Kinetic Theory of Gas: Class 11 Physics Notes, Definition, Mean Free Path Formula & More

Physics Kinetic Theory 2025

Syed Aquib Ur Rahman
Updated on Jun 12, 2025 19:23 IST

By Syed Aquib Ur Rahman, Assistant Manager

You have started studying Kinetic Theory again in Class 11 for your annuals, perhaps, brushing through the molecular nature of matter and the behaviour of gases. You may still feel clear about the kinetic interpretation of macroscopic properties.  But, the equations and derivations in the NCERT Solutions for Kinetic Theory are becoming a bit hazy. And when you shift gears to preparing for JEE Mains, the struggle intensifies more, seeing that mounting pile of study material at your desk. 

You aren’t alone in this. Every student finds it challenging right before exams to visualise atoms and molecules moving in perpetual motion, as well as understanding when and why ideal gases deviate. 

Let’s simplify those complex concepts and derivations in the Kinetic Theory of Gases with this overview-style guide, prepared for a quick brushup before you begin to move on to the nitty-gritty of this chapter and tackle JEE and any other competitive test with confidence. 

Table of content
  • Kinetic Theory of Gases - An Overview
  • Eight (8) Important Assumptions of Kinetic Theory
  • Pressure of an Ideal Gas
  • Ideal Gas Equation for Kinetic Theory
  • Vander Waal's Gas Equation for Kinetic Theory of Non-Ideal Gas
  • Molecular Speeds
  • Kinetic Energy of Gas Molecules
  • Degrees of Freedom
  • Law of Equipartition of Energy
  • Specific Heat of Gases
  • Mean Free Path
  • Common Mistakes in Kinetic Theory
  • Sample Problems
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Kinetic Theory of Gases - An Overview

The kinetic theory of gases (KTG) is a model to help us understand macroscopic properties of gases (pressure, temperature, volume) by showing how molecules behave at the microscopic level. It assumes gases consist of numerous tiny particles in constant random motion, undergoing elastic collisions with each other and container walls. This theory connects molecular motion to observable gas properties. 

Important Backgrounds to the Kinetic Theory

  1. Your previous chapter, Thermal Properties of Matter contains the ideal gas equation section, which you should refer to when trying to draw the connection with Gas Laws by Boyle, Charles, and Lussac. Do practice the NCERT Solutions for Physics Chapter 10 to draw connections on how ideal gas behaviour deviates. 
  2. The Kinetic Theory is applied universally to all gaseous systems where interatomic forces are negligible. Scientists, including Boyle, Newton, Boltzmann, and Maxwell developed this during the 16th-17th centuries, which in turn is a result of centuries-old observations of atoms by Kanada from India and Democritus from Greece - later concretised by Dalton. 
  3. The macroscopic properties of gases, such as pressure, temperature, and volume, are known as thermodynamic variables
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Eight (8) Important Assumptions of Kinetic Theory

The kinetic theory relies on these assumptions for an ideal gas. This section should accompany you with the Kinetic Theory of An Ideal Gas section of your Physics textbook from NCERT. 

We elaborate on the reasons for these assumptions as well, because learning about them helps us draw the link between the microscopic properties of gas molecules and their macroscopic properties, including pressure, temperature, and volume.

  1. Gases consist of many identical molecules, treated as point masses with negligible size 10 - 9 m . In physics, a point mass is a small object that has no measurable size or internal structure.
  2. Molecules move randomly in all directions with a range of velocities. This ensures that gas properties, such as pressure, are uniform with no preferred direction of molecular motion. 

    1. Collisions between molecules and with container walls are perfectly elastic, conserving momentum and kinetic energy. Meaning, the molecules can rebound without any loss of total kinetic energy or momentum.
    2. No intermolecular forces exist except during collisions. This is because gas molecules are considered to move freely and independently between brief, instantaneous interactions.
    3. Molecular volume is negligible compared to the gas volume. For instance, it is 0.014% for oxygen at STP, i.e., standard temperature 273 K and pressure 1 atm.
    4. Collision time is negligible compared to the time between collisions. The reason for this is that the duration of an actual collision between molecules (when they are interacting) is extremely short compared to the much more extended periods they spend travelling freely through space without any interaction.
    5. Molecules follow Newtonian mechanics, moving in straight lines between collisions. This is a simplification of classical physics, or, more accurately, a framework to simplify the complex behaviour of molecules when they aren’t colliding with each other. 
    8. Gravitational forces between molecules are negligible due to their small mass and high speed. Individual gas molecules are too small and move too fast for their weak gravitational attraction to affect their paths or interactions significantly. Learn more about gravity.
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Pressure of an Ideal Gas

We know that pressure results from molecular collisions with container walls. 

From a physics perspective, we need to understand how a macroscopic property, such as pressure, arises from the random motion of individual molecules within a container. 

For a gas with N molecules, each of mass m , in a cube of side L :

  • A molecule with velocity v = v x , v y , v z hits a wall perpendicular to the x-axis, changing momentum by Δ p = - 2 m v x .
  • Time between collisions with the same wall: Δ t = 2 L v x .
  • Collisions per second: n = v x 2 L .
  • Force by one molecule: F = m v x 2 L .
  • Total force: F x = m L v x 2 .
  • Pressure: P = F x A = m V v x 2 .
  • For isotropic motion, v x 2 = v y 2 = v z 2 = 1 3 v 2 , so: P = 1 3 m N V v r m s 2 = 1 3 ρ v r m s 2 where v rms   = v 2 N and ρ = m N V .

Key Implications

  • Pressure v rms   2 .

    Pressure is directly proportional to the mean squared speed of the gas molecules

  • For constant V and T , P m N .

    For a gas kept at a constant volume (V) and temperature (T), the pressure is proportional to the total mass of the gas (mN).

  • For constant m and T , P 1 V

         This is precisely Boyle's Law, which states that for a fixed amount of gas at constant temperature, pressure and volume are inversely related. 

  • For constant m and V , P T .
  • This directly aligns with the ideal gas law, where if N and V are constant, P is directly proportional to T. It’s also Charles’ Law, when extended to constant pressure. 

Physics Kinetic Theory

Try these practice questions

Q1:

What will be the effect on the root mean square velocity of oxygen molecules if the temperagture is doubled and oxygen molecule dissociates into atomic oxygen?

Q2:

A flask contains argon and oxygen in the ratio of 3 : 2 in mass and the mixture is kept at 27°C. The ratio of their average kinetic energy per molecule respectively will be:

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Ideal Gas Equation for Kinetic Theory

The NCERT textbook mentions that the “kinetic theory of an ideal gas is completely consistent with the ideal gas equation and the various gas laws based on it.” 

As you know, an ideal gas obeys

P V = μ R T = N k T where μ is moles, R = 8.31 J / ( m o l K ) , k = R N A = 1.38 × 10 - 23 J / K , and N A = 6.02 × 10 23 .

For density ρ : P = ρ R T M 0 where M 0 is the molar mass.

 

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Vander Waal's Gas Equation for Kinetic Theory of Non-Ideal Gas

So far, we have considered kinetic theory in terms of an ideal gas, which is a theoretical model. In reality, however, there is no ‘ideal gas’, a concept corroborated by your NCERT book as well. That’s why we need to learn about the Van der Waals equation. It is a modification of the ideal gas equation, which takes into account the intermolecular forces between gas molecules and the finite volume of the gas molecules themselves. 

Real gases deviate from ideal behaviour due to molecular size and forces,

P + a μ 2 V 2 ( V - μ b ) = μ R T where a corrects for attractions, b for molecular volume.

 

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Molecular Speeds

Molecules have a distribution of speeds, where macroscopic properties, such as temperature and pressure, play a crucial role. In kinetic theory, the molecules of a gas move at various speeds when looked at from the perspective of Newtonian mechanics. This distribution of gas is described by the Maxwell-Boltzmann distribution.

The main macroscopic properties—temperature and pressure—are directly related to this distribution, and it applies to real gases. Temperature is a measure of the average kinetic energy of the molecules, and pressure results from collisions of these molecules with the walls of the container. 

Root Mean Square Speed

This is the square root of the average of the squares of the molecular speeds.

v r m s = 3 R T M = 3 k T m

Here, R is the universal gas constant, k is Boltzmann’s constant, T is the absolute temperature, M is the molar mass, and m is the molecular mass.

Most Probable Speed

v m p = 2 R T M = 2 k T m

This is the speed at which the majority of molecules are moving. It shows the peak of the Maxwell-Boltzmann distribution. 

Average Speed

v a v = 8 R T π M = 8 k T π m

This is the arithmetic mean of the speeds of all molecules. 

Now, all three of the above depend on temperature and molecular mass. But, they are not equal. Their ratio reflects the different ways in which speed is averaged. 

Relation: v rms   : v av   : v mp   = 3 : 8 π : 2 1.224 : 1.128 : 1 .

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Kinetic Energy of Gas Molecules

So far, we have seen how molecular speeds are distributed in a gas. But how does such motion convert into kinetic energy, especially when we have to quantify it?

We have taken into account how temperature is linked to the movement of molecules at measurable speeds. Now we need to determine how much kinetic energy they possess. 

Average translational kinetic energy per molecule E = 3 2 k T

E is the average kinetic energy per molecule

k is the Boltzmann constant

T is the absolute temperature

For one mole or Avogadro’s number of molecules

E = 3 2 R T

Here, R is the Universal Gas Constant.

This energy also connects to pressure through the kinetic theory equation.

P V = 2 3 E

 

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Degrees of Freedom

Degrees of freedom ( f ) determine energy storage.

  • Monoatomic (e.g., He): f = 3 (translational).
  • Diatomic (e.g., O 2 ): f = 5 ( 3 translational +2 rotational).
  • Triatomic non-linear (e.g., H 2 O ): f = 6 ( 3 translational +3 rotational).

General: f = 3 A - B , where A is atoms, B is restrictions.

Use this table to recall the concepts quickly. 

Molecule Type

Degrees of Freedom (f)

Breakdown

Why & How

Monoatomic

3

3 translational

Only moves in x, y, z directions; no rotation or vibration due to spherical shape.

Diatomic

5

3 translational + 2 rotational

Rotates about 2 perpendicular axes; rotation along bond axis is negligible.

Triatomic (non-linear)

6

3 translational + 3 rotational

Asymmetrical shape allows full 3D rotation along all axes.

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Law of Equipartition of Energy

You observed how the degrees of freedom represent how molecules move. We also need to know how the energy is shared during such motions. This is where we need to learn about the Law of Equipartition of Energy. 

Maxwell and other pioneers in statistical mechanics established this principle. 

Simply put, the Law of Equipartition of Energy states that energy is distributed equally among all available energy modes in thermal equilibrium

The condition for applying this law is only when the system reaches thermal equilibrium at a defined temperature T. 

This is valid only for translational, rotational, and vibrational energy modes, which are covered in the above section on Degrees of Freedom. 

Energy is equally distributed among degrees of freedom, each contributing 1 2 k T per molecule. For one mole: E = f 2 R T

This equation explains how thermal energy partitions among different molecular motions. 

 

Each degree of freedom contributes ½kT to total energy, with vibrational modes contributing kT due to both kinetic and potential components.

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Specific Heat of Gases

Specific heat tells us how much energy is needed to raise the temperature of a gas. This is something we need to learn after we understand how the Law of Equipartition of Energy works. 

You may also review specific heat capacity for a quick refresher on older but related concepts in Class 11 Physics. 

Here are the formulae for the specific heat of ideal gases at constant volume and constant pressure.

Specific heats C v = ( Δ Q ) v μ Δ T , C p = ( Δ Q ) p μ Δ T

At constant volume, there is no work done - a concept you may recall when you are practising the NCERT Solutions for Chapter 5.

However, at constant pressure, the gas still performs some work. So, as the gas expands while also increasing internal energy, the total heat required is higher. That’s where we have to use Mayer’s Relation. 

Mayer's formula: C p - C v = R .

So, when we are using the Law of Equipartition of Energy, we get

C v = f 2 R , C p = f 2 + 1 R , γ = C p C v = 1 + 2 f

 

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Mean Free Path

After understanding how gas molecules carry and share energy, we can also examine how frequently they collide.

Mean Free Path is the average distance a molecule travels between collisions.

λ = 1 2 π n d 2 = k T 2 π d 2 P

where n is number density, d is molecular diameter. Key: λ 1 ρ , and at constant V , λ is independent of P and T .

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Common Mistakes in Kinetic Theory

Have a quick look at some misconceptions students usually have in this chapter, which you can easily avoid. 

  1. Students often mix up v rms   , v av   , and v mp   , assuming they are equal. Remember their definitions and ratio.
  2. Incorrectly assigning f (e.g., assuming diatomic gases have f = 3 ). f = 5 Use for diatomic at room temperature.
  3. Applying ideal gas equations to real gases at high pressure or low temperature, where deviations occur.
  4. Assuming λ varies with T at constant V . At constant V , λ is independent of T and P .
  5. Assuming kinetic energy depends on pressure or volume, whereas it depends only on T for ideal gases. 
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Sample Problems

If you have come this far, test your knowledge on kinetic theory with these problems with answers. 

Example 1. Calculate the pressure of hydrogen gas with v r m s = 1840 m / s and density 8.99 × 10 - 2 k g / m 3 .

Ans: 1.02 × 10 5 N / m 2 )

Example 2. A gas at 27 C has volume V and pressure P . If pressure doubles and volume triples, find the final temperature.

Ans: 1200 K 

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