Question

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

Answer 1

For one mole, van der Waal’s equation is $=(P+\frac{a}{V^2})(V-b)=RT$
as given that $b=0$
$PV+\frac{a}{V}=RT\Rightarrow PV=RT-\frac{a}{V}$
Comparing with straight line equation:
$y=mx+c$
$\text{Intercept}\left(c\right)=RT,\text{slope}\left(m\right)=-a$
We know,
$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{20.1-21.6}{3-2}=-1.5$
Thus, $a=1.5$
Therefore, the correct answer is (C)