Assuming that water vapour is an ideal gas, the internal energy ($$\Delta U$$) when 1 mol of water is vapourised at 1 bar pressure and $$100^\circ C$$, (Given: Molar enthalpy of vapourization of water at 1 bar and 373 K = 41 kJ $$mol^{-1}$$ and R = 8.3 J $$mol^{-1} K^{-1}$$) will be

The relation between molar enthalpy change ($$\Delta H$$) and molar internal-energy change ($$\Delta U$$) at constant pressure is

$$\Delta H = \Delta U + \Delta (PV)$$

For 1 mol of an ideal gas, $$PV = RT$$. During vaporisation the initial state is liquid (very small volume) and the final state is water vapour (ideal gas). Hence

$$\Delta (PV) = P\,V_{\text{vapour}} - P\,V_{\text{liquid}} \approx RT - 0 = RT$$

Therefore

$$\Delta U = \Delta H - RT$$

Substitute the given data (for 1 mol, $$T = 373\ \text{K}$$, $$R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}$$, $$\Delta H = 41\ \text{kJ mol}^{-1} = 41000\ \text{J mol}^{-1}$$):

$$RT = 8.3 \times 373 = 3095.9\ \text{J mol}^{-1} \approx 3.096\ \text{kJ mol}^{-1}$$

$$\Delta U = 41.000\ \text{kJ mol}^{-1} - 3.096\ \text{kJ mol}^{-1} = 37.904\ \text{kJ mol}^{-1}$$

Hence, the molar internal-energy change for vaporising water at 1 bar and $$100^\circ\text{C}$$ is

Option C which is: 37.904 kJ $$\text{mol}^{-1}$$