Question

Calculate the value of $\lambda$ (cm) for the nitrogen molecules at ${25}^{\circ }C$ and at pressure of ${10}^{-3}$mm Hg. Given that value of $b$ for nitrogen is $39.1$ cm${}^3$mol${}^{-1}$(Round off answer to integer).

Answer 1

$b=\frac{4}{3}\pi r^3\times 4N_A$

$39.1=[\frac{4}{3}\times 3.1416\times r^3]\times 4\times 6.023\times {10}^{23}$

$r^3=$

$r=1.57\times {10}^{-8}$ cm

$\sigma =2r=3.14\times {10}^{-8}cm=3.14\times {10}^{-8}dm$

$\lambda =\frac{kT}{\sqrt{2}\pi {\sigma }^{2}P}=\frac{0.0821\times 298\times 760}{6.023\times {10}^{23}\times 3.1416(3.14\times {10}^{-9}{)}^2\times {10}^{-3}}=0.7015dm=7.015cm$

Hence answer is $7$.