Gibbs free energy for an irreversible process

Question

In an irreversible process taking place at constant T and P in which only pressure-volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS) are?

The answer to this question was given as dS > 0, dG < 0. I know that ∆G is negative and ∆S is positive for an irreversible process but ∆G gives the maximum amount of "non-expansion" work that can be extracted from a system, whereas in this question only pressure volume work is being done which is non-useful (expansion) work. So how can ∆G be negative here? Please help me to clear this confusion. Thanks!

Answer 1

For a process at constant T and P in which no non-expansion work is done, $$\Delta U=Q-P\Delta V$$ and $$\Delta S=\frac{Q}{T}+\sigma$$where $\sigma$ is the irreversibly generated entropy. Combining these equations gives $$\Delta U=T\Delta S-P\Delta V-T\sigma$$or$$\Delta G=-T\sigma$$which is negative. On the other hand, the sign of $\Delta S$ can be positive only if $$Q>-T\sigma$$