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

∆G = 0, ∆S = + ve

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∆G = 0, ∆S = -ve

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∆G = + ve, ∆S = 0

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∆G = - ve, ∆S = + ve

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Solution

The correct option is A
∆G = 0, ∆S = + ve

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

At B.P temperature of water (373k) then is a state of equilibrium in the liquid and vapour state. There fore $Δ G = 0$ Since there is increase in entropy. When water changes to the vapour state $Δ S = + v e$

$Δ G = 0, Δ S = + v e$

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Similar questions

H_{2} O (l) (1 b a r, 373 K) ⇌ H_{2} O (g) (1 b a r, 373 K)

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

H_{2} O (l) (1 bar, 373 K) ⇌ H_{2} O (g) (1 bar, 373 K)

H_{2} O_{(l)} (1 bar, 373 K) ⇌ H_{2} O_{(g)} (1 bar, 373 K)

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