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The change in enthalpy of vaporisation o...

The change in enthalpy of vaporisation of 1 mol of water is `40850 " J mol"^(-1) " at " 100^(@)C` and 1 atm pressure. What is the difference between molar entropy of water and water vapour at that temperature and pressure ?

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Verified by Experts

`DeltaS_("vaporisation")=(DeltaH_("vaporisation"))/(T_(b))= (40850)/(373)J. K^(-1) " mol"^(-1)`
`= 109.5 J. K^(-1)" mol"^(-1)`
`overline(S)_(H_(2)O(g))-overline(S)_(H_(2)O(l))= 109.5 J. K^(-1) " mol"^(-1)`
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