I have read that the equation $\Delta G = \Delta H-T\Delta S$ is valid only at constant temperature and pressure. However, $\Delta H=0$ in an isothermal process which would give $\Delta G = -T\Delta S$ which is clearly wrong. Where have I gone wrong?
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$\begingroup$ You have to be careful about the meaning of $\Delta G = \Delta H-T\Delta S$. In this equation, $\Delta G$ represents "the maximum non-expansion work which a closed system can transfer at constant temperature and pressure" as quoted in a similar way by wikipedia, en.wikipedia.org/wiki/Gibbs_free_energy $\endgroup$– Andy ChenCommented Jul 1, 2022 at 6:23
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$\begingroup$ The definition of $G$ is $G=U+PV-TS=H-TS$. Since $\delta U=T\delta S-(P\delta V+\delta W_{ne})+\sum_{j}{\mu_j\delta N_j}$ where $\delta W_{ne}$ is the non-expansion work (like magnetic, electric work) which the system does to the environment, if we take a differential on $G$, we have $\delta G=\delta U+P\delta V+V\delta P-T\delta S - S\delta T=V\delta P -S\delta T+\sum_{j}{\mu_j\delta N_j}-\delta W_{ne}$. At constant temperature and pressure, and given the system being closed, $\delta G=\delta H - T\delta S=-\delta W_{ne}$. This is what you consider to be valid. $\endgroup$– Andy ChenCommented Jul 1, 2022 at 6:36
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I have read that the equation $\Delta G = \Delta H-T\Delta S$ is valid only at constant temperature and pressure.
Not quite; $G\equiv H-TS$, so $\Delta G\equiv \Delta H-T\Delta S$ requires only constant temperature $T$.
$G$ is the potential that's minimized at constant temperature and pressure (and is frequently used in this context), so either the source was confused or there was some misinterpretation.
However, $\Delta H=0$ in an isothermal process
This is the case only for an ideal gas (and ideal-gas-like models such as the ideal elastomer). In that case, you're free to use $\Delta G = -T\Delta S$. Since interparticle bonding isn't relevant in the ideal gas, whose stiffness is purely entropic, it's not surprising that the entropy and temperature (and not the enthalpy) are the key parameters.
answered Jul 1, 2022 at 4:11
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$\begingroup$ Can we continue this discussion in chat? $\endgroup$– BosonCommented Jul 1, 2022 at 4:18
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$\begingroup$ chat.stackexchange.com/rooms/137446/… $\endgroup$– BosonCommented Jul 1, 2022 at 4:26
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$\begingroup$ I’m sorry, I don’t see any content there. $\endgroup$ Commented Jul 2, 2022 at 16:17
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You said that $\Delta G=-T\Delta S$ is valid for an isothermal process. That is right for an ideal gas. In that case, the heat exchange must be used to do work unrelated to expansion. So, I do not understand why you say that is clearly wrong. In any case, we could wonder whether such a process is indeed possible.
