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 '$a^{\prime }$ is _____.

Answer 1

Answer: 1.2

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

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

$V=0.134L$

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 ${}^{2}mo{l}^{-2}atm$