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Mean Free Path - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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

If the successive collision time interval of an ideal gas at a pressure of 2 atm and Temperature is 300 K is 6.0×10-8. What will be the new vision time if the pressure is doubled and tempercutue is set to 500k.

Consider an ideal gas continued in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increase as V9, Where V in the volume of the gas.
The value of q is r=\frac{C_p}{C_v}

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

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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}{ \pi d^2PN_A}=\frac{kT}{\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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Mean free path

Physics Part II Textbook for Class XI

Page No. : 336

Line : 3

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