A Curious (definitely wrong) result of Absolute Entropy
The Gibbs free energy change is zero in the case of reversible processes carried out at constant temperature and pressure, but that isn't the case if these conditions are not observed.
As a demonstration consider an isothermal reversible expansion of an ideal gas.
Since the temperature is constant, the free energy change is given by
$$\Delta G = \Delta H - T \Delta S$$
Because the gas is ideal and T constant, $\Delta U = 0$, and from the second law of thermodynamics it follows that
$$\Delta S = \frac{q_\text{rev}}{T}= -\frac{w_\text{rev}}{T} = \frac{\int_{V_1}^{V_2}p dV}{T} = \frac{\int_{V_1}^{V_2} \frac{nRT}{V} dV}{T} = nR\log \left( \frac{V_2}{V_1} \right)$$
It also follows that
$$\Delta H = \Delta U + \Delta (pV) = 0 + \Delta (nRT) = RT \Delta n = 0$$
since the system is closed.
But then
$$\Delta G = - nRT\log \left( \frac{V_2}{V_1} \right) \neq 0$$
The free energy change is not zero because the pressure changed during the process.
For the reversible adiabatic expansion of an ideal gas
$$\begin{align}  \Delta S &amp;= 0  \\ \Delta H &amp;= \int_{T_1}^{T_2} C_p dT  = C_p \Delta T \\ \Delta G &amp;= \int_{P_1}^{P_2} V dP - \int_{T_1}^{T_2} S dT =  (V_1P_1^{C_V/C_P})\left(\frac{C_P}{C_P-C_V}\right)( P_2^{1-C_V/C_P}-P_1^{1-C_V/C_P}) - S\Delta T  \end{align}$$
The integral over P in the expression for $\Delta G$ can be evaluated after substituting the following expression for the volume in terms of pressure:
$$V=(V_1P_1^{C_V/C_P}) P^{-C_V/C_P}$$