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Question

Calculate entropy change for vaporization of 1 mole of liquid water to steam at 100C if ΔHv=40.8 kJ mol1.


A

ΔSv=107.38 JK1 mol1

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B

ΔSv=109.48 JK1 mol1

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C

ΔSv=109.38 JK1 mol1

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D

ΔSv=107.38 JK1 mol1

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Solution

The correct option is C

ΔSv=109.38 JK1 mol1


For entropy change of vaporization can be given as, ΔSv=ΔHvT
Given values are, ΔHv=40.8×103 Jmol1; T=373K
Therefore ΔSv=(40.8×103373)
ΔSv=109.38 JK1 mol1


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