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. 1.0
• B. 4.5
• C. 1.5
• D. 3.0

Answer 1

Answer: (C)

The van der Waals equation of state is
$\left(p+\frac{n^{a}}{V^{2}}\right)(V-n b)=n R T$
For one mole and when $b=0$, the above equation condenses to $\left(p+\frac{n^{a}}{V^{2}}\right) V=R T$
$ \Rightarrow p V=R T-\frac{a}{V} \ldots(i)$
Eq. (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.
$-a=\frac{20.1-21.6}{3-2}=-1.5$
$\Rightarrow a=1.5$