For the reversible isothermal evaporation of 90.0 g of water at 100⁰C. The value of $∆\mathrm{U}$ for the reaction is:

[Assuming that water vapours behave as an ideal gas and heat of evaporation of water is 540 cal/g and R=2 cal mol^-1 K^-1]

Heat given at constant pressure:

$$\begin{gathered} ∆\mathrm{H}=\mathrm{Heat} \mathrm{of} \mathrm{evaporation}\times \mathrm{amount} \mathrm{of} \mathrm{evaporated} \\ ∆\mathrm{H}=540\times 90=48600 \mathrm{cal} \end{gathered}$$

Moles of water at 90.0 g:

$\mathrm{Moles}=\frac{90.0 \mathrm{g}}{18 \mathrm{g}/\mathrm{mol}}=5 \mathrm{mol}$

Using the expression to determine the value of $∆\mathrm{U}$:

$$\begin{gathered} ∆\mathrm{H}=∆\mathrm{U}+∆{\mathrm{n}}_{\mathrm{g}}\mathrm{RT} \\ ∆\mathrm{U}=∆\mathrm{H}-∆{\mathrm{n}}_{\mathrm{g}}\mathrm{RT} \\ ∆\mathrm{U}=48600-5\times 2\times 373 \\ ∆\mathrm{U}=44870 \mathrm{cal} \end{gathered}$$