Question

For one mole of a van der Waal's 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. $litr{e}^{2}mo{l}^{-2})$ is

• A. $1.0$
• B. $4.5$
• C. $1.5$
• D. $3.0$

Answer 1

The correct option is C $1.5$

Van der Waals' equation for 1 mole of a real gas is-

$[P+\frac{a}{V^2}][V-b]=RT$

Given that, $b=0$

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

$\therefore PV=RT-\frac{a}{V}$ ....(1)

Following $y=mx+c$ for the curve PV vs $\frac{1}{V}$

$Slope=-a$

$Slope=\frac{21.6-20.1}{3-2}=1.5$

$\therefore a=1.5$

Hence, the correct option is $C$