Mean free path

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 = nN_A

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 C_p/C_v 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.