In an irreversible process taking place at constant `T` and `P` and in which only pressure-volume work is being done, the change in Gibbs free energy `(dG)` and the change in entropy `(dS)` satisfy the criteriaA. `(dS)_(V , E) lt 0, (dG)_(T,P) lt 0`B. `(dS)_(V , E) = 0, (dG)_(T,P) gt 0`C. `(dS)_(V , E) = 0, (dG)_(T,P) = 0`D. `(dS)_(V , E) gt 0, (dG)_(T,P) lt 0`

In an irreversible process taking place at constant `T` and `P` and in

1.	In an irreversible process taking place at constant `T` and `P` and in which only pressure-volume work is being done, the change in Gibbs free energy `(dG)` and the change in entropy `(dS)` satisfy the criteriaA. `(dS)_(V , E) lt 0, (dG)_(T,P) lt 0`B. `(dS)_(V , E) = 0, (dG)_(T,P) gt 0`C. `(dS)_(V , E) = 0, (dG)_(T,P) = 0`D. `(dS)_(V , E) gt 0, (dG)_(T,P) lt 0`
Answer» Correct Answer - D According to the second law of thermodynamics, for an irreversible process (spontaneous process), `Delta S_(total) gt 0`. Since `- T Delta S_(Total) = Delta G`, we have `Delta F lt o` for a spontaneous process. Thus, `(dS)_(V, E) gt 0` and `(dG)_(T,P) lt 0`

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