Question

Van der Waals' constant ${}^{\prime }{b}^{\prime }$ for oxygen is $32c{m}^2/mol$. Assume $b$ is four times the actual volume of a mole of "billiard-ball" $O_2$ molecules and compute the diameter of an $O_2$ molecule.

Answer 1

$b=4V_m=32c{m}^3/mol$
$V_m=8c{m}^3/mol$
For one molecule,
$V=\frac{8c{m}^3/mol}{6.023\times {10}^{23}molecules/mol}$
$V=1.33\times {10}^{-23}c{m}^3/molecule$
But $V=\frac{4}{3}\pi r^3$
$\frac{4}{3}\pi r^3=1.33\times {10}^{-23}c{m}^3/molecule$
$r^3=3.17\times {10}^{-24}c{m}^3/molecule$
$r=1.47\times {10}^{-8}cm$
The diameter of an $O_2$ molecule $=2r=2\times 1.47\times {10}^{-8}cm=2.94\times {10}^{-8}cm$