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

1.	If water vapour is assumed to be perfect gas, molar enthalpy change 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) For vaporisation of water, the change is `: H_(2)O (l ) rarr H_(2)O(g)` `Delta n_(g) = 1 -0=1` ` Delta H = Delta U + Delta n_(g) RT` or `Delta U = Delta H - Delta n_(g) RT = 41.00 kJ mol^(-1) - ( 1 mol) xx ( 8.314 xx 10^(-3)kJ K^(-1) mol^(-1)) ( 373 K) ` `= 41.00 - 3.10 k J mol^(-1) = 37.90 kJ mol^(-1)` (ii) For conversion of water into ice, the change is`H_(2)O(l) rarr H_(2)O(s)` In this case, the volume change in negligible. Hence, `Delta H =Delta U = 41.00 kJ mol^(-1)`