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

△ G = 0, △ S = + v e

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△ G = 0, △ S = - v e

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△ G = + v e, △ S = 0

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△ G = - v e, △ S = + v e

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Solution

The correct option is A $△ G = 0, △ S = + v e$
At $373 K, H_{2} O (l)$ is in equilibrium with $H_{2} O (g)$ , and for an equilibrium process we know change in gibbs free energy is zero,
therefore $△ G = 0$ .

As liquid phase is changes to gaseous phase sp molecules entropy increases therefore $△ S = + i v e$ .

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

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