If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporization of 1 mol of water at 1 bar and `100^(@)`C is 41 kJ `mol^(-1)`. Calculate the internal energy, when 1 mol of water is vapourised at one bar pressure and `100^(@)`C.

Explanation

The change `H_(2)O(I) rarr H_(2)O(g)`
`Delta H = Delta U + n_(g)RT`
`or Delta U = Delta H - Delta n_(g)RT`
= 41 kJ `mol^(-1) - 1 xx "8.314 J"" " mol^(-1)K^(-1)xx 373 K`
= 41 kJ `mol^(-1)` - 3101 J `mol^(-1)`
= 41 kJ`mol^(-1)` - 3.101 kj `mol^(1)`
= 37.9 kJ `mol^(-1)`