The mean free path and the average speed of oxygen molecules at 300~K and 1~atm are 3 × 10⁻⁷~m and 600~m/s, respectively. Find the frequency of its collisions.

Solution & Explanation

Related Formula
f = (1)/(T) = vavgλ

where:

  • f = frequency of collisions
  • vavg = average speed of the molecules
  • λ = mean free path
Core Logic

Given parameters:

  • Average speed, vavg = 600~m/s
  • Mean free path, λ = 3 × 10⁻⁷~m
Step 1: Calculate Frequency

Substitute the values into the formula:

f = 6003 × 10⁻⁷ = 2 × 10⁹~s⁻¹

Hence, the collision frequency is 2 × 10⁹/s.

Pattern Recognition

Collision frequency is simply distance covered per unit time (average velocity) divided by the average distance between consecutive collisions (mean free path).

Chapter Mix

Class 11 Physics: Kinetic Theory

Reference Study Guides

Kinetic Theory

More Kinetic Theory Previous-Year Questions — Page 6

Q31 jee_main_2024_31_jan_morning Kinetic Energy Of Gas Molecules
The parameter that remains the same for molecules of all gases at a given temperature is :
  • A. kinetic energy
  • B. momentum
  • C. mass
  • D. speed

Solution

Related Formula
KE = (f)/(2)kT
Core Logic

The average translational kinetic energy of any gas molecule depends only on the absolute temperature of the gas and is independent of the nature or mass of the gas.

For 1 mole of any ideal gas, the average translational kinetic energy is (3)/(2)RT. Therefore, at a given temperature, the kinetic energy parameter is uniform across all ideal gases.

Pattern Recognition

Temperature is directly proportional to average translational kinetic energy. If T is constant, KE is constant for all gases regardless of mass.

Chapter Mix

Class 11 Physics: Kinetic Theory

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