Calculate the ratio of the mean free paths of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The mean free path $\bar{l}$ of a gas molecule is given by the formula:$\bar{l} = \frac{1}{\sqrt{2} \pi d^{2} n}$where $d$ is the molecular diamet

Vedclass · Accepted Answer

$4:1$

Vedclass · Answer

(B) The mean free path $\bar{l}$ of a gas molecule is given by the formula:$\bar{l} = \frac{1}{\sqrt{2} \pi d^{2} n}$where $d$ is the molecular diameter and $n$ is the number density.Given that the gases are under identical conditions of temperature,pressure,and volume,the number density $n$ remains constant for both gases.From the formula,we observe that $\bar{l} \propto \frac{1}{d^{2}}$.Therefore,the ratio of the mean free paths is:$\frac{\bar{l}_{1}}{\bar{l}_{2}} = \left(\frac{d_{2}}{d_{1}}\right)^{2}$Given $d_{1} = 1\,\mathring{A}$ and $d_{2} = 2\,\mathring{A}$,we substitute these values:$\frac{\bar{l}_{1}}{\bar{l}_{2}} = \left(\frac{2}{1}\right)^{2} = \frac{4}{1}$Thus,the ratio of the mean free paths is $4:1$.

Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters $1\,\mathring{A}$ and $2\,\mathring{A}$. The gases may be considered under identical conditions of temperature,pressure,and volume.

Similar Questions

If the diameter of a gas molecule is very small,what will be the mean free path?

In Van der Waals equation $\left( P + \frac{a}{V^2} \right)(V - b) = RT$,what do $a$ and $b$ represent?

$A$ real gas behaves like an ideal gas if its

What is mean free path?

Vedclass Test Series

Exam Paper Generator

Online Exam Module