Question

van der Waals' constant '$b$' of $Ar$ is $3.22\times {10}^{-5}{\text{m}}^3{\text{mol}}^{-1}$.Calculate the molecular diameter(in $nm$) of $Ar$.

Answer 1

We know that, exclused volume 'b' is given as :
$b=4\times$ 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 $r=\left(\frac{3.22\times {10}^{-5}\times 3\times 7}{4\times 6.02\times {10}^{23}\times 4\times 22}\right)$
$=0.1472\times {10}^{-9}\text{m}$
$\therefore$ Diameter $=2r=2\times 0.1472\times {10}^{-9}$
$=0.2944\times {10}^{-9}\text{m}=0.2944\text{nm}$