A Collision Problem

Let $\sigma_A , \sigma_B$ and $\sigma_C$ be the Collision Diameters of $A,B$ and $C$ .

Let $\lambda$ be the Mean Free Path i.e the average distance travelled by a molecule between two successive collisions , $P$ be the pressure in atm .

$\therefore \lambda = \dfrac{k\cdot T}{\sqrt{2}\cdot \pi\cdot \sigma^{2}\cdot P} \\ \Rightarrow \sigma \propto \sqrt{\dfrac{T}{\lambda\cdot P}}$

$\therefore \sigma_A:\sigma_B:\sigma_C \\ \\ \sqrt{\dfrac{1600}{0.16\cdot 1}} : \sqrt{\dfrac{200}{0.16\cdot 2}} : \sqrt{\dfrac{400}{0.04\cdot 4}} \\= 4:1:2$

Now using this we'll find the rest .

$Z_{11} \propto \dfrac{\sigma^{2}\cdot P^{2}}{\sqrt{m\cdot T^{3}}}$ $\Rightarrow A:B:C \\ = \dfrac{16\times 1}{\sqrt{20\cdot 1600^{3}}} : \dfrac{1\times 4}{\sqrt{40\cdot 200^{3}}} : \dfrac{4\times 16}{\sqrt{80\cdot 400^{3}}} \\ 1:4:16$ .

$\implies j=1,k=4,l=16$

$Z_1 \propto \dfrac{\sigma^{2}\cdot P}{\sqrt{MT}}$

$\Rightarrow A:B:C \\ = \dfrac{16\cdot 1}{\sqrt{20\cdot 1600}} : \dfrac{1\cdot 2}{\sqrt{40\cdot 200}} : \dfrac{4\cdot 4}{\sqrt{80\cdot 400}} \\= 4:1:4$

$\implies m=4,n=1,o=4$

The final answer is $\frac{m}{j} + \frac{k}{n} + \frac{l}{o} \\= \frac{4}{1} + \frac{4}{1} + \frac{16}{4} = 12$ .

The collision frequency for single molecule $Z_1$ , for all molecules $Z_{11}$ , mean free path $\lambda$ and average speed of molecule $\overline{c}$ are given below:

\(\begin{cases} Z_1 = \sqrt{2}\pi d^2 \overline{c} \left(\dfrac{N}{V}\right) &...(1) \\ Z_{11} = \dfrac{1}{2} Z_1 \left(\dfrac{N}{V}\right) &...(2) \\ \lambda = \dfrac{1}{\sqrt{2}d^2 N} & ...(3) \\ \overline{c} = \sqrt{\dfrac{8k_BT}{\pi m}} & ...(4) \end{cases} \quad \begin{array} {} \text{where } & d = \text{the diameter of molecule} \\ & N = \text{the number of molecules in volume }V \\ & k_B = \text{Boltzmann constant} \\ & T = \text{absolute temperature} \\ & m = \text{molecular mass} \end{array} \)

$\begin{aligned} (1): \quad Z_1 & \propto d^2 \overline{c} N \quad \quad \small \color{#3D99F6}{\text{Since } (3): \space d^2 \propto \frac{1}{\lambda N} \quad (4): \space \overline{c} \propto \sqrt{\dfrac{T}{m}}} \\ & \propto \frac{1}{\lambda}\sqrt{\dfrac{T}{m}} \end{aligned}$

$\Rightarrow m : n : o = \frac{1}{0.16}\sqrt{\frac{1600}{20}} : \frac{1}{0.16}\sqrt{\frac{200}{40}} : \frac{1}{0.04}\sqrt{\frac{400}{80}} = 4: 1 : 4$

$\begin{aligned} (2): \quad Z_{11} & \propto Z_1 N \quad \quad \small \color{#3D99F6}{\text{Since } pV = NRT, \text{ where }p \text{ is pressure and } R \text{, the universal gas constant}} \\ & \propto \frac{Z_1 p}{T} \end{aligned}$

$\Rightarrow j : k : l = \frac{m}{1600} : \frac{2n}{200} : \frac{4o}{400} = \frac{4}{1600} : \frac{2}{200} : \frac{16}{400} = 1: 4 : 16$

$\Rightarrow \dfrac{m}{j} + \dfrac{k}{n} + \dfrac{l}{o} = \dfrac{4}{1} + \dfrac{4}{1} + \dfrac{16}{4} = \boxed{12}$