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So $\Delta G = \Delta H - T\Delta S - S\Delta T$.

Is there a way to express $\Delta G$ for a process where T, V, and P change simultaneously only in terms of the initial and final T, V, P? Assume this process is for an ideal gas.

Since $G$ is a state function I can calculate it separately for an isothermal process and an isochoric process, but in that case $\Delta G$ for the isochoric process has dependence on the entropy of one of the states.

Auguste Baudin
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  • To 'change simultaneously' means we could also be looking at instantaneous change, which was previously addressed on this forum here https://chemistry.stackexchange.com/questions/50569/what-is-the-difference-between-%CE%94g-and-%CE%94rg . – AJKOER May 07 '20 at 16:30

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Your original equation is incorrect. It should read $\Delta G=\Delta H -\Delta (TS)$, and $\Delta (TS)\neq T\Delta S+S\Delta T$ unless the changes are infinitesimal.

That said, for an ideal gas, irrespective of path, $$\Delta S=S_f-S_i=C_p\ln{(T_f/T_i)}-R\ln{(P_f/P_i)}=C_v\ln{(T_f/T_i)}+R\ln{(V_f/V_i)}$$and$$\Delta H=C_p(T_f-T_i)$$So, $$\Delta G=G_f-G_i=C_p\Delta T-\frac{(T_f+T_i)}{2}\Delta S-\frac{(S_f+S_i)}{2}\Delta T$$Clearly the 3rd term on the rhs depends on either $S_i$, $S_f$, or both, so $\Delta G$ depends on these. It then all comes down to how you assign your reference state values for G, H, and S, such that $G_{ref}=H_{ref}-T_{ref}S_{ref}$.

Chet Miller
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