Question

For one mole of a van der Waals' gas when $b=0$ and $T=300K$, the $PVvs1/V$ plot is shown below. The value of the van der Waals' constant $a\left(atmlitr{e}^{2}mo{l}^{-2}\right)$ is :

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

Answer 1

Answer: (C)

$1.5$

Van der Waal's equation is $(P+\frac{a}{V^2})(V-b)=RT$

As given that $b=0$

$PV+\frac{a}{V}=RT\Rightarrow PV=RT-\frac{a}{V}$

Comparing with $y=mx+c$, Intercept $\left(c\right)=RT$, Slope $\left(m\right)=-a$

$Slope=\frac{y_2-y_1}{x_2-x_1}=\frac{20.1-21.6}{3-2}=-1.5$.

Thus, $a=1.5$.

Therefore, the correct option is C.