1. Given data

• Mass of water film $= 36\,\text{g}$
• Molar mass of water $= 18\,\text{g mol}^{-1}$
• Therefore moles of water: $n = \frac{36}{18} = 2\,\text{mol}$
• Enthalpy of vaporization of water at $373\,\text{K}$ and $1\,\text{bar}$: $\Delta H_{\text{vap}}^\Theta = 41.1\,\text{kJ mol}^{-1}$
• Gas constant: $R = 8.31\,\text{J K}^{-1}\text{mol}^{-1}$
• Temperature: $T = 373\,\text{K}$

2. Relation between enthalpy and internal energy

For vaporization, $\Delta H = \Delta U + \Delta n_g RT$

Here, liquid water changes to steam. For 1 mol of water:

• Initial gaseous moles $=0$
• Final gaseous moles $=1$

So, $\Delta n_g = 1$

Hence, $\Delta U_{\text{vap}} = \Delta H_{\text{vap}} - RT$

3. Calculate $RT$

$RT = 8.31 \times 373\,\text{J mol}^{-1}$ $RT = 3099.63\,\text{J mol}^{-1} = 3.10\,\text{kJ mol}^{-1}$

4. Calculate internal energy of vaporization

$\Delta U_{\text{vap}} = 41.1 - 3.10 = 38.0\,\text{kJ mol}^{-1}$

5. Nearest integer

$\boxed{38}$

6. Note on the 36 g water film

Since the question asks for internal energy for vaporization in kJ mol$^{-1}$, the answer is on a per mole basis. The 36 g corresponds to $2$ mol, but the molar value remains the same.

Thus the required integer is 38.