Question

For one mole of a van der Waal's gas when b =o and T = 300 K, the PV vs $\frac{1}{V}$ plot is shown below. The value of the Van der Waal's 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

​

$⟮P+\frac{a}{V^2}⟯⟮V⟯=RT$

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

y = RT –a(V)

y = RT –a(x)

So, slope = a = $\frac{21.6–20.1}{3-2}=\frac{1.5}{1}=1.5$