$10$ moles of a van der Waals' gas are subjected to a process from $2\mathrm{L}, 300 \mathrm{K}$to $2\mathrm{I}, 350 \mathrm{K}$.

Given : $C_{\mathrm{ym}}=20\mathrm{J}/\mathrm{mole}-\mathrm{K},b=0.04\mathrm{L}/\mathrm{mole}$

$C_{\mathrm{pm}}=30\mathrm{J}/\mathrm{mole}-\mathrm{K},a=\text{ negligible }$

Calculate $∆\mathrm{H}$ (in KJ ).

[GIven : $R=0.08$ litre atm ${\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$ I litre aim = 100 J]

$\mathrm{\Delta }U=n{C}_{vm}\mathrm{\Delta }T$ for real gas at constant volume.

$\mathrm{\Delta }U=10\times 20\times 50$

$\mathrm{\Delta }U=10\mathrm{kJ}$

$\text{ For van der Waals' gas }H=U+PV$

$\mathrm{\Delta }H=\mathrm{\Delta }U+\mathrm{\Delta }\left(PV\right)\text{ as volume is constant. }$

$\mathrm{\Delta }H=\mathrm{\Delta }U+V\mathrm{\Delta }P$

To find  change in pressure 

$\left(P+\frac{a{n}^2}{V^2}\right)(V-nb)=nRT$

a=0

$P(V-nb)=nRT$

$P=\frac{nRT}{V-nb}$

$dP=\frac{nRdt}{(V-nb)}$

$=\frac{10\times 0.08\times 50}{2-10\times 0.04}$

$=25\mathrm{atm}$

$\mathrm{\Delta }H=10\mathrm{kJ}+2\times 25\text{ litre atm }$

$\mathrm{\Delta }H=15\mathrm{kJ}$