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

$Δ G = 0$

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$Δ S = - v e$

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$Δ G = - v e$

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$Δ S = + v e$

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Solution

The correct option is D
$Δ S = + v e$

For $H_{2} O (I) (1 b a r, 373 K) - H_{2} O (g) (1 b a r, 373 K)$

At 1 atm and (373 K), this transormation will be at equilibrium (vapour - liquid) since this is the boiling point.
Therefore $Δ G = 0$ Water changes to the vapour state increasing the entropy of the system
$Δ S = + v e$

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