For a molecule of an ideal gas,the number den | vedclass

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

Question

(A) The formula for the mean free path $\lambda$ of a gas molecule is given by:$\lambda = \frac{1}{\sqrt{2} \pi n d^2}$where $n$ is the number density

Vedclass · Accepted Answer

$5 	imes 10^{-4} 	ext{ cm}$

Vedclass · Answer

(A) The formula for the mean free path $\lambda$ of a gas molecule is given by:$\lambda = \frac{1}{\sqrt{2} \pi n d^2}$where $n$ is the number density and $d$ is the diameter of the molecule.Given:$n = 2 \sqrt{2} 	imes 10^8 	ext{ cm}^{-3}$$\lambda = \frac{10^{-2}}{\pi} 	ext{ cm}$Substituting these values into the formula:$\frac{10^{-2}}{\pi} = \frac{1}{\sqrt{2} \pi (2 \sqrt{2} 	imes 10^8) d^2}$$\frac{10^{-2}}{\pi} = \frac{1}{\pi (2 	imes 2 	imes 10^8) d^2}$$\frac{10^{-2}}{\pi} = \frac{1}{\pi (4 	imes 10^8) d^2}$$d^2 = \frac{1}{4 	imes 10^8 	imes 10^{-2}} = \frac{1}{4 	imes 10^6}$$d = \sqrt{\frac{1}{4 	imes 10^6}} = \frac{1}{2 	imes 10^3} = 0.5 	imes 10^{-3} 	ext{ cm} = 5 	imes 10^{-4} 	ext{ cm}$.

For a molecule of an ideal gas,the number density is $2 \sqrt{2} \times 10^8 \text{ cm}^{-3}$ and the mean free path is $\frac{10^{-2}}{\pi} \text{ cm}$. The diameter of the gas molecule is

Similar Questions

For matter to exist simultaneously in gas and liquid phases:

For a van der Waals gas,if $P_c, V_c$,and $T_c$ are the critical pressure,volume,and temperature respectively,then the value of $P_cV_c/T_c$ is:

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?

Write the equation for the mean free path of gas molecules.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module