For one mole of a van der Waal’s gas when $$b = 0$$  and $$T = 30K,$$  the $$PV$$ vs, $$\frac{1}{V}$$ plot is shown below. The value of the van der Waal’s constant $$a\left( {atm.\,lite{r^2}mo{l^{ - 2}}} \right)$$     is:

Solution :

$$\eqalign{ & \left( {P + \frac{a}{{{V^2}}}} \right)\left( V \right) = RT \cr & PV + \frac{a}{V} = RT;\,PV = RT - a\left( V \right) \cr & y = RT - a\left( x \right) \cr & {\text{So, slope}} = a = \frac{{21.6 - 20.1}}{{3 - 2}} = \frac{{1.5}}{1} = 1.5 \cr} $$