States of Matter 1 Question 13
13. For one mole of a van der Waals’ gas when $b=0$ and $T=300 \mathrm{~K}$, the $p V v s 1 / V$ plot is shown below. The value of the van der Waals’ constant $a$ (atm $\left.\mathrm{L} \mathrm{mol}^{-2}\right)$
(2012)
(a) 1.0
(b) 4.5
(c) 1.5
(d) 3.0
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Solution:
- The van der Waals’ equation of state is
$$ \left(p+\frac{n^{2} a}{V^{2}}\right)(V-n b)=n R T $$
For one mole and when $b=0$, the above equation condenses to
$$ \left(p+\frac{a}{V^{2}}\right) V=R T $$
$$ \Rightarrow \quad p V=R T-\frac{a}{V} $$
Eq. (i) is a straight equation between $p V$ and $\frac{1}{V}$ whose slope is c $-a^{\prime}$. Equating with slope of the straight line given in the graph.
$$ \begin{array}{rlrl} & & -a & =\frac{20.1-21.6}{3-2}=-1.5 \ \Rightarrow & a & =1.5 \end{array} $$
