Mean Free Path MCQ Quiz - Objective Question with Answer for Mean Free Path - Download Free PDF

We need to determine the rate at which a molecule has collisions with other molecules for different ideal gases given their average speeds and molecular diameters.

The rate of collisions (Z) for a molecule in an ideal gas is given by the formula:

Z = nσv

Where:

• n is the number of molecules per unit volume,
• σ is the collision cross-section, which is proportional to the square of the molecular diameter (σ ∝ d²),
• v is the average speed of the molecules.

Given that the number of molecules per unit volume (n) is the same for all the gases, we can focus on the product of the collision cross-section and the average speed to compare the rates of collision.

Let's analyze each option:

Option 1: v = v₀ and d = d₀

Option 2: v = 2v₀ and d = d₀/2

Option 3: v = v₀ and d = 2d₀

Option 4: v = 4v₀ and d = d₀/2

For each option, the collision rate Z can be expressed as:

Option 1: Z₁ ∝ v₀ × (d₀)² = v₀d₀²

Option 2: Z₂ ∝ 2v₀ × (d₀/2)² = 2v₀ × (d₀²/4) = (1/2)v₀d₀²

Option 3: Z₃ ∝ v₀ × (2d₀)² = v₀ × 4d₀² = 4v₀d₀²

Option 4: Z₄ ∝ 4v₀ × (d₀/2)² = 4v₀ × (d₀²/4) = v₀d₀²

Comparing the values:

Z₁ = v₀d₀²

Z₂ = (1/2)v₀d₀²

Z₃ = 4v₀d₀²

Z₄ = v₀d₀²

It is clear that the rate of collision is greatest for Option 3, where Z₃ = 4v₀d₀².

Final Answer: The rate at which a molecule has collisions with other molecules is greatest for Option 3.

With increase in the collision frequency, the molecular speed decreases then the expected so the distance covered will be less.

Concept:

Mean Free Path and Molecular Diameter:

• The mean free path (λ) of a gas molecule is the average distance a molecule travels before colliding with another molecule. It is inversely proportional to the number density of molecules and the cross-sectional area, which depends on the molecular diameter.
• The formula for the mean free path is given by:
  λ ∝ 1 / (n × d²)
  Where n is the number density (number of molecules per unit volume) and d is the molecular diameter.
• Since the number of molecules per unit volume is the same for both gases, the ratio of the mean free paths is inversely proportional to the square of the ratio of their molecular diameters.

Calculation:

Given that the molecular diameters of the two gases are in the ratio 1 : 3, the ratio of their mean free paths will be:
λ₁ / λ₂ = (d₂ / d₁)² = (3 / 1)² = 9

∴ The ratio of their mean free paths is 9 : 1.

∴ Hence, the correct answer is: Option 4.

CONCEPT:

Mean Free Path (λ): 

• The distance travelled by a gas molecule between two successive collisions is known as a free path.
  \(λ = \frac{{Total\;distance\;travelled\;by\;a\;gas\;molecule\;between\;successive\;collisions\;\;}}{{Total\;number\;of\;collisions}}\)
• During two successive collisions, a molecule of gas moves in a straight line with constant velocity, and the mean free path of a gas molecule is given by
  \(λ = \frac{1}{{\sqrt 2 \pi n{d^2}}}\)

Where n = number of molecules per unit volume and d = diameter of a molecule

EXPLANATION:

• The mean free path of a gas molecule is given by

\(⇒ λ = \frac{1}{{\sqrt 2 \pi n{d^2}}}\)

⇒ λ ∝ d^-2

Concept:

• The mean free path ($\lambda$) is the average distance a molecule travels between two successive collisions.
• It depends on the number density (" id="MathJax-Element-29-Frame" role="presentation" style="position: relative;" tabindex="0"> n) of the molecules and the diameter (dUnknown node type: span" id="MathJax-Element-30-Frame" role="presentation" style="position: relative;" tabindex="0">Unknown node type: span<merror><mtext>Unknown node type: span</mtext></merror> <span style="font-family: var(--bs-body-font-family); font-size: var(--bs-body-font-size); font-weight: var(--bs-body-font-weight); text-align: var(--bs-body-text-align);">) of the molecules.</span>
• The mean free path is given by the following formula
• λ = \(\frac{1}{\sqrt{2} \pi \mathrm{d}^2 n}\)

​Calculation:

Thus by concept.

n = number of molecule per unit volume 

d = diameter of the molecule 

λ = \(\frac{1}{\sqrt{2} \pi \mathrm{d}^2 n}\) 

∴ The correct option is 3

CONCEPT:

• Mean Free Path (λ): The distance travelled by a gas molecule between two successive collisions is known as a free path.

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

• During two successive collisions, a molecule of gas moves in a straight line with constant velocity and mean free path of a gas molecule is given by

\(\lambda = \frac{1}{{\sqrt 2 π n{d^2}}}\)

Where n = number of molecules per unit volume and d = diameter of a molecule

• The mean free path depends on 
  1. Density 
  2. 2.Number of molecules 
  3. 3.Temperature and Pressure

EXPLANATION:

• Suppose d is the diameter of each molecule of the gas.
• A particular molecule will suffer a collision with any molecule that comes within a distance d between the centers of two molecules.
• If \(\bar c\) is the average speed of the molecules.
• The volume swept by the molecules in small time Δt, in which molecules will collide with it is 

\(\Rightarrow V=\pi d^2̅ v\Delta t\)

​

CONCEPT:

• Mean Free Path (λ): The distance travelled by a gas molecule between two successive collisions is known as a free path.

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

• During two successive collisions, a molecule of gas moves in a straight line with constant velocity and mean free path of a gas molecule is given by

\(\lambda = \frac{1}{{\sqrt 2 π n{d^2}}}\)

Where n = number of molecules per unit volume and d = diameter of a molecule

EXPLANATION:

• The volume swept by the molecules in small time Δt, in which molecules will collide with it is 

\(\Rightarrow V=\pi d^2\bar v\Delta t\)

Where d = distance between the center of two molecules,  v = average speed of the molecules

• If n is the number of molecules per unit volume of the gas, then the collision suffered by the molecule in time  Δt is \(\pi d^2 v\Delta t\times n\)
• The number of collisions per sec is

\(\Rightarrow Number\, of \,collisions/sec=\frac{n\pi d^2 v\Delta t}{\Delta t}=n\pi d^2 v\)

• The average time between two successive collisions is 

\(\Rightarrow \tau=\frac{1}{n\pi d^2 v}\)

CONCEPT:

• Mean Free Path (λ): The distance travelled by a gas molecule between two successive collisions is known as a free path.

\(\Rightarrow \lambda = \frac{{Total\;distance\;travelled\;by\;a\;gas\;molecule\;between\;successive\;collisions\;\;}}{{Total\;number\;of\;collisions}}\)

• During two successive collisions, a molecule of gas moves in a straight line with constant velocity, and the mean free path of a gas molecule is given by

\(\Rightarrow \lambda = \frac{1}{{\sqrt 2 \pi n{d^2}}}\)

Where n = number of molecules per unit volume and d = diameter of a molecule

EXPLANATION:

• From the above, it is clear that the mean free path of a gas molecule depends on the number of molecules per unit volume and diameter of a molecule. Therefore option 3 is correct.

CONCEPT:

Mean Free Path (λ): 

• The distance travelled by a gas molecule between two successive collisions is known as a free path.
  \(λ = \frac{{Total\;distance\;travelled\;by\;a\;gas\;molecule\;between\;successive\;collisions\;\;}}{{Total\;number\;of\;collisions}}\)
• During two successive collisions, a molecule of gas moves in a straight line with constant velocity, and the mean free path of a gas molecule is given by
  \(λ = \frac{1}{{\sqrt 2 \pi n{d^2}}}\)

Where n = number of molecules per unit volume and d = diameter of a molecule

EXPLANATION:

• The mean free path of a gas molecule is given by

\(⇒ λ = \frac{1}{{\sqrt 2 \pi n{d^2}}}\)

⇒ λ ∝ d^-2

CONCEPT:

• Mean Free Path (λ): The distance travelled by a gas molecule between two successive collisions is known as a free path.

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

• During two successive collisions, a molecule of gas moves in a straight line with constant velocity and mean free path of a gas molecule is given by

\(\lambda = \frac{1}{{\sqrt 2 π n{d^2}}}\)

Where n = number of molecules per unit volume and d = diameter of a molecule

EXPLANATION:

• The volume swept by the molecules in small time Δt, in which molecules will collide with it is 

\(\Rightarrow V=\pi d^2\bar v\Delta t\)

Where d = distance between the centre of two molecules, ̅ c = average speed of the molecules

• If n is the number of molecules per unit volume of the gas, then the collision suffered by the molecule in time  Δt is \(\pi d^2\bar v\Delta t\times n\)
• Number of collisions per sec is

\(\Rightarrow Number\, of \,collisions/sec=\frac{n\pi d^2\bar v\Delta t}{\Delta t}=n\pi d^2\bar v\)

CONCEPT:

Mean Free Path (λ): 

The distance travelled by a gas molecule between two successive collisions is known as a free path.
\(\rm \lambda = \frac{{Distance\;travelled\;by\;a\;gas\;molecule\;between\;successive\;collisions\;\;}}{{Total\;number\;of\;collisions}}\)

During two successive collisions, a molecule of gas moves in a straight line with constant velocity, and the mean free path of a gas molecule is given by
\(\lambda = \frac{1}{{\sqrt 2 \pi n{d^2}}}\)

Where n = number of molecules per unit volume and d = diameter of a molecule

EXPLANATION:

From the above, it is clear that the average distance between two successive collisions called the mean free path is \(\lambda = \frac{1}{{ \pi n{d^2}}}\).

CONCEPT:

• Mean Free Path (λ): The distance traveled by a gas molecule between two successive collisions is known as a free path.
  \(\lambda = \frac{{Total\;distance\;travelled\;by\;a\;gas\;molecule\;between\;successive\;collisions\;\;}}{{Total\;number\;of\;collisions}}\)

During two successive collisions, a molecule of gas moves in a straight line with constant velocity, and the mean free path of a gas molecule is given by
\(\lambda = \frac{1}{\sqrt{2}n\pi a^2}\)

Where n = number of molecules per unit volume and a = diameter of a molecule

• The mean free path depends on 
  1. Density 
  2. Number of molecules 
  3. Temperature and Pressure

EXPLANATION:

\(\lambda = \frac{1}{{\sqrt (2 \pi n{a^2})}}\) is the mean free path given by Maxwell. So option 2 is correct.

• In the given options, m is the mass of the molecules and T is the temperature of the gas.

EXTRA POINTS:

• Suppose d is the diameter of each molecule of the gas.
• A particular molecule will suffer a collision with any molecule that comes within a distance d between the centers of two molecules.
• If \(\bar c\) is the average speed of the molecules.
• The volume swept by the molecules in small time Δt, in which molecules will collide with it is 

\(\Rightarrow V=\pi d^2̅ v\Delta t\)

Concept:

The mean free path of a gas molecule is given by:

\(λ =\frac{1}{\sqrt{2}\pi d^2n'}\)

where d is the diameter of the gas molecule and n' is the number of molecules per unit volume of the gas.

Calculations:

Given:

From the ideal gas equation

PV = nRT

where 'n' is the number of the moles

\(PV=(nN_A)\left(\frac{R}{N_A}\right)T\)

Here nNA is the number of molecules in n moles.

\(P=\left(\frac{nN_A}{V}\right)\left(\frac{R}{N_A}\right)T\)

Here \(\left(\frac{nN_A}{V}\right)=n'\;\)is the number of molecules per unit volume of the gas and \(\left(\frac{R}{N_A}\right)=k\;\)is the Boltzmann constant.

\(\Rightarrow n'=\frac{P}{kT}\)

The mean free path of a gas molecule is given by:

\(λ =\frac{1}{\sqrt{2}\pi d^2n'}\)

∴ λ ∝ (1/n')

∴ with an increase in the number of molecules per unit volume, the mean free path decreases.

∴ λ ∝ (1/d²) (Option 4 is not true)

∴ with a decrease in the size of molecules, the mean free path increases. (Option 3 is true)

After replacing \(\Rightarrow n'=\frac{P}{kT}\), we will get;

\(λ =\frac{1}{\sqrt{2}\pi d^2n'}\)

\(λ =\frac{kT}{\sqrt{2}\pi d^2P}\)

At constant temperature T, the mean free path will decrease on increasing P. (Option 1 is not true)

Similarly, 

At constant pressure P, the mean free path will increase on increasing T. (Option 4 is not true)

Concept:

Mean Free Path (λ):

 The distance traveled by a gas molecule between two successive collisions is known as a free path.
\(\lambda = \frac{{Total\;distance\;travelled\;by\;a\;gas\;molecule\;between\;successive\;collisions\;\;}}{{Total\;number\;of\;collisions}}\)

During two successive collisions, a molecule of gas moves in a straight line with constant velocity, and the mean free path of a gas molecule is given by
\(\lambda = \frac{1}{{\sqrt 2 \pi n{d^2}}}\) -- (1)

n is the number of molecules per unit volume.

\(n = \frac{\rho}{m} = \frac{P}{K_B T}\) --- (2)

Here K_B is Boltzman Constant, P is pressure, T is temperature

Calculation:

Now if we put (2) in (1) we get

\(\lambda = \frac{1}{{\sqrt 2 \pi n{d^2}}}\)

\(\implies \lambda = \frac{1}{{\sqrt 2 \pi (\frac{P}{K_B T}){d^2}}}\)

\(\implies \lambda = \frac{K_B T}{{\sqrt 2 \pi (P){d^2}}}\)

K is constant, T is the temperature which is constant for isothermal Gas.

\( \lambda ∝ \frac{1}{P}\)

Explanation:

Mean free path- The mean free path for gas is defined as the average distance of an object that will move in between collisions and it is directly proportional to the temperature and inversely proportional to the pressure and the diameter of the molecule. It is written as;

\(\lambda = \frac{kT}{{\sqrt 2 n\pi {d^2}}}\)

⇒ \(\lambda \propto \frac{1}{{\sqrt 2 n\pi {d^2}}}\)

Where d is the diameter and n is the molecular density of the gas.

Hence, option 4) is the correct answer.