For the process: $H_{2} O (l) (1 b a r, 373 K) ⇌ H_{2} O (g) (1 b a r, 373 K)$ the correct set of thermodynamic parameter is :

Δ G = 0, Δ S = + v e

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Δ G = 0, Δ S = - v e

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Δ G = + v e, Δ S = 0

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Δ G = - v e, Δ S = + v e

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Solution

The correct option is A $Δ G = 0, Δ S = + v e$
Here liq. and vap. phase are in equilibrium therefore $Δ U = 0$ . Degree of randomness increases hence $Δ S > 0$

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