Standard enthalpy of vaporization  ${\Delta }_{vap}H°$ for water at $100°C$  is $40.66kJmo{l}^{-1}$.
The internal energy of vaporization of water at $100°C \left(inkJmo{l}^{-1}\right)$ is :

$$\begin{gathered} {\Delta }_{vap}H°=40.66kJmo{l}^{-1} \\ T=100+273=373K,\Delta E=? \\ \Delta H=\Delta E+\Delta n_{g}RT\Rightarrow \Delta E=\Delta H-\Delta n_{g}RT \\ {H}_2{O}_{\left(l\right)}\rightleftharpoons H_2{O}_{\left(g\right)}\Delta n_g=1-0=1 \\ \Delta E=\Delta H-RT=(40.66\times {10}^3)-(8.314\times 373)=37559J/molor37.56kJ/mol \end{gathered}$$