Class 11 Physics Kinetic Theory | Mean free path |

**Mean free path**

- Mean free path is the average distance between the two successive collisions.
- Inside the gas there are several molecules which are randomly moving and colliding with each other.
- The distance which a particular gas molecule travels without colliding is known as mean free path.

__Expression for mean free path__

- Consider each molecule of gas is a sphere of diameter (d).The average speed of each molecule is<v>.
- Suppose the molecule suffers collision with any other molecule within the distance (d). Any molecule which comes within the distance range of its diameter this molecule will have collision with that molecule.
- The volume within which a molecule suffer collision =<v>Δtπd
^{2}. - Let number of molecules per unit volume =n
- Therefore the total number of collisions in time Δt =<v>Δtπd
^{2}xn - Rate of collision =<v>Δtπd
^{2}xn/Δt=<v>πd^{2}n - Suppose time between collision τ =1/<v>πd
^{2}n - Average distance between collision = τ<v> = 1/πd
^{2} - 1/πd
^{2}n this value was modified and a factor was introduced. - Mean free path(l) =
**1/√2 π d**^{2}n

**Conclusion**: - Mean free path depends inversely on:

- Number density (number of molecules per unit volume)
- Size of the molecule.

The volume swept by a molecule in time Δtin which any molecule will collide with it.

**Problem:- **Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N

**Answer:- **Mean free path = 1.11 × 10^{–7} m

Collision frequency = 4.58 × 10^{9} s^{–1}

Successive collision time ≈ 500 × (Collision time)

Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2.026 × 10^{5} Pa

Temperature inside the cylinder, T = 17°C =290 K

Radius of a nitrogen molecule, r = 1.0 Å = 1 × 1010 m

Diameter, d = 2 × 1 × 1010 = 2 × 1010 m

Molecular mass of nitrogen, M = 28.0 g = 28 × 10^{–3} kg

The root mean square speed of nitrogen is given by the relation:

v_{rms}=√3RT/M

Where,

R is the universal gas constant = 8.314 J mole^{–1} K^{–1}

Therefore v_{rms}=√ (3x8.314x290)/ (28x10^{-3})

=508.26m/s

The mean free path (l) is given by the relation:

l= KT/√2xd^{2}xP

Where, k is the Boltzmann constant = 1.38 × 10^{–23} kgm^{2} s^{–2}K^{–1}

Therefore l= 1.38x10^{-23}x290/√2x 3.14x (2x10^{-10})^{2}x2.026x10^{5}

= 1.11 × 10^{–7} m

Collision frequency=v_{rms}/l

=508.26/1.11x10^{-7}

= 4.58 × 10^{9} s^{–1}

Collision time is given as:

T=d/v_{rms}

= 2x10^{-10}/508.26 = 3.93 × 10^{–13} s

Time taken between successive collisions:

T’=l/v_{rms} = 1.11x10^{-7}m/508.26m/s= 2.18 × 10^{–10} s

T’/T= 2.18 × 10^{–10} s/3.93x10^{-13} = 500

Hence, the time taken between successive collisions is 500 times the time taken for a collision.

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