Question
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 change in entropy $$(dS),$$ satisfy the criteria
A.
$${\left( {dS} \right)_{V.E}} > 0,{\left( {dG} \right)_{T.P}} < 0$$
B.
$${\left( {dS} \right)_{V.E}} = 0,{\left( {dG} \right)_{T.P}} = 0$$
C.
$${\left( {dS} \right)_{V,E}} = 0,{\left( {dG} \right)_{T.P}} > 0$$
D.
$${\left( {dS} \right)_{V,E}} < 0,{\left( {dG} \right)_{T.P}} < 0$$
Answer :
$${\left( {dS} \right)_{V.E}} > 0,{\left( {dG} \right)_{T.P}} < 0$$
Solution :
For spontaneous reaction, $$dS > 0$$ and $$dG$$ should be negative ie. $$<0.$$