Question

For one mole of a Van der Waals gas when $b=0$ and $T=300K$, the $PVvs.1/V$ plot is shown below. The value of the Van der Waals constant $a\left(atm{L}^{2}mo{l}^{-2}\right):$

Answer 1

We know,
Vander Waal's equation, for one mole of the gas,
$(p+\frac{a}{V^2})(V-b)=RT$
Since,
$b=0$
$(P+\frac{a}{V^2})\left(V\right)=RT$
$PV+a/V=RT$
$PV=RT-\frac{a}{V}$
This is an equation of a straight line.
$y=c+m\left(x\right)$

slope $m=-a$

So slope $-a=\frac{20.1-21.6}{3-2}=\frac{1.5}{1}=-1.5$
$a=1.5$