IIT-JEE 2012 · Paper 1 · Chemistry Q22

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ChemistrySingle correct+3 / -1Question 22 of 60

For one mole of a van der Waals gas when $b=0$ and $T=300$ K, the $PV$ vs. $1/V$ plot is shown below. The value of the van der Waals constant $a$ (atm.liter² mol⁻²) is

PV (liter-atm mol⁻¹) against 1/V (mol liter⁻¹), not to scale: a straight line falling from an intercept of 24.6 at 1/V = 0, through 21.6 at 1/V = 2.0 and 20.1 at 1/V = 3.0 (23.1 also marked on the axis).

1.0

4.5

1.5

3.0

Answer and worked solution

Official answer: C

Van der Waals equation with $b=0$ . For one mole,
$\left(P+\frac a{V^2}\right)V=RT\implies PV=RT-a\cdot\frac1V.$
So the plot of $PV$ against $1/V$ is a straight line of slope $-a$ .

Slope from the graph. $PV$ falls from 24.6 to 21.6 as $1/V$ goes from 0 to 2.0 (and to 20.1 at 3.0):
$a=\frac{24.6-21.6}{2.0}=1.5\ \mathrm{atm\,L^2\,mol^{-2}}.$
(The intercept $24.6=RT$ at 300 K checks.)

(C)

Topics:Real gases and van der Waals equation

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IIT-JEE 2012 · Paper 1, Chemistry question 22