Question

Assuming the water vapour is an ideal gas, the internal energy change $\left(\Delta U\right)$ in kJ when 1 mole of water is vapourised at ${177}^{\circ }\text{C}$. (Given molar enthalpy of vapourisation of water ${177}^{\circ }\text{C}$ = 37 kJ/mol and $R=\frac{25}{3}\text{J/molK}$)

Answer 1

$H_{2}O\left(l\right)\to H_{2}O\left(g\right)$
$\Delta H=37\text{kJ/mol}$
$\Delta H=\Delta U+\Delta n_{g}RT$
$\Delta n_g=1$
$37=\Delta U+1\times \frac{25}{3}\times {10}^{-3}\times 450$
$37=\Delta U+3.75$
or, $\Delta U=37-3.75=33.25\text{kJ/mol}$