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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 '21 at 8:40
  • $\begingroup$ Yes, you are correct. $\endgroup$ Jun 1 '21 at 10:27
  • $\begingroup$ Maybe my answer here would help. $\endgroup$
    – Buraian
    Jun 1 '21 at 15:15
  • $\begingroup$ @Poutnik ok got it now $\endgroup$ Jun 2 '21 at 4:31
  • $\begingroup$ @Buraian thank you ,answer helped $\endgroup$ Jun 2 '21 at 4:32
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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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