For the vaporisation of water: $H_2O(l) \rightarrow H_2O(g)$ The change in the number of moles of gaseous species, $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - 0 = 1$. The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) is given by : $\Delta H = \Delta U + \Delta n_g RT$ Rearranging the formula to find $\Delta U$: $\Delta U = \Delta H - \Delta n_g RT$ Given data: $\Delta H = 40.66 \text{ kJ mol}^{-1}$ $T = 100^{\circ}\text{C} = 373 \text{ K}$ $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1} = 8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{mol}^{-1}$ Substituting the values: $\Delta U = 40.66 \text{ kJ mol}^{-1} - (1 \times 8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{mol}^{-1} \times 373 \text{ K})$ $\Delta U = 40.66 - 3.101 \text{ kJ mol}^{-1}$ $\Delta U = +37.559 \text{ kJ mol}^{-1} \approx +37.56 \text{ kJ mol}^{-1}$.