Question

Calculate the volume occupied by $0.2$ mole of a Vander Waal gas at ${27}^o$C and $0.0821$ atm. $[a=4.105L^2$ atm $mo{l}^{-2},b=\frac{1}{6}Lmo{l}^{-1}]$.

Answer 1

$n=0.2$ mol, temp $=T=298k$ $P=0.0821$ atm

$a=4.105$ and $b=\frac{1}{6}L{mol}^{-1}$

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

$\Rightarrow [0.0821+\frac{0.04}{V^2}\left(4.105\right)](V-0.2\times \frac{1}{6})=0.2\times 0.082\times 298$

$\Rightarrow [0.0821+\frac{0.1642}{V^2}](V-0.033)=4.8872$

$\Rightarrow$ Volume required $V=59.5268$ Litres.