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Question

For H2O(l) (1bar,373K) H2O(g) (1bar,373K).The correct set of thermodynamic parameter is


A

∆G = 0, ∆S = + ve

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B

∆G = 0, ∆S = -ve

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C

∆G = + ve, ∆S = 0

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D

∆G = - ve, ∆S = + ve

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Solution

The correct option is A

∆G = 0, ∆S = + ve


For H2O(I) (1bar,373K) H2O(g) (1bar,373K)

At B.P temperature of water (373k) then is a state of equilibrium in the liquid and vapour state. There fore ΔG = 0 Since there is increase in entropy. When water changes to the vapour state ΔS = +ve

ΔG = 0,ΔS = +ve


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