# Variable Molecular Weight

1. Mar 29, 2005

### Clausius2

Hi friends.

I am wondering about the next calculation.

Imagine there is a hole of some diameter, through which it is discharging a jet of an ideal gas of molecular weight $$W_1$$ (Kg/mol). Outside, there is an atmosphere at rest of molecular weight $$W_2$$ (Kg/mol). The external atmosphere has not got any volume definited, it can be assumed to have an infinite extension. Both, jet gas and atmosphere are at the same pressure $$P$$ and temperature $$T$$. There are no gradients of pressure nor temperature.

My question is, how can I calculate the average molecular weight $$\overline{W}$$ of the binary mixture?.

Well, this maybe a kinda easy question, but I don't see it. I wait your suggestions to continue with it.

Thanks.

2. Mar 29, 2005

### SpaceTiger

Staff Emeritus
Well, if you know the relative fractions of each gas at a given point, the mean molecular weight is given by a density-weighted average of $$W_1$$ and $$W_2$$, but calculating the relative densities seems to me a complicated fluid flow problem. If you assume it continues expelling gas, after a long time and at large distances, you could probably treat it as a point source and use simple diffusion arguments to get the relative fractions as a function of radius from the hole... is that what you want?

3. Mar 30, 2005

### Clausius2

Thanks for your explanation, but is not exactly what I am looking for. I mean, I have calculated it as:

-The mixture has to yield ideal gas state equation:

$$\overline{\rho}=\frac{P\overline{W}}{RT}$$

Now is when I am not too sure; mixture molecular weight can be calculated as a function of molar fractions $$X_i$$ of each gas (sure these molar fractions are a function of spatial coordinates):

$$\overline{W}=\sum^2_{i=1}X_iW_i$$

but where this formula comes from?

I am not sure about what are the assumptions to derivate it.

4. Mar 30, 2005

### SpaceTiger

Staff Emeritus
Ok, sorry, I guess I took your question too far.

For the purposes of the gas pressure, you can think of a multi-component gas with mean molecular weight W as being a single-component gas of particles all with weight W. That's pretty much just by definition. Then, when combining mean molecular weights, you can just treat it the same way as you did the original multi-component gas. I was bored during lecture, so I did the full derivation:

For a gas with N components, each with its own mass, $$m_i$$, the total pressure is just the sum of the component pressures:

$$P = \frac{\sum_{i=1}^{N}n_i~RT}{V}$$

The total mass density is given by:

$$\rho = \frac{\sum_{i=1}^{N}n_im_i}{V}$$

so

$$P=\rho RT\frac{\sum_{i=1}^{N}n_i}{\sum_{i=1}^{N}n_im_i}=\frac{\rho RT}{W}$$
$$W=\frac{\sum_{i=1}^{N}n_im_i}{\sum_{i=1}^{N}n_i}$$

That defines the mean molecular weight. Then, for two gases (denoted "A" and "B"), each with a different mean molecular weight:

$$P = \frac{(\sum_{i=1}^{N_1}n_i^A+\sum_{i=1}^{N_2}n_i^B)~RT}{V}$$
$$\rho = \frac{\sum_{i=1}^{N_1}n_i^Am_i^A+\sum_{i=1}^{N_2}n_i^Bm_i^B}{V}$$
$$P=\rho RT\frac{\sum_{i=1}^{N_1}n_i^A+\sum_{i=1}^{N_2}n_i^B}{\sum_{i=1}^{N_1}n_i^Am_i^A+\sum_{i=1}^{N_2}n_i^Bm_i^B}=\frac{\rho RT}{\bar{W}}$$
$$\bar{W}=\frac{\sum_{i=1}^{N_1}n_i^Am_i^A+\sum_{i=1}^{N_2}n_i^Bm_i^B}{\sum_{i=1}^{N_1}n_i^A+\sum_{i=1}^{N_2}n_i^B}=\frac{(\frac{\sum_{i=1}^{N}n_i^Am_i^A}{\sum_{i=1}^{N}n_i^A})\sum_{i=1}^{N}n_i^A+(\frac{\sum_{i=1}^{N}n_i^Bm_i^2}{\sum_{i=1}^{N}n_i^B})\sum_{i=1}^{N}n_i^B}{n_{tot}}=W_AX_A+W_BX_B$$

Once you've gone through that derivation, the generalization to arbitrary numbers of components (with $$W_i$$) is obvious. I just wanted to avoid too much cumbersome notation.

Last edited: Mar 30, 2005
5. Mar 30, 2005

### Gokul43201

Staff Emeritus
How is this self-evident ?

6. Mar 31, 2005

### SpaceTiger

Staff Emeritus
If you assume that there are no intermolecular forces and the collisions are elastic (as in an ideal gas), then the presence of other components shouldn't change the amount of pressure each one exerts. Thus, it should just be additive.

I was having trouble putting it into words, so I did a quick google search. Turns out it's true and is often referred to as Dalton's Law. I think the web sites turned up on a google search would do a better job of explaining it than I. The law isn't always true in non-ideal conditions, however.

Last edited: Mar 31, 2005
7. Mar 31, 2005

### Clausius2

Right. You hit the head of the nail. I knew how to derivate it too from Dalton's Law. (But I acknowledge your effort by the way!).

The point here is if, with assumptions made previously of an isothermic and isobaric mixture this derivation is valid. You are assuming: i) both components have the same temperature, ii) each component has its partial pressure given by Dalton's Law.

The question is: if jet gas is discharging at the same pressure of the ambient and there is not any pressure gradient, does it has sense to talk about "partial pressures"??? i.e the total pressure P is the sum of partial pressures $$P_i$$??? Wouldn't be each partial pressure equal to P?.

Maybe It is a misconception what I am saying here.

8. Mar 31, 2005

### Clausius2

I have been thinking of it.

I don't see a practical difference between making this three distinctions:

-case a) I am adding pressures under a unique volume:

$$P_iV=n_i RT$$

then $$\sum P_iV=PV=\sum n_iRT=\sum\Big(\frac{m_i}{W_i}\Big)RT=\frac{\sum m_i}{\overline{W}}RT$$

working out the average molecular weight from the last equal:

$$\overline{W}=\sum X_iW_i$$

-case b) adding volumes under a unique pressure:

$$PV_i=n_i RT$$

then $$\sum PV_i=PV=\sum n_iRT=\sum\Big(\frac{m_i}{W_i}\Big)RT=\frac{\sum m_i}{\overline{W}}RT$$

working out the average molecular weight from the last equal:

$$\overline{W}=\sum X_iW_i$$

$$P_i V_i=n_i RT$$

then $$\sum P_iV_i=PV=\sum n_iRT=\sum\Big(\frac{m_i}{W_i}\Big)RT=\frac{\sum m_i}{\overline{W}}RT$$

working out the average molecular weight from the last equal:

$$\overline{W}=\sum X_iW_i$$

In all cases the result is the same.

But it arises another question: In what practical-physical situation will be add pressures, volumes or energy of a mixture?. In cases b) and c) the result of $$P_i=X_i P$$ to denote the partial pressure is non sense. But I don't know what does it physically mean.

Last edited: Mar 31, 2005
9. Apr 1, 2005

### SpaceTiger

Staff Emeritus
These are both assumptions you stated in the original problem, the second one being a consequence of them being ideal gases. I don't think they're bad assumptions.

Partial pressure refers to the pressure of the individual subcomponents of the gas. I'll respond to your other post and hopefully make this clearer.

10. Apr 1, 2005

### SpaceTiger

Staff Emeritus
In this case, you're talking about the pressure from each individual component of a gas in a given volume. On average, they all occupy the same space, but they each apply their own force on a unit area. If you put a hypothetical unit area inside the volume, the total force on its surface will the sum of all the forces from the individual gas components.

In this case, you're splitting the volume into subvolumes. The equation implies that each component of the gas is present in a different subunit, so describing it all as a single ideal gas doesn't make sense, even though the equation works out.

This is an acceptable way to think about it only if Vi=V. Otherwise, you have the same subvolume problem I described above.

11. Apr 1, 2005

### Clausius2

Thanks for your aid, by the way.

Let's think of this case b). Don't you think that this assumption of unique pressure and different volumes is more accurated for my original problem?.

Jet gas is totally expanded when discharging, and external atmosphere has not a definited volume, so I don't see how can I imagine a partial pressure of some component. On the other hand I would see some volume occupied by one gas and another volume by the other. That's make sense if you imagine some imaginary control volume.

Also I haven't understood when you said: "so describing it all as a single ideal gas doesn't make sense, even though the equation works" . Why doesn't it make sense?.

12. Apr 1, 2005

### SpaceTiger

Staff Emeritus
The best way to treat your problem is with densities, not with moles and volumes. Honestly, I don't know why anybody bothers with that formalism anymore, it's only applicable to special cases in which the density is constant. Like I said in my first response, the real trick would be in figuring out the mean molecular weight as a function of position. Once you have that, the other things follow.

I'm not sure how to explain it in another way without drawing a figure. Perhaps someone coming from an engineering background would be able to explain it better...

But it wouldn't be right. The gases are mixed and the relative mixing is a continuous function of position. You can't just choose a non-infinitesmal volume and say that it's composed of only one gas. This is why people use differential equations to solve fluid flows.
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