Question

For oxygen at 25 ${}^{o}C$, the collision diameter is 0.361 nm. What is the mean free path for oxygen molecules at:
(a) 1 atm pressure and (b) 0.1 $Pa$ pressure

Answer 1

Collision diameter, $d=0.361nm=3.61\times {10}^{-10}m$

We know that $PV=\frac{N}{N_A}RT$

(a) $\therefore \frac{N}{V}=\frac{P{N}_A}{RT}=\frac{1atm\times 6.022\times {10}^{23}mo{l}^{-1}}{0.08206Latm{K}^{-1}mo{l}^{-1}\times 298K}$

$=2.46\times {10}^{22}{L}^{-1}=2.46\times {10}^{25}{m}^{-3}$

The mean free path= $\lambda =\frac{1}{\sqrt{2}\Pi d^2\left(\frac{N}{V}\right)}$

$=\frac{1}{\sqrt{2}\Pi (3.61\times {10}^{-10}m{)}^2\times 2.46\times {10}^{25}{m}^{-3}}$

$=6.983\times {10}^{-8}m$

(b) $0.1Pa=0.987\times {10}^{-6}atm$

$\frac{N}{V}=\frac{PNa}{RT}=\frac{6.022\times {10}^{23}mo{l}^{-1}\times 0.987\times {10}^{-6}atm}{0.08206Latm{K}^{-1}mo{l}^{-1}\times 298K}$

$=2.428\times {10}^{19}{m}^{-3}$

$\therefore$ The mean free path= $\lambda =\frac{1}{\sqrt{2}\Pi d^2\left(\frac{N}{V}\right)}$

$=\frac{1}{\sqrt{2}\times \Pi (3.6\times {10}^{-10}m{)}^2\times 2.428\times {10}^{19}{m}^{-3}}$

$=0.071m$