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 'a' (in $L^{2}mo{l}^{-1}atm$)

Answer 1

Given,
$\text{Moles of gas}=1\text{mol}$
$T=0{}^{\circ }\text{C}=273\text{K}$
$P=100\text{atm}$
$Z=0.5$
We know that,
$Z=\frac{PV}{RT}$
$0.5=\frac{100\times V}{0.082\times 273}$
V = 0.112 L
Accoording to van der Waals' equation,
$(P+\frac{a}{V^2})(V-b)=RT$ for 1 mole
As volume of a gas moecule is negligible so $b=0$
$[100+\frac{a}{(0.112{)}^2}][0.112-0]=0.0821\times 273$
On solving, we get $a=1.253{\text{L}}^2{\text{mol}}^{-2}\text{atm}$