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enter image description here I'm somewhat confused by this derivation (in Schroeder's Book of Thermal Physics) of the fact that in a closed system, the change in the Gibbs Energy is always less than the Non-Expansion work done on the system.

If we take the change in the entropy of the universe (system + ideal surroundings maintained at constant temperature and pressure also assuming that reactions are taking place in system only i.e. the composition of various species present in surroundings remain constant), then it turns out (in this derivation) that it equals to negative of change in Gibbs Energy divided by Temp of system (which is same as that of surroundings).

Now since entropy of Universe always increases it turns out that Gibbs Energy of system will always decrease. But this is only true in absence of Non-Expansion work.

However during this derivation nowhere it's mentioned that 'no non expansion work assumed' yet the result only holds true in absence of non expansion work.

I think that maybe it was assumed somewhere and I'm not able to spot where. Please help me to clear my doubt.

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The whole point of Gibbs free energy is to quantify a bound on possible non-expansion work at constant p and T. That non-expansion work is implicit as part of the term "dU" which in the reversible case can be written as

$$dU = -pdV + TdS + dw_{\text{non-exp,rev}}$$

The $-pdV$ term is the expansion work if the process is carried out reversibly. You remove that contribution by adding a $pdV$ term:

$$dU + pdV= -pdV + TdS + dw_{\text{non-exp,rev}} + pdV = TdS + dw_{\text{non-exp,rev}}$$

Finally if you subtract the entropy term you are left with the maximum possible non-expansion work (obtained when the process is performed in a reversible fashion):

$$dU + pdV - TdS= dw_{\text{non-exp,rev}}$$

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  • $\begingroup$ I think you did not understand my question. What I mean to say that in general the change in Gibbs energy is less than the non expansion work done on system, and in the absence of non expansion work the Gibbs energy of system falls. However during this derivation the final result is the latter one. That's the point which confused me. Because to get the latter result one has to explicitly assume the absence of non expansion work, which I dont think is assumed in this proof. Can you please point out where in the derivation they assumed absence of non expansion work? $\endgroup$ – Shivansh J May 2 '20 at 7:40
  • $\begingroup$ @ShivanshJ please see my edit. Hopefully clearer now that I understand your question better. $\endgroup$ – Buck Thorn♦ May 2 '20 at 8:52
  • $\begingroup$ Buck Thorn: Thanks! Now it's much clearer! $\endgroup$ – Shivansh J May 2 '20 at 9:00
  • $\begingroup$ A small question (to make sure I understand correctly). In equation 5.27 (In the photo I attached with my question). This thermodynamic identity assumes no non expansion work done on system. Am I right? $\endgroup$ – Shivansh J May 2 '20 at 9:02
  • $\begingroup$ @ShivanshJ Yes, and in hindsight you are right, that derivation is not clear. $\endgroup$ – Buck Thorn♦ May 2 '20 at 9:07

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