The excluded volume $b$ in the van der Waals equation is related to the molecular diameter $d$ by $b = \frac{2}{3} π N_{A} d^{3}$ . Which statements are correct?

Answer: A, B, D

Actual volume of

N_{A}

molecules is

N_{A} \cdot \frac{4}{3} π (d /2)^{3} = \frac{1}{6} π N_{A} d^{3}

, and

b = \frac{2}{3} π N_{A} d^{3}

is four times this (A true).

b \propto d^{3}

, so doubling

d

multiplies

b

by 8 (B true). Volume of

N_{A}

spheres of diameter

2 d

N_{A} \cdot \frac{4}{3} π d^{3} = \frac{4}{3} π N_{A} d^{3}

, which is twice

b

(C false). The two centres cannot approach closer than

d

, so the excluded sphere per pair has radius

d

and volume

\frac{4}{3} π d^{3}

(D true).