NCERT Solutions for Class 11 Physics Chapter 12: Kinetic Theory

These Class 11 Physics Chapter 12 solutions cover Kinetic Theory with every NCERT exercise solved step by step, units carried throughout and final answers cross-checked against the official key. The kinetic theory explains the behaviour of gases by treating them as a large number of rapidly moving molecules, and links measurable quantities like pressure and temperature to molecular motion. Updated for session 2026–27.

Class: 11 Subject: Physics Chapter: 12 Title: Kinetic Theory Exercises: 12.1–12.10 Session: 2026–27
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Class 11 Physics Chapter 12 – Overview

Chapter 12, Kinetic Theory, builds a molecular picture of gases. Starting from the molecular nature of matter, it explains the behaviour of gases through the ideal-gas equation PV = μRT and derives an expression for the pressure of an ideal gas, P = (1/3) n m v̅². This leads to the kinetic interpretation of temperature: the average translational kinetic energy of a molecule is (3/2)kBT, independent of the nature of the gas. The chapter then introduces the law of equipartition of energy, uses it to predict the molar specific heats of monatomic, diatomic and polyatomic gases, and finishes with the idea of mean free path — the average distance a molecule travels between collisions. These ideas connect microscopic molecular parameters to macroscopic, measurable properties.

Key Concepts & Definitions

Ideal gas: a gas that obeys PV = μRT exactly at all pressures and temperatures. Real gases approach ideal behaviour at low pressure and high temperature.

Avogadro’s law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. NA = 6.022 × 1023 mol−1.

Dalton’s law of partial pressures: the total pressure of a mixture of non-reactive ideal gases is the sum of the partial pressures of the constituents.

RMS speed: the square root of the mean of the squared molecular speeds; at the same temperature, lighter molecules have a larger rms speed.

Degrees of freedom: the number of independent ways a molecule can store energy — 3 translational for monatomic gases; 3 translational + 2 rotational for rigid diatomic gases.

Law of equipartition of energy: in thermal equilibrium, energy is shared equally among all energy modes, each quadratic term contributing ½kBT per molecule (a vibrational mode contributes kBT, having kinetic + potential parts).

Mean free path: the average distance a molecule travels between two successive collisions.

Important Formulas

Ideal gas equation: PV = μRT = kBNT, with R = 8.314 J mol−1 K−1, kB = R/NA = 1.38 × 10−23 J K−1

Pressure (kinetic theory): P = (1/3) n m v̅², where n = number density

Kinetic interpretation of temperature: ½ m v̅² = (3/2) kBT

RMS speed: vrms = √(v̅²) = √(3kBT/m) = √(3RT/M0)

Average thermal energy per molecule: E̅ = (3/2) kBT

Molar specific heats: monatomic Cv = (3/2)R, Cp = (5/2)R; rigid diatomic Cv = (5/2)R, Cp = (7/2)R; always Cp − Cv = R

Mean free path: l = 1 / (√2 · nπd²), where d = molecular diameter

NCERT Exercises — Full Solutions

Questions are reproduced verbatim from the NCERT textbook (Reprint 2026–27). Answers are original, worked step by step, with units, and verified against the NCERT key.

12.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.

SOLUTION Given: diameter d = 3 Å = 3 × 10−10 m, so radius r = 1.5 × 10−10 m. At STP, 1 mole of gas occupies V = 22.4 litres = 22.4 × 10−3 m³ and contains NA = 6.022 × 1023 molecules. Volume of one molecule (treated as a sphere) = (4/3)πr³ = (4/3)(3.14)(1.5 × 10−10)³ = 1.41 × 10−29 m³. Total molecular volume in 1 mole = NA × (4/3)πr³ = 6.022 × 1023 × 1.41 × 10−29 = 8.5 × 10−6 m³. Fraction = (8.5 × 10−6) / (22.4 × 10−3) = 3.8 × 10−4 ≈ 4 × 10−4. This tiny fraction shows that the actual space occupied by the molecules of a gas is negligibly small compared with the volume of the gas — the molecules are far apart. (Matches NCERT answer.)

12.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.

SOLUTION Given (STP): P = 1 atm = 1.013 × 105 Pa, T = 273.15 K, μ = 1 mol, R = 8.314 J mol−1 K−1. From the ideal gas equation PV = μRT, the molar volume is V = μRT / P. V = (1 × 8.314 × 273.15) / (1.013 × 105) = 2271 / (1.013 × 105) = 2.24 × 10−2 m³. Converting: 2.24 × 10−2 m³ × 1000 L/m³ = 22.4 litres. Hence the molar volume of an ideal gas at STP is 22.4 L. (Proved.)

12.3 Figure 12.8 shows plot of PV/T versus P for 1.00 × 10−3 kg of oxygen gas at two different temperatures. (a) What does the dotted plot signify? (b) Which is true: T1 > T2 or T1 < T2? (c) What is the value of PV/T where the curves meet on the y-axis? (d) If we obtained similar plots for 1.00 × 10−3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot)? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mol−1 K−1.)

SOLUTION (a) The dotted line is parallel to the P-axis (PV/T = constant = μR, independent of P). It signifies ideal-gas behaviour, for which PV/T stays constant at all pressures. (b) The real-gas curve closer to the dotted (ideal) line corresponds to the higher temperature, since a real gas behaves more like an ideal gas at higher temperature. As the curve for T1 is nearer the ideal line, T1 > T2. (c) Where the curves meet the y-axis (P → 0), the gas is ideal and PV/T = μR. For oxygen, μ = mass/M0 = (1.00 × 10−3 kg) / (32.0 × 10−3 kg mol−1) = 3.125 × 10−2 mol. PV/T = μR = 3.125 × 10−2 × 8.31 = 0.26 J K−1. (d) No. PV/T = μR depends on the number of moles μ, which depends on the mass. For 1.00 × 10−3 kg of H2, μ would be different, so PV/T would differ. To get the same PV/T = 0.26 J K−1, we need the same number of moles, μ = 3.125 × 10−2 mol. Required mass of H2 = μ × M0 = 3.125 × 10−2 × 2.02 × 10−3 kg = 6.3 × 10−5 kg. So 6.3 × 10−5 kg of H2 yields the same value. (Matches NCERT answer.)

12.4 An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol−1 K−1, molecular mass of O2 = 32 u).

SOLUTION Initial state: V1 = 30 L = 30 × 10−3 m³, P1 = 15 atm = 15 × 1.013 × 105 = 1.52 × 106 Pa, T1 = 27 + 273 = 300 K. Number of moles initially: μ1 = P1V1 / (RT1) = (1.52 × 106 × 30 × 10−3) / (8.31 × 300) = 45600 / 2493 = 18.3 mol. Final state: V2 = 30 × 10−3 m³, P2 = 11 atm = 1.114 × 106 Pa, T2 = 17 + 273 = 290 K. Number of moles finally: μ2 = P2V2 / (RT2) = (1.114 × 106 × 30 × 10−3) / (8.31 × 290) = 33430 / 2410 = 13.9 mol. Moles withdrawn = μ1 − μ2 = 18.3 − 13.9 = 4.4 mol. Mass withdrawn = (μ1 − μ2) × M0 = 4.4 × 32 = 141 g = 0.14 kg. (Matches NCERT answer.)

12.5 An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?

SOLUTION At the bottom: V1 = 1.0 cm³ = 1.0 × 10−6 m³, T1 = 12 + 273 = 285 K. Pressure = atmospheric + water column = P0 + hρg. P1 = 1.013 × 105 + (40 × 1000 × 9.8) = 1.013 × 105 + 3.92 × 105 = 4.93 × 105 Pa. At the surface: P2 = P0 = 1.013 × 105 Pa, T2 = 35 + 273 = 308 K. Using (P1V1)/T1 = (P2V2)/T2, so V2 = (P1V1T2) / (P2T1). V2 = (4.93 × 105 × 1.0 × 10−6 × 308) / (1.013 × 105 × 285) = (0.1518) / (2.887 × 107) = 5.3 × 10−6 m³ (= 5.3 cm³). The bubble grows to about 5.3 times its original volume as it rises. (Matches NCERT answer.)

12.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27 °C and 1 atm pressure.

SOLUTION Given: V = 25.0 m³, T = 27 + 273 = 300 K, P = 1 atm = 1.013 × 105 Pa, kB = 1.38 × 10−23 J K−1. Using PV = kBNT, the number of molecules N = PV / (kBT). N = (1.013 × 105 × 25.0) / (1.38 × 10−23 × 300) = (2.53 × 106) / (4.14 × 10−21). N = 6.10 × 1026 molecules. (Matches NCERT answer.)

12.7 Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).

SOLUTION Average thermal energy of an atom = (3/2) kBT, with kB = 1.38 × 10−23 J K−1. (i) T = 27 + 273 = 300 K. E̅ = (3/2)(1.38 × 10−23)(300) = 6.21 × 10−21 J ≈ 6.2 × 10−21 J. (ii) T = 6000 K. E̅ = (3/2)(1.38 × 10−23)(6000) = 1.24 × 10−19 J. (iii) T = 107 K. E̅ = (3/2)(1.38 × 10−23)(107) = 2.07 × 10−16 J ≈ 2.1 × 10−16 J. (Matches NCERT answer.)

12.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest?

SOLUTION Number of molecules: Yes. By Avogadro’s law, equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. So all three vessels contain the same number of molecules. RMS speed: No, it is not the same. Since vrms = √(3kBT/m), at the same temperature vrms is inversely proportional to √m. The molecular masses are Ne ≈ 20.2 u, Cl2 ≈ 70.9 u, UF6 ≈ 352 u. The lightest molecule has the largest rms speed. Therefore vrms is largest for neon (the lightest of the three gases). (Matches NCERT answer.)

12.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at −20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).

SOLUTION Given: He at THe = −20 + 273 = 253 K, MHe = 4.0 u; Ar with MAr = 39.9 u; find TAr. RMS speed vrms = √(3RT/M). Setting the two rms speeds equal: √(3RTAr/MAr) = √(3RTHe/MHe) ⇒ TAr/MAr = THe/MHe. TAr = THe × (MAr/MHe) = 253 × (39.9 / 4.0) = 253 × 9.975 = 2524 K. So TAr ≈ 2.52 × 103 K. (Matches NCERT answer.)

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 °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 N2 = 28.0 u).

SOLUTION Given: P = 2.0 atm = 2 × 1.013 × 105 = 2.026 × 105 Pa, T = 17 + 273 = 290 K, radius = 1.0 Å so diameter d = 2.0 × 10−10 m, M0 = 28.0 × 10−3 kg mol−1. Number density: n = N/V = P / (kBT) = (2.026 × 105) / (1.38 × 10−23 × 290) = (2.026 × 105) / (4.0 × 10−21) = 5.10 × 1025 m−3. Mean free path: l = 1 / (√2 · nπd²) = 1 / (1.414 × 5.10 × 1025 × 3.14 × (2.0 × 10−10)²). Denominator = 1.414 × 5.10 × 1025 × 3.14 × 4.0 × 10−20 = 9.06 × 106. So l = 1 / (9.06 × 106) = 1.0 × 10−7 m. RMS speed: vrms = √(3RT/M0) = √[(3 × 8.31 × 290) / (28.0 × 10−3)] = √(2.58 × 105) = 5.1 × 102 m s−1. Collision frequency: ν = vrms / l = (5.1 × 102) / (1.0 × 10−7) = 5.1 × 109 s−1. Time between successive collisions: τ = l / vrms = (1.0 × 10−7) / (5.1 × 102) = 2 × 10−10 s. Collision time (time to cross molecular diameter) = d / vrms = (2.0 × 10−10) / (5.1 × 102) = 4 × 10−13 s. Comparison: τ / (collision time) = (2 × 10−10) / (4 × 10−13) = 500. The time between successive collisions is about 500 times the time of a collision, so a molecule moves essentially free for most of the time. (Matches NCERT answer.)

Extra Practice Questions

Short Answer Type Questions

Q1. State Avogadro’s law.

ANSWEREqual volumes of all gases, under the same conditions of temperature and pressure, contain equal numbers of molecules. The number per unit volume is the same for all gases at a fixed temperature and pressure.

Q2. On what factors does the average kinetic energy of a gas molecule depend?

ANSWERThe average translational kinetic energy of a molecule = (3/2)kBT depends only on the absolute temperature T. It is independent of the pressure, the volume and the nature of the gas.

Q3. Why does the rms speed of hydrogen molecules exceed that of oxygen molecules at the same temperature?

ANSWERBecause vrms = √(3RT/M0) is inversely proportional to √M0 at a fixed temperature. Hydrogen (M = 2 u) is much lighter than oxygen (M = 32 u), so its molecules move faster (by a factor √16 = 4).

Q4. State the law of equipartition of energy.

ANSWERIn thermal equilibrium at temperature T, the total energy of a system is distributed equally among all its energy modes, each quadratic (squared) term in the energy contributing an average of ½kBT per molecule. A vibrational mode contributes kBT because it has both a kinetic and a potential term.

Q5. Define mean free path and state how it depends on number density and molecular size.

ANSWERThe mean free path is the average distance a molecule travels between two successive collisions, l = 1/(√2 nπd²). It is inversely proportional to the number density n and to the square of the molecular diameter d, so it increases at low pressure (small n).

Long Answer Type Questions

Q1. Derive the expression for the pressure exerted by an ideal gas on the walls of its container using kinetic theory.

ANSWERConsider a gas in a cube of side l with N molecules of mass m. A molecule moving with x-velocity vx hits a wall (area A = l²) and rebounds elastically, so its momentum change is −2mvx; the wall receives +2mvx. In time Δt, only molecules within a distance vxΔt reach the wall, and on average half move toward it, so the number hitting is ½ n A vx Δt (n = number density). Total momentum transferred Q = (2mvx)(½ n A vx Δt). Pressure P = Q/(AΔt) = n m vx². Averaging over the velocity distribution and using isotropy, v̅x² = v̅y² = v̅z² = (1/3)v̅². Hence P = (1/3) n m v̅², the kinetic theory expression for gas pressure.

Q2. Using kinetic theory, show that the average kinetic energy of a molecule is (3/2)kBT and hence interpret temperature.

ANSWERFrom kinetic theory, P = (1/3) n m v̅². For a volume V containing N molecules, n = N/V, so PV = (1/3) N m v̅² = (2/3) N (½ m v̅²). Comparing with the ideal gas law PV = N kBT (since PV = μRT = N kBT), we get (2/3)(½ m v̅²) = kBT, i.e. ½ m v̅² = (3/2)kBT. Thus the average translational kinetic energy of a molecule is directly proportional to the absolute temperature, and is independent of pressure, volume and the nature of the gas. Temperature is therefore a measure of the average molecular kinetic energy.

Q3. Using the law of equipartition of energy, obtain the molar specific heats and ratio γ for monatomic and rigid diatomic gases.

ANSWERMonatomic gas: 3 translational degrees of freedom, so energy per molecule = 3 × ½kBT = (3/2)kBT; internal energy of one mole U = (3/2)RT. Then Cv = dU/dT = (3/2)R, and Cp = Cv + R = (5/2)R, giving γ = Cp/Cv = 5/3 ≈ 1.67. Rigid diatomic gas: 3 translational + 2 rotational = 5 degrees of freedom, so U = (5/2)RT, Cv = (5/2)R, Cp = (7/2)R, γ = 7/5 = 1.40. (Note Cp − Cv = R holds for any ideal gas.)

MCQs & Assertion–Reason

1. The pressure of an ideal gas according to kinetic theory is:

(a) P = n m v̅²    (b) P = (1/2) n m v̅²    (c) P = (1/3) n m v̅²    (d) P = (2/3) n m v̅²

2. The average translational kinetic energy of a gas molecule depends only on:

(a) pressure    (b) volume    (c) absolute temperature    (d) the nature of the gas

3. The rms speed of a gas molecule is proportional to:

(a) T    (b) √T    (c) 1/T    (d) T²

4. At a given temperature, which gas has the highest rms speed?

(a) O2    (b) N2    (c) H2    (d) CO2

5. The number of degrees of freedom of a rigid diatomic molecule is:

(a) 3    (b) 5    (c) 6    (d) 7

6. The value of γ (= Cp/Cv) for a monatomic ideal gas is:

(a) 1.40    (b) 1.33    (c) 1.67    (d) 1.00

7. For any ideal gas, Cp − Cv equals:

(a) R/2    (b) R    (c) 2R    (d) 3R/2

8. The mean free path of a gas molecule is inversely proportional to:

(a) d    (b) d²    (c) √d    (d) 1/d²

9. According to the law of equipartition, each vibrational mode of a molecule contributes an average energy of:

(a) ½kBT    (b) kBT    (c) (3/2)kBT    (d) 2kBT

10. Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. This is:

(a) Boyle’s law    (b) Charles’ law    (c) Avogadro’s law    (d) Dalton’s law

Answer key: 1-(c), 2-(c), 3-(b), 4-(c), 5-(b), 6-(c), 7-(b), 8-(b), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: At the same temperature, lighter gas molecules have a greater rms speed than heavier ones.

Reason: At a fixed temperature, all gas molecules have the same average kinetic energy, so vrms ∝ 1/√m.

A-R 2. Assertion: The internal energy of an ideal gas depends only on its temperature.

Reason: For an ideal gas there are no intermolecular forces, so its internal energy is purely kinetic.

A-R 3. Assertion: The molar specific heat of a diatomic gas is greater than that of a monatomic gas.

Reason: A diatomic molecule has rotational degrees of freedom in addition to translational ones.

A-R 4. Assertion: A real gas behaves most like an ideal gas at high pressure and low temperature.

Reason: At high pressure and low temperature the molecules are far apart and intermolecular forces are negligible.

A-R 5. Assertion: The mean free path of gas molecules increases as the pressure is reduced.

Reason: Reducing the pressure at constant temperature decreases the number density of molecules.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Forgetting to convert temperatures to kelvin (T = t °C + 273) before using gas equations.
  • Using gauge pressure directly in gas-law numericals where absolute pressure (gauge + atmospheric) is required — check what the question gives.
  • Confusing v̅² (mean of squared speed) with (v̅)² (square of mean speed); they are not equal.
  • Dropping the √2 factor in the mean free path formula l = 1/(√2 nπd²).
  • Mixing up radius and diameter when the molecular radius is given but d² is needed.
  • Treating a vibrational degree of freedom as contributing only ½kBT — it contributes kBT (kinetic + potential).
  • Forgetting that average kinetic energy depends only on temperature, not on molecular mass or the nature of the gas.

Exam tips for scoring full marks

Write every numerical with a clear Given → Formula → Substitution → Answer with units layout — marks are awarded for each step. Memorise the standard constants (R = 8.31 J mol−1 K−1, kB = 1.38 × 10−23 J K−1, NA = 6.022 × 1023, molar volume 22.4 L at STP). Learn the degrees-of-freedom table for monatomic (3), diatomic (5) and the resulting Cv, Cp, γ values — they appear in 1-mark and assertion-reason questions. For derivations (pressure of a gas, kinetic interpretation of temperature) practise stating each assumption of kinetic theory.

Frequently Asked Questions

What is Class 11 Physics Chapter 12 Kinetic Theory about?

Chapter 12 explains the behaviour of gases by treating them as a large number of rapidly moving molecules. It covers the ideal-gas equation, the kinetic theory expression for pressure, the kinetic interpretation of temperature, the law of equipartition of energy, specific heats of gases, and the mean free path.

How many exercises are there in Class 11 Physics Chapter 12?

The NCERT textbook has 10 exercise questions, numbered 12.1 to 12.10. All of them are reproduced verbatim and solved step by step on this page, with numerical answers verified against the NCERT key.

What is the kinetic interpretation of temperature?

It is the result that the average translational kinetic energy of a gas molecule equals (3/2)kBT. This means the absolute temperature of a gas is a direct measure of the average kinetic energy of its molecules, independent of pressure, volume or the nature of the gas.

Are these Class 11 Physics Chapter 12 solutions free?

Yes. All solutions are free and follow the official NCERT Physics textbook for session 2026–27.

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