A molecule is travelling in air at 300 , K an | 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 $\lambda$ is given by $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$, where $d$ is the diameter of the molecule and $

Vedclass · Accepted Answer

$\frac{0.2}{\pi} 	imes 10^{-5} \, m$

Vedclass · Answer

(A) The formula for the mean free path $\lambda$ is given by $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$, where $d$ is the diameter of the molecule and $n$ is the number density.Given:Radius $r = 0.6 	imes 10^{-10} \, m$, so diameter $d = 2r = 1.2 	imes 10^{-10} \, m$.Number density $n = 2.44 	imes 10^{25} \, 	ext{molecules}/m^3$.Substituting the values:$\lambda = \frac{1}{\sqrt{2} 	imes \pi 	imes (1.2 	imes 10^{-10})^2 	imes 2.44 	imes 10^{25}}$$\lambda = \frac{1}{1.414 	imes \pi 	imes 1.44 	imes 10^{-20} 	imes 2.44 	imes 10^{25}}$$\lambda = \frac{1}{4.97 	imes \pi 	imes 10^5} \approx \frac{0.2}{\pi} 	imes 10^{-5} \, m$.

$A$ molecule is travelling in air at $300 \, K$ and $1 \, atm$, and the radius of the molecule is $0.6 \times 10^{-10} \, m$. Calculate the approximate mean free path of the molecule. (The number density is $2.44 \times 10^{25} \, \text{molecules}/m^3$)

Similar Questions

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

The vapour of a substance behaves as a gas:

If the volume of the gas containing $n$ number of molecules is $V,$ then the pressure will decrease due to the force of intermolecular attraction in the proportion:

For a gas,$C_{p} - C_{V} = R$ in state $P$ and $C_{p} - C_{V} = 1.10 R$ in state $Q$. If $T_{P}$ and $T_{Q}$ are the temperatures in states $P$ and $Q$ respectively,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module