Answer:
153273.68816 m
Explanation:
k = Boltzmann constant = [tex]1.3\times 10^{-23}\ J/K[/tex]
T = Temperature = 304 K
P = Pressure = [tex]3.8\times 10^{-13}\ atm[/tex]
r = Radius = [tex]2\times 10^{-10}\ m[/tex]
d = Diameter= 2r = [tex]2\times 2\times 10^{-10}\ m=4\times 10^{-10}\ m[/tex]
Mean free path is given by
[tex]\lambda=\frac{kT}{\sqrt2\pi d^2P}\\\Rightarrow \lambda=\frac{1.38\times 10^{-23}\times 304}{\sqrt2 \pi (4\times 10^{-10})^2\times 3.8\times 10^{-13}\times 101325}\\\Rightarrow \lambda=153273.68816\ m[/tex]
The mean free path of the air molecule is 153273.68816 m
