For one mole of a van der Waal's gas when $b=0$ and $\mathrm{T}=300 \mathrm{~K}$, the PV vs, $1 / \mathrm{V}$ plot is shown below. The value of the van der Waal's constant $a$ (atm. liter ${ }^{2} \mathrm{~mol}^{-2}$ ) is :

• A

  $1.0$
• B

  $4.5$
• Correct answer
  C

  $1.5$
• D

  $3.0$

Solution

$\begin{array}{l} \left(P+\frac{\mathrm{a}}{V^{2}}\right)(V)=R T \\ P V+a / V=R T ; P V=R T-a(V) \\ y=R T-a(x) \end{array}$ So, slope $=a=\frac{21.6-20.1}{3-2}=\frac{1.5}{1}=1.5$

Hence, the correct answer is option: 'C'.