Question

Van der Waals' constant ${}^{\prime }{b}^{\prime }$ of $Ar$ is $3.22\times {10}^{-5}{m}^{3}mo{l}^{-1}$. Calculate the molecular diameter of $Ar$.

Answer 1

$b=4\times$ volume occupied by the molecules in one mole of a gas
$=4\times N_0\times \left(\frac{4}{3}\pi r^3\right)$
$3.22\times {10}^{-5}=4\times 6.02\times {10}^{23}\times \frac{4}{3}\times \frac{22}{7}\times r^3$
or $r={\left(\frac{3.22\times {10}^{-5}\times 3\times 7}{4\times 6.02\times {10}^{23}\times 4\times 22}\right)}^{1/3}$
$=0.1472\times {10}^{-9}m$
$\therefore Diameter=2r=2\times 0.1472\times {10}^{-9}$
$=0.2944\times {10}^{-9}m$