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!