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 thermodynamics parameters 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 B $Δ G = 0, Δ S = + v e$
$H_{2} O (l) ⇌ H_{2} O (g)$
It is an equilibrium process.
Therefore, $Δ G = 0$
Now, $Δ G = Δ H - T Δ S$
where $Δ H$ is the enthalpy of vaporization & $Δ H > 0$
Therefore, for $Δ G = 0$ , the value of $Δ S$ should be positive. Also, the $Δ S$ is always when a molecule is energized. Here, the liquid $H_{2} O$ is energized to gaseous $H_{2} O$ . So, $Δ S$ is +ve.

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