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JEE Main 2025 Syllabus PDF - Subject-Wise Detailed Syllabus

Mean Free Path - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • 12 Questions around this concept.

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25 \times 10^{-3}m^{3} volume cylinder is filled with 1 mol of O_{2} gas  at room temperature (300 K). The molecular diameter of O_{2}, and its root mean square speed are found to be 0.3 nm and 200 m/s respectively. What is the average collision rate (per second) for an O_{2} molecule?

The number of air molecules in cm^3 is increased from 3 \times 10^{19} \text{ to }12 \times 10^{19}. The ratio of collision frequency of air molecules before and after the increase in the number respectively is :

The mean free path of molecules of a certain gas at STP is 1500d, where d is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at 373 K is approximately :

The mean free path of molecules of a gas,(radius r ) is inversely proportional to

Concepts Covered - 1

Mean free path

Mean Free Path - 

On the basis of kinetic theory of gases, it is assumed that the molecules of a gas are continuously colliding against each other. So, the distance travelled by a gas molecule between any two successive collisions is known as free path.

There are assumption for this theory that during two successive collisions, a molecule of a gas moves in a straight line with constant velocity. Now, let us discuss the formula of mean free path - 

                                                                          

Let  \lambda _1,\lambda _2.......\lambda_n be the distance travelled by a gas molecule during n collisions respectively, then the mean free path of a gas molecule is defined as - 

                       \lambda=\frac{\text { Total distance travelled by a gas molecule between successive collisions }}{\text { Total number of collisions }}

Here, \lambda is the mean free path.

It can also be written as -        \boldsymbol{\lambda=\frac{\lambda_{1}+\lambda_{2}+\lambda_{3}+\ldots+\lambda_{n}}{n}}

Now, let us take d = Diameter of the molecule,
                          N = Number of molecules per unit volume.

 

Also, we know that,  PV = nRT 

So, Number of moles per unit volume  =  \frac{n}{V }=\frac{P}{RT}

Also we know that number of molecules per unit mole  = N_A = 6.023 \times10^{23} 

So, the number of molecules in 'n' moles = nNA

So the number of molecules per unit volume is N =  \frac{PN_A}{RT}

                                                                 So, \dpi{100} \mathbf{\lambda = \frac{RT}{\sqrt{2} \pi d^2PN_A}=\frac{kT}{\sqrt{2}\pi d^2P}}

If all the other molecules are not at rest then, \boldsymbol{\lambda=\frac{1}{\sqrt{2} \pi N d^{2}}}\mathbf{= \frac{RT}{\sqrt{2}\lambda d^2PN_A}=\frac{kT}{\sqrt{2}\pi d^2P}}

 

Now, if  \boldsymbol{\lambda=\frac{1}{\sqrt{2} \pi N d^{2}}}  and m = mass of each molecule then we can write - \lambda=\frac{1}{\sqrt{2} \pi N d^{2}}=\frac{m}{\sqrt{2} \pi(m N) d^{2}}=\frac{m}{\sqrt{2} \pi d^{2} \rho}

                                                                        So, \lambda \propto \frac{1}{\rho} \text { and } \lambda \propto m

 

                                         

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Mean free path

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