Enthalpy of vapourization of liquid at 127° c is 50kJ/mole. Change in Internal energy during this process would be

Let the liquid be ‘X’

$\mathrm{X}\left(\mathrm{l}\right) \to \mathrm{X}\left(\mathrm{g}\right)$

$\mathrm{\Delta n} = 1-0 = 1$

$\mathrm{\Delta H} = \mathrm{\Delta E} +\mathrm{\Delta nRT}$

$50 = \mathrm{\Delta E} +\left(1\right)\left(8.3\times {10}^{-3}\mathrm{kJ} {\mathrm{K}}^{-1}{\mathrm{mole}}^{-1}\right)\left(400\right)$

$50 = \mathrm{\Delta E} +3.32$

$\mathrm{\Delta E} = 46.68 \mathrm{kJ}$