If water vapour is assumed to be a perfect gas, molar enthalpy change for vapouration of 1 mole of water at 1 bar and `100^(@)C` is 41 kJ `mol^(-1)`. Calculate the internal energy change when
a) 1 mol of water is vapourised at 1 bar and `100^(@)C`
b) 1 mol of water liquid is converted into ice.

Explanation

a) `H_(2)O(I)to H_(2)O(g)`
`Deltan-1`
`DeltaH=DeltaU+Deltan_(g)RT`
`41=DeltaU+(1xx8.314xx10^(-3)xx373)`
`DeltaE=41-3.1 =37.9 kJ.`
`H_(2)O(I)hArrH_(2)O(g)`
`Deltan=0`
`DeltaH=DeltaU+Deltan_(g)RT`
`therefore DeltaH=DeltaU(becausen_(g)RT=0)`
So, `DeltaU=41`