# How can the Gibbs free energy equation be at constant temperature and pressure?

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?

• 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 Commented Jul 1, 2022 at 6:23
• 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. Commented Jul 1, 2022 at 6:36

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.

• Can we continue this discussion in chat? Commented Jul 1, 2022 at 4:18
• chat.stackexchange.com/rooms/137446/… Commented Jul 1, 2022 at 4:26
• I’m sorry, I don’t see any content there. Commented Jul 2, 2022 at 16:17

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.