As Second Law says, the entropy must increase for universe

$$\Delta S_\textrm{universe}\ge 0$$

Now, we know, $$\Delta G= -T\Delta S_\textrm{universe}\le 0^\dagger$$

For reversible process, $$\Delta S_\textrm{reversible, universe} = 0$$

That would mean $$\Delta G_\textrm{reversible}= 0$$

But does it happen so?

Peter Atkins, in his book writes:

At constant temperature and pressure, for a reversible process: $\mathrm dG = \mathrm dw'_\rm{rev}\;.$

Now, isn't it contradicting that $\mathrm dG\ne 0$ for reversible process? If $\Delta G= -T\Delta S_\textrm{universe}$, then wouldn't $\mathrm dG= 0$ for $\mathrm dS_\textrm{universe}= 0$ during reversible process?

\begin{align}^\dagger \Delta S_\textrm{universe}&=\Delta S_\textrm{system}+ \Delta S_\textrm{surroundings} \\ &= \Delta S_\text{system}+ \frac{-\Delta H_\text{system}}{T} \\ \implies -T\Delta S_\text{universe} &= \Delta H_\text{system} - T\Delta S_\text{system} \\&=\Delta G \end{align}

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    $\begingroup$ Forgot about enthalpy H? $\endgroup$ – Mithoron Feb 23 '16 at 17:21
  • $\begingroup$ @Mithoron: Huh? $\endgroup$ – user5764 Feb 23 '16 at 17:22
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    $\begingroup$ G=H-TS so your equations look wrong... $\endgroup$ – Mithoron Feb 23 '16 at 17:28
  • $\begingroup$ @Mithoron: Edited. $\endgroup$ – user5764 Feb 23 '16 at 17:41
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    $\begingroup$ Think more about it and you'll see why you're wrong. $\endgroup$ – Mithoron Feb 23 '16 at 18:10

The equation in Atkins' book refers to the system only, and not to the universe, and, moreover, in this equation, w' specifically refers to non-PV work, like electrical.

  • $\begingroup$ G at constant temperature and pressure doesn't talk about the entropy of the universe? $\endgroup$ – user5764 Feb 23 '16 at 20:14
  • $\begingroup$ I thought dG would be zero as $\mathrm dG= -T\mathrm dS_\rm{universe}$ and $\mathrm dS_\textrm{universe}= 0$ for reversible process? $\endgroup$ – user5764 Feb 23 '16 at 20:18
  • $\begingroup$ Doesn't Atkins' book specify these constraints when it talks about the change in G for a system experiencing a process at constant temperature and pressure? What exactly does the book say leading up to your highlighted statement? How exactly does it define w' (with the prime)? You seem to have omitted important details in your question. $\endgroup$ – Chet Miller Feb 23 '16 at 20:32
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – user5764 Feb 24 '16 at 16:53
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    $\begingroup$ @user36790 In the future, it's better to formulate some of these into new questions rather than having a back and forth in chat between just two people. $\endgroup$ – jonsca Feb 25 '16 at 0:08

The proper equation is $$\mathrm dG=\mathrm dH-\mathrm d(TS)$$ Since $H=U+pV$, we rewrite the above equation as $$\mathrm dG=\mathrm dU+(p\mathrm dV+V\mathrm dp)-T\mathrm dS-S\mathrm dT$$ Furthermore, we know from the first law that $$\mathrm dU=\delta Q + \delta W$$ and $$\Delta S=\int{\frac{\delta Q_\text{rev}}{T}}$$ which is equivalent to writing $$T\mathrm dS=\delta Q_\text{rev}$$ so, putting it all together and remembering that temperature is constant, we have $$\mathrm dG=T\mathrm dS+\delta W_\text{rev}-T\mathrm dS$$ Then, $$\mathrm dG=\delta W_\text{rev}$$ where the pressure-volume work went to zero because, as you point out in one of your comments, all of this is only valid for zero expansion work.

I hope this kind of answers your question, but you seem to have a pretty good grasp on this already.

Also, it probably shouldn't be written that $$\mathrm dG=\mathrm dW_\text{rev}$$ because this seems to imply that integrating this expression will have you take an integral with respect to work, but this cannot be done because work is not a state function. You would figure this out, however, when trying to figure out what values of work to use as the boundaries in that integral.

  • $\begingroup$ $\delta Q\ne 0\,\,\textrm{but}\,\,= T\mathrm dS$ $\endgroup$ – user5764 Feb 25 '16 at 3:14
  • $\begingroup$ For a reversible process the heat transfer does equal zero. $\endgroup$ – jheindel Feb 25 '16 at 6:07
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    $\begingroup$ No, no; this does happen only for adiabatic process. $\endgroup$ – user5764 Feb 25 '16 at 6:20
  • $\begingroup$ Ya you're right about that. I think there shouldn't be heat transfer in this case because I believe this is how Atkins gets to the result you're talking about. $\endgroup$ – jheindel Feb 25 '16 at 7:33
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    $\begingroup$ @jheindel It is not correct to say that "for a reversible process, there must be no heat loss from the system." What about an isothermal reversible process? $\endgroup$ – Chet Miller Feb 25 '16 at 13:11

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