Question

The compressibility factor for $1$ mole of a van der Waal's gas at $273K$ and $100atm$ 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 $L^{2}mo{l}^{-2}$ ).

Answer 1

Answer: 1.26

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