Question

Two flask A and B have equal volume at 100K and 200K and have pressure 4 atm and 1 atm respectively. The flask A contains $H_2$ gas and B contains ${CH}_4$ gas. The collision diameter of ${CH}_4$ is twice that of $H_2$. Calculate ratio of mean free path of ${CH}_4$ to $H_2$.

Answer 1

Formula for mean free path is
$\lambda =\frac{RT}{\sqrt{2}\pi d^2{N}_{A}P}$
R=gas constant
T=temperature
$N_A$=Avogadro's constant
P=Pressure
d=collision diameter
Let collision diameter of $H_2=d$
Then collision diameter of ${CH}_4=2d$
Given:
$A\Rightarrow H_2\Rightarrow 100K\to 4atm\to d$
$B\Rightarrow {CH}_4\Rightarrow 200K\to 1atm\to 2d$
$\frac{\lambda \left({CH}_4\right)}{\lambda \left(H_2\right)}=\frac{\left(\frac{P\times 200}{\sqrt{2}\times \pi \times {\left(2d\right)}^2{\times N}_A\times 1}\right)}{\left(\frac{P\times 100}{\sqrt{2}\times \pi \times {\left(d\right)}^2{\times N}_A\times 4}\right)}$
ratio=2:1