Test: Kinetic theory- Mean free path (27 Sep) - JEE MCQ

The expression for the mean free path is given by:

Here, the area of the cross-section is πd².

The volume of the cylindrical path is given by πd² x vt, where t is the time, and v is the molecule’s velocity and the number of molecules per unit volume as (N/V).

Thus we can say that;

λ ∝ 1/d² or d^-2

∴ k = -2

Given : 
T = 27^oC = 27 + 273.15 = 300.15 K
P = 1.01 x 10⁵ Pa
d = 0.3 nm = 0.3 x 10^-9 m 
k = 1.38 x 10^-23 JK^-1
We know that PV = NkT then after rearranging we get,

Mean free path for oxygen molecules

Given:

From the ideal gas equation

PV = nRT

where 'n' is the number of the moles

Here nNA is the number of molecules in n moles.

Here  is the number of molecules per unit volume of the gas and  is the Boltzmann  constant.

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

∴ 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)

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)

From the ideal gas equation -

PV = nRT

Where n is the number of the moles

Here nN_A is the number of molecules in n moles.

Here,  is the number of the molecules per unit volume of the gas and  = k, the Boltzmann constant

Here we can see that,
λ ∝ T and λ ∝ (1/P)
∴ On increasing T and P two times, the net effect is zero on λ.
Hence λ₂ = λ₁
⇒ λ₁:λ₂ = 1 : 1

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

 

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 πd²vΔt x n
• The number of collisions per sec is
  
• The average time between two successive collisions is 
  

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  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 

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.

The mean free path of a gas molecule is given by

 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.

Mean Free Path (λ):

• The distance traveled by a gas molecule between two successive collisions is known as a free path.
  
• 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
  

n is the number of molecules per unit volume.

Here K_B is Boltzman Constant, P is pressure, T is temperature
Now if we put (2) in (1) we get

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