Question

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

• A. 1
• B. 4.5
• C. 1.5
• D. 3

Answer 1

Answer: (C)

1.5

Answer 2

The van der Waals equation of state is
$\, \, \, \, \, \, \, \, \, \, \, \, \, \Bigg(p+\frac{n^ a}{V^2}\Bigg)(V-nb)=nRT$
For one mole and when b = 0, the above equation condenses to
$\, \, \, \, \, \, \, \, \, \, \, \, \, \Bigg(p+\frac{n^ a}{V^2}\Bigg)V=RT$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, pV=RT-\frac{a}{V}\, \, \, \, \, \, \, \, \, ...(i)$
E (i) is a straight equation between pV and$\frac{1}{V}$ whose slope is -
- a'. Equating with slope of the straight line given in the graph.
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, -a=\frac{20.1-21.6}{3-2}=-1.5$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, a=1.5$