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.

If water vapour is assumed to be a perfect gas, molar enthalpy change

1.	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.
Answer» Solution :(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)`