KVPY SX SB Exam 2016 Question Paper

As partition has hole of diameter d, mean pressure between both sections is equal.
$\because$ Mean force path $\left(\mathit{\lambda }\right)$ is given by $=\frac{1}{\sqrt{2\mathit{\pi }{\mathit{r}}^2{\mathit{n}}_{\mathit{v}}}}$
$\mathit{r}=$ radius of the gas particle
${\mathit{n}}_{\mathit{V}}=$ number density of gas i.e.
$\because$ From ideal gas equation $\mathit{P}\mathit{V}=\mathit{n}\mathit{R}\mathit{T}=\mathit{N}∕{\mathit{N}}_{\mathit{A}}\mathit{R}\mathit{T}={\mathit{N}\mathit{K}}_{\mathit{B}}\mathit{T}$
${\mathit{N}}_{\mathit{A}}=$ Avogadro number; ${\mathit{K}}_{\mathit{B}}=$ Boltzmann constant
${\mathit{N}}_{\mathit{A}}=$ Avogadro number; ${\mathit{K}}_{\mathit{B}}=$ Boltzmann constant
$\Rightarrow {\mathit{n}}_{\mathit{y}}=\frac{\mathit{N}}{\mathit{V}}=\frac{\mathit{P}}{{\mathit{k}}_{\mathit{B}}\mathit{T}}$
$\therefore \mathit{\lambda }=\frac{1}{\sqrt{2}\mathit{\pi }{\mathit{r}}^2}\frac{{\mathit{K}}_{\mathit{B}}\mathit{T}}{\mathit{P}}$
$\frac{{\mathit{\lambda }}_1}{{\mathit{\lambda }}_2}=\frac{{\mathit{T}}_1}{{\mathit{T}}_2}\times \frac{{\mathit{P}}_2}{{\mathit{P}}_1}\Rightarrow \frac{{\mathit{\lambda }}_1}{{\mathit{\lambda }}_2}=\frac{{\mathit{T}}_1}{{\mathit{T}}_2}\times \frac{\sqrt{{\mathit{T}}_2}}{\sqrt{{\mathit{T}}_1}}$
$\frac{{\mathit{\lambda }}_1}{{\mathit{\lambda }}_2}=\sqrt{\frac{{\mathit{T}}_1}{{\mathit{T}}_2}}=\sqrt{\frac{150}{300}}=0.7$