Question

The compression factor (compressibility factor) for one mole of a van der Waal's gas at $0^{\circ} C$ and 100 atmosphere pressure is found to be $0.6$ . Assuming that the volume of a gas molecule is negligible, the van der Waal's constant '$a'$ is _____.

Answer 1

$Z = \frac{PV}{RT} = 0.6$

$\therefore \frac{100 \times V }{0.082 \times 273}=0.6$

$V =0.134 \,L$

Now, using van der Waal's equation

$\left[ P +\frac{ a }{ V ^{2}}\right][ V ]= RT$

or $\left[ P +\frac{ a }{ V ^{2}}\right]=\frac{ RT }{ V }$

$(\because b$ is negligible $)$

$\therefore \left[100+\frac{ a }{(0.134)^{2}}\right]$

$=\frac{0.082 \times 273}{0.134}=167$

$\therefore \frac{ a }{(0.134)^{2}}=67$

$a =1.2$ litre ${ }^{2} mol ^{-2} atm$