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 :

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Aicharya · Accepted Answer

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

For one mole of a van der Waal's gas when $b=0$ and $\mathrm{T}=300 \mathrm{~K}$, the PV vs, $1 /…

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