Question
The diameter of an oxygen molecule is approximately $3 \text Å $. Calculate the mean free path of oxygen molecules and the mean time taken between collisions at normal temperature and pressure. At normal temperature and pressure the number of molecules per cubic is $3 \times 10^{19}$.

Answer

$ \begin{array}{l} d =3 \textÅ =3 \times 10^{-8} cm \\ n _0  =3 \times 10^{19}\end{array}$
Mean free path
$\begin{aligned}\lambda & =\frac{1}{\sqrt{2} n_0 \pi d^2} \\& =\frac{1}{1.414 \times 3 \times 10^{19} \times 3.14 \times\left(3 \times 10^{-8}\right)^2} \\& =\frac{1}{1.414 \times 3 \times 3.14 \times 9 \times10^{19} \times 10^{-16}} \\& =\frac{1}{119.88 \times 10^3}=8.33 \times 10^{-6} cm\end{aligned}$
Mean time taken in collisions
$\begin{aligned} t & =\frac{\lambda}{\overline{ C }}=\lambda\left[\frac{ M }{3 RT }\right]^{\frac{1}{2}} \\ & =8.33 \times 10^{-6} \times\left[\frac{32}{3 \times 8.3 \times 10^7 \times 273}\right]^{\frac{1}{2}} \\ & =1.75 \times 10^{-10} sec .\end{aligned}$

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