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

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · States of Matter

For one mole of a van der Waals gas when

b = 0

and

T = 300 K

, the

P V

vs.

1/ V

plot is shown below. The value of the van der Waals constant

a

(atm.liter

^{2} mol^{- 2}

) is

(A)1.0
(B)4.5
(C)1.5✓ Correct
(D)3.0

Answer: (C)

Solution

For a van der Waals gas with

b = 0

, the equation becomes

(P + \frac{a}{V ^{2}}) V = R T

, which rearranges to

P V = - a \times \frac{1}{V} + R T

. This is a linear equation in

P V

vs.

1/ V

with slope

= - a

and intercept

= R T

. From the graph, the slope is

tan (π - θ) = - a

, where

θ

is the angle the line makes with the positive

x

-axis. Reading from the graph:

tan θ = \frac{21.6 - 20.1}{3 - 2} = 1.5

tan θ = \frac{24.6 - 20.1}{3 - 0} = 1.5

. Therefore,

a = 1.5

atm·L²·mol⁻².

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Single correct option

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