1.

Molar enthalpy change for vaporisation of 1 mole of water at 1 bar and 100^(@)C is 41 kJ mol^(-1) (If water vapour is assumed to be perfect gas). Find out of the internal energy change. If 1 mole of water is vaporised at 1 bar pressure and 100^(@)C.

Answer»

`+37.904kJmol^(-1)`
`-37.904kJmol^(-1)`
`44.096kJmol^(-1)`
`-44.096kJmol^(-1)`

SOLUTION :`H_(2)O(l)toH_(2)O(G)`
`DeltaH=DeltaU+Deltan_(g)RT`
`DeltaU=41.00" KJ "mol^(-1)-8.3JK^(-1)mol^(-1)xx373K`
`=41.00-3.096" kJ "mol^(-1)=37.904" kJ "mol^(-1)`


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