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Class 11 Physics · NCERT Chapter 12

Kinetic Theory Class 11 Physics Notes

Complete, exam-ready notes on the kinetic theory of gases: the assumptions and model, the ideal gas equation and its laws, the pressure exerted by an ideal gas, the kinetic interpretation of temperature, root-mean-square speed, degrees of freedom, the law of equipartition of energy and specific heats of gases, and the Maxwell speed distribution — every NCERT topic with MCQs, mark-wise questions and solved numericals for CBSE, JEE and NEET.

ChapterKinetic TheoryClassClass 11SubjectPhysicsBoardCBSEExamsCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is the kinetic interpretation of temperature?

The average kinetic energy of a gas molecule is directly proportional to its absolute temperature: (½mv²)_avg = (3/2)k_B T. Temperature is thus a measure of the average translational kinetic energy of the molecules — at higher temperature molecules move faster on average.

Ideal Gas and Its Assumptions

Kinetic theory assumptions

The kinetic theory models a gas as a large number of identical molecules. The assumptions are: (1) molecules are in random motion, (2) the size of molecules is negligible compared with the distance between them, (3) collisions are perfectly elastic and the time of collision is negligible, and (4) the only interactions occur during collisions (no intermolecular forces otherwise).

Ideal gas

An ideal gas obeys the ideal gas equation PV = nRT exactly. Its molecules occupy no volume and exert no intermolecular forces, so its internal energy depends only on temperature. Real gases approach ideal behaviour at low pressure and high temperature.

PV=nRT=NNART=NkBTPV = nRT = \frac{N}{N_A}RT = N k_B T
Ideal gas equation

The ideal gas equation combines Boyle's law (P ∝ 1/V at constant T), Charles' law (V ∝ T at constant P) and Avogadro's law (V ∝ n at constant P and T). R = 8.31 J mol⁻¹ K⁻¹ is the universal gas constant, and k_B = R/N_A ≈ 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant.

  • Boyle's law: PV = constant at constant temperature.
  • Charles' law: V/T = constant at constant pressure.
  • Avogadro's law: equal volumes of gases at the same P and T contain equal numbers of molecules.

Real gases become ideal

At low pressure molecules are far apart and their own volume and mutual attractions become negligible, so real gases obey PV = nRT closely. At high pressure or low temperature they deviate.

MCQ Questions

1In the kinetic theory model, collisions between gas molecules are —

2The ideal gas equation is —

3Boyle's law states that at constant temperature, —

4A real gas behaves most like an ideal gas at —

1 Mark Questions

  1. State the main assumptions of the kinetic theory of gases.
  2. Write the ideal gas equation and define each symbol.
  3. State Boyle's law.

2 Mark Questions

  1. What is an ideal gas? Under what conditions does a real gas approach ideal behaviour?
  2. State Boyle's law, Charles' law and Avogadro's law and show how they combine into the ideal gas equation.
  3. Find the number of moles in 44 g of CO₂. (Molar mass = 44 g mol⁻¹.)

3 Mark Questions

  1. State the postulates of the kinetic theory of gases and explain why the ideal gas equation holds at low pressure and high temperature.
  2. A gas occupies 2 m³ at a pressure of 10⁵ Pa and a temperature of 300 K. Find the number of moles and the number of molecules. (R = 8.31 J mol⁻¹ K⁻¹, N_A = 6.02 × 10²³.)

Pressure Exerted by an Ideal Gas

Pressure of a gas

The pressure of a gas arises from the collisions of its molecules with the walls of the container. Each collision transfers momentum 2mv to a wall (for a molecule moving perpendicular to it), and the rate of momentum transfer per unit area is the pressure.

P=13NmVv2‾=13ρv2‾P = \frac{1}{3}\frac{Nm}{V}\overline{v^2} = \frac{1}{3}\rho \overline{v^2}
Pressure of an ideal gas in terms of molecular speed

Using the average of the square of the molecular speed v², the pressure is P = (1/3)ρv²_avg, where ρ = Nm/V is the density. This derivation connects the macroscopic quantity pressure with the microscopic motion of the molecules.

  • Pressure depends on the mean-square speed of the molecules, not their individual speeds.
  • The same formula holds in three dimensions because each of the three velocity components contributes equally on average.
  • Doubling the mean-square speed doubles the pressure at constant volume.

P = ⅓ρv² is rigorous for an ideal gas

This exact result for an ideal gas follows purely from momentum transfer during elastic collisions, without any assumption about the distribution of molecular speeds.

MCQ Questions

1The pressure of an ideal gas is directly proportional to its —

2The pressure of a gas arises from —

3For an ideal gas, P = —

4If the mean-square speed of gas molecules is doubled at constant volume, the pressure —

1 Mark Questions

  1. Explain how pressure arises in the kinetic theory of gases.
  2. Write the expression P = (1/3)ρv²_avg and identify the symbols.
  3. On which molecular quantity does gas pressure depend?

2 Mark Questions

  1. Derive the relation P = (1/3)ρv²_avg from the kinetic theory model.
  2. The density of a gas is 1.5 kg m⁻³ and its molecules have a mean-square speed of 10⁶ m² s⁻². Find the pressure it exerts.
  3. Show that the pressure of a gas is directly proportional to its density at constant mean-square speed.

3 Mark Questions

  1. Use the kinetic theory to derive an expression for the pressure exerted by an ideal gas in terms of the mass, number density and mean-square speed of the molecules.
  2. A gas has a density of 2 kg m⁻³ and exerts a pressure of 2 × 10⁵ Pa. Find the root-mean-square speed of its molecules.

Kinetic Interpretation of Temperature

Temperature and molecular energy

Combining the ideal gas equation with P = (1/3)ρv²_avg shows that the average translational kinetic energy per molecule is (1/2)mv²_avg = (3/2)k_B T. Thus the temperature of a gas is a direct measure of the average kinetic energy of its molecules.

12mv2‾=32kBT,KEavg=32NkBT=32nRT\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_B T, \qquad \text{KE}_{\text{avg}} = \frac{3}{2}N k_B T = \frac{3}{2}nRT
Kinetic interpretation of temperature

At a given temperature all gases, regardless of molecular mass, have the same average translational kinetic energy per molecule (3/2)k_B T. Lighter molecules move faster to have the same energy — at the same temperature, hydrogen molecules are much faster than oxygen molecules.

  • Temperature is proportional to the mean translational kinetic energy, not the total energy.
  • At absolute zero all translational motion ceases in the classical model.
  • Two gases at the same temperature have equal average translational KE per molecule.

Same T, same average KE

At the same temperature, a light gas (H₂) and a heavy gas (O₂) have the same average translational kinetic energy per molecule — but the light molecules move faster.

MCQ Questions

1The average translational kinetic energy of a gas molecule is —

2Two different gases at the same temperature have —

3The temperature of a gas is a measure of the —

4At a given temperature, which gas molecules move fastest on average? —

1 Mark Questions

  1. State the kinetic interpretation of temperature.
  2. Write the expression for the average kinetic energy of a gas molecule.
  3. Two gases at the same temperature: what is equal for their molecules?

2 Mark Questions

  1. Show that the average kinetic energy per molecule of a gas equals (3/2)k_B T.
  2. Find the average translational kinetic energy of a molecule at 300 K. (k_B = 1.38 × 10⁻²³ J K⁻¹.)
  3. Why is the average translational kinetic energy of all gases the same at a given temperature?

3 Mark Questions

  1. Derive the kinetic interpretation of temperature, starting from P = (1/3)ρv²_avg and the ideal gas equation.
  2. Compute the total translational kinetic energy of 2 moles of an ideal gas at 400 K. (R = 8.31 J mol⁻¹ K⁻¹.)

Root-Mean-Square Speed

RMS speed

The root-mean-square speed is the square root of the mean of the squares of the molecular speeds: v_rms = √(v²_avg). From the kinetic interpretation, v_rms = √(3k_B T/m) = √(3RT/M), where M is the molar mass.

vrms=3kBTm=3RTMv_{\text{rms}} = \sqrt{\frac{3k_B T}{m}} = \sqrt{\frac{3RT}{M}}
Root-mean-square speed of gas molecules

The RMS speed is proportional to the square root of temperature and inversely proportional to the square root of molar mass. At 300 K, the RMS speed of nitrogen molecules (M = 28 g/mol) is about 517 m/s. For a given gas, raising the temperature increases the RMS speed, and for a given temperature, lighter gases have higher RMS speeds.

  • v_rms ∝ √T at fixed mass.
  • v_rms ∝ 1/√M at fixed temperature.
  • The RMS speed is slightly larger than the mean speed and the most probable speed.

Remember the ordering of speeds

For a gas, v_rms > v_mean > v_mp (most probable). The RMS value emphasises the faster molecules because it averages the squares.

MCQ Questions

1The RMS speed of gas molecules is given by —

2The RMS speed of gas molecules is proportional to —

3At a given temperature, a lighter gas has —

4For an ideal gas, the ordering of the three speeds is —

1 Mark Questions

  1. Define the root-mean-square speed of gas molecules.
  2. Write the expression for v_rms in terms of the molar mass M.
  3. How does v_rms depend on temperature?

2 Mark Questions

  1. Find the RMS speed of oxygen molecules at 300 K. (M = 32 × 10⁻³ kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹.)
  2. Compare the RMS speeds of hydrogen and oxygen at the same temperature.
  3. If the temperature of a gas is raised from 300 K to 1200 K, by what factor does its RMS speed increase?

3 Mark Questions

  1. Derive an expression for the root-mean-square speed of gas molecules in terms of T and molar mass.
  2. At what temperature will the RMS speed of oxygen molecules equal that at which hydrogen has an RMS speed of 2000 m/s at 300 K? Derive your answer.

Degrees of Freedom and Specific Heats

Degrees of freedom

The number of degrees of freedom of a gas molecule is the number of independent ways it can possess energy. A monatomic molecule has 3 translational degrees of freedom. A diatomic molecule has 3 translational, 2 rotational and (at high temperature) 2 vibrational degrees of freedom.

Law of equipartition of energy

The law of equipartition of energy states that in thermal equilibrium, energy is shared equally among all degrees of freedom, each contributing (1/2)k_B T per molecule (or (1/2)RT per mole). Hence a monatomic gas has internal energy (3/2)nRT and C_V = (3/2)R.

energy per degree=12kBT,CV=f2R,CP=(f2+1)R\text{energy per degree} = \frac{1}{2}k_B T, \qquad C_V = \frac{f}{2}R, \qquad C_P = \left(\frac{f}{2}+1\right)R
Equipartition energy and specific heats in terms of f

For a monatomic gas C_V = 3R/2 and γ = 5/3. For a diatomic gas (translational + rotational) C_V = 5R/2 and γ = 7/5. At very high temperatures the vibrational degrees are also excited and C_V rises.

  • Monatomic (He, Ar): f = 3, C_V = 3R/2, γ = 5/3.
  • Diatomic (N₂, O₂) without vibration: f = 5, C_V = 5R/2, γ = 7/5.
  • Each degree of freedom contributes (1/2)R per mole to C_V.

Which degrees are active?

At ordinary temperatures only the translational and rotational degrees of a diatomic gas are fully active; vibrational degrees contribute at high temperatures. This is why γ is 7/5 at room temperature rather than a smaller value.

MCQ Questions

1The number of translational degrees of freedom of a monatomic gas is —

2According to the equipartition theorem, energy per degree of freedom per molecule is —

3For a monatomic gas, C_V is —

4For a diatomic gas excluding vibration, γ = C_P/C_V is —

1 Mark Questions

  1. Define the degrees of freedom of a gas molecule.
  2. State the law of equipartition of energy.
  3. Write the value of C_V for a monatomic gas.

2 Mark Questions

  1. Find the number of degrees of freedom of (a) monatomic and (b) diatomic gas molecules.
  2. State the law of equipartition of energy and hence write C_V and C_P for a monatomic gas.
  3. For a diatomic gas, using f = 5, find C_V, C_P and γ.

3 Mark Questions

  1. State the law of equipartition of energy and use it to obtain the specific heats and γ for a monatomic and a diatomic gas.
  2. Compute the internal energy of 3 moles of a monatomic gas at 400 K. (R = 8.31 J mol⁻¹ K⁻¹.)

The Maxwell Speed Distribution

Maxwell distribution

In a gas, molecules do not all move at one speed; their speeds follow the Maxwell speed distribution. The fraction of molecules with speed between v and v + dv depends on temperature and molecular mass. The distribution has a maximum at the most probable speed v_mp = √(2k_B T/m).

vmp=2kBTm,vmean=8kBTπm,vrms=3kBTmv_{\text{mp}} = \sqrt{\frac{2k_B T}{m}}, \qquad v_{\text{mean}} = \sqrt{\frac{8k_B T}{\pi m}}, \qquad v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}
Most probable, mean and RMS speeds

The three speeds are related by v_mp : v_mean : v_rms = √2 : √(8/π) : √3 ≈ 1 : 1.128 : 1.225. As temperature rises, the distribution broadens and shifts to higher speeds, and the most probable, mean and RMS speeds all increase.

  • The most probable speed is the speed possessed by the largest number of molecules.
  • The distribution spreads out with increasing temperature.
  • At the same temperature, lighter gases have their distribution peak at higher speeds.

The three characteristic speeds

v_mp = √(2kT/m), v_mean = √(8kT/πm), v_rms = √(3kT/m). They are all proportional to √(T/m) and differ by small numerical factors.

MCQ Questions

1The most probable speed of gas molecules is —

2As the temperature of a gas rises, its Maxwell speed distribution —

3For a gas, the correct order of the three speeds is —

4The Maxwell distribution describes —

1 Mark Questions

  1. What is the Maxwell speed distribution?
  2. Define the most probable speed.
  3. Write the expressions for the three characteristic speeds of a gas.

2 Mark Questions

  1. Find the most probable speed of nitrogen molecules at 300 K. (M = 28 × 10⁻³ kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹.)
  2. Explain why the Maxwell distribution broadens as the temperature increases.
  3. State the ordering v_mp < v_mean < v_rms and explain it briefly.

3 Mark Questions

  1. Describe the Maxwell speed distribution and explain the physical meaning of the most probable, mean and RMS speeds.
  2. Compute the most probable, mean and RMS speeds of argon atoms at 300 K. (M = 40 × 10⁻³ kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹.)

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Ideal gas equation

PV=nRT=NkBTPV = nRT = Nk_B T

Pressure of a gas

P=13NmVv2‾=13ρv2‾P = \frac{1}{3}\frac{Nm}{V}\overline{v^2} = \frac{1}{3}\rho \overline{v^2}

Kinetic interpretation

12mv2‾=32kBT\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_B T

RMS speed

vrms=3kBTm=3RTMv_{\text{rms}} = \sqrt{\frac{3k_B T}{m}} = \sqrt{\frac{3RT}{M}}

Equipartition

CV=f2R,CP=(f2+1)RC_V = \frac{f}{2}R, \quad C_P = \left(\frac{f}{2}+1\right)R

Most probable speed

vmp=2kBTmv_{\text{mp}} = \sqrt{\frac{2k_B T}{m}}

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • PV = nRT; k_B = R/N_A ≈ 1.38 × 10⁻²³ J K⁻¹.
  • P = (1/3)ρv²_avg: pressure from molecular collisions.
  • Average translational KE per molecule = (3/2)k_B T.
  • v_rms = √(3RT/M); v_rms ∝ √T, ∝ 1/√M.
  • v_mp : v_mean : v_rms ≈ 1 : 1.13 : 1.22.
  • Equipartition: ½k_B T per degree of freedom.
  • Monatomic C_V = 3R/2, γ = 5/3; diatomic C_V = 5R/2, γ = 7/5.

Solved problems

Worked examples

JEE / NEET-style numericals, solved step by step.

1

RMS speed of oxygen

Find the RMS speed of oxygen molecules at 300 K. (M = 32 × 10⁻³ kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹.)

  1. v_rms = √(3RT/M) = √(3 × 8.31 × 300 / 0.032).
  2. v_rms = √(7479 / 0.032) = √233718 ≈ 483 m/s.

Answer

≈ 483 m/s

2

Average kinetic energy at a temperature

Find the average translational kinetic energy of a gas molecule at 300 K. (k_B = 1.38 × 10⁻²³ J K⁻¹.)

  1. KE = (3/2)k_B T = 1.5 × 1.38 × 10⁻²³ × 300.
  2. KE = 6.21 × 10⁻²¹ J.

Answer

6.21 × 10⁻²¹ J

3

Pressure from molecular speed

A gas has density 2 kg m⁻³ and exerts pressure 3 × 10⁵ Pa. Find the RMS speed of its molecules.

  1. P = (1/3)ρv²_rms, so v²_rms = 3P/ρ.
  2. v²_rms = 3 × 3 × 10⁵ / 2 = 4.5 × 10⁵.
  3. v_rms = √(4.5 × 10⁵) ≈ 671 m/s.

Answer

≈ 671 m/s

FAQ

Common questions

What does the kinetic theory assume about gas molecules?

It assumes a large number of identical molecules in random motion, negligible molecular size compared with intermolecular distances, perfectly elastic collisions of negligible duration, and no intermolecular forces except during collisions.

What is the kinetic interpretation of temperature?

The average translational kinetic energy per molecule equals (3/2)k_B T, so temperature is a direct measure of the average kinetic energy of the molecules. At a fixed temperature, all gases have the same average translational KE per molecule.

Why do lighter gases move faster than heavier ones at the same temperature?

Since average kinetic energy (3/2)k_B T is the same for all gases at a given temperature, lighter molecules must move faster to have the same energy: v_rms = √(3RT/M) is inversely proportional to √M.

What is the law of equipartition of energy?

In thermal equilibrium, energy is shared equally among all degrees of freedom, each contributing (1/2)k_B T per molecule. This gives a monatomic gas C_V = 3R/2 and a diatomic gas C_V = 5R/2 at ordinary temperatures.

Why do molecules in a gas have many different speeds?

Collisions continually redistribute energy among the molecules, so their speeds are distributed rather than uniform. The Maxwell distribution describes this, with a most probable, a mean and an RMS speed, all of which shift higher as temperature rises.

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