Assuming the water vapour to be a perfect gas,calculate the internal energy change when $1 \ mol$ of water at $100^{\circ} C$ and $1 \ bar$ pressure is converted to ice at $0^{\circ} C$. Given the enthalpy of fusion of ice is $6.00 \ kJ \ mol^{-1}$ and heat capacity of water is $4.2 \ J \ g^{-1} {\circ} C^{-1}$.

The change takes place as follows:
Step-$1$: $1 \ mol \ H_2O(l, 100^{\circ} C) \rightarrow 1 \ mol \ H_2O(l, 0^{\circ} C)$,Enthalpy change $\Delta H_1 = n \times C_p \times \Delta T = 1 \ mol \times 18 \ g \ mol^{-1} \times 4.2 \ J \ g^{-1} {\circ} C^{-1} \times (0^{\circ} C - 100^{\circ} C) = -7560 \ J \ mol^{-1} = -7.56 \ kJ \ mol^{-1}$.
Step-$2$: $1 \ mol \ H_2O(l, 0^{\circ} C) \rightarrow 1 \ mol \ H_2O(s, 0^{\circ} C)$,Enthalpy change $\Delta H_2 = -6.00 \ kJ \ mol^{-1}$ (freezing is the reverse of fusion).
Total enthalpy change: $\Delta H = \Delta H_1 + \Delta H_2 = -7.56 \ kJ \ mol^{-1} - 6.00 \ kJ \ mol^{-1} = -13.56 \ kJ \ mol^{-1}$.
Since the change in volume during the phase transition from liquid to solid is negligible,$P\Delta V \approx 0$.
Using the relation $\Delta H = \Delta U + P\Delta V$,we get $\Delta U = \Delta H = -13.56 \ kJ \ mol^{-1}$.