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The latent heat of vaporisation of water...

The latent heat of vaporisation of water at `100 ^(@)C` is `540 cal g^(-1)`. Calculate the entropy increase when one mole of water at `100^(@)C` is evaporated

A

`26 cal K^(-1) mol^(-1)`

B

`1.82 cal K^(-1) mol^(-1)`

C

`367 cal K^(-1) mol^(-1)`

D

`540xx18 cal K^(-1) mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaS=(DeltaH_(vap))/(T)=(540xx18)/(373)`
`=26 cal K^(-1) mol^(-1)`
`(DeltaH_(vap) "for" 1 "mole"=540xx18)`
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DINESH PUBLICATION-CHEMICAL THERMODYNAMICS AND CHEMICAL ENERGETICS -Exercise
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