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The derivation of $$\mathrm{d}G=\delta w_{non.exp. max.}$$ $$G=H-TS$$ $$ \mathrm{d}G=\mathrm{d}H-T\mathrm{d}S-S\mathrm{d}T$$ $$\mathrm{d}G=\mathrm{d}U+P\mathrm{d}V+V\mathrm{d}P-T\mathrm{d} S-S\mathrm{d} T$$ $$dG=\delta w_{exp}+\delta w_{non.exp}+\delta Q+P\mathrm{d}V+V\mathrm{d}P-T\mathrm{d}S-S\mathrm{d}T$$ Now the term $\delta Q$ and the term $-T\mathrm{d} S$ are cancelled, Note that this means process is reversible($ \mathrm{d} Q_{rev}=TdS$).Also the term $\delta w_{exp}$ and the term $P\mathrm{d}V$are cancelled.So the only terms remaining are$$dG=\delta w_{non.exp.}+V\mathrm{d}P-S\mathrm{d}T$$ when pressure and temperature are kept constant then $$dG=\delta w_{non.exp.}$$ So looking at this implies that Gibbs free energy is equal to the non expansion work only when a process is reversible and when the pressure and temperature are constant. Is this correct? If no then what am I missing here? If yes then what does Gibbs free energy imply(Any kind of work?) when process is irreversible?

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  • $\begingroup$ Gibbs energy change at p,T=const is the maximum nonvolume work (negative as done by the system ), achievable at reversible conditions. For irreversible conditions, this available work is lower. Imagine e.g. an ideal reversible rechargable cell versus a real cell with Ohmic ( and some other ) loses. $\endgroup$
    – Poutnik
    Jun 1, 2021 at 8:40
  • $\begingroup$ Yes, you are correct. $\endgroup$ Jun 1, 2021 at 10:27
  • $\begingroup$ Maybe my answer here would help. $\endgroup$ Jun 1, 2021 at 15:15
  • $\begingroup$ @Poutnik ok got it now $\endgroup$ Jun 2, 2021 at 4:31

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We are assuming in this analysis that the system is in contact with a single constant temperature reservoir at temperature T, the same temperature as the initial temperature of the system, and a single external pressure at pressure P, the same pressure as the initial pressure of the system (see Denbigh). From the first law, for such a process, we have $$\Delta U=Q-P\Delta V -W_{nonPV}$$and, from the 2nd law, we have $$\Delta S=\frac{Q}{T}+\sigma$$where $\sigma$ is the entropy generated within the system due to irreversibility (an always-positive entity). So, if we combine these two equations, we obtain: $$\Delta G=\Delta U+P\Delta V-T\Delta S=-W_{nonPV}-T\sigma$$or equivalently, $$W_{nonPV}=-\Delta G-T\sigma$$So, for a reversible path between the initial and final states of the system, the nonPV work is maximum and equal to $-\Delta G$ and, for an irreversible path between the same two end states, the nonPV work is diminished by T times the amount of entropy generated due to irreversibility (and $-\Delta G$ exceeds the amount of nonPV work done by the system).

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  • $\begingroup$ Hey Chet! I wanted to ask you something. I read your answer here and was wondering if $\Delta A$ is also diminished in irreversible path? I applied your method you wrote here, and came to a conclusion that just like $\Delta G$, $\Delta A$ is also diminished for irreversible process. What do you think about it? Is it correct? $\endgroup$
    – Natasha J
    Jan 28, 2023 at 5:26
  • $\begingroup$ @NatashaJ Yes. It is worked out in detail in Denbigh, Principles of Chemical Equilibrium. $\endgroup$ Jan 28, 2023 at 12:39
  • $\begingroup$ Thank you for reply! I read in a book that though the work involved is different from maximum work, yet $\Delta A$ has the same value in both, reversible and irreversible path. I think it may be a misprint. Thank you. $\endgroup$
    – Natasha J
    Jan 28, 2023 at 12:50
  • $\begingroup$ @Natasha Why do you think that? $\endgroup$ Jan 28, 2023 at 15:17
  • $\begingroup$ @NatashaJ What Denbigh shows is that, if the initial and final thermodynamic equilibrium temperatures of a process in a closed system are the same value T, and the process takes place with all the heat transfer to the system solely involving an ideal constant temperature reservoir also at T, then the work done by the system is less than the reversible work equal to $-Delta A$ between the initial and final states. $\endgroup$ Jan 28, 2023 at 15:45

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