If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1 bar and 100°C is 41 kJ `"mol"^(-1)`. Calculate the internal energy change, when
(i) 1 mol of water is vaporised at 1 bar pressure and 100°C.
(ii) 1 mol of water is converted into ice.

Explanation

(i) The change `H_(2) O_((l)) to H_(2) O_((g))`
`DeltaH= Delta U + Deltan_(g) RT`
OR `Delta U = Delta H - Deltan_(g) RT`, substituting the values we gat.
`Delta U = 41.00 "kJ mol"^(-1) - 1xx 8.3 "J mol"^(-1) K^(-1) xx 373 K`
`= 41.00 "kJ mol"^(-1) -3.096 "kJ mol"^(-1)`
`= 37.904 "kJ mol"^(-1)`
(ii) The change `H_(2) O_((l)) to H_(2) O_((s))`
There is negligible change in volume
So, we can put `pDelta V=Deltan_(g) RT~~0` in this case,
`Delta H ~=Delta U`
So, `Delta U = 41.00 "kJ mol"^(-1)`