Mean free path

Mean free path

Edited By Vishal kumar | Updated on Sep 10, 2024 08:37 PM IST

The mean free path is a fundamental concept in physics that describes the average distance a particle travels before colliding with another particle. This concept is crucial in understanding the behaviour of gases, where molecules are in constant motion and frequently collide with one another. In simpler terms, the mean free path helps us predict how far a molecule can move freely without interruption. In real life, the mean free path can be related to everyday experiences, such as walking through a crowded room. Imagine navigating through a bustling crowd; the distance you can move freely before bumping into someone is akin to the mean free path. Similarly, in the air we breathe, the molecules of gases like oxygen and nitrogen move in all directions, with their paths continuously interrupted by collisions, determining the overall properties of the gas, such as pressure and temperature. Understanding the mean free path is essential for applications ranging from designing efficient engines to predicting atmospheric phenomena.

This Story also Contains
  1. Mean Free Path
  2. Solved Examples Based on Mean Free Path
  3. Summary

Mean Free Path

On the basis of the 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 the free path.

There are assumptions 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 the mean free path.

Let $\lambda_1, \lambda_2 \ldots \ldots \lambda_n$ 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 $\lambda=\frac{\lambda_1+\lambda_2+\lambda_3+\ldots+\lambda_n}{n}$

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 = $\frac{n}{V}=\frac{P}{R T}$

Also, we know that the number of molecules per unit mole $=N_A=6.023 \times 10^{23}$

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

So the number of molecules per unit volume is $N=\frac{P N_A}{R T}$
So, $\lambda=\frac{\text { RT }}{\sqrt{2} \pi \mathrm{d}^2 \mathbf{P N}_{\mathrm{A}}}=\frac{\mathrm{kT}}{\sqrt{2} \pi \mathrm{d}^2 \mathbf{P}}$

If all the other molecules are not at rest then, $\quad \lambda=\frac{1}{\sqrt{2} \pi N d^2}=\frac{R T}{\sqrt{2} \lambda d^2 \mathrm{PN}_{\mathrm{A}}}=\frac{\mathrm{kT}}{\sqrt{2} \pi \mathrm{d}^2 \mathrm{P}}$

Now, if $\lambda=\frac{1}{\sqrt{2} \pi N d^2}$ 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, $\lambda \propto \frac{1}{\rho}$ and $\lambda \propto m$

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Solved Examples Based on Mean Free Path

Example 1: For a diatomic gas, the ratio of two specific heat Cp/Cv is equal to

1) 1.66

2) 1.4

3) 1.28

4) 2

Solution:

Value of degree of freedom for diatomic gas

$
f=5
$

So
$
\begin{aligned}
& \frac{C_p}{C_v}=1+\frac{2}{f} \\
& \frac{C_p}{C_v}=1+\frac{2}{5}=\frac{7}{5}
\end{aligned}
$

Hence, the answer is the option (2).

Example 2: The molecules of an ideal gas at 270 C have a mean velocity $v$. At what temperature (in oC) will the mean velocity be $\frac{3}{2} v$?

1) 350

2) 450

3) 402

4) 315

Solution:

The mean velocity of the gas molecule is

$
\bar{v}=\sqrt{\frac{8 R T}{\pi M}}
$
for same gas
$
\begin{aligned}
& \bar{v}=\sqrt{T} \\
& \frac{\overline{v_1}}{\overline{v_2}}=\sqrt{\frac{T_1}{T_2}}=\sqrt{\frac{(273+27)}{T_2}} \\
& \frac{2 v}{3 v}=\sqrt{\frac{300}{T_2}} \Rightarrow \frac{300}{T_2}=\frac{4}{9} \\
& \quad T_2=\frac{2700}{4}=675 \mathrm{~K} \\
& 675-373=402^{\circ} \mathrm{C}
\end{aligned}
$

Hence, the answer is the option (3).

Example 3: For a triatomic ( non-linear) gas, the total no. of degrees of freedom is

1) 4

2) 6

3) 7

4) 5

Solution:

Value of degree of freedom for triatomic gas

$f=6$

A triatomic gas can have three translational degrees of freedom and three rotational degrees of freedom.

3(translational) + 3(rotational) = 6

Hence, the answer is the option (2).

Example 4: For monoatomic gas, the incorrect statement is:

(i) It has all translational degrees of freedom

(ii) examples He, H2, Ne

(iii) It can have a maximum of 6 degrees of freedom 3 translational and 3 rotational

1) Only (ii)

2) only (i)

3) (ii) and (iii)

4) (i) and (iii)

Solution

Value of degree of freedom for monoatomic gas

$f=3$

wherein

A monoatomic gas can only have a translational degree of freedom.

* H2 is an example of a diatomic gas

* monoatomic gas can have a maximum of three degrees of freedom and is all translational.

Hence, the answer is the option of (3).

Example 5: Which statement is true for the term degree of freedom?

(i) Monatomic gas has all translational degrees of freedom

(ii) There can be a maximum of two rotational degrees of freedom

(iii) Total degree of freedom is of three types

1) (i), (ii), (iii)

2) (i) and (ii)

3) (i) and (iii)

4) (ii) and (iii)

Solution:

Degree of freedom

It is the number of directions in which a particle can move freely or the number of independent coordinates required to describe the system completely.

It is denoted by $f$.

$\begin{aligned} & f=3 N-R \\ & N=n o . \text { of particle } \\ & R=\text { no. of relation }\end{aligned}$

* monatomic gas has only three degrees of freedom and is all translational.

* Total degree of freedom is of three types

(i) Translational

(ii) Rotational

(iii) Vibrational

Hence, the answer is the option (3).

Summary

The mean free path is the average distance a gas molecule travels between collisions, crucial in understanding gas behaviour. It is calculated using the formula $\lambda=\frac{\mathrm{RT}}{\sqrt{2} \pi d^2 \mathrm{PN}_A}$, where $\lambda$ is the mean free path, $d$ is the molecule's diameter, and $N$ is the number of molecules per unit volume This concept is foundational in predicting gas properties, like pressure and temperature, and is essential in various scientific and engineering applications.

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