For one mole of a van der Waals' gas when b = 0 and T = 300 K, the pV vs $\frac{1}{V}$ plot is shown below. The value of the van der Waals' constant a (atm L ${\mathrm{mol}}^{-2}$)

 

van der Waals' equation of state is

$\left(\mathrm{p}+\frac{{\mathrm{n}}^2\mathrm{a}}{{\mathrm{V}}^2}\right)(\mathrm{V}-\mathrm{nb})=\mathrm{nRT}$

For one mole and b = 0 the above equation condenses to

$\left(\mathrm{p}+\frac{\mathrm{a}}{{\mathrm{V}}^2}\right)\mathrm{V}=\mathrm{RT}$

$\Rightarrow$$\mathrm{pV}=\mathrm{RT}-\frac{\mathrm{a}}{\mathrm{V}}$----------(i)

Eq. (i) is a straight line equation between $\mathrm{pV}\text{ and }\frac{1}{\mathrm{V}}$ whose slope is $-\mathrm{a}$

Equating with slope of the straight line given in the graph

$-\mathrm{a}=\frac{20.1-216}{3-2}=-1.5$

$\Rightarrow$a = 1.5