Question

For one mole of a van der Waals gas when $b=0$ and $T=300K$, the $PV$ vs. $1/V$ plot is shown below. The value of the van der Waals constant $a$ (atm.litre ${}^{2}mo{l}^{-2}$ ) is

• A. 1.0
• B. 4.5
• C. 1.5
• D. 3.0

Answer 1

Answer: (C)

van der Waals equation for 1 mole of real gas is,

$\left(P+\frac{a}{V^2}\right)(V-b)=RT$

but, $b=0$ (given)

$\Rightarrow \left(P+\frac{a}{V^2}\right)(V)=RT$.....(i)

$\therefore PV=-a\times \frac{1}{V}+RT$

$y=mx+c$

Slope $=\tan (\pi -\theta )=-a$

So, $\tan \theta =a=\frac{21.6-20.1}{3-2}=1.5$

or, $\tan \theta =\frac{24.6-20.1}{3-0}=1.5$