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Question

For the process:
H2O(l) (1 bar, 373 K)H2O(g) (1 bar, 373 K)
the correct set of thermodynamic parameters is:

A
ΔG=0, ΔS=+ve
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B
ΔG=0, ΔS=ve
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C
ΔG=+ve, ΔS=0
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D
ΔG=ve, ΔS=+ve
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Solution

The correct option is A ΔG=0, ΔS=+ve
H2O(l)(1 bar, 373 K)H2O(g)(1 bar, 373 K)
At 100 C, H2O(l) is in equilibrium with H2O(g) therefore, ΔG=0. Because liquid molecules are converting into gaseous molecules, therefore randomness increases and hence, ΔS=+ve

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