Question

The compression factor (compressibility factor) for $1$ mole of a van der Waals' gas at $0^{\circ }C$ and $100$ atmospheric pressure is found to be $0.5$. Assuming that the volume of a gas molecule is negligible, calculate the van der Waals' constant ${}^{\prime }{a}^{\prime }$.

Answer 1

We know that,

$Z=\frac{PV}{RT}$

$0.5=\frac{100\times V}{0.0821\times 273}$

$V=0.112litre$

According to van der Waals' equation,

$(P+\frac{a}{V^2})(V-b)=RT$ for $1$ mole

$[100+\frac{a}{(0.112{)}^2}][0.112-0]-0.0821\times 273$

On solving, we get $a=1.253L^{2}mo{l}^{-2}atm$.