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JEE Main levelSingle correct optionDifficulty 2 of 3rg016

The van der Waals constant $b$ is related to the radius $r$ of a spherical molecule by

A $b = N_{A} \cdot \frac{4}{3} π r^{3}$
B $b = 4 N_{A} \cdot \frac{4}{3} π r^{3}$
C $b = 2 N_{A} \cdot \frac{4}{3} π r^{3}$
D $b = \frac{N _{A}}{4} \cdot \frac{4}{3} π r^{3}$

Answer: B

Two molecules of radius

r

cannot approach closer than

2 r

, so the excluded volume per pair is

\frac{4}{3} π (2 r)^{3} = 8 \times \frac{4}{3} π r^{3}

, or

4 \times \frac{4}{3} π r^{3}

per molecule. Per mole,

b = 4 N_{A} \cdot \frac{4}{3} π r^{3}

, four times the actual molecular volume.

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