Question

Calculate the value of $\sigma ,$ for nitrogen molecules at ${25}^{\circ }C$ and at pressure of ${10}^{-3}mmHg$. Given that b for nitrogen is 39.1 $c{m}^{3}mo{l}^{-1}.$ (in pm)

Answer 1

$b=39.1c{m}^{3}mol{e}^{-1}=39.1\times {10}^{-6}{m}^{3}mol{e}^{-1}=4V\times N_A$
$39.1\times {10}^{-6}=4\times \frac{4}{3}\pi r^3\times 6.023\times {10}^{23}$
$r=1.57\times {10}^{-10}m$
$\sigma =2r=3.14\times {10}^{-10}m=314pm$
${\mu }_{avg}=\sqrt{\frac{8RT}{\pi M}}=\sqrt{\frac{8\times 8.314\times 298}{3.14\times 28\times {10}^{-3}}}=474.8m/sec$.
$N^{\ast }=\frac{P}{kT}=\frac{{10}^{-3}\times 101325}{760\times 1.38\times {10}^{-23}\times 298}=3.24\times {10}^{19}$
${\lambda }_1=\frac{1}{\sqrt{2}\pi {\sigma }^2{N}^{\ast }}=\frac{1}{\sqrt{2}\times 3.14\times (3.14\times {10}^{-10}{)}^2\times 3.24\times {10}^{19}}$
$=0.0705m=7.05cm$
$Z_1=\sqrt{2}\pi {\sigma }^{2}U{N}^{\ast }=\sqrt{2}\times 3.14\times (3.14\times {10}^{-10}{)}^2\times 474.8\times 3.24\times {10}^{19}=6739.4se{c}^{-1}$
$Z_{11}=\frac{1}{\sqrt{2}}\pi {\sigma }^{2}U{N}^{\ast }=\frac{Z_1{N}^{\ast }}{2}=\frac{6739.4\times 3.24\times {10}^{19}}{2}$
$=1.09\times {10}^{23}{m}^{-3}se{c}^{-1}=1.09\times {10}^{17}c{m}^{-1}se{c}^{-1}$