Consider a gas for which the diameter of mole | vedclass

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

Question

(C) The mean free path $\lambda$ for a gas is given by the formula $\lambda = \frac{1}{\sqrt{2} \pi \sigma^2 n}$,where $n$ is the number density of mo

Vedclass · Accepted Answer

$\frac{RT}{\sqrt{6} P N_a \pi \sigma^2}$

Vedclass · Answer

(C) The mean free path $\lambda$ for a gas is given by the formula $\lambda = \frac{1}{\sqrt{2} \pi \sigma^2 n}$,where $n$ is the number density of molecules.Using the ideal gas equation $PV = n_{mol} RT$,where $n_{mol} = \frac{N}{N_a}$,we have $P = \frac{N}{V} \frac{RT}{N_a} = n \frac{RT}{N_a}$.Thus,$n = \frac{P N_a}{RT}$.Substituting $n$ into the expression for $\lambda$,we get $\lambda = \frac{RT}{\sqrt{2} \pi \sigma^2 P N_a}$.Since the motion of molecules is isotropic,the mean square free path is related to its components by $\lambda^2 = \lambda_x^2 + \lambda_y^2 + \lambda_z^2$. Due to symmetry,$\lambda_x = \lambda_y = \lambda_z$,so $\lambda^2 = 3 \lambda_x^2$,which implies $\lambda_x = \frac{\lambda}{\sqrt{3}}$.Substituting the value of $\lambda$,we get $\lambda_x = \frac{RT}{\sqrt{3} \sqrt{2} P N_a \pi \sigma^2} = \frac{RT}{\sqrt{6} P N_a \pi \sigma^2}$.

Consider a gas for which the diameter of molecules is $\sigma$. The gas is at a pressure $P$ and temperature $T$,and $N_a$ is Avogadro's number. What is the mean free path along the $x$-axis?

Similar Questions

The value of critical temperature in terms of van der Waals' constants $a$ and $b$ is given by

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

What is mean free path?

The diameter of an oxygen molecule is $2.94 \times 10^{-10} \ m$. The van der Waals gas constant '$b$' in $m^3/mol$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module