Question

For the process H2O (l) (1 bar, 373 K) →H2O (G) (1 bar, 373 K), the correct set of thermodynamic parameters is

(A) ∆G = 0. ∆S = + ve

(B) ∆G = 0. ∆S = - ve

(C) ∆G = +ve, ∆S = 0

(D) ∆G = -ve, ∆S = + ve

Answer 1

Correct option (A) ∆G = 0. ∆S = + ve

Explanation:

At transition point (373 K, 1.0 bar), liquid *** in equilibrium with vapour phase, therefore ∆G =0. As vaporisation occur, degree of randomness increase, hence ∆S > 0