Question

Calculate the mean free path in $C{O}_2$ at ${27}^{\circ }C$ and a pressure of ${10}^{-9}$ bar. Molecular diameter of $C{O}_2$ is $500$ pm.

[ Given : R $=\frac{25}{3}$ $J.mo{l}^{-1}{K}^{-1},\sqrt{2}=1.4,\pi =\frac{22}{7},N_A=6\times {10}^{23}$]

• A. $3.788\times {10}^3$ cm
• B. $7.47\times {10}^3$ cm
• C. $6.43\times {10}^5$ cm
• D. None of these

Answer 1

Answer: (A)

$3.788\times {10}^3$ cm

The number of molecules per unit volume $N=\frac{n}{V}\times N_A=\frac{P}{RT}\times N_A=\frac{{10}^{-9}}{1.01325\times 0.0821\times 300}\times 6.023\times {10}^{23}$

$=2.4\times {10}^{13}{\text{molecules/dm}}^3=2.4\times {10}^{10}{\text{molecules/cm}}^3$

The mean free path is $\frac{1}{\sqrt{2}\times \pi \times {\sigma }^2\times N}=\frac{1}{1.414\times 3.1416\times (500\times {10}^{-10}{)}^2\times 2.4\times {10}^{+10}}=3.788\times {10}^{3}cm$