Question

For water at ${100}^{\circ }\text{C}$ and 1 bar,
$\Delta H_{\text{vap}}-\Delta U_{\text{vap}}$is $x\times {10}^2{\text{J mol}}^{-1}$
The value of x (Round off to the Nearest Integer) is
[Use : $R=8.31{\text{mol}}^{-1}{\text{K}}^{-1}$]
[Assume volume of $H_{2}O\left(l\right)$ is much smaller than volume of $H_{2}O\left(g\right)$. Assume $H_{2}O\left(g\right)$ can be treated as an ideal gas]

Answer 1

$\Delta H-\Delta U=\Delta n_{g}RT$
$\Delta n_g=1-0=1$
$\Delta H-\Delta U=1\times 8.31\times 373$
$\Delta H-\Delta U=30.99\times {10}^2{\text{J mol}}^{-1}$
$\Delta H-\Delta U\approx 31\times {10}^2{\text{J mol}}^{-1}$
$x=31$