NCERT solution Class 11 kinetic theory

NCERT Solutions

Class 11 Physics

Kinetic Theory

NCERT Questions

Q. 12.1
Q. 12.2
Q. 12.3
Q. 12.4
Q. 12.5
Q. 12.6
Q. 12.7
Q. 12.8
Q. 12.9
Q. 12.10

Q. 12.10

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 0C. 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 N₂ = 28.0 u).

Given:

Mean free path = 1.11 × 10^–7 m

Collision frequency = 4.58 × 10⁹ s^–1

Successive collision time ≈ 500 × (Collision time)

Pressure inside the cylinder containing nitrogen, P = 2.0 atm

= (2.026 × 10⁵) 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=

Where,R is the universal gas constant = 8.314 J mole^–1 K^–1

Therefore v_rms=

=508.26m/s

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

Where, k is the Boltzmann constant = (1.38 × 10^–23) kgm² s^–2K^–1

Therefore l= = 1.11 × 10^–7 m

Collision frequency=

= (4.58 × 10⁹) s^–1

Collision time is given as:

= = 3.93 × 10^–13 s

Time taken between successive collisions:

T’= = m/s = 2.18 × 10^–10 s

= = 500

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

Video solution:

Kinetic Theory Class 11 Physics NCERT Chapter 12 NCERT Solutions

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NCERT solutions of Kinetic Theory Class 11 Physics - all questions of all exercises are solved. Q. No. 1, 2, 3, and so on. Complete explanation and detailed solution makes them the best videos for Class 11 Physics Kinetic Theory NCERT solutions.