If water vapour is assumed to be a perfect gas,the molar enthalpy change for the vaporisation of $1 \ mol$ of water at $1 \ bar$ and $100^{\circ} C$ is $41 \ kJ \ mol^{-1}$. Calculate the internal energy change when $1 \ mol$ of water is vaporised at $1 \ bar$ pressure and $100^{\circ} C$.

(N/A) The process is $H_2O_{(l)} \rightarrow H_2O_{(g)}$.
The relationship between enthalpy change and internal energy change is given by $\Delta H = \Delta U + \Delta n_g RT$.
Rearranging for internal energy change: $\Delta U = \Delta H - \Delta n_g RT$.
Here,$\Delta H = 41 \ kJ \ mol^{-1}$,$R = 8.314 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1}$,$T = 373 \ K$,and $\Delta n_g = 1 - 0 = 1$.
Substituting the values:
$\Delta U = 41 \ kJ \ mol^{-1} - (1 \times 8.314 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1} \times 373 \ K)$
$\Delta U = 41 \ kJ \ mol^{-1} - 3.101 \ kJ \ mol^{-1}$
$\Delta U = 37.899 \ kJ \ mol^{-1} \approx 37.9 \ kJ \ mol^{-1}$.