Assumption behind Gibbs energy and maximum work

Question

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.

Answer 1

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):

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