Class 11 Physics NCERT Solutions

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Kinetic Theory Class 11 Physics NCERT Solutions

The complete NCERT exercise solutions for Chapter 12, Kinetic Theory — 10 questions from 12.1 to 12.10, each worked through step by step in the CBSE marking pattern. The ideal gas equation, molecular speeds, degrees of freedom and equipartition of energy.

Class:11Subject:PhysicsChapter:12

3 Key Formulas25 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 11 Physics Chapter 12?

Chapter 12 carries 10 exercise questions, numbered 12.1 to 12.10. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Table of Contents11 sections

Revision & Practice

Chapter Overview

Kinetic theory links the microscopic world of molecules to the gas laws: PV = nRT, the rms speed (1/2)mv² = (3/2)kT, mean free path, and the distribution of molecular energy. Nearly every board question here is a clean ideal-gas calculation — moles, volumes, bubbles, cylinders, mean free paths — with one reading of the PV/T-versus-P plot. Every question below is from the NCERT Class 11 textbook (rationalised edition), solved line by line in the board pattern.

Board pattern

Work in absolute temperatures (K), add 1 atm to gauge pressures, and quote the two workhorses: PV = NkT = nRT and v_rms = √(3kT/m) = √(3RT/M). For the mean free path remember λ = kT/(√2 π d² P). Express molecular speeds the same way every time, and relate the PV/T intercept to nR.

$Ex 12.1 — Fraction of molecular volume to actual volume for oxygen at STP$
$Ex 12.2 — Showing molar volume is 22.4 litres at STP$
$Ex 12.3 — Reading the PV/T versus P plot for oxygen$
$Ex 12.4 — Mass of oxygen drawn from a cylinder$
$Ex 12.5 — Growth of an air bubble rising from a lake floor$
$Ex 12.6 — Number of air molecules in a room$
$Ex 12.7 — Average thermal energy of a helium atom$
$Ex 12.8 — Equal molecules but unequal rms speeds$
$Ex 12.9 — Temperature at which argon matches helium’s rms speed$
$Ex 12.10 — Mean free path and collision frequency of nitrogen$

NCERT Exercise 12.1 — Fraction of Molecular Volume to Actual Volume for Oxygen at STP

1Exercise question

Step-by-step solution

1d = 3 Å, so r = 1.5 Å = 1.5 × 10⁻¹⁰ m.
2Actual volume of 1 mole of O₂ at STP = 22.4 L = 22.4 × 10⁻³ m³.
3Molecular volume of one mole = (4/3)πr³ × N = (4/3) × 3.14 × (1.5 × 10⁻¹⁰)³ × 6.02 × 10²³.
4= 8.51 × 10⁻⁶ m³ (≈ 8.5 cm³).
5Fraction = 8.5 × 10⁻⁶/22.4 × 10⁻³ = 3.8 × 10⁻⁴.

Final answer

The fraction is about 3.8 × 10⁻⁴.

NCERT Exercise 12.2 — Showing Molar Volume Is 22.4 Litres at STP

1Exercise question

Step-by-step solution

1Ideal gas law: PV = nRT with n = 1 mol.
2V = RT/P = 8.31 × 273/(1.013 × 10⁵).
3V = 0.0224 m³ = 22.4 L.

Final answer

V = RT/P = 0.0224 m³ = 22.4 litres at STP.

NCERT Exercise 12.3 — Reading the PV/T Versus P Plot for Oxygen

1Exercise question

Step-by-step solution

1(a) The dotted line is the ideal-gas plot: PV/T = nR, a constant independent of pressure.
2(b) A real gas approaches ideal behaviour as temperature rises. The T₁ curve lies closer to the dotted line than T₂, so T₁ > T₂.
3(c) At the y-axis intercept the gases are ideal: PV/T = nR = (1.00 × 10⁻³/0.032) × 8.31 = 0.26 J K⁻¹.
4(d) No — the intercept equals nR, and 1.00 × 10⁻³ kg of H₂ is a different number of moles, so PV/T differs.
5To get the same value 0.26 J K⁻¹: n = PV/RT = 0.26/8.31 = 0.0313 mol, giving m = 0.0313 × 0.00202 = 6.3 × 10⁻⁵ kg of hydrogen.

Final answer

(a) ideal-gas behaviour (b) T₁ > T₂ (c) PV/T = 0.26 J K⁻¹ (d) no; 6.3 × 10⁻⁵ kg of hydrogen gives the same value.

NCERT Exercise 12.4 — Mass of Oxygen Drawn From a Cylinder

1Exercise question

Step-by-step solution

1Absolute pressures: P₁ = 16 atm, P₂ = 12 atm; V = 30 L = 0.030 m³.
2n₁ = P₁V/RT₁ = 16 × 1.013 × 10⁵ × 0.030/(8.31 × 300) = 19.5 mol.
3n₂ = P₂V/RT₂ = 12 × 1.013 × 10⁵ × 0.030/(8.31 × 290) = 15.1 mol.
4Moles withdrawn = 19.5 − 15.1 = 4.4 mol.
5Mass = 4.4 × 32 × 10⁻³ = 1.4 × 10⁻¹ kg ≈ 0.14 kg.

Final answer

About 0.14 kg of oxygen is taken out of the cylinder.

NCERT Exercise 12.5 — Growth of an Air Bubble Rising From a Lake Floor

1Exercise question

Step-by-step solution

1Bottom pressure: P₁ = 1 atm + ρgh = 1.013 × 10⁵ + 10³ × 9.8 × 40 = 4.93 × 10⁵ Pa.
2Surface pressure P₂ = 1.013 × 10⁵ Pa; T₁ = 285 K, T₂ = 308 K, V₁ = 1.0 × 10⁻⁶ m³.
3P₁V₁/T₁ = P₂V₂/T₂.
4V₂ = V₁ × (P₁/P₂) × (T₂/T₁) = 1.0 × 10⁻⁶ × (4.93/1.013) × (308/285).
5V₂ = 5.26 × 10⁻⁶ m³ = 5.3 cm³.

Final answer

The bubble grows to about 5.3 cm³.

NCERT Exercise 12.6 — Number of Air Molecules in a Room

1Exercise question

Step-by-step solution

1PV = NkT, so N = PV/kT.
2N = 1.013 × 10⁵ × 25.0/(1.38 × 10⁻²³ × 300).
3N = 6.11 × 10²⁶ molecules.

Final answer

There are about 6.11 × 10²⁶ air molecules in the room.

NCERT Exercise 12.7 — Average Thermal Energy of a Helium Atom

1Exercise question

Step-by-step solution

1Average thermal energy = (3/2)kT with k = 1.38 × 10⁻²³ J K⁻¹.
2(i) T = 300 K: (3/2) × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J.
3(ii) T = 6000 K: (3/2) × 1.38 × 10⁻²³ × 6000 = 1.24 × 10⁻¹⁹ J.
4(iii) T = 10⁷ K: (3/2) × 1.38 × 10⁻²³ × 10⁷ = 2.07 × 10⁻¹⁶ J.

Final answer

(i) 6.21 × 10⁻²¹ J (ii) 1.24 × 10⁻¹⁹ J (iii) 2.07 × 10⁻¹⁶ J.

NCERT Exercise 12.8 — Equal Molecules but Unequal rms Speeds

1Exercise question

Step-by-step solution

1Equal P, V, T means equal n; by Avogadro's law each vessel holds the same number of molecules (equal to N_A per mole of gas).
2v_rms = √(3kT/m): the same T but different molecular masses give different rms speeds.
3Neon (monatomic, mass 20 u) is the lightest of the three molecules, so its v_rms is the largest.

Final answer

Yes, equal numbers of molecules; v_rms differs, and it is largest for neon, the lightest.

NCERT Exercise 12.9 — Temperature at Which Argon Matches Helium's rms Speed

1Exercise question

Step-by-step solution

1v_rms ∝ √(T/M), so equal speeds give T_Ar/M_Ar = T_He/M_He.
2T_He = −20 °C = 253 K.
3T_Ar = 253 × 39.9/4.0.
4T_Ar = 2.52 × 10³ K.

Final answer

Approximately 2.52 × 10³ K (2523 K).

NCERT Exercise 12.10 — Mean Free Path and Collision Frequency of Nitrogen

1Exercise question

Step-by-step solution

1v_rms = √(3RT/M) = √(3 × 8.31 × 290/0.028) = 508 m s⁻¹.
2Mean free path λ = kT/(√2 π d² P) with d = 2 × 10⁻¹⁰ m, P = 2.026 × 10⁵ Pa.
3λ = 1.38 × 10⁻²³ × 290/(1.414 × 3.14 × (2 × 10⁻¹⁰)² × 2.026 × 10⁵) = 1.11 × 10⁻⁷ m.
4Collision frequency = v_rms/λ = 508/1.11 × 10⁻⁷ = 4.58 × 10⁹ s⁻¹.
5Collision time ≈ d/v_rms = 2 × 10⁻¹⁰/508 = 3.9 × 10⁻¹³ s; free time = λ/v_rms = 2.2 × 10⁻¹⁰ s.
6The free time is about 500 times the collision time.

Final answer

Mean free path = 1.11 × 10⁻⁷ m; collision frequency = 4.58 × 10⁹ s⁻¹; free time ≈ 500 × collision time.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Rms speed

v_{r m s} = \frac{3 R T}{M}

Mean kinetic energy per molecule

\overline{E_{k}} = \frac{3}{2} k_{B} T

Degrees of freedom

f = \frac{2}{3} C_{V} = \frac{5}{3} C_{P}

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Rms speed is the square root of the mean of the squared speeds — not the square root of the mean speed.
A diatomic gas has f = 5 at ordinary temperatures, not 3; the extra two come from rotation.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Physics Chapter 12 (Kinetic Theory)?

There are 10 exercise questions in this chapter, numbered 12.1 to 12.10. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Kinetic Theory Class 11 Physics?

The formulas this chapter's questions actually turn on are: Rms speed, Mean kinetic energy per molecule, Degrees of freedom. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Kinetic Theory important for JEE Main and NEET?

Mostly formula-based and quick — molecular speeds and degrees of freedom are reliable two-mark questions in the boards and appear regularly in NEET.

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