frac{1}{P}\right) dP & \approx \int^{P^*}_0 \left(\frac{\overline{V^{\textrm{⦁}}_A}}{RT}-\frac{1}{P}\right)dP \ ~ & \approx 0 \end{aligned} \nonumber
where the approximation becomes exact in the limit as $P^*\to 0$. Simplifying the natural logarithm terms and expanding the integral, we obtain
\begin{aligned} \frac{{\mu }_A-{\mu }^o_A}{RT} & = \ln \left[\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)}\right] \ ~ & = \ln \frac{x_AP}{P^o} +\int^P_0 \left(\frac{\overline{V}_A}{RT}-\frac{1}{P}\right) dP-\int^{P^*}_0 \left(\frac{\overline{V}_A}{RT}-\frac{1}{P}\right)dP \ ~ & = \ln \frac{x_AP}{P^o} +\int^P_0 \left(\frac{\overline{V}_A}{RT}-\frac{1}{P}\right)dP \end{aligned} \nonumber
Defining the fugacity coefficient for $A$ in this mixture, ${\gamma }_A$, by ${\gamma }_A=\frac{f_A\left(P\right)}{x_AP} \nonumber$
and recalling that $f_A\left({HIG}^o\right)=P^o$, we use this result to find ${ \ln {\gamma }_A\ }={ \ln \left(\frac{f_A\left(P\right)}{x_AP}\right)=\ }\int^P_0{\left(\frac{{\overline{V}}_A}{RT}-\frac{1}{P}\right)}dP \nonumber$
This differs from the corresponding relationship for the fugacity of a pure gas only in that the partial molar volume is that of gas $A$ in a mixture with other gases. This is a trifling difference in principle, but a major difference in practice. To find the fugacity of pure $A$, we use the partial molar volume of the pure gas, which is readily calculated from any empirical pure-gas equation of state. However, to experimentally obtain the partial molar volume of gas $A$ in a gas mixture, we must collect pressure–volume–temperature data as a function of the composition of the system. If we contemplate creating a catalog of such data for the mixtures of even a modest number of compounds, we see that an enormous amount of data must be collected. Just the number of systems involving only binary mixtures is large. For $N$ compounds, there are $N{\left(N-1\right)}/{2}$ binary mixtures—each of which would have to be studied at many compositions in order to develop good values for the partial molar volumes.
Fortunately, practical experience shows that a simple approximation often gives satisfactory results. In this approximation, we assume that the partial molar volume of gas $A$—present at mole fraction $x_A$ in a system whose pressure is $P$— is equal to the partial molar volume of the pure gas at the same pressure. That is, for a binary mixture of gases $A$ and $B$, we assume
${\overline{V}}_A\left(P,x_A,x_B\right)=\overline{V^{\textrm{⦁}}_A}\left(P\right) \nonumber$
In this approximation, we have
$\frac{{\mu }_A-{\mu }^o_A}{RT}={ \ln \frac{x_AP}{P^o}\ }+\int^P_0{\left(\frac{\overline{V^{\textrm{⦁}}_A}}{RT}-\frac{1}{P}\right)}dP \nonumber$ and ${ \ln {\gamma }_A\ }={ \ln \left(\frac{f_A\left(P\right)}{x_AP}\right)=\ }\int^P_0{\left(\frac{\overline{V^{\textrm{⦁}}_A}}{RT}-\frac{1}{P}\right)}dP \nonumber$
We make the same assumption for gas $B$. From Euler’s theorem on homogeneous functions, we have $\overline{V}=x_A{\overline{V}}_A+x_B{\overline{V}}_B$. Therefore, in this approximation, we have
$\overline{V}=x_A{\overline{V}}_A+x_B{\overline{V}}_B\approx x_A\overline{V^{\textrm{⦁}}_A}+x_B\overline{V^{\textrm{⦁}}_B} \nonumber$
The last sum is the Amagat’s law representation of the molar volume of the gas mixture. We see that our approximation is equivalent to assuming that the system obeys Amagat’s law. Physically, this assumes that the gas mixture is an ideal (gaseous) solution. We discuss ideal solutions in Chapter 16. In an ideal solution, the intermolecular interactions between an $A$ molecule and a $B$ molecule are assumed to have the same effect as the interactions between two $A$ molecules or between two $B$ molecules. This differs from the ideal-gas assumption; there is no effect from the interactions between any two ideal-gas molecules.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/14%3A_Chemical_Potential_-_Extending_the_Scope_of_the_Fundamental_Equation/14.11%3A_Back_to_the_Fugacity-_the_Fugacity_of_A_.txt
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While we develop the fugacity concept by thinking about the Gibbs free energy of a pure real gas, our definition means that the fugacity of a substance in any system depends only on the difference between the chemical potential in that system and its chemical potential in the standard state. There is no reason to confine the definition of fugacity to gaseous systems. We generalize it: Our defining relationship specifies the fugacity of any substance in any system as a function of the difference between its chemical potential in that system and its chemical potential in its hypothetical ideal gas standard state.
Since the Gibbs free energy of an arbitrary system depends on the pressure, temperature, and composition of the system, the fugacity of any component also depends on these variables. However, there is an additional constraint on the fugacity. At a sufficiently low system pressure, the gases in any system behave ideally, and they obey Boyle’s law of partial pressures; the integral in the fugacity-coefficient equation becomes zero, and the fugacity coefficient becomes unity. In the limit as $P\to 0$, $f_A\to x_AP$, where $x_A$ is the mole fraction of $A$ in the gas phase. At a sufficiently low pressure, any gas mixture behaves ideally, and the fugacity of a constituent species becomes equal to its gas-phase partial pressure. (For a mixture of ideal gases, the fugacity of a component is always equal to its partial pressure.)
In Section 14.10, we define the activity of component A in an arbitrary system by the relationship
${\mu }_A={\widetilde{\mu }}^o_A+RT{ \ln \left[{\tilde{a}}_A\left(P,T,c_A,c_B,c_C,\dots \right)\right]\ } \nonumber$
where ${\widetilde{\mu }}^o_A$ is the chemical potential of $A$ in an activity standard state in which we stipulate that the activity of $A$ is unity. Since the defining equations for activity and for fugacity are formally identical, the distinction between activity and fugacity lies in our choices of standard states and in the facts that activity is dimensionless while fugacity has the units of pressure. If we use the hypothetical ideal-gas standard state for activity and measure the concentration in bars, the practical distinction between activity and fugacity vanishes. We can view the fugacity as a specialization of the activity concept.
In summary: The fugacity function proves to be a useful way to express the difference between the chemical potential of a substance in two different states. Fugacity is measured in bars with the hypothetical ideal gas standard state as the reference state. We add chemical activity to our list of useful thermodynamic properties because it extends the advantages of the fugacity representation to non-volatile components of systems that contain condensed phases. The standard state for activity can be any particular state of any convenient system that contains the substance. We define the activity of the substance in this reference system to be unity. (As with the hypothetical ideal gas standard state, we often find it useful to define a hypothetical system as the standard state for the activity.) The chemical potential of the substance in the standard-state system is, by definition, the standard chemical potential, ${\widetilde{\mu }}^o_A$, for this particular activity scale.
In the remainder of this chapter and in Chapters 15-17, we consider additional properties and applications of fugacity and chemical activity. Before doing so however, we digress to observe that we can choose an alternate definition of activity: chemical activity can be defined as a ratio of fugacities. Let us consider three systems that contain substance $A$.
The first is pure gas $A$ in its hypothetical ideal gas standard state at temperature $T$, $A\left({HIG}^o,T\right)$. In this state, the fugacity of $A$ is unity, $f_A\left({HIG}^o,T\right)=1\ \mathrm{bar}$. The chemical potential of $A$ is the standard-state chemical potential, which we equate to the Gibbs free energy of formation:
${\overline{G}}_A\left({HIG}^o,T\right)={\Delta }_fG^o\left(A,{HIG}^o,T\right)={\mu }^o_A \nonumber$
The second is a system that we define as the standard state for the activity of $A$ at temperature $T$. Apart from convenience, there is no reason to prefer any particular system for this role. We denote this system as $A\left(activity\ standard\ state,T\right)$ or, for short, $A\left(ss,T\right)$. By definition, the activity of $A$ in this state is unity, ${\tilde{a}}^o_A\left(ss\right)=1$, and the chemical potential of $A$ in this system is the standard chemical potential for this particular activity scale: ${\overline{G}}_A\left(ss,T\right)={\widetilde{\mu }}^o_A$. We denote the fugacity of $A$ in this state as $f_A\left(ss\right)$. By our definition of fugacity, we have
${\widetilde{\mu }}^o_A={\mu }^o_A+RT{ \ln \left[\frac{f_A\left(ss\right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$
The third is an arbitrary system that we denote as $A\left(P,T,c_A,c_B,c_C,\dots \right)$. We denote the chemical potential, fugacity, and activity of $A$ in this system as ${\overline{G}}_A\left(P,T,c_A,c_B,c_C,\dots \right)={\mu }_A\left(P,T,c_A,c_B,c_C,\dots \right)$, $f_A\left(P,T,c_A,c_B,c_C,\dots \right)$, and ${\tilde{a}}_A\left(P,T,c_A,c_B,c_C,\dots \right)$, respectively. By our definition of fugacity, we have
${\mu }_A\left(P,T,c_A,c_B,c_C,\dots \right)={\mu }^o_A+RT{ \ln \left[\frac{f_A\left(P,T,c_A,c_B,c_C,\dots \right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$
and by our definition of activity,
${\mu }_A\left(P,T,c_A,c_B,c_C,\dots \right)={\widetilde{\mu }}^o_A+RT{ \ln \left[{\tilde{a}}_A\left(P,T,c_A,c_B,c_C,\dots \right)\right]\ } \nonumber$
Figure 2 summarizes the relationships among the Gibbs free energies of these three states of substance $A$. From the cycle in Figure 2, we have ${\Delta }_1G+{\Delta }_3G={\Delta }_2G$, so that
$RT{ \ln \left[\frac{f_A\left(ss,T\right)}{f_A\left({HIG}^o\right)}\right]+RT{ln \left[\frac{f_A\left(P,T,c_A,c_B,\dots .\right)}{f_A\left(ss,T\right)}\right]\ }\ } \nonumber$ $=RT{ \ln \left[\frac{f_A\left(P,T,c_A,c_B,c_C,\dots \right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$ $=RT{ \ln \left[{\tilde{a}}_A\left(P,T,c_A,c_B,\dots .\right)\right]\ } \nonumber$
and ${\tilde{a}}_A\left(P,T,c_A,c_B,c_C,\dots \right)=\frac{f_A\left(P,T,c_A,c_B,c_C,\dots \right)}{f_A\left(ss,T\right)} \nonumber$
That is, the activity of a substance in a particular system is always equal to its fugacity in that system divided by its fugacity in the standard state for activity. From this relationship, it is evident that activity is always a dimensionless function of concentrations.
14.13: Relating the Differentials of Chemical P
Let us write ${\left(d{\mu }_A\right)}_{PT}$ to represent the differential of ${\mu }_A$ at constant pressure and temperature. From the general expression for $d{\mu }_A$ and the definition of activity, we can write the total differential of the chemical potential of substance $A$ in a particular system in several equivalent ways
\begin{aligned} d{\mu }_A & = \left(\frac{\partial {\mu }_A}{\partial P}\right)_TdP+ \left(\frac{\partial {\mu }_A}{\partial T}\right)_PdT+ \left(d{\mu }_A\right)_{PT} \ ~ & = \left(\frac{\partial {\mu }_A}{\partial P}\right)_TdP+ \left(\frac{\partial {\mu }_A}{\partial T}\right)_PdT+\sum^{\omega }_{j=1} \left(\frac{\partial {\mu }_A}{\partial n_j}\right)_{PT}dn_j \ ~ & =\overline{V}_AdP-\overline{S}_AdT+RT \left(d \ln \tilde{a}_A \right)_{PT} \ ~ & =RT \left(\frac{\partial \ln \tilde{a}_A}{\partial P}\right)_TdP+ \left(\frac{\partial {\mu }_A}{\partial T}\right)_PdT + RT \left(d \ln \tilde{a}_A \right)_{PT} \end{aligned} \nonumber
In short, we have developed several alternative notations for the same physical quantities. From the dependence of chemical potential on pressure, and because ${\widetilde{\mu }}^o_A$ is not a function of pressure, we have a very useful relationship:
${\left(\frac{\partial {\mu }_A}{\partial P}\right)}_T=RT{\left(\frac{\partial { \ln {\tilde{a}}_A\ }}{\partial P}\right)}_T={\overline{V}}_A \nonumber$
From the definition of activity and the dependence of chemical potential on temperature, we have: ${\left(\frac{\partial {\mu }_A}{\partial T}\right)}_P={\left(\frac{\partial {\widetilde{\mu }}^o_A}{\partial T}\right)}_P+R{ \ln {\tilde{a}}_A\ }+RT{\left(\frac{\partial { \ln {\tilde{a}}_A\ }}{\partial T}\right)}_P=-{\overline{S}}_A \nonumber$ From the dependence of chemical potential on the composition of the system, we have
$\sum^{\omega }_{j=1}{{\left(\frac{\partial {\mu }_A}{\partial n_j}\right)}_{PT}dn_j}={\left(d{\mu }_A\right)}_{PT}=RT{\left(d{ \ln {\tilde{a}}_A\ }\right)}_{PT} \nonumber$
This last equation shows explicitly that the activity of component $A$ depends on all of the species present. The effects of interactions between $A$ molecules and $B$ molecules are represented in this sum by the term ${\left({\partial {\mu }_A}/{\partial n_B}\right)}_{PT}$. When the effects of intermolecular interactions on the chemical potential are independent of the component concentrations, ${\left({\partial {\mu }_A}/{\partial n_B}\right)}_{PT}=0$, and the only surviving term is ${\left({\partial {\mu }_A}/{\partial n_A}\right)}_{PT}$. If the interactions between $A$ molecules and the rest of the system are constant over a range of concentrations of $A$, ${\gamma }_A$ is constant over this range.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/14%3A_Chemical_Potential_-_Extending_the_Scope_of_the_Fundamental_Equation/14.12%3A_Relating_Fugacity_and_Chemical_Activity.txt
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Having found the activity of a component at one temperature, we want to be able to find it at a second temperature. The equation developed in Section 14.13 does not provide a practical way to find the temperature dependence of ${\tilde{a}}_A$ or ${ \ln {\tilde{a}}_A\ }$. We can obtain a useful equation by rearranging the defining equation, taking the partial derivative of ${ \ln {\tilde{a}}_A\ }$ with respect to temperature, and making use of the Gibbs-Helmholtz equation:
\begin{aligned} \left(\frac{\partial \ln \tilde{a}_A}{\partial T} \right)_P & =\left[\frac{\partial }{\partial T}\left(\frac{{\mu }_A}{RT}-\frac{ \widetilde{\mu }^o_A}{RT}\right)\right]_P \ ~ & =\frac{1}{R}\left[ \left(\frac{\partial \left({\mu }_A/T \right)}{\partial T}\right)_P- \left(\frac{\partial \left( \widetilde{\mu }^o_A/T\right)}{\partial T}\right)_P\right] \ ~ & =-\frac{\overline{H}_A}{RT^2}+\frac{\tilde{H}^o_A}{RT^2} \end{aligned} \nonumber
${\overline{H}}_A$ is the partial molar enthalpy of component $A$ as it is present in the system. ${\tilde{H}}^o_A$ is the partial molar enthalpy of component $A$ in its activity standard state. Since this standard state need not correspond to any real system, ${\tilde{H}}^o_A$ can be the partial molar enthalpy of the substance in a hypothetical state.
In general, it is not possible to find $-\left({\overline{H}}_A-{\tilde{H}}^o_A\right)$ by experimental measurement of the heat exchanged in a single-step process in which $A$ passes from its state in the system of interest to its activity standard state. Instead, we devise a multi-step cycle in which we can determine the enthalpy change for each step. This cycle includes yet another state of substance $A$, which we call the reference state and whose molar enthalpy we designate as ${\overline{H}}^{ref}_A$. We devise this cycle to find two enthalpy changes. One is the enthalpy change that occurs when one mole of $A$ passes from the system of interest to the reference state; this enthalpy change is represented by the difference $-\left({\overline{H}}_A-{\overline{H}}^{ref}_A\right)$. The other is the enthalpy change that occurs when one mole of A passes from the activity standard state to the reference state. This enthalpy change is represented by the difference $-\left({\tilde{H}}^o_A-{\overline{H}}^{ref}_A\right)$. The enthalpy change we seek is the difference between these two differences:
$-\left({\overline{H}}_A-{\tilde{H}}^o_A\right)=-\left({\overline{H}}_A-{\overline{H}}^{ref}_A\right)+\left({\tilde{H}}^o_A-{\overline{H}}^{ref}_A\right) \nonumber$
Because we explicitly choose the reference state so that these differences are experimentally measurable, it is useful to introduce still more terminology. We define the relative partial molar enthalpy of substance $A$, ${\overline{L}}_A$, as the difference between the partial molar enthalpy of $A$ in the state of interest, ${\overline{H}}_A$, and the partial molar enthalpy of A in the reference state, ${\overline{H}}^{ref}_A$; that is, ${\overline{L}}_A={\overline{H}}_A-{\overline{H}}^{ref}_A$ and ${\tilde{L}}^o_A={\tilde{H}}^o_A-{\overline{H}}^{ref}_A$ where ${\tilde{L}}^o_A$ is the relative partial molar enthalpy of substance $A$ in the activity standard state. Clearly, the values of ${\overline{L}}_A$ and ${\tilde{L}}^o_A$ depend on the choice of reference state. We choose the reference state so that ${\overline{L}}_A$ and ${\tilde{L}}^o_A$ can be measured directly. The quantity that we must evaluate experimentally in order to find ${\left({\partial { \ln {\tilde{a}}_A\ }}/{\partial T}\right)}_P$ becomes
$-\left({\overline{H}}_A-{\tilde{H}}^o_A\right)=-\left({\overline{L}}_A-{\tilde{L}}^o_A\right) \nonumber$
To see how these ideas and definitions can be given practical effect, let us consider a binary solution that comprises a relatively low concentration of solute, $A$, in a solvent, $B$, at a fixed pressure, normally one bar. We suppose that pure $A$ is a solid and pure $B$ is a liquid in the temperature range of interest. We need to choose an activity standard state and an enthalpy reference state for each substance. The most generally useful choices use the concept of an infinitely dilute solution. In an infinitely dilute solution, $A$ molecules are dispersed so completely that they can interact only with $B$ molecules. Consequently, the energy of the $A$ molecules cannot change if additional pure solvent (initially at the same temperature and pressure) is added. Operationally then, we can recognize an infinitely dilute solution by mixing it with additional pure solvent; if no heat must be exchanged with the surroundings in order to keep the temperature constant, the original solution is infinitely dilute.
For the solvent, the concept of an infinitely dilute solution gives rise to the following choices, which are shown schematically in Figure 3:
For the activity standard state of the solvent, $B$, we choose pure liquid $B$. Then, the activity of pure liquid $B$ is unity at any temperature. Letting ${\tilde{H}}^o_B$ be the partial molar enthalpy of $B$ in the activity standard state and $H^{\textrm{⦁}}_B$ be the molar enthalpy of pure liquid $B$, we have ${\tilde{H}}^o_B=H^{\textrm{⦁}}_B$.
For the enthalpy reference state of the solvent, we choose $B$ in an infinitely dilute solution. We represent the partial molar enthalpy of $B$ in this infinitely dilute solution by ${\overline{H}}^{ref}_B$.
For the solute, the infinitely dilute solution involves the following choices, which are shown schematically in Figure 4:
For the activity standard state of the solute, we choose the hypothetical solution in which the concentration of $A$ is one molal, and the activity of $A$ is unity, but all of the effects of intermolecular interactions are the same as they are in an infinitely dilute solution. We represent the partial molar enthalpy of $A$ in the activity standard state by ${\tilde{H}}^o_A$.
For the enthalpy reference state of the solute, we choose the infinitely dilute solution and designate the partial molar enthalpy of $A$ in this reference state by ${\overline{H}}^{ref}_A$. Since all of the intermolecular interactions are the same in the enthalpy reference state as they are in the activity standard state, there can be no energy change when one mole of $A$ goes from one of these states to the other. It follows that ${\tilde{H}}^o_A={\overline{H}}^{ref}_A$.
It follows from these choices that the relative partial molar enthalpies of $A$ and $B$ in their activity standard states are ${\tilde{L}}^o_A={\tilde{H}}^o_A-{\overline{H}}^{ref}_A=0$ and
${\tilde{L}}^o_B={\tilde{H}}^o_B-{\overline{H}}^{ref}_B={\mathop{\mathrm{lim}}_{n_A\to 0} {\left(-\frac{\partial {\Delta }_{mix}H}{\partial n_B}\right)}_{P,T,n_A}\ } \nonumber$
The relevant process for which we can measure an enthalpy change is the isothermal mixing of $n_A$ moles of pure solid $A$ with $n_B$ moles of pure liquid $B$ to form a solution:
${n_AA\left(\mathrm{pure\ solid}\right)+n}_BB\left(\mathrm{pure\ liquid}\right) \begin{array}{c} {\Delta }_{mix}H \ \to \ \end{array} {A}/{B\left(\mathrm{solution},\ n_A,n_B\right)} \nonumber$
Letting the enthalpy of mixing be ${\Delta }_{mix}H$ and the enthalpy of the resulting solution be $H$, we have
$H={\Delta }_{mix}H+n_AH^{\textrm{⦁}}_A+n_BH^{\textrm{⦁}}_B \nonumber$
Then
${\overline{H}}_A={\left(\frac{\partial H}{\partial n_A}\right)}_{PT}={\left(\frac{\partial {\Delta }_{mix}H}{\partial n_A}\right)}_{PT}+H^{\textrm{⦁}}_A \nonumber$
and
${\overline{H}}_B={\left(\frac{\partial H}{\partial n_B}\right)}_{PT}={\left(\frac{\partial {\Delta }_{mix}H}{\partial n_B}\right)}_{PT}+H^{\textrm{⦁}}_B \nonumber$ where, of course, $H$, ${\Delta }_{mix}H$, ${\overline{H}}_A$, and ${\overline{H}}_B$ are all functions of $n_A$ and $n_B$. The partial molar enthalpies in the reference states are the limiting values of ${\overline{H}}_A$ and ${\overline{H}}_B$ as $n_A\to 0$. That is,
${\overline{H}}^{ref}_A={\mathop{\mathrm{lim}}_{n_A\to 0} {\left(\frac{\partial {\Delta }_{mix}H}{\partial n_A}\right)}_{P,T,n_B}\ }+H^{\textrm{⦁}}_A \nonumber$ and ${\overline{H}}^{ref}_B={\mathop{\mathrm{lim}}_{n_B\to 0} {\left(\frac{\partial {\Delta }_{mix}H}{\partial n_B}\right)}_{P,T,n_A}\ }+H^{\textrm{⦁}}_B \nonumber$
When we base the enthalpy reference states on the infinitely dilute solution, we find for the solute
$-\left({\overline{H}}_A-{\tilde{H}}^o_A\right)=-\left({\overline{H}}_A-{\overline{H}}^{ref}_A\right)+\left({\tilde{H}}^o_A-{\overline{H}}^{ref}_A\right) \nonumber$ $=-\left({\overline{L}}_A-{\tilde{L}}^o_A\right)=-{\overline{L}}_A=-{\left(\frac{\partial {\Delta }_{mix}H}{\partial n_A}\right)}_{P,T,n_B}+{\mathop{\mathrm{lim}}_{n_A\to 0} {\left(\frac{\partial {\Delta }_{mix}H}{\partial n_A}\right)}_{P,T,n_B}\ } \nonumber$
and for the solvent
$-\left({\overline{H}}_B-{\tilde{H}}^o_B\right)=-\left({\overline{H}}_B-{\overline{H}}^{ref}_B\right)+\left({\tilde{H}}^o_B-{\overline{H}}^{ref}_B\right)=-\left({\overline{L}}_B-{\tilde{L}}^o_B\right)=-{\left(\frac{\partial {\Delta }_{mix}H}{\partial n_B}\right)}_{P,T,n_A} \nonumber$
The temperature dependence of the activities becomes
${\left(\frac{\partial { \ln {\tilde{a}}_A\ }}{\partial T}\right)}_P=\frac{1}{RT^2}\left[-{\left(\frac{\partial {\Delta }_{mix}H}{\partial n_A}\right)}_{P,T,n_B}+{\mathop{\mathrm{lim}}_{n_A\to 0} {\left(\frac{\partial {\Delta }_{mix}H}{\partial n_A}\right)}_{P,T,n_B}\ }\right] \nonumber$
and ${\left(\frac{\partial { \ln {\tilde{a}}_B\ }}{\partial T}\right)}_P=-\frac{1}{RT^2}{\left(\frac{\partial {\Delta }_{mix}H}{\partial n_B}\right)}_{P,T,n_A} \nonumber$
To a good first approximation, we can measure ${\Delta }_{mix}H$ as a function of composition at a single temperature, determine ${\overline{L}}_A$ and ${\overline{L}}_B-{\tilde{L}}^o_B$ at that temperature, and assume that these values are independent of temperature. For a more exact treatment, we can measure ${\Delta }_{mix}H$ as a function of composition at several temperatures and find ${\overline{L}}_A$ and ${\overline{L}}_B-{\tilde{L}}^o_B$ as functions of temperature. It proves to be useful to define the relative partial molar heat capacity of $A$, to which we give the symbol, ${\overline{J}}_A$, as the temperature derivative of ${\overline{L}}_A$:
${\overline{J}}_A={\left(\frac{\partial {\overline{L}}_A}{\partial T}\right)}_P \nonumber$
To illustrate the use of these ideas, let us suppose that we measure the enthalpy of mixing of solute $A$ in 1 kg water (solvent $B$). We make this measurement for several quantities of $A$ at each of several temperatures between 273.15 K and 293.15 K. For each experiment in this series, $n_B$ is 55.51 mole and $n_A$ is equal to the molality of $A$, $\underline{m}$, in the solution. We fit the experimental data to empirical equations. Let us suppose that the enthalpy of mixing data at any given temperature are described adequately by the equation ${\Delta }_{mix}H={\alpha }_1\underline{m}+{\alpha }_2{\underline{m}}^2$ and that ${\alpha }_1$ and ${\alpha }_2$ depend linearly on $T$ according to
${\alpha }_1={\beta }_{11}+{\beta }_{12}\left(T-273.15\right) \nonumber$ and
${\alpha }_2={\beta }_{21}+{\beta }_{22}\left(T-273.15\right) \nonumber$
Then
${\left(\frac{\partial {\Delta }_{mix}H}{\partial n_A}\right)}_{P,T,n_B}={\left(\frac{\partial {\Delta }_{mix}H}{\partial \underline{m}}\right)}_{P,T,n_B}={\alpha }_1+2{\alpha }_2\underline{m} \nonumber$
and
${\mathop{\mathrm{lim}}_{n_A\to 0} {\left(\frac{\partial {\Delta }_{mix}H}{\partial n_A}\right)}_{P,T,n_B}\ }={\alpha }_1 \nonumber$
so that
${\overline{L}}_A={\left(\frac{\partial {\Delta }_{mix}H}{\partial n_A}\right)}_{P,T,n_B}-{\mathop{\mathrm{lim}}_{n_A\to 0} {\left(\frac{\partial {\Delta }_{mix}H}{\partial n_A}\right)}_{P,T,n_B}\ }=2{\alpha }_2\underline{m} \nonumber$
and
$\left(\frac{\partial \ln \tilde{a}_A}{\partial T}\right)_P=-\frac{\overline{L}_A}{RT^2}=-\frac{2{\alpha }_2\underline{m}}{RT^2} \nonumber$
and
${\overline{J}}_A={\left(\frac{\partial {\overline{L}}_A}{\partial T}\right)}_P=2\underline{m}{\left(\frac{\partial {\alpha }_2}{\partial T}\right)}_P=2\underline{m}{\beta }_{22} \nonumber$
In the experiments of this illustrative example, $n_B$ is constant. This might make it seem that we would have to do an additional set of experiments, a set in which $n_B$ is varied, in order to find ${\overline{L}}_B$, ${\left({\partial { \ln {\tilde{a}}_B\ }}/{\partial T}\right)}_P$, and ${\overline{J}}_B$. However, this is not the case. Since ${\overline{L}}_A$ and ${\overline{L}}_B$ are partial molar quantities, we have
$n_Ad{\overline{L}}_A+n_Bd{\overline{L}}_B=0 \nonumber$ so that $d{\overline{L}}_B=-\left(\frac{n_A}{n_B}\right)d{\overline{L}}_A \nonumber$
and we can find ${\overline{L}}_B$ from
$\overline{L}_B\left(\underline{m}\right)-\overline{L}_B\left(0\right)=\overline{L}_B\left(\underline{m}\right)=\int^{\underline{m}}_0 -\left(\frac{2\underline{m}{\alpha }_2}{55.51}\right)d\underline{m}=-\frac{\underline{m}^2{\alpha }_2}{55.51} \nonumber$
and
$\overline{J}_B= \left(\frac{\partial \overline{L}_B}{\partial T}\right)_P=-\frac{\underline{m}^2}{55.51} \left(\frac{\partial {\alpha }_2}{\partial T} \right)_{P,n_A}=-\frac{\underline{m}^2{\beta }_{22}}{55.51} \nonumber$
14.15: Problems
Problems
1. When we express the energy of a system as a function of entropy, volume, and composition, we have $E=E\left(S,V,n_1,n_2,\dots ,\ n_{\omega }\right)$. Since $S$ and $V$ are extensive variables, we have $\lambda E=E\left(\lambda S,\lambda V,{\lambda n}_1,\lambda n_2,\dots ,\ \lambda n_{\omega }\right)$. Find ${\left({\partial \left(\lambda E\right)}/{\partial \lambda }\right)}_{SV}$. From this result, show that $G=\sum^{\omega }_{j=1}{{\mu }_jn_j} \nonumber$
2. When we express the energy of a system as a function of pressure, temperature, and composition, we have $E=E\left(P,T,n_1,n_2,\dots ,\ n_{\omega }\right)$. Because P and T are independent of $\lambda$, $\lambda E=E\left(P,T,{\lambda n}_1,\lambda n_2,\dots ,\ \lambda n_{\omega }\right)$. Show that
$E=\sum^{\omega }_{j=1} \overline{E}_jn_j \nonumber$
3. From $E\mathrm{=E}\left(P,T,n_{\mathrm{1}},n_{\mathrm{2}}\mathrm{,\dots ,\ }n_{\omega }\right)$ and the result in problem 2, show that
$\left[{\left(\frac{\partial H}{\partial T}\right)}_P + P{\left(\frac{\partial V}{\partial T}\right)}_P\right]dT + \left[P{\left(\frac{\partial V}{\partial P}\right)}_T + T{\left(\frac{\partial V}{\partial T}\right)}_P\right]dP = \sum^{\omega }_{j\mathrm{=1}}{n_j}d{\overline{E}}_j \nonumber$
Note that at constant pressure and temperature,
$\sum^{\omega }_{j\mathrm{=1}}{n_j}d{\overline{E}}_j\mathrm{=0} \nonumber$
4. If pressure and temperature are constant, $E=E\left(n_1,n_2,\dots ,\ n_{\omega }\right)$ and $\lambda E=E\left({\lambda n}_1,\lambda n_2,\dots ,\ \lambda n_{\omega }\right)$. Show that $\sum^{\omega }_{j\mathrm{=1}}{n_j}d{\overline{E}}_j\mathrm{=0}$ follows from these relationships.
5. A solution contains $n_1$ moles of component 1, $n_2$ moles of component 2, $n_3$ moles of component 3, etc. Let $n=n_1+n_2+n_3+...$ The mole fraction of component $j$ is $x_j={n_j}/{n}$. Show that $\left(\frac{\partial x_j}{\partial n_j}\right)=\frac{n-n_j}{n^2} \nonumber$ and, for $j\neq k$, $\ \left(\frac{\partial x_j}{\partial n_k}\right)=\frac{-n_j}{n^2} \nonumber$ What are $\left(\frac{\partial x_1}{\partial n_1}\right) \nonumber$ and $\left(\frac{\partial x_2}{\partial n_2}\right) \nonumber$ if the solution has only two components?
6. For any extensive state function, $Y\left(P,T,n_1,n_2,\dots ,\ n_{\omega }\right)$, the arguments developed in this chapter lead, at constant $P$ and$\ T$, to the equations
$Y=n_1{\overline{Y}}_1+n_2{\overline{Y}}_2+\dots +n_{\omega }{\overline{Y}}_{\omega } \nonumber$ and $0=n_1d{\overline{Y}}_1+n_2d{\overline{Y}}_2+\dots +n_{\omega }d{\overline{Y}}_{\omega } \nonumber$
Where ${\overline{Y}}_j$ is the partial molar quantity ${\left({\partial Y}/{\partial n_j}\right)}_{P,T,n_{m\neq j}}$.
(a) Prove that $0=x_1d{\overline{Y}}_1+x_2d{\overline{Y}}_2+\dots +x_{\omega }d{\overline{Y}}_{\omega }$
(b) Prove that $0=n_1\left(\frac{\partial {\overline{Y}}_1}{\partial n_1}\right)+n_2\left(\frac{\partial {\overline{Y}}_2}{\partial n_2}\right)+\dots +n_{\omega }\left(\frac{\partial {\overline{Y}}_{\omega }}{\partial n_{\omega }}\right) \nonumber$ (c) Prove that $0=x_1\left(\frac{\partial {\overline{Y}}_1}{\partial x_1}\right)+x_2\left(\frac{\partial {\overline{Y}}_2}{\partial x_2}\right)+\dots +x_{\omega }\left(\frac{\partial {\overline{Y}}_{\omega }}{\partial x_{\omega }}\right) \nonumber$
7. The enthalpy of mixing is measured in a series of experiments in which solid solute, $A$, dissolves to form an aqueous solution. These enthalpy data are represented well by empirical equations ${\Delta }_{mix}H={\alpha }_1\underline{m}+{\alpha }_2{\underline{m}}^2$, ${\alpha }_1={\beta }_{11}+{\beta }_{12}\left(T-273.15\right)$ and
${\alpha }_2={\beta }_{21}+{\beta }_{22}\left(T-273.15\right)$ with ${\beta }_{11}=10.0\ \mathrm{kJ}\ {\mathrm{molal}}^{-1} \nonumber$ ${\beta }_{12}=-0.14\ \mathrm{kJ}\ {\mathrm{molal}}^{-2}\ K^{-1} \nonumber$ ${\beta }_{21}=-3.00\ \mathrm{kJ}\ {\mathrm{molal}}^{-1} \nonumber$ ${\beta }_{22}=-0.040\ \mathrm{kJ}\ {\mathrm{molal}}^{-2}\ K^{-1} \nonumber$ Find ${\overline{L}}_A$, ${\overline{L}}_{H_2O}$, ${\overline{J}}_A$, and ${\overline{J}}_{H_2O}$ as functions of ${\underline{m}}_A$ and $T$. Find ${\overline{L}}_A$, ${\overline{L}}_{H_2O}$, ${\overline{J}}_A$, and ${\overline{J}}_{H_2O}$ for a one molal solution at 209 K. What is the value of
${ \ln \frac{{\tilde{a}}_A\left(1\mathrm{\ molal},290\mathrm{\ K}\right)}{{\tilde{a}}_A\left(1\mathrm{\ molal},273.15\mathrm{\ K}\right)}\ } \nonumber$
Notes
${}^{1}$ We can make other assumptions. It is possible to describe an inhomogeneous system as a collection of many macroscopic, approximately homogeneous regions.
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In Chapters 11-14, we define chemical potential, fugacity, and activity. We find numerous relationships among these quantities. In Sections 15.1 and 15.2, we summarize the principal relationships between chemical potential and fugacity and between chemical potential and activity. Thereafter, we introduce some basic ideas about the chemical potentials, fugacities, and activities of liquids, solids, solvents, and solutes. We use these ideas to relate standard Gibbs free energy changes to fugacities and activities in systems at equilibrium.
• 15.1: The Chemical Potential and Fugacity of a Gas
The fugacity of a gas in any system is a measure of the difference between its chemical potential in that system and its chemical potential in its hypothetical ideal-gas standard state at the same temperature. The chemical potential of A in a particular system, μA , is the change in the Gibbs free energy when the amounts of the elements that form one mole of A pass from their standard states as elements into the (very large) system as one mole of substance A .
• 15.2: The Chemical Potential and Activity of a Gas
To make predictions about processes involving substance A, we need information about the chemical potential of A. Introducing the fugacity does not introduce new information; the fugacity is merely a convenient way to relate the chemical potential to the composition of the system. The fugacity relationship is valid whether we can actually measure the chemical potential or not. To use the relationship for practical calculations, we must know both, of course.
• 15.3: The Pressure-dependence of the Fugacity and Activity of a Condensed Phase
So far, we have investigated fugacity and activity only for gases. Let us now consider a system that consists entirely of substance A present as either a pure liquid or a pure solid. We assume that the temperature is fixed and that the pressure on this condensed phase can be varied. For our present discussion, it does not matter whether the condensed phase is a liquid or a solid.
• 15.4: Standard States for the Fugacity and Activity of a Pure Solid
If substance is a liquid at one bar and the temperature of interest, pure liquid is the standard state for the calculation of the enthalpy and Gibbs energy of formation. From thermal measurements, we can find the standard Gibbs energy of formation of this liquid. If we can measure the vapor pressure of the substance and find an equation of state that describes the behavior of the real vapor, we can also find its fugacity and the standard Gibbs energy of formation of its hypothetical ideal gas
• 15.5: The Chemical Potential, Fugacity, and Activity of a Pure Solid
The relationship between the standard Gibbs free energy of formation of a substance whose standard state is a solid and the Gibbs free energy of the substance in its hypothetical ideal-gas standard state is essentially the same as described in the previous section for a liquid. In each case, to find the fugacity of the condensed phase in its standard state, it is necessary to find a reversible path that takes the condensed-phase substance to its hypothetical ideal-gas standard state.
• 15.6: Chemical Potential, Fugacity, and Equilibrium
In practice, a great many substances are non-volatile. The Gibbs free energy of formation of their hypothetical ideal-gas standard states and their fugacities cannot be measured. For such substances, we have recourse to other standard states and use activities to express the equilibrium constant.
• 15.7: Chemical Potential, Activity, and Equilibrium
The relationship between the equilibrium constant and the standard Gibbs free energy change for a reaction is extremely useful. If we can calculate the standard Gibbs free energy change from tabulated values, we can find the equilibrium constant and predict the position of equilibrium for a particular system. Conversely, if we can measure the equilibrium constant, we can find the standard Gibbs free energy change.
• 15.8: The Rate of Gibbs Free Energy Change with Extent of Reaction
If a reacting system is not at equilibrium, the extent of reaction is time-dependent.
• 15.9: Problems
15: Chemical Potential Fugacity Activity and Equilibrium
The third law and the fugacity of a pure real gas.
In Chapter 11, we introduce the fugacity as a measure of the difference between the molar Gibbs free energy of a real gas, $\overline{G}\left(P,T\right)$ at pressure $P$, and that of the pure gas in its hypothetical ideal-gas standard state at the same temperature. We choose the standard Gibbs free energy of formation, ${\Delta }_fG^o\left({HIG}^o,T\right)$, to be the Gibbs free energy of the real gas in its hypothetical ideal-gas standard state. Letting the gas be $A$, we find
$\overline{G}\left(P,T\right)={\Delta }_fG^o\left({HIG}^o,T\right)+RT{ \ln \left[\frac{f^{\textrm{⦁}}_A\left(P\right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$
(real gas)
where the fugacity depends on pressure according to
${ \ln \left[\frac{f^{\textrm{⦁}}_A\left(P\right)}{f_A\left({HIG}^o\right)}\right]\ }={ \ln \left[\frac{P}{P^o}\right]\ }+\int^P_0{\left[\frac{{\overline{V}}^{\textrm{⦁}}_A}{RT}-\frac{1}{P}\right]}dP \nonumber$ (real gas)
and ${\overline{V}}^{\textrm{⦁}}_A$ is the molar volume of the pure real gas. (In Chapter 14, we introduce a solid-bullet superscript to indicate that a particular property is that of a pure substance.) Given ${\Delta }_fG^o\left({HIG}^o,T\right)$ and an equation of state for the real gas, we can calculate the fugacity and molar Gibbs free energy of the real gas at any pressure.
The fugacity of a pure ideal gas
For a pure ideal gas, we have
$\frac{{\overline{V}}^{\textrm{⦁}}_A}{RT}-\frac{1}{P}=0 \nonumber$ (ideal gas)
The fugacity becomes equal to the ideal-gas pressure $f^{\textrm{⦁}}_A\left(P\right)=P \nonumber$
(ideal gas)
and the Gibbs free energy relationship becomes
${\overline{G}}_A\left(P,T\right)={\Delta }_fG^o\left(A,P^o,T\right)+RT{ \ln \left[\frac{P}{P^o}\right]\ } \nonumber$ (ideal gas)
For pure gases, the system pressure that appears in these equations, $P$, is the same thing as the pressure of the gas.
The fugacity of an ideal gas in a mixture
In Chapter 13, we find that the molar Gibbs free energy of a component of an ideal gas mixture is unaffected by the presence of the other gases. For an ideal gas, $A$, present at mole fraction $x_A$, in a system whose pressure is $P$, the partial pressure is $P_A=x_AP$. Since the partial pressure is the pressure that the system would exhibit if only ideal gas $A$ were present, the molar Gibbs free energy of an ideal gas in a mixture is
${\overline{G}}_A\left(x_A,P,T\right)={\Delta }_fG^o\left(A,P^o,T\right)+RT{ \ln \left[\frac{x_AP}{P^o}\right]\ } \nonumber$ (ideal gas)
The chemical potential and fugacity of real gases
In Chapter 14, we introduce the chemical potential as the partial molar Gibbs free energy. The defining relationship is ${\mu }_A={\overline{G}}_A={\left(\frac{\partial G}{\partial n_A}\right)}_{P,T,n_{i\neq A}} \nonumber$ (any substance in any system)
When the system is a pure substance, the chemical potential is identical to the Gibbs free energy per mole of the pure substance at the same temperature and pressure. For the chemical potential of $A$ in a system comprised of pure $A$, we can write
${\mu }^{\textrm{⦁}}_A={\overline{G}}^{\textrm{⦁}}_A=\frac{G^{\textrm{⦁}}}{n_A}=\frac{dG^{\textrm{⦁}}}{dn_A} \nonumber$ (any system comprised of pure A)
From Euler’s theorem, we find that the Gibbs free energy of any system is the composition-weighted sum of the chemical potentials of the substances present:
$G=\sum^{\omega }_{i=1}{n_i{\mu }_i} \nonumber$
For a pure real gas, the partial molar Gibbs free energy and the molar Gibbs free energy are the same thing; we also write
${\mu }^{\textrm{⦁}}_A\left(P,T\right)={\Delta }_fG^o\left(A,{HIG}^o,T\right)+RT{ \ln \left[\frac{f^{\textrm{⦁}}_A\left(P\right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$ (pure real gas A)
and introducing ${\mu }^o_A\left(T\right)={\Delta }_fG^o\left(A,{HIG}^o,T\right)$, we write
${\mu }^{\textrm{⦁}}_A\left(P,T\right)={\mu }^o_A\left(T\right)+RT{ \ln \left[\frac{f^{\textrm{⦁}}_A\left(P\right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$ (pure real gas A)
Since ${\mu }^o_A$, ${\Delta }_fG^o\left(A\right)$, and $f_A\left({HIG}^o\right)$ are defined to be properties of one mole of pure $A$, it is not necessary to include either the solid-bullet superscript or the solid over-bar in these symbols.
In Section 14.11, we find that the partial molar Gibbs free energy of a component of a real-gas mixture is
${\mu }_A\left(P,T\right)={\mu }^o_A\left(T\right)+RT{ \ln \left[\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$ (real gas A in a mixture)
where the fugacity of $A$, present at mole fraction $x_A$ in a system whose pressure is $P$, is given by
$RT{ \ln \left[\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)}\right]\ }={ \ln \left[\frac{x_AP}{P^o}\right]\ }+\int^P_0{\left[\frac{{\overline{V}}_A}{RT}-\frac{1}{P}\right]}dP \nonumber$ (real gas A in a mixture)
where $f_A\left({HIG}^o\right)=P^o=1\ \mathrm{bar}$. The partial molar volume is a function of the system’s pressure, temperature, and composition; that is,
${\overline{V}}_A\left(P\right)={\overline{V}}_A\left(P,T,x_A,x_B,\dots ,x_{\omega }\right) \nonumber$
and the fugacity depends on the same variables,
$f_A\left(P\right)=f_A\left(P,T,x_A,x_B,\dots ,x_{\omega }\right) \nonumber$
If the system is a mixture of ideal gases,
$V=\left(n_A+n_B+\dots +n_{\omega }\right){RT}/{P} \nonumber$ and ${\overline{V}}_A={\left(\partial {V}/{\partial n_A}\right)}_{PTn_{m\neq A}}={RT}/{P} \nonumber$
The integrand becomes zero, and the fugacity relationship reduces to the ideal-gas fugacity equation introduced in Chapter 13 and repeated above.
The fugacity of a gas in any system is a measure of the difference between its chemical potential in that system and its chemical potential in its hypothetical ideal-gas standard state at the same temperature. The chemical potential of $A$ in a particular system, ${\mu }_A$, is the change in the Gibbs free energy when the amounts of the elements that form one mole of $A$ pass from their standard states as elements into the (very large) system as one mole of substance $A$.
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To make predictions about processes involving substance $A$, we need information about the chemical potential of $A$. Introducing the fugacity does not introduce new information; the fugacity is merely a convenient way to relate the chemical potential to the composition of the system. The fugacity relationship is valid whether we can actually measure ${\mu }_A$ and ${\mu }^o_A\left({HIG}^o\right)$ or not. To use the relationship for practical calculations, we must know both, of course.
We introduce the activity function to cope with situations in which we cannot measure the fugacity. For volatile liquids—or solids—we can obtain the Gibbs free energy of formation for both the condensed phase and for the hypothetical ideal-gas standard state. In Section 15.4, we consider the relationship between the two.
The chemical activity of substance $A$ measures the change in the chemical potential when one mole of $A$ in some arbitrarily chosen standard state passes into a very large system of specified composition. We introduce
${\mu }_A={\widetilde{\mu }}^o_A+RT{ \ln \left[\frac{{\tilde{a}}_A}{{\tilde{a}}_A\left(ss\right)}\right]\ } \nonumber$
where, as always,
${\mu }_A={\left(\frac{\partial G}{\partial n_A}\right)}_{P,T,n_{i\neq A}} \nonumber$
We let the activity of $A$ in the arbitrarily chosen standard state, designated “$ss$”, be unity, so that ${\tilde{a}}_A\left(ss\right)=1$ and the chemical potential of $A$ in this standard state is ${\widetilde{\mu }}^o_A$. The activity, ${\tilde{a}}_A$, is a function of the pressure, temperature, and composition of the system.
While we are free to choose any standard state we please for the activity of a gas, the hypothetical ideal-gas standard state is the most practical. In this case, the activity of a gas is given by
${\tilde{a}}_A\left(P\right)=\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)} \nonumber$ and ${\widetilde{\mu }}^o_A={\mu }^o_A\left({HIG}^o\right) \nonumber$
Then the only difference between fugacity and activity is that fugacity has the units of pressure, whereas activity is dimensionless. For any gas in any state, we have
$\frac{{\mu }_A-{\mu }^o_A\left({HIG}^o\right)}{RT}={ \ln \left[\frac{f_A\left(P\right)}{f_A\left({HIG}^o\right)}\right]\ }={ \ln \left[{\tilde{a}}_A\left(P\right)\right]\ } \nonumber$
For an ideal-gas mixture whose pressure is $P$ and in which the mole fraction of $A$ is $x_A$, we have
${\tilde{a}}_A\left(x_A,P\right)=\frac{f_A\left(x_A,P\right)}{f_A\left({HIG}^o\right)}=\frac{x_AP}{P^o} \nonumber$
(ideal gas mixture)
15.03: The Pressure-dependence of the Fugacity and Activity of
So far, we have investigated fugacity and activity only for gases. Let us now consider a system that consists entirely of substance $A$ present as either a pure liquid or a pure solid. We assume that the temperature is fixed and that the pressure on this condensed phase can be varied. For our present discussion, it does not matter whether the condensed phase is a liquid or a solid. For specificity, let us assume that it is a liquid. We can imagine that the pressure changes are effected with the pure substance contained in a cylinder that is sealed by a frictionless piston. We ask how the fugacity and activity vary when the system pressure changes. Let the molar volume of the pure condensed phase be ${\overline{V}}^{\textrm{⦁}}_A$. Since the coefficient of isothermal compression is small for condensed phases, it is often adequate to assume that ${\overline{V}}^{\textrm{⦁}}_A$ is a constant. We do so here.
We have developed several designations for the molar Gibbs free energy of the condensed phase. For pure liquid $A$ at pressure $P$, we can write
${\overline{G}}^{\textrm{⦁}}\left(A,\ell ,P\right)={\overline{G}}^{\textrm{⦁}}_A\left(\ell ,P\right)={\mu }^{\textrm{⦁}}_A\left(\ell ,P\right) \nonumber$ (pure liquid)
For the pure liquid at a pressure of one bar, it is convenient to let the molar Gibbs free energy of the liquid be equal to the molar Gibbs free energy of formation. We let
${\overline{G}}^{\textrm{⦁}}\left(A,\ell ,P^o\right)={\overline{G}}^{\textrm{⦁}}_A\left(\ell ,P^o\right)={\mu }^{\textrm{⦁}}_A\left(\ell ,P^o\right)={\Delta }_fG^o\left(A,\ell ,P^o\right) \nonumber$ (pure liquid, $P=1\ \mathrm{bar}$)
We also have several designations for the pressure dependence of the Gibbs free energy of this liquid. Since the system consists entirely of the pure liquid, we have
\begin{aligned} \left(\partial \overline{G}^{\textrm{⦁}}\left(A,\ell ,P\right)/\partial P\right)_T & =\left( \partial \overline{G}^{\textrm{⦁}}_A\left(\ell ,P\right)/ \partial P\right)_T \ ~ & = \left(\partial {\mu }^{\textrm{⦁}}_A\left(\ell ,P\right)/ \partial P\right)_T \ ~ & =RT\ \left( \partial \left\{ \ln \left[f^{\textrm{⦁}}_A\left(\ell ,P\right)\right] \right\} /\partial P \right)_T \ ~ & =RT \left( \partial \left\{ \ln \left[ \tilde{a}^{\textrm{⦁}}_A\left(\ell ,P\right)\right] \right\}/\partial P \right)_T \ ~ & =\overline{V}^{\textrm{⦁}}_A\left(\ell ,P\right)dP \end{aligned} \nonumber
We are free to choose any state of any system that contains $A$ as the standard state for the activity of substance $A$. Often, it is convenient to let this standard state be the pure liquid (or the pure solid) at a pressure of one bar. The activity of $A$ is unity in the standard state. Taking the partial molar volume to be constant, ${\overline{V}}^{\textrm{⦁}}_A\left(\ell ,P\right)={\overline{V}}^{\textrm{⦁}}_A$, and integrating between one bar and an arbitrary pressure, $P$, we can express the pressure dependence of the Gibbs free energy of the pure liquid in several equivalent ways:
\begin{aligned} \overline{G}^{\textrm{⦁}}_A\left(\ell ,P\right) & =\overline{G}^{\textrm{⦁}}\left(A,\ell ,P\right)-\overline{G}^{\textrm{⦁}}\left(A,\ell ,P^o\right) \ ~ & = \overline{G}^{\textrm{⦁}}\left(A,\ell ,P\right)-{\Delta }_fG^o\left(A,\ell \right) \ ~ & ={\mu }^{\textrm{⦁}}_A\left(\ell ,P\right)-{\mu }^{\textrm{⦁}}_A\left(\ell ,P^o\right) \ ~ & =RT \ln \left[\frac{f^{\textrm{⦁}}_A\left(\ell ,P\right)}{f^{\textrm{⦁}}_A\left(\ell ,P^o\right)}\right] \ ~ & =RT \ln \left[\tilde{a}^{\textrm{⦁}}_A\left(\ell ,P\right)\right] \ ~ & =\overline{V}^{\textrm{⦁}}_A\left(P-P^o\right) \end{aligned} \nonumber
From the last equations, we see that the activity and fugacity of the pure liquid vary with the system pressure as
${\tilde{a}}^{\textrm{⦁}}_A\left(\ell ,P\right)=\frac{f^{\textrm{⦁}}_A\left(\ell ,P\right)}{f^{\textrm{⦁}}_A\left(\ell ,P^o\right)}=\mathrm{exp}\left[\frac{{\overline{V}}^{\textrm{⦁}}_A\left(P-P^o\right)}{RT}\right] \nonumber$
At ordinary temperatures and pressures, ${\overline{V}}^{\textrm{⦁}}_A\left(P-P^o\right)\ll RT$. In consequence, the system pressure must become much greater than one bar before the exponential term becomes significantly different from one. Thus, the activity of a condensed phase is approximately unity until the system reaches high pressures. At pressures near one bar, the fugacity of a condensed phase is only a weak function of pressure.
This argument provides rigorous justification for treating the activities (or concentrations) of pure solids and liquids as constants when we use equilibrium constant data to calculate the compositions of systems that are at equilibrium. (We address this issue previously in Section 5.17 and Section 13.8.)
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If substance $A$ is a liquid at one bar and the temperature of interest, pure liquid $A$ is the standard state for the calculation of the enthalpy and Gibbs free energy of formation. From thermal measurements, we can find the standard Gibbs free energy of formation of this liquid, ${\Delta_fG}^o\left(A,\ell \right)=\mu^{\textrm{⦁}}_A\left(\ell ,P^o\right)$. If we can measure the vapor pressure of the substance and find an equation of state that describes the behavior of the real vapor, we can also find its fugacity and the standard Gibbs free energy of formation of its hypothetical ideal gas, ${\Delta_fG}^o\left(A,{HIG}^o\right)=\mu^o_A\left({HIG}^o\right)$. From the principle that the chemical potential of substance $A$ is the same in any two phases that are at equilibrium, it follows that the fugacity is the same in each phase.
If we choose the hypothetical ideal-gas standard state as the standard state for the activity of $A$, then the activity and fugacity are the same thing, and the standard state chemical potential is the same thing as the Gibbs free energy of formation of the hypothetical ideal gas.
$\mu^o_A={\widetilde\mu}^o_A=\Delta_fG^o\left(A,{HIG}^o\right) \nonumber$
(activity standard state is the hypothetical ideal gas)
Alternatively, we can choose the pure liquid as the standard state for the activity of $A$. In this case, there are two further options: We can choose the pure liquid either at one bar pressure, $P^o$, or at its equilibrium vapor pressure, $P^{\textrm{⦁}}_{vp}$. If we choose the pure liquid at $P^o$, we have
${\widetilde\mu}^o_A=\mu^{\textrm{⦁}}_A\left(\ell ,P^o\right)=\Delta_fG^o\left(A,{\ell ,P}^o\right) \nonumber$
(activity standard state is the pure liquid at $P^o$)
In Chapter 16, we see that the pure liquid at $P^{\textrm{⦁}}_{vp}$ proves to be the most generally useful choice for $A$ in a solution. For the pure liquid at $P^{\textrm{⦁}}_{vp}$, we have
${\widetilde\mu}^o_A=\mu^{\textrm{⦁}}_A\left(\ell ,P^{\textrm{⦁}}_{vp}\mathrm{\ }\right) \nonumber$
(activity standard state is the pure liquid at $P^{\textrm{⦁}}_{vp}$)
Evidently, it is useful to be able to relate the quantities $\Delta_fG^o\left(A,{HIG}^o\right)$, $\mu^{\textrm{⦁}}_A\left(\ell ,P^o\right)=\Delta_fG^o\left(A,{\ell ,P}^o\right)$, and $\mu^{\textrm{⦁}}_A\left(\ell ,P^{\textrm{⦁}}_{vp}\mathrm{\ }\right)$ to one another.
The difference between the Gibbs free energy of formation of the ideal gas in its hypothetical ideal-gas state and the Gibbs free energy of formation of the liquid in its standard state is a quantity that we can call the standard Gibbs free energy of vaporization, ${\Delta_{vap}G}^o\left(A,P^o\right)$, because both the initial and final states are at a pressure of one bar. At an arbitrary temperature, a liquid at one bar is not at equilibrium with its own ideal gas at one bar, and the standard Gibbs free energy of vaporization is not zero. We have
\begin{align*} \Delta_{vap}G^o\left(A,P^o\right) &= \Delta_fG^o\left(A,{HIG}^o\right)-\Delta_fG^o\left(A,{\ell ,P}^o\right) \[4pt] &= \mu^o_A-\mu^{\textrm{⦁}}_A\left(\ell ,P^o\right) = RT \ln \left[\frac{f_A\left({HIG}^o\right)}{f^{\textrm{⦁}}_A\left(\ell ,P^o\right)}\right] \end{align*}
Figure 1 describes a reversible process that takes one mole of $A$ from the pure liquid state at one bar to its hypothetical ideal-gas standard state. We first reversibly decrease the pressure applied to pure liquid A until we reach its equilibrium vapor pressure, $P^{\textrm{⦁}}_{vp}$, at the temperature of interest. The molar Gibbs free energy change for this process is
$\Delta_{press}G\left(A,\ell \right) = \mu^{\textrm{⦁}}_A \left(\ell ,P^{\textrm{⦁}}_{vp}\right)-\Delta_fG^o\left(A,\ell ,P^o\right) = \mu^{\textrm{⦁}}_A\left(\ell ,P^{\textrm{⦁}}_{vp}\right)-\mu^{\textrm{⦁}}_A\left(\ell ,P^o\right) = \int^{P^{\textrm{⦁}}_{vp}}_{P^o}{\overline{V}^{\textrm{⦁}}_A}\left(\ell \right)\ dP \nonumber$
To reach the hypothetical ideal-gas standard state, we then reversibly and isothermally evaporate one mole of the liquid to its real gas. The Gibbs free energy change for this reversible process is zero, $\Delta_{vap}G\left(A,P^{\textrm{⦁}}_{vp}\right)=0$. Finally, we isothermally and reversibly expand the real gas to an arbitrarily low pressure, $P^*$, conceptually convert the real gas to an ideal gas, and compress this ideal gas from $P^*$ to one bar. The Gibbs free energy change for these latter steps is
\begin{align*} \Delta_{press}G\left(A,g\right) &= RT \ln \left[\frac{f_A\left({HIG}^o\right)}{f^{\textrm{⦁}}_A\left(RG,P^{\textrm{⦁}}_{vp}\right)}\right] \[4pt]&= -RT{ \ln \left[\frac{P^{\textrm{⦁}}_{vp}}{P^o}\right]-\ }RT\int^{P^{\textrm{⦁}}_{vp}}_0{\left[\frac{\overline{V}^{\textrm{⦁}}_A\left(g\right)}{RT}-\frac{1}{P}\right]}dP \end{align*}
so that we can also express the standard Gibbs free energy of vaporization as
\begin{align*} \Delta_{vap}G^o\left(A,P^o\right) &= \Delta_{press}G\left(A,\ell \right)+\Delta_{press}G\left(A,g\right) \[4pt]&=\int^{P^{\textrm{⦁}}_{vp}}_{P^o}{\overline{V}^{\textrm{⦁}}_A}\left(\ell \right)\ dP-RT{ \ln \left[\frac{P^{\textrm{⦁}}_{vp}}{P^o}\right]-\ }RT\int^{P^{\textrm{⦁}}_{vp}}_0{\left[\frac{\overline{V}^{\textrm{⦁}}_A\left(g\right)}{RT}-\frac{1}{P}\right]}dP \end{align*}
Equating expressions for ${\Delta_{vap}G}^o\left(A,P^o\right)$, we find
\begin{align*} \Delta_fG^o\left(A,{HIG}^o\right)-{\Delta_fG}^o\left(A,\ell ,P^o\right) &= RT \ln \left[\frac{f_A\left({HIG}^o\right)}{f^{\textrm{⦁}}_A\left(\ell ,P^o\right)}\right] \[4pt]&= \int^{P^{\textrm{⦁}}_{vp}}_{P^o}{\overline{V}^{\textrm{⦁}}_A}\left(\ell \right)\ dP-RT{ \ln \left[\frac{P^{\textrm{⦁}}_{vp}}{P^o}\right]-\ }RT\int^{P^{\textrm{⦁}}_{vp}}_0{\left[\frac{\overline{V}^{\textrm{⦁}}_A\left(g\right)}{RT}-\frac{1}{P}\right]}dP \end{align*}
Thus, we can find the standard Gibbs free energy of formation of the hypothetical ideal gas from the standard Gibbs free energy of formation of the liquid, the equilibrium vapor pressure of the pure substance, and the equation of state of the pure gas:
\begin{align*} \Delta_fG^o\left(A,{HIG}^o\right)-\Delta_fG^o\left(A,\ell ,P^o\right) &=\Delta_fG^o\left(A,{HIG}^o\right)-\mu^{\textrm{⦁}}_A\left(\ell ,P^o\right) \[4pt] &=\int^{P^{\textrm{⦁}}_{vp}}_{P^o}{\overline{V}^{\textrm{⦁}}_A}\left(\ell \right)\ dP-RT{ \ln \left[\frac{P^{\textrm{⦁}}_{vp}}{P^o}\right]-\ }RT\int^{P^{\textrm{⦁}}_{vp}}_0{\left[\frac{\overline{V}^{\textrm{⦁}}_A\left(g\right)}{RT}-\frac{1}{P}\right]}dP \end{align*}
If the vapor of the substance behaves as an ideal gas, the last integral vanishes. If we also neglect the integral of the molar volume of the liquid, we have
${\Delta_fG}^o\left(A,{HIG}^o\right)={\Delta_fG}^o\left(A,\ell ,P^o\right)-RT{ \ln \left[\frac{P^{\textrm{⦁}}_{vp}}{P^o}\right]\ }=\mu^{\textrm{⦁}}_A\left(\ell ,P^o\right)-RT{ \ln \left[\frac{P^{\textrm{⦁}}_{vp}}{P^o}\right]\ } \nonumber$ (ideal gas)
and $f^{\textrm{⦁}}_A\left(\ell ,P^o\right)=P^{\textrm{⦁}}_{vp}$.
In this development, we suppose that we know ${\Delta_fG}^o\left(A,\ell ,P^o\right)$ from thermal measurements. We can calculate the Gibbs free energy difference between the liquid and hypothetical ideal-gas standard states if we have an equation of state for the vapor. The chemical potential in the hypothetical ideal-gas standard state is
$\mu^o_A\left({HIG}^o\right)={\Delta_fG}^o\left(A,{HIG}^o\right) \nonumber$
and the chemical potential of the pure liquid
$\mu^{\textrm{⦁}}_A\left(\ell ,P^o\right)={\Delta_fG}^o\left(A,\ell ,P^o\right) \nonumber$
is expressed as a function of the fugacity of the pure liquid:
$\mu^{\textrm{⦁}}_A\left(\ell ,P^o\right)=\mu^o_A\left({HIG}^o\right)+RT{ \ln \left[\frac{f^{\textrm{⦁}}_A\left(\ell ,P^o\right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$
The activity formalism provides an alternative way to express the same information. When we choose the pure liquid at one bar as the activity standard state; we set ${\tilde{a}}_A\left(\ell ,P^o\right)=1$. For this activity scale, the standard chemical potential becomes the Gibbs free energy of formation of the pure liquid, ${\widetilde\mu}^o_A\left(\ell \right)={\Delta_fG}^o\left(A,\ell ,P^o\right)$. Since the chemical potential of the hypothetical ideal-gas standard state is $\mu^o_A\left({HIG}^o\right){{=\Delta }_fG}^o\left(A,{HIG}^o\right)$, the activity relationship becomes
${\Delta_fG}^o\left(A,{HIG}^o\right)-{\Delta_fG}^o\left(A,\ell ,P^o\right) =\mu^o_A\left({HIG}^o\right)-{\widetilde\mu}^o_A\left(\ell \right)=-\left(\mu^{\textrm{⦁}}_A\left({\ell ,P}^o\right)-\mu^o_A\left({HIG}^o\right)\right) =RT{ \ln \left[{\tilde{a}}_A\left({HIG}^o\right)\right]\ } \nonumber$
Comparison with the previous equation shows that ${ \ln \left[{\tilde{a}}_A\left({HIG}^o\right)\right]\ }={ \ln \left[\frac{f_A\left({HIG}^o\right)}{f^{\textrm{⦁}}_A\left(\ell ,P^o\right)}\right]\ } \nonumber$
If the pure liquid is at equilibrium with a gas mixture at $P^o$ in which the mole fraction of $A$ is $x^{eq}_A$, the fugacity of the pure liquid is equal to the fugacity of the gas in the mixture; that is,
$f^{\textrm{⦁}}_A\left(\ell ,P^o\right)=f_A\left(RG,x^{eq}_A,P^o\right) \nonumber$
so that
$\ln \left[{\tilde{a}}_A\left({HIG}^o\right)\right] = \ln \left[\frac{f_A\left({HIG}^o\right)}{f_A\left(RG,x^{eq}_A,P^o\right)}\right] \nonumber$
and
${\tilde{a}}_A\left({HIG}^o\right)=\frac{f_A\left({HIG}^o\right)}{f_A\left(RG,x^{eq}_A,P^o\right)} \nonumber$
Finally, let us take the activity standard state to be pure liquid $A$ at 1 bar and find the activity of $A$ in an arbitrary real-gas mixture whose pressure is $P$ and in which the mole fraction of $A$ is $x_A$. Let us represent this state as $A\left(RG,x_A,P\right)$. The activity of real gas $A$ in this state is
${\tilde{a}}_A\left(RG,x_A,P\right)=\frac{f_A\left(RG,x_A,P\right)}{f^{\textrm{⦁}}_A\left(\ell ,P^o\right)} = \frac{f_A\left(RG,x_A,P\right)}{f_A\left(RG,x^{eq}_A,P^o\right)} \nonumber$
and the chemical potential is
$\mu_A\left(RG,x_A,P\right)={\widetilde\mu}^o_A\left(\ell \right)+RT{ \ln \left[{\tilde{a}}_A\left(RG,x_A,P\right)\right]\ } \nonumber$
If the real gas that is present at mole fraction $x_A$ in a system whose pressure is $P$ can be treated approximately as an ideal gas, these gas-fugacity terms can be approximated as
$f_A\left(RG,x_A,P\right)\approx f_A\left(IG,x_A,P\right)=x_AP \nonumber$ and $f_A\left(RG,x^{eq}_A,P^o\right)\approx f_A\left(IG,x^{eq}_A,P^o\right)=f_A\left(IG,P^{\textrm{⦁}}_{vp}\right)=P^{\textrm{⦁}}_{vp} \nonumber$
The activity becomes
${\tilde{a}}_A\left(RG,x_A,P\right)\approx {\tilde{a}}_A\left(IG,x_A,P\right)\approx {x_AP}/{P^{\textrm{⦁}}_{vp}} \nonumber$
and the chemical potential becomes
$\mu_A\left(RG,x_A,P\right)\approx \mu_A\left(IG,x_A,P\right)={\widetilde\mu}^o_A\left(\ell \right)+RT{ \ln \left(\frac{x_AP}{P^{\textrm{⦁}}_{vp}}\right)\ } \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/15%3A_Chemical_Potential_Fugacity_Activity_and_Equilibrium/15.04%3A_Standard_States_for_the_Fugacity_and_Activity_of_a_Pure_.txt
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The relationship between the standard Gibbs free energy of formation of a substance whose standard state is a solid and the Gibbs free energy of the substance in its hypothetical ideal-gas standard state is essentially the same as described in the previous section for a liquid. In each case, to find the fugacity of the condensed phase in its standard state, it is necessary to find a reversible path that takes the condensed-phase substance to its hypothetical ideal-gas standard state. If the solid substance has a significant vapor (sublimation) pressure, the paths described for a liquid in the previous section are also available for the solid. Otherwise, it may be possible to determine the Gibbs free energy change along some more complicated path.
Of course, whether the Gibbs free energy of formation for the hypothetical ideal-gas standard state can be evaluated or not, the fugacity and activity relationships remain valid. For substance $A$ in another state—in which $A$ need not be pure and the pressure is generally not one bar—we have
$\mu_A=\mu^o_A+RT{ \ln \left[\frac{f_A\left(P,T,x_A,x_B,\dots .\right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$
where $\mu^o_A={\Delta }_fG^o\left(A,{HIG}^o,T\right)$, and the fugacity of substance $A$, $f_A\left(P,T,x_A,x_B,\dots .\right)$, is simply an alternative expression of the difference between the chemical potential of the substance, as it occurs in the system, and its chemical potential in the hypothetical ideal-gas standard state. We write $f_A\left(P,T,x_A,x_B,\dots .\right)$ to indicate that the state of the system is specified by its pressure, temperature, and composition.
When the fugacity is difficult to measure, the activity function becomes essential. Choosing the standard state for the activity to be pure solid A in the same standard state that we use for the Gibbs free energy of formation, we have
${\widetilde\mu}^o_A\left(s\right)={\Delta }_fG^o\left(A,s\right) \nonumber$ and
${\widetilde\mu}^o_A\left(s\right)=\mu^o_A+RT{ \ln \left[\frac{f^{\textrm{⦁}}_A\left(s,P^o\right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$
Then, for substance $A$ in an arbitrary state at the temperature of interest:
\begin{align*} \mu_A &= \mu^o_A+RT{ \ln \left[\frac{f_A\left(P,T,x_A,x_B,\dots .\right)}{f_A\left({HIG}^o\right)}\right]\ } \[4pt] &= {\widetilde\mu}^o_A\left(s\right)-RT{ \ln \left[\frac{f^{\textrm{⦁}}_A\left(s,P^o\right)}{f_A\left({HIG}^o\right)}\right]+\ }RT{ \ln \left[\frac{f_A\left(P,T,x_A,x_B,\dots .\right)}{f_A\left({HIG}^o\right)}\right]\ } \[4pt] &= {\widetilde\mu}^o_A\left(s\right)+RT{ \ln \left[\frac{f_A\left(P,T,x_A,x_B,\dots .\right)}{f^{\textrm{⦁}}_A\left(s,P^o\right)}\right]\ } \[4pt] &={\widetilde\mu}^o_A\left(s\right)+RT{ \ln {\tilde{a}}_A\ } \end{align*}
From one perspective, the activity function is simply a mathematical device that expresses the chemical potential relative to an arbitrarily chosen reference state. If we can measure this difference experimentally, we can find ${\tilde{a}}_A$ whether we can measure the fugacity of $A$ or not. For clarity, we designate the chemical potential in the reference state as ${\widetilde\mu}^o_A\left(s\right)$. When we let ${\widetilde\mu}^o_A\left(s\right)$ be the Gibbs free energy of formation of the substance in the arbitrarily chosen reference state and let the activity in this reference state be unity, $\mu_A$ is the chemical potential difference between the substance in the state of interest and the chemical potential of its constituent elements in their standard states at the same temperature.
15.06: Chemical Potential Fugacity and Equilibrium
In Chapter 13, we develop the relationship between the standard Gibbs free energy change for a reaction and the equilibrium constant for that reaction, under the assumption that all of the substances involved in the reaction behave ideally. In the gas phase, they behave as ideal gases; when dissolved in a solution, their concentrations are proportional to their mole fractions in a gas phase at equilibrium with the solution.
We can now repeat this development using the fugacities instead of the pressures of the reacting species. Let the reaction be
$aA+bB\to cC+dD. \nonumber$
We introduce the reactions that create the reactants and the products in their hypothetical ideal-gas standard states from their elements in their standard states. The Gibbs free energy change for creating the reactants in their hypothetical ideal-gas standard states from the elements in their standard states is
$a{\Delta }_fG^o\left(A,{HIG}^o\right)+b{\Delta }_fG^o\left(B,{HIG}^o\right) \nonumber$
The Gibbs free energy change for creating the products in their standard states, from the same set of elements, is
$c{\Delta }_fG^o\left(C,{HIG}^o\right)+d{\Delta }_fG^o\left(D,{HIG}^o\right) \nonumber$
Next, we introduce a set of processes, each of which adds a further quantity of a reactant or product to a very large system at equilibrium. That is, we transfer an additional $a$ moles of pure $A$ from its hypothetical ideal-gas standard state into a very large system in which its fugacity is $f_A\left(P,T,x_A,x_B,\dots .\right)$. The Gibbs free energy change for this process is
$a\ RT{ \ln \left[\frac{f_A\left(P,T,x_A,x_B,\dots .\right)}{f_A\left({HIG}^o\right)}\right]\ } \nonumber$
Corresponding processes add $b$ moles of pure $B$ at fugacity $f_B\left(P,T,x_A,x_B,\dots .\right)$, etc. These processes are diagrammed in Figure 2. Since the fugacity of a substance in any state is a rigorous measure of the difference between its chemical potential in that state and its chemical potential in its hypothetical ideal-gas standard state, these Gibbs free energy changes are exact. Since the very large system is at equilibrium, there is no Gibbs free energy change when a moles of $A$ and $b$ moles of $B$ react according to $aA+bB\to cC+dD$.
Let us compute the Gibbs free energy change in a clockwise direction around the reversible cycle in Figure 2. The elements in their standard states are converted first to isolated products, then to components of the large equilibrium system, then to separated reactants, and finally back to the elements in their standard states. We have
$0=c\ {\Delta }_fG^o\left(C,{HIG}^o\right)+d\ {\Delta }_fG^o\left(D,{HIG}^o\right)-a\ {\Delta }_fG^o\left(A,{HIG}^o\right)-b\ {\Delta }_fG^o\left(B,{HIG}^o\right)+c\ RT{ \ln \left[\frac{f_C\left(P,T,x_A,x_B,\dots .\right)}{f_C\left({HIG}^o\right)}\right]\ }+d\ RT{ \ln \left[\frac{f_D\left(P,T,x_A,x_B,\dots .\right)}{f_D\left({HIG}^o\right)}\right]\ }-a\ RT{ \ln \left[\frac{f_A\left(P,T,x_A,x_B,\dots .\right)}{f_A\left({HIG}^o\right)}\right]\ }-b\ RT{ \ln \left[\frac{f_B\left(P,T,x_A,x_B,\dots .\right)}{f_B\left({HIG}^o\right)}\right]\ } \nonumber$
To express the fugacity ratios more compactly, we let
$f_A=\frac{f_A\left(P,T,x_A,x_B,\dots .\right)}{f_A\left({HIG}^o\right)} \nonumber$
etc. Letting
${\Delta }_rG^o=c\ {\Delta }_fG^o\left(C,{HIG}^o\right)+d\ {\Delta }_fG^o\left(D,{HIG}^o\right)-a\ {\Delta }_fG^o\left(A,{HIG}^o\right)-b\ {\Delta }_fG^o\left(B,{HIG}^o\right) \nonumber$
the Gibbs free energy around this reversible cycle simplifies to ${\Delta }_rG^o=-RT{ \ln \frac{f^c_Cf^d_D}{f^a_Af^b_B}\ } \nonumber$
We can express the criterion for equilibrium as
$K_f=\frac{f^c_Cf^d_D}{f^a_Af^b_B} \nonumber$
where
$K_f=\mathrm{exp}\left(\frac{-{\Delta }_rG^o}{RT}\right) \nonumber$
We introduce the subscript, “$f$”, to indicate that the equilibrium constant, $K_f$, is a function of the fugacities of the reacting substances. These relationships parallel those that we found for equilibrium among ideal gases, with real-gas fugacities replacing ideal-gas pressures.
As we did when we considered ideal-gas equilibria, let us suppose that the very large equilibrium system contains a liquid phase in which the reactants and products are soluble. The reaction can also occur in this liquid phase, and this liquid-phase reaction must be at equilibrium. Since the system is at equilibrium, each chemical species must have the same chemical potential in the solution as it does in the gas phase. Hence, the fugacities and the fugacity-based equilibrium constant are the same in both phases.
At this point, we have obtained—in principle—a complete solution to the problem of predicting the equilibrium position for any reaction. If we can find the Gibbs free energy of formation of each substance in its hypothetical ideal-gas standard state, and we can find its fugacity as a function of the composition and pressure of the system in which the reaction occurs, we can find the equilibrium constant and the equilibrium composition of the system.
In practice, a great many substances are non-volatile. The Gibbs free energy of formation of their hypothetical ideal-gas standard states and their fugacities cannot be measured. For such substances, we have recourse to other standard states and use activities to express the equilibrium constant.
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When we choose a standard state for the activity of a substance, we want the chemical potential that we calculate from the measured activity of a substance in a particular system to be the Gibbs free energy difference for the formation of the substance in that system from its constituent elements in their standard states.
Thus we must be able to measure the Gibbs free change for the process that produces the substance in its activity standard state from its constituent elements; this is the quantity that we designate as ${\Delta }_f{\tilde{G}}^o\left(A,ss\right)={\widetilde{\mu }}^o_A$. We must also be able to measure the difference between the chemical potential of the substance in this standard state and its chemical potential in the system of interest; this is the quantity that we designate as $RT{ \ln {\tilde{a}}_A\ }$. The sum ${\mu }_A={\widetilde{\mu }}^o_A+RT{ \ln {\tilde{a}}_A\ } \nonumber$
is then the desired Gibbs free energy change for formation of the substance in the system of interest.
When it is convenient to choose the activity standard state to be the standard state of the pure liquid or the pure solid, we have
${\Delta }_f{\tilde{G}}^o\left(A,ss\right)={\widetilde{\mu }}^o_A={\Delta }_fG^o\left(A,\ell \right) \nonumber$ or ${\Delta }_f{\tilde{G}}^o\left(A,ss\right)={\widetilde{\mu }}^o_A={\Delta }_fG^o\left(A,s\right) \nonumber$
For other choices, we may be unable to measure the difference between the chemical potential of the substance in the activity standard state and that of its constituent elements in their standard states; that is, the value of ${\Delta }_f{\tilde{G}}^o\left(A,ss\right)={\widetilde{\mu }}^o_A$ may be unknown. Nevertheless, we can describe the Gibbs free energy changes in a cycle that goes from the elements in their standard states to the chemical species in the equilibrium system and back.
For the reaction $aA+bB\to cC+dD$, this cycle is shown in Figure 3. Summing the Gibbs free energy changes in a clockwise direction around this cycle, we have
$0=c\ {\Delta }_f{\tilde{G}}^o\left(C,ss\right)+d\ {\Delta }_f{\tilde{G}}^o\left(D,ss\right)-a\ {\Delta }_f{\tilde{G}}^o\left(A,ss\right)-b\ {\Delta }_f{\tilde{G}}^o\left(B,ss\right) \nonumber$ $+c\ RT{ \ln {\tilde{a}}^c_C\ }+d\ RT{ \ln {\tilde{a}}^d_D\ }-a\ RT{ \ln {\tilde{a}}^a_A\ }-b\ RT{ \ln {\tilde{a}}^b_B\ } \nonumber$
Writing ${\Delta }_r{\tilde{G}}^o$ to emphasize that the standard Gibbs free energy change for the reaction is now a difference between the chemical potentials of reactants and products in their activity standard states, we have
${\Delta }_r{\tilde{G}}^o=c\ {\Delta }_f{\tilde{G}}^o\left(C,ss\right)+d\ {\Delta }_f{\tilde{G}}^o\left(D,ss\right)-a\ {\Delta }_f{\tilde{G}}^o\left(A,ss\right)-b\ {\Delta }_f{\tilde{G}}^o\left(B,ss\right)={\Delta }_r{\widetilde{\mu }}^o \nonumber$
and the equation for the Gibbs free energy change around the cycle becomes
${\Delta }_r{\tilde{G}}^o={\Delta }_r{\widetilde{\mu }}^o=-RT{ \ln \frac{{\tilde{a}}^c_C{\tilde{a}}^d_D}{{\tilde{a}}^a_A{\tilde{a}}^b_B}\ } \nonumber$
Since the chemical reaction is at equilibrium, the activities term is constant. We have
$K_a=\frac{{\tilde{a}}^c_C{\tilde{a}}^d_D}{{\tilde{a}}^a_A{\tilde{a}}^b_B} \nonumber$
and $K_a=\mathrm{exp}\left(\frac{-{\Delta }_r{\tilde{G}}^o}{RT}\right)=\mathrm{exp}\left(\frac{-{\Delta }_r{\widetilde{\mu }}^o}{RT}\right) \nonumber$
We append the subscript, “$a$”, to indicate that the equilibrium constant, $K_a$, is expressed in terms of the activities of the reacting substances. This cycle demonstrates the underlying logic for our calculation of $K_a$ from our definition of activity, ${\mu }_A={\widetilde{\mu }}^o_A+RT{ \ln {\tilde{a}}_A\ }$, our definition of ${\Delta }_r\mu$, and the fact that
${\Delta }_r\mu =\sum^{\omega }_{j=1}{{\mu }_j{\nu }_j=0} \nonumber$
is a criterion for equilibrium.
Introducing activity coefficients, we have
$K_a=\frac{{\tilde{a}}^c_C{\tilde{a}}^d_D}{{\tilde{a}}^a_A{\tilde{a}}^b_B}=\left(\frac{c^c_Cc^d_D}{c^a_Ac^b_B}\right)\left(\frac{{\gamma }^c_C{\gamma }^d_D}{{\gamma }^a_A{\gamma }^b_B}\right) \nonumber$
When we know $K_a$ and can estimate the activity coefficients, we can predict the thermodynamic properties of a system whose component concentrations are known. In general, to determine $K_a$ and the activity coefficients is more difficult than to determine the equilibrium concentrations. Over narrow ranges of concentrations it is often adequate to assume that the activity-coefficient function,
$K_{\gamma }=\frac{{\gamma }^c_C{\gamma }^d_D}{{\gamma }^a_A{\gamma }^b_B} \nonumber$
is approximately constant. When this is the case, the concentration function,
$K_c=\frac{c^c_Cc^d_D}{c^a_Ac^b_B} \nonumber$
is approximately constant. Then, a direct measurement of $K_c$ in one equilibrium system makes it possible to predict the position of equilibrium in other similar systems.
The relationship between the equilibrium constant and the standard Gibbs free energy change for a reaction is extremely useful. If we can calculate the standard Gibbs free energy change from tabulated values, we can find the equilibrium constant and predict the position of equilibrium for a particular system. Conversely, if we can measure the equilibrium constant, we can find the standard Gibbs free energy change, ${\Delta }_r{\tilde{G}}^o$.
This brings us back to a central challenge in our development. We now have rigorous relationships between the Gibbs free energy and the fugacity or activity of a substance. To use these relationships, we must be able to relate the fugacity or activity to the concentration of a substance in any particular system that we want to study. For many common systems, this is difficult. In Chapter 16, we discuss some basic approaches to accomplishing it.
15.08: The Rate of Gibbs Free Energy Change with Extent of Reac
In Section 13.5, we demonstrate that the Gibbs free energy change for a reaction among ideal gases is the same thing as the rate at which the Gibbs free energy of the system changes with the extent of reaction. That is, for an ideal-gas reaction at constant temperature and pressure, we find ${\Delta }_rG={\left({\partial G}/{\partial \xi }\right)}_{TP}$. We can now show that this conclusion is valid for any reaction.
With the introduction of the activity function, we have developed a very general expression for the Gibbs free energy of any substance in any system. For substance $A$ at a fixed temperature, we have
${\overline{G}}_A={\mu }_A={\widetilde{\mu }}^o_A+RT{ \ln {\tilde{a}}_A\ } \nonumber$
For a reaction that we describe with generalized substances and stoichiometric coefficients as
$\left|{\nu }_1\right|X_1+\left|{\nu }_2\right|X_2+\dots +\left|{\nu }_i\right|X_i\to \ \left|{\nu }_j\right|X_j+\left|{\nu }_k\right|X_k+\dots +\left|{\nu }_{\omega }\right|X_{\omega } \nonumber$
we can write the Gibbs free energy change in several equivalent ways:
\begin{align*} {\Delta }_rG &= \sum^{\omega }_{j=1}{\nu_j{\overline{G}}_j} \[4pt] &={\Delta }_r\mu \[4pt] &=\sum^{\omega }_{j=1}{{\nu }_j{\mu }_j\le 0} \end{align*}
The Gibbs free energy of the system is a function of temperature, pressure, and composition,
$G=G\left(P,T,n_1,n_2,\dots ,n_{\omega }\right) \nonumber$
To introduce the dependence of the Gibbs free energy of the system on the extent of reaction, we use the stoichiometric relationships $n_j=n^o_j+{\nu }_j\xi$. ($n_j$ is the number of moles of the $j^{th}$ reacting species; $n^o_j$ is the number of moles of the $j^{th}$ reacting species when $\xi$ =0. If the $k^{th}$ substance does not participate in the reaction, ${\nu }_k=0$.) Then,
$G=G\left(P,T,n^o_1+{\nu }_1\xi ,{\ n}^o_2+{\nu }_2\xi ,\dots ,\ n^o_{\omega }+{\nu }_{\omega }\xi \right) \nonumber$
At constant temperature, pressure, and composition, the dependence of the Gibbs free energy on the extent of reaction is
\begin{align*} \left(\frac{\partial G}{\partial \xi }\right)_{PTn_m} &= \sum^{\omega}_{j=1} \left(\frac{\partial G}{\partial \left(n^o_j + \nu_j\xi \right)}\right)_{PTn_{m\neq j}} \left(\frac{\partial \left(n^o_j + \nu_j \xi \right)}{\partial \xi }\right)_{PTn_m} \[4pt] &=\sum^{\omega}_{j=1} \nu_j \mu_j \end{align*}
It follows that
${\left(\frac{\partial G}{\partial \xi }\right)}_{PTn_m}={\Delta }_rG\le 0 \nonumber$
expresses the thermodynamic criteria for change when the process is a chemical reaction.
If a reacting system is not at equilibrium, the extent of reaction is time-dependent. We see that the Gibbs free energy of a reacting system depends on time according to
$\frac{dG}{dt}={\left(\frac{\partial G}{\partial \xi }\right)}_{PTn_m}\left(\frac{d\xi }{dt}\right)={\Delta }_r\mu \left(\frac{d\xi }{dt}\right) \nonumber$
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Problems
Use data from the table below to find the thermodynamic properties requested in problems 1 to 7.
Properties $\boldsymbol{C}{\boldsymbol{H}}_{\boldsymbol{3}}\boldsymbol{OH}$ $\boldsymbol{C}{\boldsymbol{H}}_{\boldsymbol{3}}\boldsymbol{C}{\boldsymbol{H}}_{\boldsymbol{2}}\boldsymbol{OH}$
Density, $\mathrm{g}\ {\mathrm{cm}}^{-3}$ at 20 C 0.7914 0.7893
Mol mass, $\mathrm{\ g}\ {\mathrm{mol}}^{-1}$ 32.04 46.07
bp, C 64.6 78.2
${\Delta }_fG^o\left(300\mathrm{K},\ {HIG}^o\right)$$\mathrm{kJ}\ {\mathrm{mol}}^{-1}$ –159.436 –162.934
Vapor pressure at320 K, bar 0.5063 0.2764
Virial coefficient, $B$, $\mathrm{\ }{\mathrm{m}}^3\ {\mathrm{mol}}^{-1}$ $-1.421\times {10}^{-3}$ $-2.710\times {10}^{-3}$
1. Find the chemical potentials of the pure gases, assuming that they are ideal, taking the hypothetical ideal gas standard state as the standard state for activity ($f=\tilde{a}=P$).
2. Find the chemical potentials of the mixed gases, assuming that they are ideal, taking the hypothetical ideal gas standard state as the standard state for activity ($f={\tilde{a}}_A=x_AP)$.
3. Find the chemical potentials of the mixed gases, assuming them to obey the Virial equation, ${PV}/{RT=1+\left({BP}/{RT}\right)}, \nonumber$ assuming that the partial molar volumes in the mixture are equal to the partial molar volumes of the pure gases at the same pressure, and taking the hypothetical ideal gas standard state as the standard state for activity.
4. Find the standard chemical potentials of the pure liquids at 320 K, assuming that the gases behave ideally.
5. Find the standard chemical potentials of the pure liquids at 320 K, assuming that the gases obey the Virial equation.
6. Find the chemical potential of the pure liquids as a function of pressure, assuming that the partial molar volumes of the pure liquids are constant and that the gases obey the Virial equation.
7. Find the activity and chemical potential of the pure liquids at 101 bar, taking the pure liquids at 1 bar as the standard state for activity.
8. A system is created by mixing one mole of gas $A$ with one mole of gas $C$. Reaction occurs according to the stoichiometry $A+C\ \ \rightleftharpoons \ D$. Assume that the behaviors of these gases in their equilibrium mixture are adequately approximated by the Virial equations ${P{\overline{V}}_A}/{RT=1+\left({B_AP}/{RT}\right)}$, etc.
(a) Show that the fugacity of gas A is given by ${ \ln f_A={ \ln \left(\frac{x_AP}{P^o}\right)\ }\ }+\frac{B_AP}{RT} \nonumber$ (b) Write an equation for ${\mu }_A={\mu }_A\left(x_A,P\right)$ at constant temperature, $T$.
(c) Write an equation for ${\Delta }_r\mu$.
(d) Assume that
${\Delta }_r{\mu }^o={\mu }^o_D-{\mu }^o_A-{\mu }^o_C=1000\ \mathrm{J}\ {\mathrm{mol}}^{-1}$. If all three gases behave ideally ($B_A=B_C=B_D=0$), what is $K_P$ for this reaction at 300 K? At equilibrium at 300 K and 1 bar, what are the mole fractions of $A$, $C$, and $D$?
(e) Under the assumptions in part (d), what are the equilibrium mole fractions of $A$, $C$, and $D$ at 300 K and 10 bar?
(f) Suppose that, contrary to the assumptions in (d) and (e), the Virial coefficients are not zero and that $B_A=B_C=B_D=-{10}^{-3}\ {\mathrm{m}}^3\ {\mathrm{mol}}^{-1}$. At equilibrium at 300 K and 1 bar, what are the mole fractions of $A$, $C$, and $D$?
(g) Under the assumptions in part (f), what are the equilibrium mole fractions of $A$, $C$, and $D$ at 300 K and 10 bar?
9. Suppose that the reaction $A+C\ \ \rightleftharpoons \ D$ occurs in an inert solvent and that it is convenient to express concentrations as molarities. A frequently convenient choice of activity standard state for a solute is a hypothetical one-molar solution in which the chemical potential of the solute is equal to the chemical potential of the solute in a very (“infinitely”) dilute solution in the same solvent. Then ${\tilde{a}}_A=\left[A\right]{\gamma }_A$, and ${\gamma }_A\to 1$ as $\left[A\right]\to 0$. The thermodynamic equilibrium constant becomes $K_a=\frac{{\tilde{a}}_D}{{\tilde{a}}_A{\tilde{a}}_C}=\frac{\left[D\right]}{\left[A\right]\left[C\right]}\frac{{\gamma }^d_D}{{\gamma }^a_A{\gamma }^c_C}=K_cK_{\gamma } \nonumber$ where we introduce $K_c=\frac{\left[D\right]}{\left[A\right]\left[C\right]} \nonumber$ and $K_{\gamma }=\frac{{\gamma }^d_D}{{\gamma }^a_A{\gamma }^c_C} \nonumber$ In a very dilute solution, $K_{\gamma }\to 1$ and $K_c=K_a$. Therefore, we can estimate $K_a$ by finding the limiting value of $K_c$ as all of the concentrations become very small. From values of $K_c$ at higher concentrations, we can develop an empirical equation for $K_{\gamma }$. The form of this equation can be anything that can adequately represent the experimental data. Note, however, that finding an empirical model for $K_{\gamma }$ does not solve the problem of finding empirical models for ${\gamma }_A$, ${\gamma }_C$, and ${\gamma }_D$ individually.
(a) Given that the hypothetical one-molar solution is chosen to be the activity standard state for all three species, what is the physical significance of ${\Delta }_r{\widetilde{\mu }}^o$?
(b) A simple function that has the properties required of ${\gamma }_A$ is ${\gamma }_A={\alpha }^{\left[A\right]}$, where $\alpha$ is a constant. Represent , ${\gamma }_C$ and , ${\gamma }_D$ by similar functions and show that this leads to ${ \ln K_{\gamma }={\beta }_A\ }\left[A\right]+{\beta }_C\left[C\right]+{\beta }_D\left[D\right]$, where ${\beta }_A$, ${\beta }_C$, and ${\beta }_D$ are constants.
(c) A series of solutions is prepared. The equilibrium concentrations of $A$, $B$, and $D$ in these solutions are given below. Calculate $K_c$ for each solution. Estimate $K_a$ and the parameters ${\beta }_A$, ${\beta }_C$, and ${\beta }_D$ in the equation of part (b).
$\left[\boldsymbol{A}\right]$ [B] [C]
1.96 x 10${}^{-3}$ 1.96 x 10${}^{-3}$ 3.84 x 10${}^{-5}$
7.85 x 10${}^{-3}$ 1.85 x 10${}^{-3}$ 1.45 x 10${}^{-4}$
3.94 x 10${}^{-2}$ 1.44 x 10${}^{-3}$ 5.57 x 10${}^{-4}$
1.99 x 10${}^{-1}$ 7.15 x 10${}^{-4}$ 1.29 x 10${}^{-3}$
9.98 x 10${}^{-1}$ 2.84 x 10${}^{-4}$ 1.72 x 10${}^{-3}$
1.85 x 10${}^{-3}$ 7.85 x 10${}^{-3}$ 1.45 x 10${}^{-4}$
1.44 x 10${}^{-3}$ 3.94 x 10${}^{-2}$ 5.60 x 10${}^{-4}$
6.97 x 10${}^{-4}$ 1.99 x 10${}^{-1}$ 1.30 x 10${}^{-3}$
2.38 x 10${}^{-4}$ 9.98 x 10${}^{-1}$ 1.76 x 10${}^{-3}$
9.32 x 10${}^{-3}$ 7.32 x 10${}^{-3}$ 6.78 x 10${}^{-4}$
3.24 x 10${}^{-2}$ 3.04 x 10${}^{-2}$ 9.57 x 10${}^{-3}$
1.06 x 10${}^{-1}$ 1.04 x 10${}^{-1}$ 9.59 x 10${}^{-2}$
(d) Using the values you find in part (c), estimate the equilibrium concentrations of $A$, $B$, and $D$ when a solution is prepared by mixing one mole of $A$ with one mole of $C$ and sufficient solvent to make 1 L of solution at equilibrium.
10. An ester, $RCO_2R'$, undergoes hydrolysis in an ether solvent: $\ce{RCO2R^{'} + H2O <=> RCO2H + HOR^{'}} \nonumber$ We can express the activity of any of these species as the product of a concentration (in any convenient units) and an activity coefficient. When all of the reactants and products are present at low concentrations, the activity coefficients are approximately unity. The standard state for each species becomes a hypothetical solution of unit concentration in which the chemical potential (per mole) of that species is the same as its chemical potential in an arbitrarily (infinitely) dilute solution. A solution is prepared by mixing $2\times {10}^{-3}$ mole of the ester and ${10}^{-1}$ mole water in sufficient ether to make 1 L of solution. When equilibrium is reached, the acid and alcohol concentrations are $9.66\times {10}^{-3}$ molar.
(a) What is the equilibrium constant for this reaction?
(b) A solution is prepared by mixing $2\times {10}^{-2}$ mole of the acid, 3$\times {10}^{-2}$ mole of the alcohol, and ${10}^{-1}$ mole of water in sufficient ether to make 1 L of solution. When equilibrium is reached, what is the concentration of the ester?
(c) With this choice of the standard states, what physical process does the standard chemical-potential change, $\Delta {\widetilde{\mu }}^o$, describe?
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/15%3A_Chemical_Potential_Fugacity_Activity_and_Equilibrium/15.09%3A_Problems.txt
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We are frequently interested in equilibrium processes that occur in a solution at a constant temperature. If we are able to find the activities of the species making up the solution, we can describe the thermodynamics of such processes. Many experimental methods have been developed for the measurement of the activities of species in solution. In general, the accurate measurement of chemical activities is experimentally exacting. In this chapter, we consider some of the basic concepts involved. We focus primarily on molecular solvents and solutes; that is, neutral molecules that exist as such in solution. We introduce a simplified model, called the ideal solution model, which is often a useful approximation, particularly for dilute solutions. In Sections 16.16-16.18, we touch on the special issues that arise when we consider the activities of dissolved ions.
Thumbnail: The Effect of Solution Formation on Entropy. (CC BY-SA-NC; anonymous by request)
16: The Chemical Activity of the Components of a Solution
One way to find activities is to find the composition and pressure of the gas phase that is in equilibrium with the solution. If the gases are not ideal, we also need experimental data on the partial molar volumes of the components in the gas phase. Collecting such data is feasible for solutions of volatile molecular liquids. For solutions of electrolytes or other non-volatile components, other methods are required.
The curves sketched in Figure 1 describe a system containing components $A$ and $B$. The mole fractions in the solution and the mole fractions in the gas are related in a non-linear way. Let the mole fractions in the gas be $x_A$ and $x_B$; let those in the solution be $y_A$ and $y_B$. We have $x_A+x_B=1$ and $y_A+y_B=1$. At equilibrium, both phases are at the same pressure, $P$.
We imagine obtaining the data we need about this system by preparing many mixtures of $A$ and $B$. Beginning with an entirely liquid system at some applied pressure, we slowly decrease the applied pressure until the applied pressure becomes equal to the equilibrium pressure, $P$, and the liquid begins to vaporize. Figure 2 shows this system schematically.
We determine the compositions of the gas and liquid phases by chemical analysis; for each system, we determine $P$, $x_A$, $x_B$, $y_A$, and $y_B$. From these data we can develop empirical equations that express $P$, $x_A$, and $x_B$as functions of $y_A$; that is, we have $P=P\left(y_A\right)$, $x_A=x_A\left(y_A\right)$, and $x_B=x_B\left(y_A\right)$. Finally, we can find the value of the products $x_AP$ and $x_BP$. Figure 3 illustrates a possible function $P=P\left(y_A\right)$ and the products $x_AP$, and $x_BP$, when the gas-phase mole fractions depend on $y_A$ as shown in Figure 1.
With the hypothetical ideal gas standard state as the standard state for $A$ in the gas phase, we see in Section 14.11 that the chemical potential of $A$ in the gas phase is
${\mu }_A\left(g,P,x_A,x_B\right)={\Delta }_fG^o\left(A,{HIG}^o\right)+RT{ \ln \left[\frac{x_AP}{P^o}\right]+RT\int^P_0{\left(\frac{{\overline{V}}_A\left(g\right)}{RT}-\frac{1}{P}\right)}\ }dP \nonumber$ (any gas; activity standard state is HIG${}^{o}$)
where ${\overline{V}}_A\left(g\right)$ is the partial molar volume and $x_A$ is the mole fraction of $A$ in the gaseous mixture. The fugacity and activity of $A$ in the gas phase are given by
${ \ln \left[{\tilde{a}}_A,P,x_A,x_B\right]\ }={ \ln \left[\frac{f_A\left(g,P,x_A,x_B\right)}{f_A\left({HIG}^o\right)}\right]\ }={ \ln \left[\frac{x_AP}{P^o}\right]+\int^P_0{\left(\frac{{\overline{V}}_A\left(g\right)}{RT}-\frac{1}{P}\right)}\ }dP \nonumber$
and the standard state chemical potential is
${\mu }^o_A={\Delta }_fG^o\left(A,{HIG}^o\right) \nonumber$
We want to express the chemical potential of $A$ in the liquid solution using the properties of the solution. To do so, we introduce the chemical activity of component $A$. We write ${\tilde{a}}_A\left(P,y_A,y_B\right)$ to represent the activity of $A$ in a solution at pressure $P$ and in which the composition is specified by the mole fractions $y_A$ and $y_B$. If it suits our purposes, we are free to choose a standard state for the activity of $A$ in the liquid solution that is different from the standard state we choose for $A$ in the gas phase. For reasons that become apparent below, it is often useful to choose the standard state for the activity of $A$ in the liquid solution to be pure liquid $A$ at its equilibrium vapor pressure, $P^{\textrm{⦁}}_A$. We represent the chemical potential of $A$ in this standard state by ${\widetilde{\mu }}^o_A\left(\ell ,P^{\textrm{⦁}}_A\right)$. Note that this state is not identical to the standard state for the pure liquid, for which the pressure is one bar and the chemical potential is ${\Delta }_fG^o\left(A,\ell \right)$. The chemical potential and the activity of A in the solution are related by
${\mu }_A\left(P,y_A,y_B\right)={\widetilde{\mu }}^o_A\left(\ell ,P^{\textrm{⦁}}_A\right)+RT{ \ln \left[{\tilde{a}}_A\left(P,y_A,y_B\right)\right]\ } \nonumber$ (any solution; activity standard state for $A$ in solution is pure liquid $A$ at its equilibrium vapor pressure)
Since the system is at equilibrium, we have
${\mu }_A\left(g,P,x_A,x_B\right)={\mu }_A\left(P,y_A,y_B\right) \nonumber$
Equating our equations for these quantities, we find
${ \ln \left[{\tilde{a}}_A\left(P,y_A,y_B\right)\right]\ }=\frac{{\Delta }_fG^o\left(A,{HIG}^o\right)-{\widetilde{\mu }}^o_A\left(\ell ,P^{\textrm{⦁}}_A\right)}{RT}+{ \ln \left[\frac{x_AP}{P^o}\right]+\int^P_0{\left(\frac{{\overline{V}}_A\left(g\right)}{RT}-\frac{1}{P}\right)}\ }dP \nonumber$
This equation gives the activity of $A$ in a liquid solution whose state is specified by the liquid-phase mole fraction $y_A$. In particular, it must give the activity of $A$ in the “solution” for which $y_A=1$ and $y_B=0$; of course, this “solution” is pure liquid $A$. At equilibrium with pure liquid $A$, the gas phase contains pure gaseous $A$; therefore, we have $x_A=1$ and $P=P^{\textrm{⦁}}_A$. The gas-phase partial molar volume is that for the pure gas, ${\overline{V}}^{\textrm{⦁}}_A\left(g\right)$. Moreover, this “solution” is the standard state for the activity of component $A$, for which the activity of $A$ is unity; that is,
${\tilde{a}}_A\left(P,y_A,y_B\right)={\tilde{a}}_A\left(P^{\textrm{⦁}}_A,1,0\right)=1$. Making these substitutions into our equation for ${ \ln \left[{\tilde{a}}_A\left(P,y_A,y_B\right)\right]\ }$ and rearranging, we find
$\frac{{\Delta }_fG^o\left(A,{HIG}^o\right)-{\widetilde{\mu }}^o_A\left(\ell ,P^{\textrm{⦁}}_A\right)}{RT}=-{ \ln \left[\frac{P^{\textrm{⦁}}_A}{P^o}\right]-\int^{P^{\textrm{⦁}}_A}_0{\left(\frac{{\overline{V}}^{\textrm{⦁}}_A\left(g\right)}{RT}-\frac{1}{P}\right)}\ }dP \nonumber$
Substituting this result into our general equation for $\ln \left[ \tilde{a}_A\left(P,y_A,y_B\right)\right]$ we find a completely general function for the activity of component $A$.
$\ln \left[ \tilde{a}_A \left(P, y_A, y_B \right) \right] = \ln \left[ \frac{x_A P}{P^{\cdot}_A} \right] - \int_0^{P_A^{cdot}} \left( \frac{ \overline{V}_A^{\cdot} \left( g \right)}{RT} - \frac{1}{P} \right) dP + \int_)^P \left( \frac{ \overline{V}_A \left( g \right)}{RT} - \frac{1}{P} \right) dP \nonumber$
(any solution; $x_A$ is the mole fraction in the gas; the standard state for $A$ in solution is the pure liquid at its equilibrium vapor pressure)
In such a system, the roles of solute and solvent are interchangeable. Interchanging the labels “$A$” and “B” gives an equation for the activity of component $B$.
As circumstances warrant, several approximations can be applied to this result. When the partial molar volume of $A$ in the gas, ${\overline{V}}_A\left(g\right)$, is not available, some approximation is required. Perhaps the least drastic approximation is that introduced in Section 14.11. We equate the unknown partial molar volume to the partial molar volume of the pure real gas at the same system pressure. Setting ${\overline{V}}_A\left(g\right)={\overline{V}}^{\textrm{⦁}}_A\left(g\right)$, we have
${ \ln \left[{\tilde{a}}_A\left(P,y_A,y_B\right)\right]\ }={ \ln \left[\frac{x_AP}{P^{\textrm{⦁}}_A}\right]+\int^P_{P^{\textrm{⦁}}_A}{\left(\frac{{\overline{V}}^{\textrm{⦁}}_A\left(g\right)}{RT}-\frac{1}{P}\right)}\ }dP \nonumber$
Other approximations lead to greater simplifications. In the following sections, we discuss several. All of them assume that the components behave ideally in the gas phase. In this case, the integrals in our general equation for ${ \ln \left[{\tilde{a}}_A\left(\ P,y_A,y_B\right)\right]\ }$ vanish. Then,
${ \ln \left[{\tilde{a}}_A\left(P,y_A,y_B\right)\right]\ }={ \ln \left[\frac{x_AP}{P^{\textrm{⦁}}_A}\right]\ } \nonumber$
(solution; $x_A$ is the mole fraction in the ideal gas that is at equilibrium with the solution)
Our general result gives the activity of component $A$ in solution using the mole fraction of its own vapor in the equilibrium system. When we have an empirical function, $x_A=x_A\left(y_A\right)$, that relates the mole fraction in the gas to that in the solution, we can make this substitution and express the activity of $A$ in the solution using its concentration in the solution. In the next several sections, we develop some basic methods for finding $x_A=x_A\left(y_A\right)$.
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An ideal solution is a homogeneous liquid solution that is at equilibrium with an ideal-gas solution in which the vapor pressure of each component satisfies Raoult’s law${}^{1}$. Since the gas is ideal, the partial pressure of $A$ is $P_A=x_AP$. Raoult’s law asserts a relationship among the gas- and solution-phase mole fractions of $A$, the vapor pressure of the pure liquid, and the pressure of the system:
$P_A=x_AP=y_AP^{\textrm{⦁}}_A \nonumber$
(Raoult’s law)
For a binary mixture of $A$ and $B$ that satisfies Raoult’s law, we have also that $P_B=x_BP=y_BP^{\textrm{⦁}}_B$, and the total pressure becomes $P=P_A+P_B=y_AP^{\textrm{⦁}}_A+y_BP^{\textrm{⦁}}_B$. The lines sketched in Figure 4 show how $P_A$, $P_B$, and $P$ vary with the solution-phase composition when the solution is ideal.
When the standard state for $A$ in solution is taken to be pure liquid $A$ at its equilibrium vapor pressure, substitution of Raoult’s law into the results in Section 16.1 gives the activity of component $A$ in an ideal solution as
${ \ln \left[{\tilde{a}}_A\left(P,y_A,y_B\right)\right]\ }={ \ln \left[\frac{x_AP}{P^{\textrm{⦁}}_A}\right]\ }={ \ln \left[\frac{y_AP^{\textrm{⦁}}_A}{P^{\textrm{⦁}}_A}\right]\ }={ \ln y_A\ } \nonumber$ and
${\tilde{a}}_A\left(P,y_A,y_B\right)=y_A \nonumber$
(ideal solution, Raoult’s law)
In general, the activity and chemical potential of a component depend on pressure. If the solution is ideal, we see that the system pressure is fixed by $P=y_AP^{\textrm{⦁}}_A+y_BP^{\textrm{⦁}}_B$, and the pure-component vapor pressures depend only on temperature. Since for the binary solution, $y_B=1-y_A$, we can write the chemical potential of component $A$ as
${\mu }_A\left(P,y_A,y_B\right)={\mu }_A\left(y_A\right)={\widetilde{\mu }}^o_A\left(\ell ,P^{\textrm{⦁}}_A\right)+RT{ \ln y_A\ } \nonumber$ (ideal solution)
We can also use relationships we develop earlier to find another representation for ${\widetilde{\mu }}^o_A\left(\ell ,P^{\textrm{⦁}}_A\right)$. The chemical potential of $A$ in the liquid phase is the same as in the gas. Using the chemical potential for $A$ in the gas phase that we find in Section 16.1, we have
${\mu }_A\left(P,y_A,y_B\right)={\mu }_A\left(g,P,x_A,x_B\right)={\Delta }_fG^o\left(A,{HIG}^o\right)+RT{ \ln \left[\frac{x_AP}{P^o}\right]\ }={\Delta }_fG^o\left(A,{HIG}^o\right)+RT{ \ln \left[\frac{P^{\textrm{⦁}}_A}{P^o}\right]\ }+RT{ \ln y_A\ } \nonumber$ and hence, ${\widetilde{\mu }}^o_A\left(\ell ,P^{\textrm{⦁}}_A\right)={\Delta }_fG^o\left(A,{HIG}^o\right)+RT{ \ln \left[\frac{P^{\textrm{⦁}}_A}{P^o}\right]\ } \nonumber$
In Section 15.4, we find, for an ideal gas,
${\Delta }_fG^o\left(A,{HIG}^o\right)+RT{ \ln \left[\frac{P^{\textrm{⦁}}_A}{P^o}\right]\ }={\Delta }_fG^o\left(A,\ell \right)+\int^{P^{\textrm{⦁}}_A}_{P^o}{{\overline{V}}^{\textrm{⦁}}_A\left(\ell \right)}dP \nonumber$
so that the chemical potential of the pure liquid at its vapor pressure is also given by ${\widetilde{\mu }}^o_A\left(\ell ,P^{\textrm{⦁}}_A\right)={\Delta }_fG^o\left(A,\ell \right)+\int^{P^{\textrm{⦁}}_A}_{P^o}{{\overline{V}}^{\textrm{⦁}}_A\left(\ell \right)}dP \nonumber$
The integral is the difference between the Gibbs free energy of the pure liquid at its vapor pressure and that of the pure liquid at $P^o=1\ \mathrm{bar}$. Note that we can obtain the same result much more simply by integrating ${\left(dG^{\textrm{⦁}}_A\right)}_T={\overline{V}}^{\textrm{⦁}}_AdP$ between the same two states. In Section 15.3, we see that the value of the integral is usually negligible. To a good approximation, we have
${\widetilde{\mu }}^o_A\left(\ell ,P^{\textrm{⦁}}_A\right)\approx {\Delta }_fG^o\left(A,\ell \right) \nonumber$
(ideal solution)
16.03: Expressing the Activity Coefficient as a Deviation from
If the components behave ideally in the gas phase and if pure liquid $A$ at its equilibrium vapor pressure, $P^{\textrm{⦁}}_A$, is the standard state for the activity of $A$ in solution, we find in Section 16.2 that the activity of component $A$ is ${\tilde{a}}_A\left(P,y_A,y_B\right)={x_AP}/{P^{\textrm{⦁}}_A}$. Experimentally, we determine a relationship between the mole fractions of $A$ in the gas and liquid phases. By expressing this relationship as the function $x_A=x_A\left(y_A\right)$, we can express the activity as a function of $y_A$. If Raoult’s law is obeyed, we have seen that this function is $x_A\left(y_A\right)={y_AP^{\textrm{⦁}}_A}/{P}$, and the activity is ${\tilde{a}}_A\left(P,y_A,y_B\right)=y_A$.
If Raoult’s law is not obeyed, we must find an alternative function that adequately describes the experimentally observed relationship between $x_A$ and $y_A$. As we note in Section 14.10, we want to construct this function so that it approaches $y_A$ whenever the behavior of the solution approaches the behavior of an ideal solution. We can accomplish this by defining the activity coefficient for component $A$, ${\gamma }_A={\gamma }_A\left(P,y_A,y_B\right)$, by the equation
${\tilde{a}}_A\left(P,y_A,y_B\right)=y_A{\gamma }_A\left(P,y_A,y_B\right) \nonumber$ (Raoult’s law activity)
where the argument lists serve to emphasize that the activity and the activity coefficient are functions of the same thermodynamic variables.
Dropping the argument lists and equating the two activity relationships, we have
$\frac{x_AP}{P^{\textrm{⦁}}_A}=y_A{\gamma }_A \nonumber$
so that the activity coefficient is
${\gamma }_A=\frac{x_AP}{y_AP^{\textrm{⦁}}_A} \nonumber$ (Raoult’s Law activity coefficient)
Since we are using pure liquid $A$ at its equilibrium vapor pressure as the standard state for component $A$, the chemical potential can be expressed as
${\mu }_A\left(P,y_A,y_B\right)={\widetilde{\mu }}^o_A\left(\ell ,P^{\textrm{⦁}}_A\right)+RT{ \ln y_A{\gamma }_A\ } \nonumber$
Introduction of the activity coefficient adds nothing to our store of information about the system. It merely provides a convenient way to recast the available information, so that the solute mole fraction, $y_A$, becomes the independent variable in the chemical-potential equation. (For a one-phase two-component system, ${\gamma }_A$ is completely determined by the temperature, system pressure, and $y_A$. Then $y_A$ is the concentration variable in the chemical-potential equation. If there are more than two components, additional concentration variables are required to specify the composition of the system and the values of ${\tilde{a}}_A$ and ${\gamma }_A$.)
In summary, since the gas is ideal, the partial pressure of $A$ above the solution is $P_A=x_AP$, whether the solution is ideal or not. If the solution is ideal, we have $y_AP^{\textrm{⦁}}_A=x_AP$, so that ${y_AP^{\textrm{⦁}}_A}/{P^o}={x_AP}/{P^o}$. If the solution is not ideal, we introduce the activity coefficient, ${\gamma }_A$, as the “fudge factor” that makes $\gamma_Ay_AP^{\textrm{⦁}}_A=x_AP$ true. That is, the activity coefficient is just the actual value of the partial pressure of ideal gas $A$, $x_AP$, divided by the value it would have if the solution were ideal, $y_AP^{\textrm{⦁}}_A$. The activity coefficient corrects for the departure of the real solution from the behavior that Raoult’s law predicts for the ideal solution. When we define the activity coefficient so that ${\tilde{a}}_A={y_A\gamma }_A$, we have ${\tilde{a}}_AP^{\textrm{⦁}}_A=x_AP$, so that
$\tilde{a}_AP^{\textrm{⦁}}_A/{P^o}=\dfrac{x_AP}{P^o} \nonumber$
thus preserving the form of the ideal solution result—with ${\tilde{a}}_A$ replacing $y_A$. For an ideal solution, ${\gamma }_A=1$, and our result for the real solution activity reduces to the ideal solution activity.
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In describing the activities of solution components, we have taken the standard state of component $A$ to be the pure liquid at its equilibrium vapor pressure, $P^{\textrm{⦁}}_A$, at the temperature of the solution. We can also express the activity using Henry’s law. Henry’s law states that the partial pressure of a component above its solution is directly proportional to the concentration of the component. We can choose any convenient unit to express the solute concentration. The value of the proportionality constant depends on this choice. Using mole fraction as the unit of concentration, Henry’s law is
$P_A=x_AP={\kappa }_Ay_A \nonumber$
(Henry’s law)
where the proportionality constant, ${\kappa }_A$, is called the Henry’s-law constant. (When we write $P_A=x_AP$, we implicitly assume that the gas-phase components of the equilibrium system behave as ideal gases.) Henry’s law is more general than Raoult’s law. Indeed, Raoult’s law is a special case of Henry’s law; if the solute obeys Raoult’s law, the Henry’s-law constant is $P^{\textrm{⦁}}_A$.
The value of the Henry’s-law constant depends on the components and the temperature. Experimentally, the value of the constant is determined by finding the slope of a plot of $P_A$ versus $y_A$ in the limit as $y_A\to 0$. The sketch in Figure 5 illustrates the relationship between the $P_A$ curve and the Henry’s-law tangent to it at $y_A=0$. Note that the slope of the tangent line at $y_A=0$ is equal to its intercept at $y_A=1$.
In practice, a sufficiently low concentration of any non-electrolyte component, in any liquid solution, obeys Henry’s law. For this reason, we refer to the component that obeys Henry’s law as the solute. We designate the higher concentration component as the solvent. The universal validity of Henry’s law as a low-concentration approximation has a simple physical interpretation. The solute vapor pressure depends upon the net effects of solute–solute, solute–solvent, and solvent–solvent intermolecular forces.
If all of these intermolecular forces are the same, the intermolecular interactions that determine the gas-phase composition are the same for solvent molecules as they are for solute molecules; the vapor pressures of the solvent and the solute are the same; the gas above the solution has the same composition as the solution; the partial pressure of the solute is proportional to the solute concentration, $y_A$; the proportionality constant is the pure-solvent vapor pressure, $P^{\textrm{⦁}}_A$; and the solution obeys Raoult’s law. However, if the intermolecular forces are not all the same, their net effect changes as the solute concentration changes. As the solute concentration increases, the effects of solute–solute interactions become increasingly important. If these are different from the effects of solute–solvent and solvent–solvent interactions, the solute partial pressure is not proportional to the solute concentration.
Conversely, at some sufficiently low concentration, solute molecules are so far apart that the effects of solute–solute interactions become negligible. Only solute–solvent and solvent–solvent interactions affect the solute vapor pressure. Because these remain constant as the solute concentration decreases further, the solute partial pressure is proportional to the solute concentration in this low-concentration regime. However, if the effects of solute–solvent interactions are different from those of solvent–solvent interactions, the pure-solvent vapor pressure, $P^{\textrm{⦁}}_A,$ is not the proportionality constant.
We can assume the existence of a hypothetical ideal gas state for any substance, even a substance that has no measureable vapor pressure at any attainable temperature. We can assume that the substance has a well-defined Gibbs free energy of formation in this state even though there may be no possibility of measuring its value. Likewise, we can assume that any solute exerts some partial pressure over its solution. We consider that this partial pressure has some finite value, even if it is much too small to measure. It follows that the fugacity of the solute has a finite value. Henry’s law implies that the fugacity is proportional to the solute concentration, at least in the limit of arbitrarily low concentration. Expressing the concentration of solute A as its mole fraction, we have
$f_A\left(P,y_A,y_B\right)={P_A}/{P^o}={x_AP}/{P^o}={{\kappa }_Ay_A}/{P^o} \nonumber$
and the chemical potential of the solute is
${\mu }_A\left(P,y_A,y_B\right)={\Delta }_fG^o\left(A,{HIG}^o\right)+RT{ \ln f_A\ }\left(P,y_A,y_B\right) \nonumber$
If the solute behaves ideally in the gas phase, Henry’s law leads to simple expressions for its chemical potential and activity. The development of these expressions is very similar to the corresponding development from Raoult’s law. The essential differences arise from the introduction of a different standard state for the activity of the solute. We begin with our basic equation for the chemical potential of$\ A$ in the gas phase: ${\mu }_A\left(g,P,x_A,x_B\right)={\Delta }_fG^o\left(A,{HIG}^o\right)+RT{ \ln \left[\frac{x_AP}{P^o}\right]\ }+\int^P_0{\left(\frac{{\overline{V}}_A\left(g\right)}{RT}-\frac{1}{P}\right)}dP \nonumber$
The integral term vanishes because we assume ideal gas behavior. We write ${\tilde{a}}_A\left(\mathrm{solute},P,y_A,y_B\right)$ to represent the activity of $A$ in a solution at pressure $P$ and in which the composition is specified by the mole fractions $y_A$ and $y_B$.
We choose a hypothetical liquid to be the standard state for the activity of the solute. This hypothetical liquid is pure liquid $A$ at the vapor pressure it would exhibit if it followed Henry’s law over the entire range of possible system compositions. This pressure is equal to its Henry’s law constant ${\kappa }_A$. (See Figure 5.) Let us denote the chemical potential of this standard state by ${\widetilde{\mu }}^o_A\left(Hyp\ \ell ,{\kappa }_A\right)$. The chemical potential of solute $A$ and the activity of $A$ in the solution are related by
${\mu }_A\left(P,y_A,y_B\right)={\widetilde{\mu }}^o_A\left(Hyp\ \ell ,{\kappa }_A\right)+RT{ \ln {\tilde{a}}_A\ }\left(P,y_A,y_B\right) \nonumber$
and since ${\mu }_A\left(g,P,x_A,x_B\right)={\mu }_A\left(P,y_A,y_B\right)$, and the gas is ideal, we have
${ \ln \left[{\tilde{a}}_A\left(P,y_A,y_B\right)\right]\ }=\frac{{\Delta }_fG^o\left(A,{HIG}^o\right)-{\widetilde{\mu }}^o_A\left(Hyp\ \ell ,{\kappa }_A\right)}{RT}+{ \ln \left[\frac{x_AP}{P^o}\right]\ } \nonumber$
Except for the solute standard state, this is same as the equation that we develop in Section 16.1.
This equation must give the activity of solute $A$ in its standard state, which is pure hypothetical liquid at a pressure equal to the Henry’s law constant: $P=P_A={\kappa }_A$. In this state, $x_A=y_A=1$ and $x_B=y_B=0$. In its standard state, the activity of solute $A$ is unity; we have ${\tilde{a}}_A\left(P,y_A,y_B\right)={\tilde{a}}_A\left({\kappa }_A,1,0\right)=1$. Making these substitutions and rearranging, we find
$\frac{{\Delta }_fG^o\left(A,{HIG}^o\right)-{\widetilde{\mu }}^o_A\left(Hyp\ \ell ,{\kappa }_A\right)}{RT}=-{ \ln \left[\frac{{\kappa }_A}{P^o}\right]\ } \nonumber$
Substituting this result into our general equation for ${ \ln \left[{\tilde{a}}_A\left(\mathrm{solute},P,y_A,y_B\right)\right]\ }$, we find that the activity of solute $A$ is ${\tilde{a}}_A\left(P,y_A,y_B\right)=\frac{x_AP}{{\kappa }_A} \nonumber$ (any solution, ideal gas)
The chemical potential of the standard-state hypothetical pure liquid whose vapor pressure at $T$ is ${\kappa }_A$ is
${\widetilde{\mu }}^o_A\left(Hyp\ \ell ,{\kappa }_A\right)={\Delta }_fG^o\left(A,{HIG}^o\right)+RT{ \ln \left[\frac{{\kappa }_A}{P^o}\right]\ } \nonumber$
This is a general result for the activity of solute $A$ when the standard state is the hypothetical pure liquid whose pressure is ${\kappa }_A$, and $A$ behaves ideally in the gas phase. If Henry’s law is obeyed, we have $x_AP=y_A{\kappa }_A$. Substituting, we find
${\tilde{a}}_A\left(P,y_A,y_B\right)=y_A \nonumber$ (Henry’s law is obeyed)
The Henry’s law development and the Raoult’s law development give the same value for the chemical activity. However, the standard states are different.
If the solute obeys Raoult’s law, the standard state we choose for the solute is the pure solute at its equilibrium vapor pressure; in this state, the pure solute is in equilibrium with its own gas. Real substances can satisfy this condition. If the solute obeys Henry’s law, the standard state we choose for the solute is a hypothetical pure-liquid solute at a pressure equal to ${\kappa }_A$. In this state, we assume that the hypothetical liquid is in equilibrium with its own gas, also at pressure ${\kappa }_A$. The hypothetical standard-state liquid is a substance in which the interactions among $A$ molecules have the same effects as the interactions, in a very dilute solution, between $A$ molecules and the $B$ molecules that comprise the solvent. If solutions of $A$ and $B$ are described poorly by Raoult’s law, the vapor pressure of pure liquid $A$, $P^{\textrm{⦁}}_A$, is likely to be very different from the vapor pressure, ${\kappa }_A$, of the hypothetical standard-state liquid that we define using Henry’s law.
Our development produces a model for the chemical potential of the solute in which the activity is equal to the solute mole fraction. At the same mole fraction, every solute has the same activity. The chemical potentials of different solutes vary because the chemical potential in its standard state is different for every solute. We conclude that we can let ${\tilde{a}}_A=y_A$ for any sufficiently dilute solute, even when it is not feasible to measure the chemical potential of the solute in its standard state experimentally.
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Even if Henry’s law is valid only for solute concentrations very close to zero, we can use it to express the activity of the real system as a function of solute concentration. Let us suppose that we have data on the mole fraction of $A$, $x_A$, in a gas whose pressure is $P$ and which is at equilibrium with a solution in which its mole fraction is $y_A$. In the preceding section, we find that we can choose the solute’s standard state so that its activity in any state is $\tilde{a}_A=x_AP/{\kappa }_A$. Introducing the activity coefficient, defined by $\tilde{a}_A=y_A{\gamma }_A$, we have $x_AP/{\kappa }_A=y_A{\gamma }_A$. The activity coefficient is
${\gamma }_A=\frac{x_AP}{y_A{\kappa }_A} \nonumber$ (Henry’s law activity coefficient)
and the chemical potential is
${\mu }_A={\widetilde{\mu }}^o_A\left(Hyp\ \ell ,{\kappa }_A\right)+RT{ \ln y_A{\gamma }_A\ } \nonumber$
Just as when we define the activity coefficient using the deviation from Raoult’s law, this development provides a way to recast the available information in a way that makes the solute mole fraction, $y_A$, the independent variable in the chemical-potential equation. In Section 16.4, we note that Raoult’s law is the special case of Henry’s law in which $P^{\textrm{⦁}}_A={\kappa }_A$. If we make this substitution into the Henry’s-law based activity coefficient, we recover the Raoult’s-law based activity coefficient.
16.06: Henry's Law and the Hypothetical One-molal Standard S
When the solution is dilute, it is often convenient to use the molality of the solute rather than its mole fraction. We define the molality as the number of moles solute, $A$, per kilogram of solvent, $B$. We use ${\underline{m}}_A$ to represent the molality of $A$ and $\overline{M}_B$ to represent the gram-molar mass of $B$. A solution that contains $n_A$ and $n_B$ moles of $A$ and $B$, respectively, contains a mass of solvent given in kilograms by
$m_B=\frac{n_B\overline{M}_B}{1000}\nonumber$
Then the molality of $A$ is
${\underline{m}}_A=\frac{n_A}{m_B}=\frac{1000n_A}{\overline{M}_Bn_B}\nonumber$
so that
$n_A=\frac{{\underline{m}}_An_B\overline{M}_B}{1000} \nonumber$
The mole fraction of A is given by
$y_A=\frac{n_A}{n_A+n_B}=\frac{{\underline{m}}_A\overline{M}_B}{{\underline{m}}_A\overline{M}_B+1000}\nonumber$
In dilute solution, where $n_A\ll n_B$ and ${\underline{m}}_A\overline{M}_B\ll 1000$, the mole fraction and molality of the solute are related by
$y_A\approx \left(\frac{\overline{M}_B}{1000}\right){\underline{m}}_A\nonumber$ (dilute solution)
Using this approximation, assuming that solute $A$ obeys Henry’s law and that gas $A$ behaves ideally, we have
\begin{align*} \mu_A\left(P, \underline{m}_A \right)=\mu_A\left(P,y_A,y_B\right)=\mu_A\left(g,P,x_A,x_B\right) &= \Delta_fG^o\left(A, {HIG}^o\right)+RT \ln \left[\frac{x_AP}{P^o}\right] \[4pt]&= \Delta_fG^o\left(A,{HIG}^o\right)+RT \ln \left[\frac{{\kappa_Ay}_A}{P^o}\right] \[4pt]&= \Delta_fG^o\left(A,{HIG}^o\right)+RT \ln \left[\frac{\kappa_A\overline{M}_B}{{1000P}^o}\right] + RT \ln \underline{m}_A\end{align*}
(solute $A$ obeys Henry’s law)
When Henry’s law is not obeyed over the composition range of interest, it is often convenient to choose the standard state of the solute to be a one-molal solution of a hypothetical substance that obeys Henry’s law with the Henry’s law constant $\kappa_A$. Then the activity of this solution is unity, and its chemical potential is the chemical potential of $A$ in this hypothetical standard state. Letting this standard-state chemical potential be ${\widetilde\mu}^o_A\left(Hyp\ 1\ \underline{m}\mathrm{\ solute},P\right)$, we have
${\widetilde\mu}^o_A\left(Hyp\ 1\ \underline{m}\mathrm{\ solute},P\right)=\Delta_fG^o\left(A,{HIG}^o\right)+RT{ \ln \left[\frac{\kappa_A\overline{M}_B}{{1000P}^o}\right]\ }\nonumber$
The chemical potential of a substance that satisfies Henry’s law is
$\mu_A\left(HL,P,{\underline{m}}_A\right)={\widetilde\mu}^o_A\left(Hyp\ 1\ \underline{m}\mathrm{\ solute},P\right)+RT{ \ln {\underline{m}}_A\ }\nonumber$
If A behaves as an ideal gas and the solution is dilute (${\underline{m}}_A\overline{M}_B\ll 1000$), but ${\underline{m}}_A$ is above the range in which Henry’s law is obeyed, we introduce the Henry’s law activity coefficient, ${\gamma }_A={{\tilde{a}}_A}/{{\underline{m}}_A}$, to measure the departure of the real solution behavior from that predicted by Henry’s law. Then the chemical potential of $A$ in any solution is
$\mu_A\left(P,{\underline{m}}_A\right)={\widetilde\mu}^o_A\left(Hyp\ 1\ \underline{m}\mathrm{\ solute},P\right)+RT{ \ln {\tilde{a}}_A\ }={\widetilde\mu}^o_A\left(Hyp\ 1\ \underline{m}\mathrm{\ solute},P\right)+RT{ \ln {\underline{m}}_A{\gamma }_A\ }={\widetilde\mu}^o_A\left(Hyp\ 1\ \underline{m}\mathrm{\ solute},P\right)+RT{ \ln {\gamma }_A\ } + RT{ \ln {\underline{m}}_A\ }\nonumber$ and the logarithm of the activity coefficient measures the difference between the chemical potential of the real solute and that of a solute that obeys Henry’s law over an extended concentration range:
${ \ln {\gamma }_A\ }=\frac{\mu_A\left(P,{\underline{m}}_A\right)-\mu_A\left(HL,P,{\underline{m}}_A\right)}{RT}\nonumber$
In Section 16.19, we consider the determination of Henry’s law-based activity coefficients further.
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We have seen that the activity of any component of an equilibrium system contains information about the activities of all the other components. From
$d\mu =-\overline{S}dT+\overline{V}P+RTd{ \ln \tilde{a}\ }\nonumber$
and the Gibbs-Duhem equation, we can find a general relationship among the activities. Substituting $d{\mu }_A$ and $d{\mu }_B$ into the Gibbs-Duhem equation, we have, for a two-component solution,
\begin{align*} -SdT+VdP &=n_Ad{\mu }_A+n_Bd{\mu }_B \[4pt]&= n_A\left(-{\overline{S}}_AdT+{\overline{V}}_AP+RTd{ \ln {\tilde{a}}_A\ }\right)+n_B\left(-{\overline{S}}_BdT+{\overline{V}}_BP+RTd{ \ln {\tilde{a}}_B\ }\right) \[4pt]&= -\left(n_A{\overline{S}}_A+n_B{\overline{S}}_B\right)dT + \left(n_A{\overline{V}}_A+n_B{\overline{V}}_B\right)dP + n_ARTd{ \ln {\tilde{a}}_A\ }+n_BRTd{ \ln {\tilde{a}}_B\ }\end{align*}
Since $S=n_A{\overline{S}}_A+n_B{\overline{S}}_B$ and $V=n_A{\overline{V}}_A+n_B{\overline{V}}_B$, this simplifies to
$0=n_Ad{ \ln {\tilde{a}}_A\ }+n_Bd{ \ln {\tilde{a}}_B\ }\nonumber$
or, dividing by $n_A+n_B$,
$0=y_Ad{ \ln {\tilde{a}}_A\ }+y_Bd{ \ln {\tilde{a}}_B\ }\nonumber$
For simplicity, let us consider a system in which a non-volatile solute, $A$, is dissolved in a volatile solvent, $B$. Measuring the pressure of the system and applying the equations that we developed in Section 16.1 for volatile component $A$ to the volatile solvent, $B$, in the present system, we can determine the activity of the solvent, $B$. Let us use mole fractions to measure concentrations and take pure liquid $B$ at its equilibrium vapor pressure as the activity standard state for both liquid- and gas-phase $B$. When $B$ is in its standard state, we have $x_A=0$, $x_B=1$, and ${\overline{V}}_B\left(g\right)={\overline{V}}^{\textrm{⦁}}_B\left(g\right)$. Then, since the solute is non-volatile, we can determine the activity of the solvent, $B$, from the pressure of the system. We have
${ \ln \left[{\tilde{a}}_B\left(P,y_A,y_B\right)\right]=\ }RT{ \ln \left[\frac{P}{P^{\textrm{⦁}}_B}\right]-\int^P_{P^{\textrm{⦁}}_B}{\left(\frac{{\overline{V}}^{\textrm{⦁}}_B \left(g\right)}{RT}-\frac{1}{P}\right)}\ }dP\nonumber$
Assuming that the integral makes a negligible contribution to the activity, we have
${\tilde{a}}_B\left(P,y_A,y_B\right)={\tilde{a}}_B=y_B\gamma_B\left(P,y_A,y_B\right)=\frac{P}{P^{\textrm{⦁}}_B}\nonumber$ (solvent)
so that
$\gamma_B\left(P,y_A,y_B\right)=\gamma_B=\frac{P}{y_BP^{\textrm{⦁}}_B}\nonumber$ (solvent)
Since the gas-phase concentration of $A$ is immeasurably small, we must determine its activity indirectly. Let the standard state for solute activity be the hypothetical pure liquid, $y_A=1$, whose equilibrium vapor pressure is equal to the Henry’s law constant of solute $A$. (We can determine the solute’s activity without measuring its Henry’s law constant.) We have
${\mu }_A\left(P,y_A,y_B\right)={\widetilde{\mu }}^o_A\left(Hyp\ \ell ,{\textrm{ҝ}}_A\right)+RT{ \ln \left[{\tilde{a}}_A\left(P,y_A,y_B\right)\right]\ }\nonumber$
where
${\tilde{a}}_A\left(P,y_A,y_B\right)={\tilde{a}}_A=y_A\gamma_A\nonumber$ (solute)
Since we are able to measure the activity of the solvent, we can determine the activity of the solute from the relationship $0=y_Ad{ \ln {\tilde{a}}_A\ }+y_Bd{ \ln {\tilde{a}}_B\ }$. Rearranging, we have
$d{ \ln {\tilde{a}}_A\ }=-\frac{y_B}{y_A}d{ \ln {\tilde{a}}_B\ }=-\left(\frac{{1-y}_A}{y_A}\right)d{ \ln {\tilde{a}}_B\ }\nonumber$
For two solutions in which the mole fractions of $A$ are $y_A$ and $y^{\#}_A$, and in which the activities of $A$ and $B$ are ${\tilde{a}}_A$, ${\tilde{a}}_B$, ${\tilde{a}}^{\#}_A$, and ${\tilde{a}}^{\#}_B$, we have
$\ln \frac{\tilde{a}_A}{\tilde{a}^{\#}_A}=-\int^{y_A}_{y^{\#}_A} \left(\frac{1-y_A}{y_A}\right)d \ln \tilde{a}_B\nonumber$
Graphically, the integral is the area under a plot of $-{\left(1-y_A\right)}/{y_A}$ versus ${ \ln {\tilde{a}}_B\ }$, from $y^{\#}_A$ to $y_A$.
Typically, we are interested in solutions for which $y_A\ll y_B$. In the limit as the solution becomes very dilute, the activity, mole fraction, and activity coefficient of the solvent, $B$, all approach unity: ${\tilde{a}}_B\to 1$, $y_B\to 1$, and $\gamma_B\to 1$. The activity of the solute, $A$, approaches the mole fraction of $A$. As a matter of experience, the approach is asymptotic: as the mole fraction approaches zero, $y_A\to 0$, the solute activity coefficient approaches unity, $\gamma_A\to 1$, and does so asymptotically, so that ${\tilde{a}}_A\to y_A$. For dilute solutions, ${ \ln {\tilde{a}}_A\to -\infty \ }$ and ${ \ln \gamma_A\to 0\ }$ asymptotically. In consequence,
${\mathop{\mathrm{lim}}_{y_A\to 0} \left(d{ \ln \gamma_A\ }\right)\ }=0\nonumber$
Because the activity coefficient approaches a finite limit while the activity does not, we can express the solute’s activity most simply by finding the solute’s activity coefficient. Since ${\tilde{a}}_A=y_A\gamma_A$ and ${\tilde{a}}_B=y_B\gamma_B=\left(1-y_A\right)\gamma_B$, we have
\begin{align*} 0 &=y_Ad{ \ln {\tilde{a}}_A\ }+y_Bd{ \ln {\tilde{a}}_B\ } \[4pt] &=y_Ad{ \ln y_A\gamma_A\ }+y_Bd{ \ln y_B\gamma_B\ }=y_Ad{ \ln y_A\ }+y_Ad{ \ln \gamma_A\ }+y_Bd{ \ln y_B\ }+y_Bd{ \ln \gamma_B\ } \[4pt]&=\left(dy_A+dy_B\right)+y_Ad{ \ln \gamma_A\ }+y_Bd{ \ln \gamma_B\ }=y_Ad{ \ln \gamma_A\ }+y_Bd{ \ln \gamma_B\ }\end{align*}
(Since $y_A+y_B=1$, we have ${dy}_A+{dy}_B=0$.) We can rearrange this to
$d{ \ln \gamma_A\ }=-\left(\frac{{1-y}_A}{y_A}\right)d{ \ln \gamma_B\ }\nonumber$
As the solute concentration approaches zero, ${\left(1-y_A\right)}/{y_A}$ becomes arbitrarily large. However, since ${\mathop{\mathrm{lim}}_{y_A\to 0} \left(d{ \ln \gamma_A\ }\right)\ }=0$, it follows that
${\mathop{\mathrm{lim}}_{y_A\to 0} d{ \ln \gamma_B\ }\ }=0\nonumber$
We see that the solvent activity coefficient also approaches unity asymptotically as the solute concentration goes to zero. The solute activity coefficient at any $y_A>0$ is then given by
$\int^{y_A}_0 d \ln \left[\gamma_A\left(y_A\right)\right] \ln \left[\gamma_A\left(y_A\right)\right] \int^{y_A}_0 -\left(\frac{1-y_A}{y_A}\right)d \ln \left[\gamma_B\left(y_A\right)\right]\nonumber$
As sketched in Figure 6, the latter integral is the area under a graph of ${-\left(1-y_A\right)}/{y_A}$ versus ${ \ln \left[\gamma_B\left(y_A\right)\right]\ }$, between $y_A=0$ and $y_A$. Since $\gamma_A\left(y_A\right)\to 1$ as $y_A\to 0$, this integral must remain finite even though ${-\left(1-y_A\right)}/{y_A}\to -\infty$ as $y_A\to 0$. This can occur, because ${\mathop{\mathrm{lim}}_{y_A\to 0} d{ \ln \gamma_B\ }\ }=0$, as we observe above. Nevertheless, the fact that the integrand is unbounded can limit the accuracy of the necessary integration. For accurate measurement of the solute activity coefficient, it is important to obtain solvent-activity data at the lowest possible solute concentration.
The most desirable situation is to collect solvent-activity data down to solute concentrations at which the solvent activity coefficient, $\gamma_B$, becomes unity. If $\gamma_B\left(y^{\#}_A\right)=1$ when the solute concentration is $y^{\#}_A$, ${ \ln \left[\gamma_A\left(y_A\right)\right]\ }$ can be evaluated with $y^{\#}_A$, rather than zero, as the lower limit of integration. In some cases, ${ \ln \left[\gamma_A\left(y^{\triangle }_A\right)\right]\ }$ may be known from some other measurement at a particular concentration, $y^{\triangle }_A$; if so, we can find ${ \ln \left[\gamma_A\left(y_A\right)\right]\ }$ by carrying out the numerical integration between the limits $y^{\triangle }_A$ and $y_A$.
If the measurement of $\gamma_B$ cannot be extended to values of $y_A$ at which $\gamma_B\left(y_A\right)=1$, we must find an empirical function, call it $f\left(y_A\right)$, that fits the experimental values of ${ \ln \left[\gamma_B\left(y_A\right)\right]\ }$, for the smallest values of $y_A$. (That is, the empirical function is ${f\left(y_A\right)\mathrm{=ln} \left[\gamma_B\left(y_A\right)\right]\ }$.) The differential of $f\left(y_A\right)$ is then a mathematical model for $d{ \ln \left[\gamma_B\left(y_A\right)\right]\ }$ over the region of low solute concentrations. Letting $y^{\#}_A$ be the smallest solute concentration for which the solvent activity can be determined, we can integrate, using the function for $d{ \ln \left[\gamma_B\left(y_A\right)\right]\ }$ that we derive from this model, to estimate ${ \ln \left[\gamma_B\left(y^{\#}_A\right)\right]\ }$. Uncertainty about the accuracy of the mathematical model becomes a significant source of uncertainty in the calculated values of $\gamma_A$.
Of course, if we can find an analytical function that provides a good mathematical model for all of the solvent-activity data, the differential of this function can be used in the integral to evaluate ${ \ln \left[\gamma_A\left(y_A\right)\right]\ }$ over the entire range of the experimental data. If necessary, the evaluation of this integral can be accomplished using numerical methods.
It is essential that any empirical function, ${f\left(y_A\right)\mathrm{=ln} \left[\gamma_B\left(y_A\right)\right]\ }$, have the correct mathematical properties over the concentration range to which it is applied. If it is to be used to extend the integration to $y_A=0$, $f\left(y_A\right)$ must satisfy $f\left(0\right)=0$ and $df\left(0\right)=0$. This is a significant condition. For example, consider the approximation
$f\left(y_A\right)={ \ln \left[\gamma_B\left(y_A\right)\right]\ }=c_By^{{\alpha }_B}_A\nonumber$
This model gives
$\frac{d{ \ln \left[\gamma_A\left(y_A\right)\right]\ }}{dy_A}={-\left(1-y_A\right)c}_B{\alpha }_By^{{\alpha }_B-2}_A\nonumber$ and ${\mathop{\mathrm{lim}}_{y_A\to 0} \frac{d{ \ln \left[\gamma_A\left(y_A\right)\right]\ }}{dy_A}\ }=0\nonumber$
requires ${\alpha }_B>2$.
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In Section 16.4, we conclude that any sufficiently dilute solute obeys Henry’s law, and define a hypothetical, pure-liquid standard state that makes the solute activity equal to its mole fraction, $\tilde{a}_A\left(P,y_A,y_B\right)=y_A$. In Section 16.7, we find that the mole fractions and activities of the components of any binary solution are related by
$y_Ad \ln \tilde{a}_A + y_B d \ln \tilde{a}_B =0. \nonumber$
For a solute that obeys Henry’s law, we have
\begin{align*} d{ \ln \tilde{a}_B\ } &=-\left(\frac{y_A}{y_B}\right)d{ \ln y_A\ } \[4pt] &=-\left(\frac{y_A}{y_B}\right)\left(d{ \ln y_B\ }\right)\left(\frac{d \ln y_A}{d \ln y_B}\right) \[4pt] &=-\left(\frac{y_A}{y_B}\right)\left(d \ln y_B\right)\left(\frac{dy_A/y_A}{dy_B/y_B}\right) \[4pt] &= -\left(\frac{y_A}{y_B}\right)\left(d \ln y_B\right)\left(\frac{-dy_B/y_A}{dy_B/y_B}\right) \[4pt] &= d \ln y_B \end{align*}
This result follows for any choice of standard state for the activity of solvent $B$. It is satisfied by $\tilde{a}_B=ky_B$, where $k$ is a constant. It is valid even if $A$ is completely nonvolatile. When gas-phase $B$ behaves as an ideal gas, and we choose the ideal gas at $P^o$ as the standard state for both gas- and solution-phase B, we have
$\tilde{a}_B\left(\mathrm{gas}\right)=f_B={P_B}/{P^o}={x_BP}/{P^o} \nonumber$
Since the standard states are the same, the fugacity and activity of $B$ in solution are the same as they are in the gas phase above it. We have $\tilde{a}_B\left(\text{solution}\right)=ky_B={x_BP}/{P^o}$. To find $k$, we consider the system comprised of pure $B$, for which $y_B=x_B=1$ and $P=P^{\textrm{⦁}}_B$. Substituting, we find $k={P^{\textrm{⦁}}_B}/{P^o}$. With this value for $k$,
$\tilde{a}_A\left(\mathrm{solution}\right)={P^{\textrm{⦁}}_By_B}/{P^o}={x_BP}/{P^o} \nonumber$
so that
$y_BP^{\textrm{⦁}}_B=x_BP. \nonumber$
This is Raoult’s law.
Thus when the solute obeys Henry’s law and the solvent behaves as an ideal gas in the gas phase above its solution, the solvent obeys Raoult’s law.
Evidently the converse is also true. If the solvent obeys Raoult’s law, $y_BP^{\textrm{⦁}}_B=x_BP$. With pure ideal gas $B$ as the standard state for $B$ in both the gas phase and the solution phase, we have
$\tilde{a}_B\left(\text{solution}\right)=\tilde{a}_B\left(\mathrm{gas}\right)=f_B={P_B}/{P^o}={x_BP}/{P^o}=y_B\left({P^{\textrm{⦁}}_B}/{P^o}\right) \nonumber$ so that $d{ \ln \tilde{a}_B\left(\mathrm{solution}\right)\ }=d{ \ln y_B\ } \nonumber$
From $y_Ad{ \ln \tilde{a}_A\ }+y_Bd{ \ln \tilde{a}_B\ }=0$ and $d{ \ln \tilde{a}_B\ }=d{ \ln y_B\ }$, we have
$d{ \ln \tilde{a}_A\ }=-\left(\frac{y_B}{y_A}\right)d{ \ln y_B\ }=-\left(\frac{y_B}{y_A}\right)\left(\frac{dy_B}{y_B}\right)=\frac{dy_A}{y_A}=d{ \ln y_A\ } \nonumber$
so that $\tilde{a}_A\left(\mathrm{solution}\right)=ky_A$, where $k$ is a constant. When we choose the standard state such that $\tilde{a}_A\left(\mathrm{ss,\ }\mathrm{solution}\right)=1$ when $y_A=1$, we find $k=1$ and $\tilde{a}_A\left(\mathrm{solution}\right)=y_A$. The activity of the solute is related to its fugacity and the fugacity of its standard state by
$\tilde{a}_A\left(\mathrm{solution}\right)=y_A=\frac{f_A\left(\mathrm{solution}\right)}{f_A\left(ss,\ \mathrm{solution}\right)} \nonumber$
When $y_A=1$, the fugacity is that of the standard state, which is a system of the hypothetical pure liquid in equilibrium with its own ideal gas. Letting the pressure of this ideal gas be ${\textrm{ĸ}}_A$, we have $f_A\left(ss,\ \mathrm{solution}\right)={\textrm{ĸ}}_A$, so that $f_A\left(\mathrm{solution}\right)={\textrm{ĸ}}_Ay_A$, which is equal to the fugacity of the gas with which it is at equilibrium. The fugacity of the ideal gas is $x_AP$, so that
$x_AP={\textrm{ĸ}}_Ay_A. \nonumber$
This is Henry’s law. Thus, if solvent $B$ obeys Raoult’s law, solute $A$ obeys Henry’s law.
16.09: Properties of Ideal Solutions
We have found the chemical potential of any component in an ideal solution. Now let us find some other thermodynamic properties of an ideal solution. The value of an extensive thermodynamic property of the solution will be the sum of the values of that property for the separate pure components plus the change that occurs when these components are mixed. (The initial state of the system comprises the pure, separate components at a particular temperature and pressure. The mixed state is a homogeneous liquid solution at the same temperature and pressure.) If the solution contains $n_A$ moles of component $A$ and $n_B$ moles of component $B$, the Gibbs free energy is
$G_{\mathrm{solution}}\left(n_A,n_B\right)=n_A{\mu }_A+n_B{\mu }_B\nonumber$
Dividing through by $n_A+n_B$ to find the Gibbs free energy of the mixture per mole of solution, we have
\begin{align*} \overline{G}_{\mathrm{solution}} \left(y_A,y_B\right) &=y_A\mu_A + y_B\mu_B \[4pt] &=y_A \Delta_f G^o\left(A\right)+y_ART \ln y_A + y_B \Delta _fG^o \left(B\right)+y_BRT \ln y_B \end{align*} \nonumber
To make this mixture, we need $y_A$ moles of $A$ and $y_B$ moles of $B$. The Gibbs free energy of these amounts of unmixed pure $A$ and $B$, each in its standard state, is
${\overline{G}}_{\mathrm{initial}}=y_A{\Delta }_fG^o\left(A\right)+y_B{\Delta }_fG^o\left(B\right)\nonumber$
For the process of mixing pure $A$ and pure $B$, each originally in its standard state, to form one mole of an ideal solution, the Gibbs free energy change is ${\Delta }_{\mathrm{mix}}\overline{G}={\overline{G}}_{\mathrm{solution}}-{\overline{G}}_{\mathrm{initial}}=y_ART{\ln y_A\ }+y_BRT{\ln y_B\ }\nonumber$
In Section 13.3, we found this same relationship for mixing ideal gases:
${\Delta }_{\mathrm{mix}}\overline{G}\left(\mathrm{gas}\right)=x_ART{\ln x_A\ }+x_BRT{\ln x_B\ }\nonumber$
From ${\left({\partial {\Delta }_{\mathrm{mix}}G}/{\partial T}\right)}_P=-{\Delta }_{\mathrm{mix}}S$, we find
${\Delta }_{\mathrm{mix}}\overline{S}=-y_AR{\ln y_A\ }-y_BR{\ln y_B\ }\nonumber$
and the entropy of the liquid solution is
${\overline{S}}_{\mathrm{solution}}=y_AS^o_A+y_BS^o_B\ -y_AR{\ln y_A\ }-y_BR{\ln y_B\ }\nonumber$
From ${\left({\partial {\Delta }_{\mathrm{mix}}G}/{\partial P}\right)}_T={\Delta }_{\mathrm{mix}}V$, we find
${\Delta }_{\mathrm{mix}}\overline{V}=0\nonumber$
and from ${\Delta }_{\mathrm{mix}}H={\Delta }_{\mathrm{mix}}G+T{\Delta }_{\mathrm{mix}}S$, we find
${\Delta }_{\mathrm{mix}}\overline{H}=0\nonumber$
Thus, ${\Delta }_{\mathrm{mix}}\overline{S}$, ${\Delta }_{\mathrm{mix}}\overline{V}$, and ${\Delta }_{\mathrm{mix}}\overline{H}$ for forming an ideal solution are identical also to the relationships we found for mixing ideal gases.
These results have an important physical interpretation. That ${\Delta }_{\mathrm{mix}}\overline{V}=0$ implies that the molecules of $A$ and the molecules of $B$ occupy the same volume in the mixture as they do in the pure state. From ${\Delta }_{\mathrm{mix}}\overline{V}=0$ and ${\Delta }_{\mathrm{mix}}\overline{H}=0$, it follows that ${\Delta }_{\mathrm{mix}}\overline{E}=0$ at constant pressure. In turn, this implies that the forces between an $A$ molecule and a $B$ molecule are the same as the forces between two $A$ molecules or between two $B$ molecules. If the force of attraction between an $A$ molecule and a $B$ molecule were stronger than that between two $A$ molecules or between two$\ B$ molecules, molecules in the mixture would be—on average—closer together in the mixture than in the separate components; we would find ${\Delta }_{\mathrm{mix}}\overline{V}<0$. Moreover, the potential energy of the mixed state would be lower than that of the separate components; the mixing process would evolve heat at constant temperature; we would find ${\Delta }_{\mathrm{mix}}\overline{E}<0$.
Conversely, if the repulsive force between an $A$ molecule and a $B$ molecule were stronger than the repulsive forces between two $A$ molecules or between two $B$ molecules, the average separation would be greater in the mixture; we would find ${\Delta }_{\mathrm{mix}}\overline{V}>0$. The potential energy of the mixed state would be greater than that of the separate components; the mixing process would consume heat at constant temperature; we would find ${\Delta }_{\mathrm{mix}}\overline{E}>0$.
In an ideal gas, molecules do not interact at all. In an ideal solution, the molecules must interact, because only their mutual attraction can keep them in the liquid state. The ideal solution behaves ideally not because the intermolecular interactions are zero but rather because the intermolecular interactions are the same for all of the kinds of molecules present in the mixture. This interpretation implies that the vapor pressures of the pure components of an ideal solution should be equal. Even for solutions that follow Raoult’s law quite closely, this expected equality is often imperfectly realized. Not surprisingly, ideal-solution behavior is best exhibited when the components are isotopically-substituted versions of the same compound.
In an ideal solution, the activities of the components are equal to their mole fractions. The activity of the solvent depends only on the solvent mole fraction. The properties of the solvent in an ideal solution are independent of the specific substance that comprises the solute; they depend only on the concentration of solute particles present. Systems in which this is a useful approximation are sufficiently common that their properties are given a special name: A colligative property of a solution is a property that depends only on the concentration of solute particles and not on the specific chemical properties of the solute. We expect this approximation to become better as the solute concentration approaches zero. When a solute obeys Raoult’s law or Henry’s law, its effects on the thermodynamic properties of the solvent depend only on the concentration of the solute. Consequently, Raoult’s law and Henry’s law prove to be useful when we seek to model colligative properties.
In Sections 16.10 through 16.14, we evaluate five colligative properties: boiling-point elevation, freezing-point depression, osmotic pressure, solid-solute solubility and gas-solute solubility. We derive the first three of these properties from the perspective that they enable us to determine the molar mass of solutes. However, boiling-point elevation, freezing-point depression, and osmotic pressure are important methods for the measurement of activity coefficients in non-ideal solutions. To illustrate the measurement of activity coefficients, we develop a more detailed analysis of freezing-point depression in Section 16.15.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/16%3A_The_Chemical_Activity_of_the_Components_of_a_Solution/16.08%3A_When_the_Solute_Obeys_Henry%27s_Law_the_Solvent_Obeys_R.txt
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The system we envision when we talk about boiling-point elevation is described schematically in Figure 7. We consider a solution of two components, $A$ and $B$. The mole fractions of $A$ and $B$, $y_A$ and $y_B$, specify the composition of the solution. We suppose that one of the components is present at a low concentration. We call this component the solute, and designate it as compound $A$. Under these assumptions, we have $y_A\approx 0$ and $y_B=1-y_A\approx 1$. We assume further that $A$ is nonvolatile, by which we mean that the vapor pressure of pure $A$, $P^{\textrm{⦁}}_A$, is very small. Then the second component, $B$, comprises most of the material of the system. We call component $B$ the solvent. We suppose that the $A$–$B$ solution is in equilibrium with a gas phase. In principle, molecules of both components are present in this gas. Since we assume that essentially no component $A$ is present in the gas phase, we have $x_A=0$ and $x_B=1$. We assume also that gas-phase $B$ behaves as an ideal gas and solute $A$ obeys Henry’s law.
When we measure the boiling point of a liquid system, we find the temperature at which the vapor pressure of the system becomes equal to a specified value. For the normal boiling point, this pressure is 1 atmosphere, or 1.01325 bars. At the boiling point, liquid-phase solvent is in equilibrium with gas-phase solvent, so that the chemical potential of liquid-phase solvent is equal to the chemical potential of gas-phase solvent. That is, we have
$\mu_{B,\mathrm{soluti}\mathrm{on}}=\mu_{B,\mathrm{gas}}.\nonumber$
We want to describe the change in the equilibrium position that occurs when there is an incremental change in the solute concentration, $dy_A$, while the pressure of the system remains constant. If the system is to remain at equilibrium, $\mu_{B,\mathrm{solution}}=\mu_{B,\mathrm{gas}}$ must remain true. It follows that the chemical potentials of the two phases must change in tandem. Continued equilibrium implies that $d\mu_{B,\mathrm{solution}}=d\mu_{B,\mathrm{gas}}$ when the solute concentration changes by $dy_A$.
We can analyze the boiling-point elevation phenomenon for any fixed pressure at which pure liquid $B$ can be at equilibrium with pure gas $B$. Let us designate the fixed pressure as $P^{\#}$. We designate the boiling-point temperature of pure solvent $B$, at $P^{\#}$, as $T_B$; thus, $P^{\#}=P^{\textrm{⦁}}_B\left(T_B\right)$, where $P^{\textrm{⦁}}_B\left(T_B\right)$ designates the equilibrium vapor pressure of pure solvent $B$ at temperature $T_B$. Our goal is to find the temperature at which a binary solution is in equilibrium with pure gas $B$ at the fixed pressure $P^{\#}$. We let $T_{bp}$ be the boiling temperature of the solution at $P^{\#}$. The composition of the solution is specified by the solute concentration, $y_A=1-y_B$.
Since we assume that the solute obeys Henry’s law, we choose the standard state for solute $A$ to be the pure hypothetical liquid $A$ whose vapor pressure is ${\textrm{ĸ}}_A$ at $T$. We suppose that ${\textrm{ĸ}}_A$ is exceedingly small. From Section 16.4, we then have $\tilde{a}_{A,\mathrm{solution}}=y_A$, so that
$d ~ { \ln \tilde{a}_{A,\mathrm{solution}}\ }=d{ \ln y_A\ }\nonumber$
at any temperature. From Section 16.8, we have
$d ~ { \ln \tilde{a}_{B,\mathrm{solution}}\ }=d{ \ln y_B\ }\nonumber$
Pure liquid-phase solvent is at equilibrium with gas-phase solvent at $P^{\#}=P^{\textrm{⦁}}_B\left(T_B\right)$ and $T_B$. We imagine that we create a solution by adding a small amount of solute $A$, making the concentrations of solute and solvent $y_A$ and $y_B=1-y_A$, respectively. We maintain the system pressure constant at $P^{\#}$, while changing the temperature to maintain equilibrium between gas-phase and solution-phase solvent $B$. The new temperature is $T_{bp}$.
The pressure of gas-phase B is constant at $P^{\#}$. The temperature goes from $T_B$ to $T_{bp}$. We choose the activity standard state to be pure gas B at $P^{\#}$ and T. This means that the activity of the pure gas is unity at every temperature, so that $d{ \ln \tilde{a}_{B,\mathrm{gas}}\ }=0$.
It is worthwhile to note that we can arrive at this conclusion from a different perspective: From Section 14.14, the incremental change in the activity is
$d ~ { \ln \tilde{a}_{B,\mathrm{gas}}\ }=\left(-\frac{\overline{H}_B}{RT^2}+\frac{\tilde{H}^o_B}{RT^2}\right)dT\nonumber$
where $\overline{H}_B$ is the partial molar enthalpy of gas-phase $B$ at $T$, and $\tilde{H}^o_B$ is the partial molar enthalpy of $B$ in its activity standard state at $\ T$. Since we assume that the gas phase is essentially pure $B$, we have $\overline{H}_B=\tilde{H}^o_B$ and, again, $d \ln \tilde{a}_{B,\mathrm{gas}}=0$.
From Section 14.3, we have the general result that
$d\mu_B={\overline{V}}_BdP-{\overline{S}}_BdT+RT\left(d{ \ln \tilde{a}_B\ }\right)\nonumber$
The system pressure and temperature are $P=P^{\#}$ and $T=T_{bp}$. For both the gas phase and the solution phase, we have $dP=0$ and $dT=dT_{bp}$. Since $d{ \ln \tilde{a}_{B,\mathrm{gas}}\ }=0$, we have
$d\mu_{B,\mathrm{gas}}=-{\overline{S}}_{B,\mathrm{gas}}dT_{bp}\nonumber$
Since $d ~ { \ln \tilde{a}_{B,\mathrm{solution}}\ }=d{ \ln y_B\ }$, we have
$d\mu_{B,\mathrm{solution}}=-{\overline{S}}_{B,\mathrm{solution}}dT_{bp}+RT_{bp}\left(d{ \ln y_B\ }\right)\nonumber$
The chemical potential of the pure, constant-pressure, gas-phase solvent depends only on temperature. The chemical potential of the constant-pressure, solution-phase solvent depends on temperature and solute concentration. Equilibrium is maintained if
$d\mu_{B,\mathrm{solution}}=d\mu_{B,\mathrm{gas}}\nonumber$
Substituting, we have
$-{\overline{S}}_{B,\mathrm{solution}}dT_{bp}+RT_{bp}\left(d{ \ln y_B\ }\right)=-{\overline{S}}_{B,\mathrm{gas}}dT_{bp}\nonumber$
Since $y_B=1-y_A\approx 1$,
$d{ \ln y_B\ }=d{ \ln \left(1-y_A\right)\ }={-dy_A}/{\left(1-y_A\right)}\approx -dy_A\nonumber$
The relationship $d\mu_{B,\mathrm{solution}}=d\mu_{B,\mathrm{gas}}$ becomes
$-{\overline{S}}_{B,\mathrm{solution}}dT_{bp}-RT_{bp}dy_A=-{\overline{S}}_{B,\mathrm{gas}}dT_{bp}\nonumber$
or
$dy_A=\left(\frac{\overline{S}_{B,\mathrm{gas}}}-\overline{S}_{B,\mathrm{solution}} {RT_{bp}} \right)dT_{bp} \nonumber$
We consider systems in which the boiling point of the solution, $T_{bp}$, is little different from the boiling point of the pure solvent, $T_B$. Then, $T_B\approx T_{bp}$, and ${T_{bp}}/{T_B}\approx 1$. We let $\Delta T=T_{bp}-T_B$, where $\left|\Delta T\right|\ll T_B$. Since the solution is almost pure $B$, the partial molar entropy of $B$ in the solution is approximately that of pure $B.$ Consequently, this partial molar entropy difference is, to a good approximation, just the entropy of vaporization of the solvent, at equilibrium, at the boiling point for the specified system pressure, $P^{\#}=P^{\textrm{⦁}}_B\left(T_B\right)$. Then, since the vaporization of pure $B$ at $P^{\#}$ and $T_B$ is a reversible process,
${\left({\overline{S}}_{B,\mathrm{gas}}-{\overline{S}}_{B,\mathrm{solution}}\right)}_{P^{\#},T_{bp}}\approx {\left({\overline{S}}^{\textrm{⦁}}_{B,\mathrm{gas}}-{\overline{S}}^{\textrm{⦁}}_{B,\mathrm{liquid}}\right)}_{P^{\#},T_B}=\Delta_{\mathrm{vap}}S_B={\Delta_{\mathrm{vap}}H_B}/{T_B}\nonumber$
so that
$dy_A=\left(\frac{\Delta_{\mathrm{vap}}H_B}{RT_{bp}T_B}\right)dT_{bp}\nonumber$
In the solution, the solute mole fraction is $y_A$; in the pure solvent, it is zero. At $P^{\#}$ and $T_B$, $\Delta_{\mathrm{vap}}H_B$ is a constant. Integrating, between the limits $\left(0,T_B\right)$ and $\left(y_A,T_{bp}\right)$, we have
$\int^{y_A}_0{dy_A}=\frac{\Delta_{\mathrm{vap}}H_B}{RT_B}\int^{T_{bp}}_{T_B}{\frac{dT_{bp}}{T_{bp}}}\nonumber$ and $y_A=\frac{\Delta_{\mathrm{vap}}H_B}{RT_B}{ \ln \frac{T_{bp}}{T_B}\ }\nonumber$
Introducing the approximation ${ \ln x\approx x-1\ }$, which is valid for $x\approx 1$, we have
$y_A=\frac{\Delta_{\mathrm{vap}}H_B}{RT_B}\left(\frac{T_{bp}}{T_B}-1\right)=\frac{\Delta_{\mathrm{vap}}H_B}{RT^2_B}\Delta T \nonumber$
Solving for $\Delta T$,
$\Delta T=\left(\frac{RT^2_B}{\Delta_{\mathrm{vap}}H_B}\right)y_A\nonumber$
Since the enthalpy of vaporization and the mole fraction are both greater than zero, $\Delta T=T_{bp}-T_B>0$; that is, the addition of a non-volatile solute increases the boiling point of a liquid system. By measuring $\Delta T$, we can find $y_A$; if we know the molar mass of the solvent, we can calculate the number of moles of solute in the solution. If we know the mass of the solute used to prepare the solution, we can calculate the molar mass of the solute.
Frequently it is useful to express the solute concentration as a molality rather than a mole fraction. Using the dilute-solution relationship between mole fraction and molality from Section 16.6, $y_A={\overline{M}_B{\underline{m}}_A}/{1000}$, the boiling-point elevation becomes:
$\Delta T=\left(\frac{RT^2_B}{\Delta_{\mathrm{vap}}H_B}\right)\left(\frac{\overline{M}_B}{1000}\right){\underline{m}}_A\nonumber$
Our theory predicts that the boiling-point elevation observed for a given solvent is proportional to the solute concentration and independent of the molecular characteristics of the solute. Experiments validate this prediction; however, its accuracy decreases as the solute concentration increases. Letting
${\textrm{ĸ}}_B=\frac{RT^2_B}{\Delta_{\mathrm{vap}}H_B}\nonumber$ and ${\textrm{ĸ}}^*_B=\frac{RT^2_B\overline{M}_B}{1000\ \Delta_{\mathrm{vap}}H_B}\nonumber$
we have $\Delta T={\textrm{ĸ}}_By_A$ and $\Delta T={\textrm{ĸ}}^*_B{\underline{m}}_A$. We call ${\textrm{ĸ}}_B$ or ${\textrm{ĸ}}^*_B$ the boiling-point (or boiling-temperature) elevation constant for solvent $B$. For practical determination of molecular weights, we usually find ${\textrm{ĸ}}_B$ or ${\textrm{ĸ}}^*_B$ by measuring the increase in the boiling point of a solution of known composition.
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The boiling point of a pure solvent, at a specified pressure, is the temperature at which the chemical potential of the pure solvent gas is equal to the chemical potential of the pure solvent liquid. The boiling point of a solution that contains a nonvolatile solute is the temperature at which the chemical potential of the pure solvent gas is equal to the chemical potential of the solvent in the solution. In the preceding section, we found that the boiling point of the solution is greater than the boiling point of the pure solvent. The temperature difference is the boiling-point elevation.
Similarly, the freezing point of a pure solvent, at a specified pressure, is the temperature at which the chemical potential of the pure-solid solvent is equal to the chemical potential of the pure-liquid solvent. The freezing point of a solution is the temperature at which the chemical potential of the pure-solid solvent is equal to the chemical potential of the solution-phase solvent. We find that the freezing point of the solution is less than the freezing point of the pure solvent. The temperature difference is the freezing-point depression.
In the boiling-point elevation case, we assume that the pure solvent gas contains no solute. In the freezing-point depression case, we assume that the pure solvent solid contains no solute. We find the relationship between composition and freezing-point depression by an argument very similar to that for boiling-point elevation. The equilibrium state in the freezing-point depression experiment is described schematically in Figure 8.
Since the equilibrium temperature decreases as the solute concentration increases, we can realize the equilibrium state experimentally by slowly cooling a solution of the specified composition. We determine the temperature at which the first, very small, crystal of solid solvent forms. Because the pure solvent freezes at a higher temperature than any solution, the first crystal formed is nearly pure solid solvent. Since this first crystal is very small, its formation does not change the composition of the solution significantly. Hence, the solution is in equilibrium with pure solid solvent at this temperature; we call this temperature the freezing point of the solution.
In practice, it is common to determine the melting point of a solid mixture rather than the freezing point of the liquid solution. The temperature of the solid mixture is increased slowly. As the mixture melts to form a homogeneous solution, the solute–solvent ratio in the melt approaches the ratio in which the mixture was prepared. When the last bit of solid melts, the composition of the solution is known from the manner of preparation. This last bit of solid melts at the highest temperature of any part of the mixture. It contains therefore the smallest proportion of solute. If this last bit of solid is in fact pure solvent, the temperature at which the last solid melts is the freezing point of the liquid solution. In the limit that the freezing-point and melting-point experiments are carried out reversibly, the state of the freezing-point system just after the first bit of solid freezes is the same as the state of the melting-point system just before the last bit of solid melts.
We again specify the composition of the solution by the mole fractions of $A$ and$\ B$. We let the solute be compound $A$, and assume that its concentration is low. We let the solute concentration be $y_A$, where $y_A=1-y_B$, $y_A\approx 0$ and $y_B=1-y_A\approx 1$. The $A$–$B$ solution is in equilibrium with pure solid $B$. We want to find the temperature at which these phases are in equilibrium. At this temperature, $\mu_{B,\mathrm{solution}}=\mu_{B,\mathrm{solid}}$, and hence $d\mu_{B,\mathrm{solution}}=d\mu_{B,\mathrm{solid}}$ for any change that takes the system to a new equilibrium state.
We can analyze the freezing-point depression phenomenon for any fixed pressure at which pure-liquid $B$ can be in equilibrium with pure-solid$\ B$. Let us designate the fixed pressure as $P^{\#}$ and the freezing-point temperature of pure-liquid B, at $P^{\#}$, as $T_F$. Our goal is to find the temperature at which a binary solution is in equilibrium with pure-solid $B$ at the fixed pressure $P^{\#}$. We let $T_{fp}$ be the freezing-point temperature of the solution at $P^{\#}$. We base our analysis on the assumption that $A$ obeys Henry’s law.
We let the pure-solid solvent be the standard state for the solid solvent (see Section 15.5). Then, at every temperature, $\mu_{B,\mathrm{solid}}={\widetilde\mu}^o_{B,\mathrm{solid}}$, and
$\mu_{B,\mathrm{solid}}-{\widetilde\mu}^o_{B,\mathrm{solid}}=RT{ \ln {\tilde{a}}_{B,\mathrm{solid}}\ }=0 \nonumber$ At every temperature, ${\tilde{a}}_{B,\mathrm{solid}}=1$ so that ${\left({ \ln {\tilde{a}}_{B,\mathrm{solid}}\ }\right)}_{PT}=0$. The system pressure is constant at $P^{\#}$, so $dP=0$. In the general expression
$d\mu_{B,\mathrm{solid}}=\overline{V}_{B,\mathrm{solid}}dP-\overline{S}_{B,\mathrm{solid}}dT+RT{\left(d{ \ln {\tilde{a}}_{B,\mathrm{solid}}\ }\right)}_{PT} \nonumber$
only the term in $dT$ is non-zero. Recognizing that $dT$ is the change in the freezing-point temperature at $P^{\#}$, for some change in the chemical potential of pure solid $B$, we have
$d\mu_{B,\mathrm{solid}}=-\overline{S}^{\textrm{⦁}}_{B,\mathrm{solid}}\ dT_{fp} \nonumber$
We let the pure-liquid solvent at its equilibrium vapor pressure be the standard state for the solution-phase solvent (see Section 16.2). We designate this equilibrium vapor pressure as $P^{\textrm{⦁}}_B\left(T_{fp}\right)$. Now, since we ultimately find $T_{fp}<t_f$>, the pure liquid freezes spontaneously at $T_{fp}$. The standard state for the liquid solvent is therefore a hypothetical state; it is a pure, super-cooled liquid. The properties of this hypothetical liquid can be estimated from our theory; however, except possibly in unusual circumstances, they cannot be measured directly. Since we assume that the solute obeys Henry’s law, we have from Section 16.8 that $d{ \ln {\tilde{a}}_{B,\mathrm{solution}}=d{ \ln y_B\ }\ }$. Thus, while the activity of the pure-solid solvent is constant, the activity of the solvent in the solution varies with the solute concentration. We have
$d\mu_{B,\mathrm{solution}}=-\overline{S}_{B,\mathrm{solution}}\ dT_{fp}+RT_{fp}{\left(d{ \ln y_B\ }\right)}_{PT} \nonumber$
Using $d{ \ln y_B\ }\approx -dy_A$, the relationship $d\mu_{B,\mathrm{solution}}=d\mu_{B,\mathrm{solid}}$ becomes
$-\overline{S}_{B,\mathrm{solution}}\ dT_{fp}-RT_{fp}dy_A=-\overline{S}^{\textrm{⦁}}_{B,\mathrm{solid}}\ dT_{fp} \nonumber$
or
$dy_A=-\left(\frac{\overline{S}_{B,\mathrm{solution}}-\overline{S}^{\textrm{⦁}}_{B,\mathrm{solid}}}RT_{fp}\right)dT_{fp} \nonumber$
We consider systems in which the freezing point of the solution, $T_{fp}$, is little different from the freezing point of the pure solvent, $T_F$. Then, $T_F\approx T_{fp}$, and ${T_{fp}}/{T_F}\approx 1$. We let $\left|\Delta T\right|=T_F-T_{fp}$, where $\left|\Delta T\right|\ll T_F$. Since the solution is almost pure $B$, the partial molar entropy of $B$ in the solution is approximately that of pure liquid $B$. Consequently, the partial molar entropy difference is, to a good approximation, just the entropy of fusion of the pure solvent, at equilibrium, at the freezing point for the specified system pressure. That is,
${\left(\overline{S}_{B,\mathrm{solution}}-\overline{S}^{\textrm{⦁}}_{B,\mathrm{solid}}\right)}_{P^{\#},T_{fp}}\approx {\left(\overline{S}^{\textrm{⦁}}_{B,\mathrm{liquid}}-\overline{S}^{\textrm{⦁}}_{B,\mathrm{solid}}\right)}_{P^{\#},T_F}=\Delta_{\mathrm{fus}}S_B={\Delta_{\mathrm{fus}}H_B}/{T_F} \nonumber$
so that $dy_A=-\left(\frac{\Delta_{\mathrm{fus}}H_B}{RT_{fp}T_F}\right)dT_{fp} \nonumber$
At $P^{\#}$ and $T_F$, $\Delta_{\mathrm{fus}}H_B$ is a constant. In the solution, the solute mole fraction is $y_A$; in the pure solid solvent, it is zero. Integrating between the limits $\left(0,T_F\right)$ and $\left(y_A,T_{fp}\right)$, we have
$\int^{y_A}_0{dy_A}=-\frac{\Delta_{\mathrm{fus}}H_B}{RT_F}\int^{T_{fp}}_{T_F}{\frac{dT_{fp}}{T_{fp}}} \nonumber$ and $y_A=-\frac{\Delta_{\mathrm{fus}}H_B}{RT_F}{ \ln \frac{T_{fp}}{T_F}\ } \nonumber$
Introducing ${ \ln x\approx x-1\ }$, we have $y_A=-\frac{\Delta_{\mathrm{fus}}H_B}{RT_F}\left(\frac{T_{fp}}{T_F}-1\right)=\frac{\Delta_{\mathrm{fus}}H_B}{RT^2_F}\Delta T \nonumber$ Solving for $\Delta T$, $\Delta T=\left(\frac{RT^2_F}{\Delta_{\mathrm{fus}}H_B}\right)y_A \nonumber$
The fusion process is endothermic, and $\Delta_{\mathrm{fus}}H_B>0$. Therefore, we find $\Delta T=T_F-T_{fp}>0$; that is, the addition of a solute decreases the freezing point of a liquid. The depression of the freezing point is proportional to the solute concentration.
Since measurement of $\Delta T$ enables us to find $y_A$, freezing-point depression—like boiling point elevation—enables us to determine the molar mass of a solute. In our discussion of boiling-point elevation, we noted that it is often convenient to express the concentration of a dilute solute in units of molality rather than mole fraction. This applies also to freezing-point depression. Likewise, for practical applications, we usually find the freezing-point depression constant by measuring the depression of the freezing point of a solution of known composition.
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The phenomena of boiling-point elevation and freezing-point depression involve relationships between composition and equilibrium temperature—at constant system pressure. We turn now to a phenomenon, osmotic pressure, which involves a relationship between composition and equilibrium pressure—at constant system temperature.
To analyze boiling-point elevation, we equate the chemical potential of the solvent in two subsystems, a solution and the gas phase above it. To analyze freezing-point depression, we equate the chemical potential of the solvent in solution and solid subsystems. Similarly, to analyze osmotic pressure, we equate the chemical potential of the pure solvent—at one pressure—to the chemical potential of the solvent in a solution—at a second pressure. We find that equilibrium can be obtained only when the pressure in the solution subsystem exceeds the pressure in the solvent subsystem. The difference between these two pressures is the osmotic pressure.
In the boiling-point elevation and freezing-point depression phenomena, the subsystems are separated by a phase boundary. In the osmotic pressure phenomenon, a pure solvent phase is separated from a solution phase by a semi-permeable membrane. A semi-permeable membrane allows free passage to solvent molecules; however, solute molecules cannot pass through it. In practice, the semi-permeable membrane is a material that is penetrated by pores, or channels, whose cross-sectional dimensions are nearly as small as typical solvent molecules. Solvent molecules can diffuse through these pores and pass from one side of the membrane to the other. With such a membrane, we can satisfy the osmotic pressure conditions by choosing a solute whose molecules are larger than the pore diameters, because large molecules will be unable to pass through the pores. In practice, the solute in osmotic pressure experiments is typically a polymer or a biologically derived molecule of high molecular weight. Osmotic pressure measurements have been an important source of data on the molar masses of such substances.
The osmotic pressure experiment is described schematically in Figure 9. The semi-permeable membrane must be sufficiently robust to support the pressure drop between the two subsystems. At constant pressure, mixing of the two subsystems is a spontaneous process. Were we to remove the membrane and the pressure drop that it supports, the subsystems would mix to form a single, more dilute solution. We see therefore that there is a tendency for net migration of solvent molecules from the solvent side of the membrane to the solution side. We can oppose this tendency by applying additional pressure on the solution side. Evidently, for any given solution composition, there will be an applied pressure at which the subsystems are in equilibrium with one another.
We let the pure-liquid solvent at its equilibrium vapor pressure be the standard state for both the pure-liquid and the solution-phase solvent (see Section 16.2). For the two subsystems to be in equilibrium, we must have ${\mu }_{B,\mathrm{soluton}}={\mu }_{B,\mathrm{solvent}}$. For any change that takes one equilibrium state to another, we have $d{\mu }_{B,\mathrm{soluton}}=d{\mu }_{B,\mathrm{solvent}}$. Since the pure solvent subsystem contains only $B$, we have ${\tilde{a}}_{B,\mathrm{so}\mathrm{lvent}}=\mathrm{constant}$ so that $d{ \ln {\tilde{a}}_{B,\mathrm{sol}\mathrm{vent}}\ }=0$. Since the temperature is constant, we have $dT=0$. For the solvent subsystem, the general expression for$\ d{\mu }_{B,\mathrm{solvent}}$ reduces to
$d{\mu }_{B,\mathrm{solvent}}={\overline{V}}^{\textrm{⦁}}_{B,\mathrm{solvent}}dP_{\mathrm{solvent}} \nonumber$
For the solution subsystem, $dT=0$. Assuming the solvent in the solution obeys Raoult’s law, we have ${\tilde{a}}_{B,\mathrm{sol}\mathrm{ution}}=y_B$. The general equation for $d{\mu }_{B,\mathrm{soluton}}$ reduces to
$d{\mu }_{B,\mathrm{soluton}}=\overline{V}_{B,\mathrm{solution}}dP_{\mathrm{solution}}+RT\left(d{ \ln y_B\ }\right) \nonumber$
Using $d{ \ln y_B\ }\approx -dy_A$, the relationship $d{\mu }_{B,\text{solution}}=d{\mu }_{B,\mathrm{solvent}}$ becomes
${\overline{V}}_{B,\mathrm{solution}}dP_{\mathrm{solution}}-RTdy_A={\overline{V}}^{\textrm{⦁}}_{B,\mathrm{solvent}}dP_{\mathrm{solvent}} \nonumber$
The molar volume of a liquid is nearly independent of the system pressure. Because the solution is nearly pure solvent, the molar volume of $B$ in the solution is approximately equal to the molar volume of pure solvent $B$. Letting $\overline{V}_{B,\mathrm{solution}}=\overline{V}^{\textrm{⦁}}_{B,\mathrm{solvent}}={\overline{V}}^{\textrm{⦁}}_B$, this becomes
$dy_A=\left(\frac{\overline{V}^{\textrm{⦁}}_B}{RT}\right)\left(dP_{\mathrm{solution}}-dP_{\mathrm{solvent}}\right)=\left(\frac{\overline{V}^{\textrm{⦁}}_B}{RT}\right)\ d\left(P_{\mathrm{solution}}-P_{\mathrm{solvent}}\right) \nonumber$
This pressure difference is the osmotic pressure; it is often represented by the Greek alphabet capital pi: $\mathit{\Pi}=P_{\mathrm{solution}}-P_{\mathrm{solvent}}$. The osmotic pressure of the pure solvent must be zero; that is, $\mathit{\Pi}=0$ when $y_A=0$. Integrating between the limits $\left(0,0\right)$ and $\left(y_A,\mathit{\Pi}\right)$, we have
$\int^{y_A}_0{dy_A}=\frac{\overline{V}^{\textrm{⦁}}_B}{RT}\int^{\mathit{\Pi}}_0{d\mathit{\Pi}} \nonumber$
and
$y_A=\frac{\overline{V}^{\textrm{⦁}}_B\mathit{\Pi}}{RT} \nonumber$
or
${\mathit{\Pi}\overline{V}}^{\textrm{⦁}}_B=y_ART \nonumber$
From this equation, we see that the osmotic pressure must be positive; that is, at equilibrium, the pressure on the solution must be greater than the pressure on the solvent: $\mathit{\Pi}=P_{\mathrm{solution}}-P_{\mathrm{solvent}}>0$.
The osmotic pressure equation can be put into an easily remembered form. For $n_A\ll n_B$, $y_A={n_A}/{\left(n_A+n_B\right)\approx {n_A}/{n_B}}$. With this substitution, $\mathit{\Pi}\left(n_B{\overline{V}}^{\textrm{⦁}}_B\right)=n_ART$, but since ${\overline{V}}^{\textrm{⦁}}_B$ is the molar volume of pure $B$, $V=n_B{\overline{V}}^{\textrm{⦁}}_B$ is just the volume of the solvent and essentially the same as the volume of the solution. The osmotic pressure equation has the same form as the ideal gas equation:
$\mathit{\Pi}\left(n_B{\overline{V}}^{\textrm{⦁}}_B\right)=\mathit{\Pi}\ V=n_ART \nonumber$
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Although the result has few practical applications, we can also use these ideas to calculate the solubility of a solid solute in an ideal solution. The arguments are similar to those we used to estimate the freezing-point depression of a solution. The freezing point of a solution is the temperature at which the solution is in equilibrium with its pure-solid solvent. The solubility of a solute is the mole fraction of the solute in a solution that is at equilibrium with pure-solid solute. In this analysis, we assume that the solid phase is pure solute. Our analysis does not apply to a solid solution in equilibrium with a liquid solution. The properties of the solvent have no role in our description of the solid–ideal-solution equilibrium state. Consequently, our analysis produces a model in which the solubility of an ideal solute depends only on the properties of the solute; for a given solute, the ideal-solution solubility is the same in every solvent.
We specify the composition of the solution by the mole fractions of $A$ and $B$, again letting the solute be compound $A$. When we consider freezing-point depression, an $A$–$B$ solution of specified composition is in equilibrium with pure solid $B$, and we want to find the equilibrium temperature. When we consider solute solubility, the $A$–$B$ solution is in equilibrium with pure solid $A$ at a specified pressure, $P^{\#}$, and temperature, $T$; we want to find the equilibrium composition. Since pure solid $A$ is present, the temperature must be less than the melting point of pure $A$. We let the melting point of the pure solute be $T_{FA}$, at the specified system pressure.
The activity of pure solid $A$ and the system pressure are both constant; we have $d{ \ln {\tilde{a}}_{A,\mathrm{solid}}=0\ }$, $dP=0$, and
$d{\mu }_{A,\mathrm{soli}\mathrm{d}}=-{\overline{S}}^{\textrm{⦁}}_{A,\mathrm{solid}}dT \nonumber$
In the saturated ideal solution in equilibrium with this solid, we have ${\tilde{a}}_{A,\mathrm{solution}}=y_A$, $dP=0$, and
$d{\mu }_{A,\mathrm{sol}\mathrm{ution}}=-{\overline{S}}_{A,\mathrm{solution}}dT+RT\left(d{ \ln y_A\ }\right) \nonumber$
The relationship $d{\mu }_{A,\mathrm{sol}\mathrm{ution}}=d{\mu }_{A,\mathrm{solid}}$ becomes
$-{\overline{S}}_{A,\mathrm{solution}}dT+RT\left(d{ \ln y_A\ }\right)=-{\overline{S}}^{\textrm{⦁}}_{A,\mathrm{solid}}dT \nonumber$
and $d{ \ln y_A\ }=\left(\frac{\overline{S}_{A,\mathrm{solution}}-{\overline{S}}^{\textrm{⦁}}_{A,\mathrm{solid}}}{RT}\right)dT \nonumber$
Now, ${\overline{S}}_{A,\mathrm{solution}}-{\overline{S}}^{\textrm{⦁}}_{A,\mathrm{solid}}$ is the entropy change for the reversible (equilibrium) process in which one mole of pure solid $A$ dissolves in a very large volume of a saturated solution; the mole fraction of $A$ in this solution is constant at $y_A$. During this process, the pressure and temperature are constant at $P^{\#}$ and $T$. Letting the heat absorbed by the system during this process be $q^{rev}_{P^{\#}}$, we have
${\left({\overline{S}}_{A,\mathrm{solution}}-{\overline{S}}^{\textrm{⦁}}_{A,\mathrm{solid}}\right)}_{P^{\#},T}={q^{rev}_{P^{\#}}}/{T} \nonumber$
The heat absorbed is also expressible as the difference between the partial molar enthalpy of $A$ in the solution and that of the pure solid; that is,
$q^{rev}_{P^{\#}}={\left({\overline{H}}_{A,\mathrm{solution}}-{\overline{H}}^{\textrm{⦁}}_{A,\mathrm{solid}}\right)}_{P^{\#},T} \nonumber$
One of the properties of an ideal solution is that the enthalpy of mixing is zero. Thus, the partial molar enthalpy of $A$ in an ideal solution is independent of $y_A$, so that the partial molar enthalpy of $A$ in an ideal solution is the same as the partial molar enthalpy of pure liquid $A$; that is, ${\overline{H}}_{A,\mathrm{solution}}={\overline{H}}^{\textrm{⦁}}_{A,\mathrm{liquid}}$, and
$q^{rev}_{P^{\#}}={\left({\overline{H}}_{A,\mathrm{solution}}-{\overline{H}}^{\textrm{⦁}}_{A,\mathrm{solid}}\right)}_{P^{\#},T}={\left({\overline{H}}^{\textrm{⦁}}_{A,\mathrm{liquid}}-{\overline{H}}^{\textrm{⦁}}_{A,\mathrm{solid}}\right)}_{P^{\#},T}={\left(\Delta_{\mathrm{fus}}H_A\right)}_{P^{\#},T}\approx {\left(\Delta_{\mathrm{fus}}H_A\right)}_{P^{\#},T_{FA}} \nonumber$
Then,
$\left(\overline{S}_{A,\mathrm{solution}}-\overline{S}^{\textrm{⦁}}_{A,\mathrm{solid}}\right)_{P^{\#},T} \approx \frac{\left(\Delta_{\mathrm{fus}}H_A\right)_{P^{\#},T_{FA}}}{T} \nonumber$
Dropping the subscript information and replacing the approximate equality, we have
$d{ \ln y_A\ }=\left(\frac{\Delta_{\mathrm{fus}}H_A}{RT^2}\right)dT \nonumber$
At $P^{\#}$ and $T_{FA}$, $\Delta_{\mathrm{fus}}H_A$ is a property of pure $A$ and is independent of the solution composition. When the pure solid solute melts at $T_{FA}$, the solute mole fraction is unity in the liquid phase with which it is in equilibrium: At $T_{FA}$, $y_A=1$. At temperature $T$, $y_A$ is the solute mole fraction in the liquid-phase solution that is at equilibrium with the pure-solid solute. Integrating between the limits $\left(1,T_{FA}\right)$ and $\left(y_A,T\right)$, we have
$\int^{y_A}_1{d{ \ln y_A\ }}=\frac{\Delta_{\mathrm{fus}}H_A}{R}\int^T_{T_{FA}}{\frac{dT}{T^2}} \nonumber$ and ${ \ln y_A\ }=\frac{-\Delta_{\mathrm{fus}}H_A}{R}\left(\frac{1}{T}-\frac{1}{T_{FA}}\right) \nonumber$
For a given solute, $\Delta_{\mathrm{fus}}H_A$ and $T_{FA}$ are fixed and are independent of the characteristics of the solvent. The mole fraction of $A$ in the saturated solution depends only on temperature. Since $\Delta_{\mathrm{fus}}H_A>0$ and $T<t_{fa}$>, we find ${ \ln y_A\ }<0$. Therefore, we find that $y_A<1$, as it must be. However, $y_A$ increases, with T, implying that the solubility of a solid increases as the temperature increases, as we usually observe.
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A similar analysis yields an equation for the solubility of a gas in a liquid solvent as a function of temperature at a fixed pressure, $P^{\#}$. We refer to the gas component as $A$ and the liquid component as $B$. We assume that solvent $B$ is nonvolatile, so that the gas phase with which it is in equilibrium is essentially pure gaseous solute $A$. We again find that the properties of the solvent have no role in our model, and the solubility of gas $A$ is the same in every solvent.
We assume that low concentrations of the solute obey Henry’s law and choose the solution-phase standard state for solute $A$ to be the pure hypothetical liquid $A$ whose vapor pressure is ${\textrm{ĸ}}_A$ at $T$. From Section 16.4, we then have ${\tilde{a}}_{A,\mathrm{solution}}=y_A$, so that $d{ \ln {\tilde{a}}_{A,\mathrm{solutio}\mathrm{n}}\ }=d{ \ln y_A\ }$ at any temperature. Substituting into the general equation
$d{\mu }_A={\overline{V}}_AdP-\overline{S}_AdT+RT\left(d{ \ln {\tilde{a}}_A\ }\right) \nonumber$ we have $d{\mu }_{A,\mathrm{solution}}=-\overline{S}_{A,\mathrm{solution}}\ dT+RT\left(d{ \ln y_A\ }\right) \nonumber$
The pressure of gas-phase $A$ is constant at $P^{\#}$, and $dP=0$. We choose the gas-phase activity standard state to be pure gas $A$ at $P^{\#}$ and $T$. Since this makes the activity of the pure gas unity at any temperature, we have $d{ \ln {\tilde{a}}_{B,gas}\ }=0$. Substituting, we have
$d{\mu }_{A,\mathrm{gas}}=-\overline{S}_{A,\mathrm{gas}}\ dT \nonumber$
Any constant-pressure process that maintains equilibrium between gas-phase $A$ and solution-phase $A$ must involve the same change in the chemical potential of $A$ in
each phase, so that $d{\mu }_{A,\mathrm{gas}}=d{\mu }_{A,\mathrm{solution}}$, and
$-\overline{S}_{A,\mathrm{gas}}\ dT=-\overline{S}_{A,\mathrm{solution}}\ dT+RT\left(d{ \ln y_A\ }\right) \nonumber$
so that $d{ \ln y_A\ }=-\frac{\left(\overline{S}_{A,\mathrm{gas}}-\overline{S}_{A,\mathrm{solution}}\right)}{RT}\ dT \nonumber$
The difference $\overline{S}_{A,\mathrm{gas}}-\overline{S}_{A,\mathrm{solution}}$ is the entropy change for the equilibrium—and hence reversible—process in which one mole of substance $A$ originally in solution vaporizes into a gas phase consisting of essentially pure gas $A$ while the system is at the constant pressure $P^{\#}$. Let us designate the enthalpy change for this reversible process at $P^{\#}$ and $T$ as $\Delta_{\mathrm{vap}}{\overline{H}}_{A,\mathrm{solution}}$. Then, we have
$\overline{S}_{A,\mathrm{gas}}-\overline{S}_{A,\mathrm{solution}}=\frac{\Delta_{\mathrm{vap}}{\overline{H}}_{A,\mathrm{solution}}}{T} \nonumber$
so that
$d{ \ln y_A\ }=-\frac{\Delta_{\mathrm{vap}}{\overline{H}}_{A,\mathrm{solution}}}{RT^2}\ dT \nonumber$
Since enthalpy changes are generally relatively insensitive to temperature, we expect that, at least over small ranges of $y_A$ and $T$, $\Delta_{\mathrm{vap}}{\overline{H}}_{A,\mathrm{solution}}$ is approximately constant. Since the vaporization process takes $A$ from a state in which it has some of the characteristics of a liquid into a gaseous state, we can be confident that $\Delta_{\mathrm{vap}}{\overline{H}}_{A,\mathrm{solution}}>0$. This conclusion implies that
$\frac{d{ \ln y_A\ }}{dT}<0 \nonumber$
Thus, our thermodynamic model leads us to the conclusion that the solubility of gas $A$ decreases as the temperature increases. That the solubilities of gases generally decrease with increasing temperature is a well-known experimental observation. It stands in contrast to the observation that the solubilities of liquid or solid—at $P^{\#}$ and $T$—substances generally increase with increasing temperature. Our analysis of gas solubility provides a satisfying theoretical interpretation for an experimental observation which otherwise appears to be counterintuitive.
The meaning of $\Delta_{\mathrm{vap}}{\overline{H}}_{A,\mathrm{solution}}$ is unambiguous. Our analysis enables us to measure it by experimentally measuring $y_A$ as a function of $T$. We can estimate $\Delta_{\mathrm{vap}}{\overline{H}}_{A,\mathrm{solution}}$ from another perspective: When we consider the “solution” in which $y_A=1$, the vaporization process is the vaporization of liquid $A$ into a gas phase of pure $A$ at $P^{\#}$ and $T$. Since we assume that $A$ is stable as a gas at $T$, the boiling point of pure liquid $A$ must be less than $T$ at $P^{\#}$ and the vaporization $A$ must be a spontaneous process at $P^{\#}$ and $T$. The enthalpy of vaporization datum which is most accessible for liquid $A$ is that for the reversible vaporization at one atmosphere and the normal boiling point, $T_B$, which we designate as $\Delta_{\mathrm{vap}}H^o_A$. If we stipulate that $P^{\#}$ is one atmosphere; assume that our solubility equation remains valid as $y_A$ increases from $y_A\approx 0$ to $y_A=1$; and assume that the enthalpy of vaporization is approximately constant between the boiling point of $A$ and $T$, we have $\Delta_{\mathrm{vap}}{\overline{H}}_{A,\mathrm{solution}}=\Delta_{\mathrm{vap}}H^o_A$. Then,
$d{ \ln y_A\ }=-\frac{\Delta_{\mathrm{vap}}H^o_A}{RT^2}\ dT \nonumber$
and
$\int^{y_A}_1{d{ \ln y_A\ }}=\int^T_{T_B}{-\frac{\Delta_{\mathrm{vap}}H^o_A}{RT^2}}\ dT \nonumber$
so that ${ \ln y_A\ }=\frac{\Delta_{\mathrm{vap}}H^o_A}{R}\left(\frac{1}{T}-\frac{1}{T_B}\right) \nonumber$
Check
Viewed critically, the accuracy of the approximation
$\Delta_{\mathrm{vap}}{\overline{H}}_{A,\mathrm{solution}}\approx \Delta_{\mathrm{vap}}H^o_A \nonumber$
is dubious. The assumptions we make to reach it are essentially equivalent to assuming that the cohesive forces in solution are about the same between $A$ molecules as they are between $A$ molecules and $B$ molecules. We expect this approximation to be more accurate the more closely the solution exhibits ideal behavior. However, if solvent $B$ is to satisfy our assumption that the solvent is nonvolatile, the cohesive interactions between $B$ molecules must be greater than those between $A$ molecules, and this not consistent with ideal-solution behavior.
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The analysis of freezing-point depression that we present in Section 16.11 introduces a number of simplifying assumptions. We now undertake a more rigorous analysis of this phenomenon. This analysis is of practical importance. Measuring the freezing-point depression of a solution is one way that we can determine the activity and the activity coefficient of the solvent component. As we see in Section 16.7, if we have activity coefficients for the solvent over a range of solute concentrations, we can use the Gibbs-Duhem equation to find activity coefficients for the solute. Freezing-point depression measurements have been used extensively to determine the activity coefficients of aqueous solutes by measuring the activity of water in their solutions.
As in our earlier discussion of freezing-point depression, the equilibrium system is a solution of solute $A$ in solvent $B$, which is in phase equilibrium with pure solid solvent $B$. Our present objective is to determine the activity of the solvent in its solutions at the melting point of the pure solvent. Having obtained this information, we can use the Gibbs-Duhem relationship to find the activity of the solute, as a function of solute concentration, at the melting point of the pure solvent. Once we have the solute activity at the melting point of the pure solvent, we can use the methods developed in Section 14.14 to find the solute activity in a solution at any higher temperature.
In Section 14.14, we find the temperature dependence of the natural logarithm of the chemical activity of a component of a solution. For a particular choice of activity standard states and enthalpy reference states, we develop a method to obtain the experimental data that we need to apply this equation. For brevity, let us refer to these choices as the infinite dilution standard states. In order to determine the activity of a solvent in its solutions at the melting point of the pure solvent, it is useful to define an additional standard state for the solvent. At temperatures below the normal melting point, which we again designate as $T_F$, we let the activity standard state of the solvent be pure solid $B$. Above the melting point, we use the infinite dilution standard state that we define in Section 14.14; that is, we let the activity standard state of the solvent be pure liquid solvent $B$.
At and below the melting point, $T_F$, the activity standard state for the solvent, $B$, is pure solid $B$. At and above the melting point, the activity standard state for the solvent is pure liquid $B$. At the melting point, pure solid solvent is in equilibrium with pure liquid solvent, which is also the solvent in an infinitely dilute solution. At $T_F$, the activity standard state chemical potentials of the pure solid solvent, the pure liquid solvent, and the solvent in an infinitely dilute solution are all the same. It follows that the value that we obtain for the activity of the solvent at $T_F$, for any particular solution, will be the same whether we determine it from measurements below $T_F$ using the pure solid standard state or from measurements above $T_F$ using the infinitely dilute solution standard state.
Now let us consider the chemical potential of liquid solvent $B$ in a solution whose composition is specified by the molality of solute $A$, ${\underline{m}}_A$, when the activity standard state is pure solid $B$. We want to find this chemical potential at temperatures in the range $T_{fp}<t_f$>, where $T_{fp}$ is the freezing point of the solution whose composition is specified by ${\underline{m}}_A$. In this temperature range, we have
${\mu }_B\left(\mathrm{solution},\ {\underline{m}}_A,T\right) \nonumber$ $={\widetilde{\mu }}^o_B\left(\mathrm{pure\ solid},\ T\right)+RT{ \ln {\tilde{a}}_B\ }\left(\mathrm{solution},\ {\underline{m}}_A,T\right) \nonumber$
Using the Gibbs-Helmholtz equation, we obtain
${\left(\frac{\partial { \ln {\tilde{a}}_B\ }\left(\mathrm{solution},\ {\underline{m}}_A,T\right)}{\partial T}\right)}_{P,{\underline{m}}_A} \nonumber$
$\ \ \ \ =\frac{-{\overline{H}}_B\left(\mathrm{solution},\ {\underline{m}}_A,T\right)}{RT^2}+\frac{{\tilde{H}}^o_B\left(T\right)}{RT^2} \nonumber$ $=\frac{-\left[{\overline{H}}_B\left(\mathrm{solution},\ {\underline{m}}_A,T\right)-{\overline{H}}^{\textrm{⦁}}_B\left(\mathrm{solid},T\right)\right]}{RT^2} \nonumber$
where we have ${\tilde{H}}^o_B\left(T\right)={\overline{H}}^{\textrm{⦁}}_B\left(\mathrm{solid},T\right)$, because pure solid $B$ is the activity standard state. Using the ideas developed in Section 14.14, we can use the thermochemical cycle shown in Figure 10 to evaluate
${\overline{H}}_B\left(\mathrm{solution},\ {\underline{m}}_A,T\right)-{\overline{H}}^{\textrm{⦁}}_B\left(\mathrm{solid},T\right) \nonumber$
In this cycle, ${\Delta }_{\mathrm{fus}}{\overline{H}}_B\left(T_F\right)$ is the molar enthalpy of fusion of pure $B$ at the melting point, $T_F$. ${\overline{L}}_B\left(\mathrm{solution},\ {\underline{m}}_A,T\right)$ is the relative partial molar enthalpy of $B$ at $T_F$ in a solution whose composition is specified by ${\underline{m}}_A$. The only new quantity in this cycle is
$\int^T_{T_f}{{\left(\frac{\partial {\overline{H}}_B\left(\mathrm{solution},\ {\underline{m}}_A,T\right)}{\partial T}\right)}_{P,{\underline{m}}_A}dT} \nonumber$ We can use the relative partial molar enthalpy of the solution to find it. By definition,
${\overline{L}}_B\left(\mathrm{solution},\ {\underline{m}}_A,T\right) \nonumber$ $={\left(\frac{\partial H\left(\mathrm{solution},\ {\underline{m}}_A,T\right)}{\partial n_B}\right)}_{P,T,n_A}-{\overline{H}}^{\mathrm{ref}}_B\left(T\right) \nonumber$
or, dropping the parenthetical information,
${\overline{L}}_B={\overline{H}}_B-{\overline{H}}^{\mathrm{ref}}_B\left(T\right) \nonumber$
so that ${\left(\frac{{\partial \overline{L}}_B}{\partial T}\right)}_P={\left(\frac{\partial {\overline{H}}_B}{\partial T}\right)}_P-{\left(\frac{\partial {\overline{H}}^{\mathrm{ref}}_B}{\partial T}\right)}_P \nonumber$
In Section 14.14, we introduce the relative partial molar heat capacity,
${\overline{J}}_B\left(T\right)={\left(\frac{{\partial \overline{L}}_B}{\partial T}\right)}_P \nonumber$
Since the infinitely dilute solution is the enthalpy reference state for $B$ in solution, we expect the molar enthalpy of pure liquid $B$ to be a good approximation to the partial molar enthalpy of liquid $B$ in the enthalpy reference state. Then, ${\left({\partial {\overline{H}}^{\mathrm{ref}}_B}/{\partial T}\right)}_P$ is just the molar heat capacity of pure liquid $B$, $C_P\left(B,\mathrm{liquid},T\right)$. (See problem 16-11.) We find
${\left(\frac{\partial {\overline{H}}_B}{\partial T}\right)}_P={\overline{J}}_B\left(T\right)+C_P\left(B,\mathrm{liquid},T\right) \nonumber$
Using this result, the enthalpy changes around the cycle in Figure 10 yield
${\overline{H}}_B\left(\mathrm{solution},\ {\underline{m}}_A,T\right)-{\overline{H}}^{\textrm{⦁}}_B\left(\mathrm{solid},T\right) \nonumber$
$=\int^{T_F}_T{C_P\left(B,\mathrm{solid},T\right)}dT+{\Delta }_{\mathrm{fus}}{\overline{H}}_B\left(T_F\right)-L^o_B\left(T_F\right)+{\overline{L}}_B\left({\underline{m}}_A,T_F\right)+\int^T_{T_F}{\left[{\overline{J}}_B\left(T\right)+\mathrm{\ }C_P\left(B,\mathrm{liquid},T\right)\right]}dT \nonumber$
$={\Delta }_{\mathrm{fus}}{\overline{H}}_B\left(T_F\right)+{\overline{L}}_B\left({\underline{m}}_A,T_F\right)-L^o_B\left(T_F\right)-\int^{T_F}_T{\left[C_P\left(B,\mathrm{liquid},T\right)-C_P\left(B,\mathrm{solid},T\right)+{\overline{J}}_B\right]}dT \nonumber$
Since we know how to determine ${\overline{J}}_B$ and the heat capacities as functions of temperature, we can evaluate this integral to obtain a function of temperature. For present purposes, let us assume that ${\overline{J}}_B$ and the heat capacities are essentially constant and introduce the abbreviation $\Delta C_P=C_P\left(B,\mathrm{liquid},T\right)-C_P\left(B,\mathrm{solid},T\right)$, so that $\int^{T_F}_T{\left[C_P\left(B,\mathrm{liquid},T\right)-C_P\left(B,\mathrm{solid},T\right)+{\overline{J}}_B\right]}dT=\left(\Delta C_P+{\overline{J}}_B\right)\left(T_F-T\right) \nonumber$
The temperature derivative of ${ \ln {\tilde{a}}_B\ }\left(\mathrm{solution},\ {\underline{m}}_A,T\right)$ becomes
${\left(\frac{\partial { \ln {\tilde{a}}_B\ }\left(\mathrm{solution},\ {\underline{m}}_A,T\right)}{\partial T}\right)}_{P,{\underline{m}}_A} \nonumber$ $=\frac{-{\Delta }_{\mathrm{fus}}{\overline{H}}_B\left(T_F\right)-{\overline{L}}_B\left({\underline{m}}_A,T_F\right)+L^o_B\left(T_F\right)}{RT^2}+\frac{\left(\Delta C_P+{\overline{J}}_B\right)\left(T_F-T\right)}{RT^2} \nonumber$
$T_{fp}$ is the freezing point of the solution whose composition is specified by ${\underline{m}}_A$. At $T_{fp}$ the solvent in this solution is at equilibrium with pure solid solvent. Hence, the chemical potential of the solution solvent is equal to that of the pure-solid solvent. Then, because the pure solid is the activity standard state for both solution solvent and pure-solid solvent at $T_{fp}$, the activity of the solution solvent is equal to that of the pure-solid solvent. Because the pure solid is the activity standard state, the solvent activity is unity at $T_{fp}$. This means that we can integrate the temperature derivative from $T_{fp}$ to $T_F$ to obtain
\begin{aligned} \int^{T_F}_{T_{fp}} d ~ \ln \tilde{a}_B \left(\mathrm{solution},\ {\underline{m}}_A,T\right) & ~ \ ~ & = \ln \tilde{a}_B \left(\mathrm{solution},\ {\underline{m}}_A,T_F\right) \ ~ & = \left( \frac{\Delta _{\text{fus}} \overline{H}_B \left(T_F \right)+ \overline{L}_B \left( \underline{m}_A, T_F \right) +L^o_B\left(T_F\right)}{R} \right) \left(\frac{1}{T_F}-\frac{1}{T_{fp}}\right) -\left(\frac{\Delta C_P+ \overline{J}_B}{R}\right) \left(1-\frac{T_F}{T_{fp}}+ \ln \left(\frac{T_F}{T_{fp}} \right) \right) \end{aligned} \nonumber
Thus, from the measured freezing point of a solution whose composition is specified by ${\underline{m}}_A$, we can calculate the activity of the solvent in that solution at $T_F$.
Several features of this result warrant mention. It is important to remember that we obtained it by assuming that $\Delta C_P+{\overline{J}}_B\left(T\right)$ is a constant. This is usually a good assumption. It is customary to express experimental results as values of the freezing-point depression, $\Delta T=T_F-T_{fp}$. The activity equation becomes
\begin{aligned} \ln \tilde{a}_B \left(\mathrm{solution}, \underline{m}_A,T_F\right) & ~ \ ~ & =-\left(\frac{\Delta _{\mathrm{fus}} \overline{H}_B\left(T_F\right)+\overline{L}_B\left( \underline{m}_A,T_F\right)+L^o_B\left(T_F\right)}{RT_FT_{fp}}\right)\Delta T \ ~ & +\left(\frac{\Delta C_P+{\overline{J}}_B}{R}\right)\left(\frac{\Delta T}{T_{fp}}-{ \ln \left(1+\frac{\Delta T}{T_{fp}}\right)\ }\right) \end{aligned} \nonumber
The terms involving ${\overline{L}}_B$, $L^o_B$, $\Delta C_P$, and ${\overline{J}}_B$ are often negligible, particularly when the solute concentration is low. When ${T_F}/{T_{fp}\approx 1}$, that is, when the freezing-point depression is small, the coefficient of $\Delta C_P+{\overline{J}}_B$ is approximately zero. When these approximations apply, the activity equation is approximated by
${ \ln {\tilde{a}}_B\ }\left(\mathrm{solution},\ {\underline{m}}_A,T_F\right)=-\left(\frac{{\Delta }_{\mathrm{fus}}{\overline{H}}_B\left(T_F\right)}{RT^2_F}\right)\Delta T \nonumber$
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/16%3A_The_Chemical_Activity_of_the_Components_of_a_Solution/16.15%3A_Solvent_Activity_Coefficients_from_Freezing-point_Depre.txt
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Thus far in our discussion of solute activities, we have assumed that the solute is a molecular species whose chemical structure is unchanged when the pure substance dissolves. This is not the case when salts dissolve in water and other polar solvents. A pure solid salt exists as a lattice of charged ions, rather than electrically neutral molecular moieties, and its solutions contain solvated ions. Since salt solutions conduct electricity, we often call them electrolytic solutions. Solutions of salts in water are extremely important from both practical and theoretical standpoints. Accordingly, we focus our discussion on aqueous solutions; however, the ideas that we develop apply to salt solutions in any solvent that supports the formation of solvated ions.
We can apply the concepts that we develop in this chapter to measure the activities of aqueous salt solutions. When we do so, we find new features. These features arise from the formation of aquated ionic species and from electrical interactions among these species. In this chapter, we consider only the most basic issues that arise when we investigate the activities of dissolved salts. We consider only strong electrolytes; that is, salts that are completely dissociated in solution. In this section, we briefly review the qualitative features of such solutions.
Departure from Henry’s law behavior begins at markedly lower concentrations when the solute is a salt than when it is a neutral molecular species. This general observation is easily explained: Departures from Henry’s law are caused by interactions among solution species. For neutral molecules separated by a distance \(r\), the variation of the interaction energy with distance is approximately proportional to \(r^{-6}\). This means that only the very closest molecules interact strongly with one another. For ions, Coulomb’s law forces give rise to interaction energies that vary as \(r^{-1}\). Compared to neutral molecules, ions interact with one another at much greater distances, so that departures from Henry’s law occur at much lower concentrations.
Our qualitative picture of an aqueous salt solution is that the cations and anions that comprise the solid salt are separated from one another in the solution. Both the cations and the anions are surrounded by layers of loosely bound water molecules. The binding results from the electrical interaction between the ions and the water-molecule dipole. The negative (oxygen) end of the water dipole is preferentially oriented toward cations and the positive (hydrogen) end is preferentially oriented toward anions.
In aqueous solution, simple metallic cations are coordinated to a first layer of water molecules that occupy well-defined positions around the cation. In this layer, the bonding can have a covalent component. Such combinations of metal and coordinated water molecules are called aquo complexes. For most purposes, we can consider that the aquo complex is the cationic species in solution. Beyond the layer of coordinated water molecules, a second layer of water molecules is less tightly bound. The positions occupied by these molecules are more variable. At still greater distances, water molecules interact progressively more weakly with the central cation. In general, when we consider the water molecules that surround a given anion, we find that even the closest solvent molecules do not occupy well-defined positions.
In any macroscopic quantity of solution, each ion has a specific average concentration. On a microscopic level, the Coulomb’s law forces between dissolved ions operate to make the relative locations of cations and anions less random. It is useful to think about a spherical volume that surrounds a given ion. We suppose that the diameter of this sphere is several tens of nanometers. Within such a sphere centered on a particular cation, the concentration of anions will be greater than the average concentration of anions; the concentration of cations will be less than the average concentration of cations. Likewise, within a microscopic sphere centered on a given anion, the concentration of cations will be greater than the average concentration of cations; the concentration of anions will be below average.
As the concentration of a dissolved salt increases, distinguishable species can be formed in which a cation and an anion are nearest neighbors. We call such species ion pairs. At sufficiently high salt concentrations, a significant fraction of the ions can be found in such ion-pair complexes. Compared to other kinds of chemical bonds, ion-pair bonds are weak. The ion-pair bond is labile; the lifetime of a given ion pair is short. At still higher salt concentrations, the formation of significant concentrations of higher aggregates becomes possible. Characterizing the species present in an electrolytic solution becomes progressively more difficult as the salt concentration increases.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/16%3A_The_Chemical_Activity_of_the_Components_of_a_Solution/16.16%3A_Electrolytic_Solutions.txt
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We can find the activity of a salt in its aqueous salt solutions. For example, we can measure the freezing point depression for aqueous solutions of sodium chloride, find the activity of water in these solutions as a function of the sodium chloride concentration, and use the Gibbs-Duhem equation to find the activity of the dissolved sodium chloride as a function of its concentration. When we do so, we find some marked differences from our observations on molecular solutes.
For molecular solutes, the activity approaches the solute concentration as the concentration approaches zero; that is, the activity coefficient for a molecular solute approaches unity as the concentration approaches zero. For sodium chloride, and other 1:1 electrolytes, we find that the activity we measure in this way approaches the square of the solute concentration as the concentration approaches zero. For other salts, the measured activity approaches other powers of the solute concentration as the concentration approaches zero.
The dissociation of the solid salt into solvated ions explains these observations. Let us consider a solution made by dissolving $n$ moles of a salt, $A_pB_q$, in $n_{\mathrm{solvent}}$ moles of solvent. (Let $A$ be the cation and $B$ the anion.) For present purposes, the cation and anion charges are not important. We use $p$ and $q$ to designate the composition of the salt. Typically, we are interested in dilute solutions, and it is convenient to use the hypothetical one-molal solution as the standard state for the activity of a solute species. We can represent the Gibbs free energy of this solution as
$G=n_{\mathrm{solvent}}{\mu }_{\mathrm{solvent}}+n{\mu }_{A_pB_q} \nonumber$
where ${\mu }_{\mathrm{solvent}}$ and ${\mu }_{A_pB_q}$ are the partial molar Gibbs free energies of the solvent and the solute in the solution. We can also write
$\mu_{A_pB_q}={\widetilde{\mu }}^o_{A_pB_q}+RT{ \ln {\tilde{a}}_{A_pB_q}\ } \nonumber$
where ${\widetilde{\mu }}^o_{A_pB_q}$ is the partial molar Gibbs free energy when ${\tilde{a}}_{A_pB_q}={\underline{m}}_{A_pB_q}{\gamma }_{A_pB_q}=1$ in the activity standard state of the salt.
We assume that $A_pB_q$ is a strong electrolyte; its solution contains $np$ moles of the cation, $A$, and $nq$ moles of the anion, $B$. In principle, we can also represent the Gibbs free energy of the solution as
$G=n_{\mathrm{solvent}}{\mu }_{\mathrm{solvent}}+np{\mu }_A+nq{\mu }_B \nonumber$
and the individual-ion chemical potentials as ${\mu }_A={\widetilde{\mu }}^o_A+RT{ \ln {\tilde{a}}_A\ }$ and ${\mu }_B={\widetilde{\mu }}^o_B+RT{ \ln {\tilde{a}}_B\ }$, where ${\widetilde{\mu }}^o_A$ and ${\widetilde{\mu }}^o_B$ are the partial molar Gibbs free energies of the ions $A$ and $B$ in their hypothetical one-molal activity standard states. Equating the two equations for the Gibbs free energy of the solution, we have
${\mu }_{A_pB_q}=p{\mu }_A+q{\mu }_B \nonumber$
and
$\widetilde{\mu}^o_{A_pB_q}+RT{ \ln {\tilde{a}}_{A_pB_q}\ } ={p\widetilde{\mu }}^o_A+RT{ \ln {\tilde{a}}^p_A\ }+{q\widetilde{\mu }}^o_B+RT{ \ln {\tilde{a}}^q_B\ } \nonumber$
While it is often experimentally challenging to do so, we can measure ${\widetilde{\mu }}^o_{A_pB_q}$ and ${\tilde{a}}_{A_pB_q}$. In principle, the meanings of the individual-ion activities, ${\tilde{a}}_A$ and ${\tilde{a}}_B$, and their standard-state chemical potentials, ${\widetilde{\mu }}^o_A$ and ${\widetilde{\mu }}^o_B$, are unambiguous; however, since we cannot prepare a solution that contains cation $A$ and no anion, we cannot make measurements of ${\tilde{a}}_A$ or ${\widetilde{\mu }}^o_A$ that are independent of the properties of $B$, or some other anion. Consequently, we must adopt some conventions to relate these properties of the ions, which we cannot measure, to those of the salt solution, which we can.
The universally adopted convention for the standard chemical potentials is to equate the sum of the standard chemical potentials of the constituent ions to that of the salt. We can think of this as assigning an equal share of the standard-state chemical potential of the salt to each of its ions; that is, we let
${\widetilde{\mu }}^o_A={\widetilde{\mu }}^o_B=\frac{\widetilde{\mu }^o_{A_pB_q}}{p+q} \nonumber$
Then,
${\widetilde{\mu }}^o_{A_pB_q}=p{\widetilde{\mu }}^o_A+q{\widetilde{\mu }}^o_B \nonumber$
and the activities of the individual ions are related to that of the salt by
${\tilde{a}}_{A_pB_q}={\tilde{a}}^p_A{\ \tilde{a}}^q_B \nonumber$
We can develop the convention for the activities of the individual ions by representing each activity as the product of a concentration and an activity coefficient. That is, we represent the activity of each individual ion in the same way that we represent the activity of a molecular solute. In effect, this turns the problem of developing a convention for the activities of the individual ions into the problem of developing a convention for their activity coefficients. Using the hypothetical one-molal standard state for each ion, we write ${\tilde{a}}_A={\underline{m}}_A{\gamma }_A$ and ${\tilde{a}}_B={\underline{m}}_B{\gamma }_B$, where ${\underline{m}}_A$, ${\gamma }_A$, ${\underline{m}}_B$, and ${\gamma }_B$ are the molalities and activity coefficients for ions $A$ and $B$, respectively. Let the molality of the salt, $A_pB_q$, be $\underline{m}$. Then ${\underline{m}}_A=p\underline{m}$ and ${\underline{m}}_B=q\underline{m}$, and
${\tilde{a}}_{A_pB_q}={\tilde{a}}^p_A{\ \tilde{a}}^q_B={\left(p\underline{m}{\gamma }_A\right)}^p{\left(q\underline{m}{\gamma }_B\right)}^q=\left(p^pq^q\right){\underline{m}}^{p+q}{\gamma }^p_A{\gamma }^q_B \nonumber$
Now we introduce the geometric mean of the activity coefficients ${\gamma }_A$ and ${\gamma }_B$; that is, we define the geometric mean activity coefficient, ${\gamma }_{\pm }$, by
$\gamma_{\pm}=\left(\gamma^p_A \gamma^q_B\right)^{1/{\left(p+q\right)}} \nonumber$
The activity of the dissolved salt is then given by
$\tilde{a}_{A_pB_q}=\left(p^pq^q\right) \underline{m}^{p+q}{\gamma_{\pm }}^{p+q} \nonumber$
The mean activity coefficient, ${\gamma }_{\pm }$, can be determined experimentally as a function of ${\underline{m}}_{A_pB_q}$, but the individual activity coefficients, ${\gamma }_A$ and ${\gamma }_B$, cannot. It is common to present the results of activity measurements on electrolytic solutions as a table or a graph that shows the mean activity coefficient as a function of the salt molality.
While we cannot determine the activity or activity coefficient for an individual ion experimentally, no principle prohibits a theoretical model that estimates individual ion activities. Debye and Hückel developed such a theory. The Debye-Hückel theory gives reasonably accurate predictions for the activity coefficients of ions for solutions in which the total ion concentration is about 0.01 molal or less. We summarize the results of the Debye-Hückel theory in Section 16.18.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/16%3A_The_Chemical_Activity_of_the_Components_of_a_Solution/16.17%3A_Activities_of_Electrolytes_-_The_Mean_Activity_Coeffici.txt
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