Question

The compressibility factor for $1$ mole of a van der Waal's gas at $273 \,K$ and $100 \,atm$ pressure is $0.5$. Assuming that the volume of a gas molecule is negligible, calculate the van der Waal's constant ' $a$ ' (in units of atm $L^{2} \,mol ^{-2}$ ).

Answer 1

$Z =\frac{ PV }{ RT }=0.5=\frac{100 \times V }{0.0821 \times 273}$
$\therefore V =0.112\, L$
Now,
$\left(P+\frac{n^{2} a}{V^{2}}\right)(V-n b)=n R T $
$\therefore \left(100+\frac{a}{0.112 \times 0.112}\right)(0.112-0)$
$=0.0821 \times 273=22.41$
$\therefore \left((100 \times 0.112)+\frac{ a }{0.112}\right)=22.41$
$\therefore 11.2+\frac{ a }{0.112}=22.41$
$\therefore \frac{ a }{0.112}=22.41-11.2$
$\therefore \frac{ a }{0.112}=11.21$
$\therefore a =1.26\, atm \,L ^{2}\, mol ^{-2}$