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in Chemical thermodynamics by (64.7k points)

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

1 Answer

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Best answer

Correct option (A) ∆G = 0. ∆S = + ve

Explanation:

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

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