Question

For one mole of a Van der Waals gas when b = 0 and T = 300 K, then pV vs. $\frac{1}{V}$ plot is shown in figure. The value of the Van der Waals constant "a" $(atm{L}^{2}mo{l}^{-2}$)is___.

• A. 1.0
• B. 4.5
• C. 1.5
• D. 3.0

Answer 1

The correct option is C

1.5

$⟮p+\frac{a}{V^2}⟯⟮V-b⟯=RT$

But it is given that b = 0, so , the equation reduces to

$⟮p+\frac{a}{V^2}⟯V=RT\Rightarrow pV=\frac{a}{V}+RT$

Comparing it with a straight line equation , we get slope as -a. Calculating the slope , we get

$\frac{24.6-20.1}{3.0-0}=1.5\Rightarrow a=1.5$