Question

Calculate the ratio of collision frequencies $\left(Z_{11}\right)$ for the gases $A,B,C$ respectively.

• A. $1:2:4$
• B. $4:2:1$
• C. $1:4:16$
• D. $16:4:1$

Answer 1

The correct option is C $1:4:16$

Let ${\sigma }_A,{\sigma }_B$ and ${\sigma }_C$ be collision diameters of $A,B$ and $C$.
$\lambda =\frac{kT}{\sqrt{2}\pi {\sigma }^{2}P}$

$i.e,\sigma \propto \sqrt{\frac{T}{\lambda P}}$
$\therefore {\sigma }_A:{\sigma }_B:{\sigma }_C$
$\Rightarrow \sqrt{\frac{1600}{0.16\times 1}}:\sqrt{\frac{200}{0.16\times 2}}:\sqrt{\frac{400}{0.04\times 4}}$
$=4:1:2$

The ratio of collision frequencies is :
$Z_{11}\propto \frac{{\sigma }^2{P}^2}{\sqrt{m{T}^{3/2}}}$
$\Rightarrow A:B:C=\frac{16\times 1}{\sqrt{20}\times (1600{)}^{3/2}}:\frac{1\times 4}{\sqrt{40}\times (200{)}^{3/2}}:\frac{4\times 16}{\sqrt{80}\times (400{)}^{3/2}}$
$=1:4:16$

Hence, option $\left(C\right)$ is correct.