Two gases-argon atomic radius 0.07 nm,

(2)

Mean free path of a gas molecule is given by

$\lambda=\frac{1}{\sqrt{2} \pi d^{2} n}$

Here, $n=$ number of collisions per unit volume

$d=$ diameter of the molecule

If average speed of molecule is $v$ then

Mean free time, $\tau=\frac{\lambda}{v}$

$\Rightarrow \tau=\frac{1}{\sqrt{2} \pi n d^{2} v}=\frac{1}{\sqrt{2} \pi n d^{2}} \sqrt{\frac{M}{3 R T}}$

$\left(\because v=\sqrt{\frac{3 R T}{M}}\right)$

$\therefore \tau \propto \frac{\sqrt{M}}{d^{2}} \quad \therefore \frac{\tau_{1}}{\tau_{2}}=\frac{\sqrt{M_{1}}}{d_{1}^{2}} \times \frac{d_{2}^{2}}{\sqrt{M_{2}}}$

$=\sqrt{\frac{40}{140}} \times\left(\frac{0.1}{0.07}\right)^{2}=1.09$