Question

For water at $100^{\circ} C$ and 1 bar,
$\Delta_{\text {vap }} H-\Delta_{\text {vap }} U=$________$\times 10^{2} J\, mol ^{-1} .$
(Round off to the Nearest Integer)
[Use: $R=8.31\, J\,mol ^{-1} K ^{-1}$ ]
[Assume volume of $H _{2} O ( l )$ is much smaller than volume of $H _{2} O ( g )$. Assume $H _{2} O ( g )$ treated as an ideal gas]

Answer 1

$H _{2} O _{(l)} \rightleftharpoons H _{2} O _{(v)}$
$\Delta H =\Delta U+\Delta n_{g} R T$
for 1 mole waters;
$\Delta n_{g}=1$
$\therefore \Delta n_{g} R T=1\, mol \times 8.31\, J / mol -k \times 373\, K$
$=3099.63 J \cong 31 \times 10^{2} J$