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Q. What is the compressibility factor $(Z)$ for $0.02$ mole of a van der Waals’ gas at pressure of $0.1$ atm. Assume the size of gas molecules is negligible.
Given: $RT = 20 \,L$ atm mol$^{-1}$ and $a = 1000$ atm $L^2\, mol^{-2}$

States of Matter

Solution:

$\left(0.1 + \frac{1000\times\left(0.02\right)^{2}}{V^{2}}\right)V = 20 \times0.02 $
$0.1 V^{2} - 0.4 V + 0.4 = 0 $
$ V^{2} - 4V + 4 = 0 $
$V = 2L$
$Z = \frac{PV}{nRT} $
$= \frac{0.1\times2}{20\times0.02} = 0.5$