A fish swimming in water body when taken out from the water body is covered with a film of water of weight $$36$$ g. When it is subjected to cooking at $$100°$$C, then the internal energy for vaporization in kJ mol$$^{-1}$$ is ______ [integer]
[Assume steam to be an ideal gas. Given $$\Delta_{vap}H^\circ$$ for water at $$373$$ K and $$1$$ bar is $$41.1$$ kJ mol$$^{-1}$$ : $$R = 8.31$$ J K$$^{-1}$$ mol$$^{-1}$$]

We need to find the internal energy of vaporization of water at 100°C.

A water film of mass 36 g is at 100°C, which corresponds to 373 K.

The standard enthalpy of vaporization is given as $$\Delta_{vap}H^\circ = 41.1$$ kJ mol$$^{-1}$$, and the gas constant is $$R = 8.31$$ J K$$^{-1}$$ mol$$^{-1}$$.

For the vaporization process $$H_2O(l) \rightarrow H_2O(g)$$, the relationship between enthalpy change and internal energy change is

$$\Delta H = \Delta U + \Delta n_g RT$$

where $$\Delta n_g$$ is the change in moles of gas, equal to 1.

Rearranging to solve for the change in internal energy gives

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

Substituting the known values:

$$\Delta U = 41.1 - (1)(8.31 \times 10^{-3})(373)$$

$$\Delta U = 41.1 - 3.0996$$

$$\Delta U = 38.0$$ kJ mol$$^{-1}$$

Thus, the internal energy of vaporization of water at 100°C is approximately 38.0 kJ mol$$^{-1}$$.

Although the mass of water (36 g = 2 mol) is provided, the question asks for the molar internal energy, so the result is expressed per mole.