This MCQ module is based on: NCERT Exercises and Solutions: Kinetic Theory
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NCERT Exercises and Solutions: Kinetic Theory
🎓 Class 11
Physics
CBSE
Theory
Ch 12 – Kinetic Theory
⏱ ~8 min
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NCERT Exercises and Solutions: Kinetic Theory
Chapter 12 Summary — Kinetic Theory
- Ideal gas equation: PV = nRT = N k_B T (k_B = R/N_A = 1.381 × 10⁻²³ J/K).
- Pressure of an ideal gas: P = (1/3) n m ⟨v²⟩ = (1/3) ρ ⟨v²⟩.
- Kinetic interpretation of T: ⟨KE⟩_translational = (3/2) k_B T per molecule. Temperature ↔ shared kinetic energy per particle.
- Three speeds: v_p = √(2RT/M), v̄ = √(8RT/πM), v_rms = √(3RT/M). Ratio 1 : 1.128 : 1.225.
- Equipartition: Each quadratic degree of freedom carries (1/2) k_B T. A molecule with f dof has ⟨E⟩ = (f/2)k_B T.
- Specific heats of ideal gas: C_v = (f/2)R, C_p = ((f+2)/2)R, γ = (f+2)/f.
- Solids: 6 dof per atom ⇒ C ≈ 3R = 25 J/mol·K (Dulong-Petit).
- Mean free path: λ = 1/(√2 π d² n). Air at STP: ~70 nm; collision freq ~10⁹–10¹⁰ /s.
| Gas type | f | γ | v_rms / v_rms(He) at same T |
|---|---|---|---|
| Monatomic (He, Ar) | 3 | 5/3 | √(M_He/M) |
| Rigid diatomic (N₂, O₂) | 5 | 7/5 | √(4/M_diatomic) |
| Non-linear triatomic (H₂O) | 6 | 4/3 | — |
NCERT Exercises — Worked Solutions
Exercise 12.1: Diameter of an oxygen molecule
Estimate the fraction of molecular volume to the actual volume occupied by one mole of oxygen at STP. Take diameter d = 3 Å.
Volume of one molecule = (4/3)π(d/2)³ = (4/3)π(1.5 × 10⁻¹⁰)³ = 1.41 × 10⁻²⁹ m³.
Total molecular volume per mole = N_A × 1.41 × 10⁻²⁹ = 8.5 × 10⁻⁶ m³ ≈ 8.5 cm³.
Molar gas volume at STP = 22.4 L = 22400 cm³.
Fraction = 8.5 / 22400 ≈ 3.8 × 10⁻⁴. The molecules occupy only ~0.04 % of available volume — gas is mostly empty space.
Total molecular volume per mole = N_A × 1.41 × 10⁻²⁹ = 8.5 × 10⁻⁶ m³ ≈ 8.5 cm³.
Molar gas volume at STP = 22.4 L = 22400 cm³.
Fraction = 8.5 / 22400 ≈ 3.8 × 10⁻⁴. The molecules occupy only ~0.04 % of available volume — gas is mostly empty space.
Exercise 12.2: Vapour at higher T but smaller V
One mole of an ideal gas is at STP (273 K, 1 atm). Show that the volume occupied is ~22.4 L using PV = nRT.
V = nRT/P = (1)(8.314)(273)/(1.013 × 10⁵) = 2269.7 / 101325 = 0.0224 m³ = 22.4 L ✓
Exercise 12.3: Density of gas
An oxygen cylinder of volume 30 L has an initial gauge pressure of 15 atm at 27 °C. After some O₂ is withdrawn, the gauge reads 11 atm at 17 °C. Find the mass of O₂ removed.
Use absolute P (gauge + 1 atm), V = 30 × 10⁻³ m³.
n₁ = P₁V/(RT₁) = (16 × 1.013 × 10⁵)(0.030)/(8.314 × 300) = 48624/2494.2 = 19.5 mol.
n₂ = P₂V/(RT₂) = (12 × 1.013 × 10⁵)(0.030)/(8.314 × 290) = 36468/2411.1 = 15.1 mol.
Δn = 4.4 mol; mass removed = 4.4 × 32 g = 140.8 g ≈ 141 g.
n₁ = P₁V/(RT₁) = (16 × 1.013 × 10⁵)(0.030)/(8.314 × 300) = 48624/2494.2 = 19.5 mol.
n₂ = P₂V/(RT₂) = (12 × 1.013 × 10⁵)(0.030)/(8.314 × 290) = 36468/2411.1 = 15.1 mol.
Δn = 4.4 mol; mass removed = 4.4 × 32 g = 140.8 g ≈ 141 g.
Exercise 12.4: Air bubble in a lake
An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at 12 °C to the surface where T = 35 °C. Find its volume at the surface. ρ_water = 10³ kg/m³, atm pressure = 1.013 × 10⁵ Pa.
P_bottom = P_atm + ρgh = 1.013 × 10⁵ + 10³ × 9.8 × 40 = 4.93 × 10⁵ Pa. T_bottom = 285 K.
P_surface = 1.013 × 10⁵ Pa. T_surface = 308 K.
P₁V₁/T₁ = P₂V₂/T₂ ⇒ V₂ = V₁(P₁/P₂)(T₂/T₁) = 1.0 × (4.93/1.013)(308/285) = 4.87 × 1.081 = 5.27 cm³.
P_surface = 1.013 × 10⁵ Pa. T_surface = 308 K.
P₁V₁/T₁ = P₂V₂/T₂ ⇒ V₂ = V₁(P₁/P₂)(T₂/T₁) = 1.0 × (4.93/1.013)(308/285) = 4.87 × 1.081 = 5.27 cm³.
Exercise 12.5: Number of molecules in a room
Estimate the total number of air molecules (oxygen + nitrogen) in a classroom of volume 25 m³ at 27 °C and 1 atm. (k_B = 1.38 × 10⁻²³ J/K).
N = PV/(k_B T) = (1.013 × 10⁵)(25)/(1.38 × 10⁻²³ × 300) = 2.53 × 10⁶ / 4.14 × 10⁻²¹ = 6.1 × 10²⁶ molecules.
Exercise 12.6: He at 6500 K
Estimate the average thermal energy of a helium atom (i) at room temperature (27 °C), (ii) at the temperature on the surface of the Sun (6000 K), (iii) at the core of a star (10⁷ K).
⟨E⟩ = (3/2) k_B T:
(i) T = 300 K: ⟨E⟩ = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J (≈ 0.039 eV).
(ii) T = 6000 K: ⟨E⟩ = 1.24 × 10⁻¹⁹ J (≈ 0.78 eV).
(iii) T = 10⁷ K: ⟨E⟩ = 2.07 × 10⁻¹⁶ J (≈ 1300 eV) — enough for thermonuclear fusion.
(i) T = 300 K: ⟨E⟩ = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J (≈ 0.039 eV).
(ii) T = 6000 K: ⟨E⟩ = 1.24 × 10⁻¹⁹ J (≈ 0.78 eV).
(iii) T = 10⁷ K: ⟨E⟩ = 2.07 × 10⁻¹⁶ J (≈ 1300 eV) — enough for thermonuclear fusion.
Exercise 12.7: Mean speeds of three gases
Three vessels of equal capacity contain neon (monatomic), oxygen (diatomic) and CO₂ (polyatomic) at the same temperature and pressure. (a) Do they have equal numbers of molecules? (b) Do they have the same v_rms? (c) Same average translational KE per molecule? Justify each.
(a) Yes. From PV = N k_B T at same P, V, T: N is the same — Avogadro's hypothesis.
(b) No. v_rms = √(3k_BT/m) depends on molecular mass m. Lighter Ne has the largest v_rms; CO₂ the smallest.
(c) Yes. ⟨KE⟩_trans = (3/2)k_BT depends only on T, not on species.
(b) No. v_rms = √(3k_BT/m) depends on molecular mass m. Lighter Ne has the largest v_rms; CO₂ the smallest.
(c) Yes. ⟨KE⟩_trans = (3/2)k_BT depends only on T, not on species.
Exercise 12.8: Temperature of a hot gas
At what temperature is the v_rms of helium atoms equal to that of N₂ molecules at 27 °C? (M_He = 4, M_N₂ = 28 g/mol).
v_rms² = 3RT/M. Setting equal: 3R T_He / M_He = 3R T_N₂ / M_N₂ ⇒ T_He = T_N₂ × M_He/M_N₂ = 300 × (4/28) = 42.86 K (≈ −230 °C).
Exercise 12.9: Volume per molecule at STP
At STP, find the average volume per molecule and the average distance between molecules of an ideal gas.
V/N = V_m/N_A = 22.4 × 10⁻³ / 6.022 × 10²³ = 3.72 × 10⁻²⁶ m³ per molecule.
Average separation ≈ (V/N)^(1/3) = (3.72 × 10⁻²⁶)^(1/3) ≈ 3.34 × 10⁻⁹ m ≈ 3.3 nm.
Compare to molecular size ~0.3 nm — the gas spacing is ~10× the molecular size.
Average separation ≈ (V/N)^(1/3) = (3.72 × 10⁻²⁶)^(1/3) ≈ 3.34 × 10⁻⁹ m ≈ 3.3 nm.
Compare to molecular size ~0.3 nm — the gas spacing is ~10× the molecular size.
Exercise 12.10: Mean free path of nitrogen
Calculate the mean free path and collision frequency of N₂ at 2.0 atm and 17 °C. The molecular diameter of N₂ is 1.0 × 10⁻¹⁰ m, M = 28 g/mol.
n = P/(k_B T) = (2 × 1.013 × 10⁵)/(1.38 × 10⁻²³ × 290) = 5.06 × 10²⁵ m⁻³.
λ = 1/(√2 π d² n) = 1/(√2 × π × (10⁻¹⁰)² × 5.06 × 10²⁵) = 1/(2.25 × 10⁶) ≈ 4.5 × 10⁻⁷ m = 450 nm.
v̄ = √(8RT/πM) = √(8 × 8.314 × 290 / (π × 0.028)) = √(2.19 × 10⁵) ≈ 468 m/s.
Collision frequency ν = v̄/λ = 468/(4.5 × 10⁻⁷) ≈ 1.04 × 10⁹ s⁻¹.
λ = 1/(√2 π d² n) = 1/(√2 × π × (10⁻¹⁰)² × 5.06 × 10²⁵) = 1/(2.25 × 10⁶) ≈ 4.5 × 10⁻⁷ m = 450 nm.
v̄ = √(8RT/πM) = √(8 × 8.314 × 290 / (π × 0.028)) = √(2.19 × 10⁵) ≈ 468 m/s.
Collision frequency ν = v̄/λ = 468/(4.5 × 10⁻⁷) ≈ 1.04 × 10⁹ s⁻¹.
Interactive: Combined PVT Solver L3 Apply
Set initial state (P₁, V₁, T₁) and any two of the final state variables; the applet finds the third using P₁V₁/T₁ = P₂V₂/T₂.
Self-Assessment Activity — Concept Audit
L6 Create
Predict: Without looking back, can you derive both the ideal gas law and the kinetic interpretation of temperature starting only from "molecules are tiny billiard balls in random motion"?
- Sketch the box-of-gas argument (Step 1–4) on paper.
- Compare with PV = NkT to identify temperature with ⟨KE⟩.
- From there, derive Boyle's law, Charles' law, Avogadro's hypothesis, and Graham's law of effusion.
Master-list of derived results:
- Boyle: P = (1/3) ρ ⟨v²⟩, so at constant ⟨v²⟩ (i.e. constant T), P ∝ ρ ∝ 1/V.
- Charles: ⟨v²⟩ ∝ T, so V ∝ T at constant P.
- Avogadro: PV = N k_B T → at given P, V, T, N is the same for all gases.
- Graham (effusion): rate ∝ v̄ ∝ 1/√M.
- Equipartition gives γ values that match measurements.
Competency-Based Questions
A scuba diver carries a 12.0 L tank of compressed air at 200 atm and 27 °C. Air is a mixture of N₂ and O₂ approximated as ideal diatomic gas (M̄ = 29 g/mol).
Q1. L1 Remember Write the ideal gas equation in terms of total molecule count N.
PV = N k_B T, where k_B = 1.381 × 10⁻²³ J/K is Boltzmann's constant.
Q2. L3 Apply How many moles of air are in the tank?
n = PV/(RT) = (200 × 1.013 × 10⁵)(12 × 10⁻³)/(8.314 × 300) = 243120 / 2494.2 ≈ 97.5 mol; mass = 97.5 × 29 ≈ 2.83 kg.
Q3. L3 Apply Estimate v_rms of an air molecule in the tank at 300 K.
v_rms = √(3RT/M) = √(3 × 8.314 × 300 / 0.029) = √(2.58 × 10⁵) ≈ 508 m/s.
Q4. L4 Analyse When the diver is 30 m below the surface, the gauge regulator delivers air at the local water pressure (~4 atm). For a normal breathing volume of 0.5 L per breath, how many breaths does the tank give? (Use n_used / n_breath ratios.)
The 12 L tank at 200 atm contains 12 × 200 = 2400 L of air at 1 atm — but at 4 atm depth pressure each breath needs 0.5 L × 4 = 2 L of "1-atm equivalent" air. So breaths ≈ 2400/2 = 1200 breaths. Real divers leave a reserve, so usable ≈ 800 breaths.
Q5. L5 Evaluate Why must the diver ascend slowly even after a deep dive?
Dissolved nitrogen in the diver's blood follows Henry's law (gas solubility ∝ partial pressure). Rapid pressure drop on ascent makes N₂ come out of solution as bubbles — "decompression sickness" or "the bends". Slow ascent allows N₂ to be exhaled gradually. This is a real-world consequence of Dalton's law plus Henry's law of gas solubility.
Assertion-Reason Questions
Assertion (A): Two gases at the same temperature have the same average translational KE per molecule.
Reason (R): Temperature is a measure of the random translational kinetic energy of molecules.
Answer: A. ⟨KE⟩ = (3/2)k_B T is the kinetic interpretation of temperature; R explains A directly.
Assertion (A): The ratio C_p/C_v is greater than 1 for any ideal gas.
Reason (R): C_p − C_v = R > 0 for ideal gases (Mayer's relation).
Answer: A. Since C_p = C_v + R > C_v, γ = C_p/C_v > 1 always for an ideal gas. R is the direct reason for A.
Assertion (A): The mean free path of a gas can exceed the size of its container.
Reason (R): At extremely low pressures, molecules collide more often with the walls than with each other.
Answer: A. In an ultra-high-vacuum chamber, λ can exceed metres — molecules travel ballistically wall-to-wall (Knudsen regime). Wall collisions dominate. R correctly explains A.
Frequently Asked Questions - NCERT Exercises and Solutions: Kinetic Theory
What are the key NCERT exercise types in Chapter 12 Kinetic Theory?
NCERT Class 11 Physics Chapter 12 Kinetic Theory exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question in the chapter, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Kinetic Theory?
For numerical problems in NCERT Class 11 Physics Chapter 12 Kinetic Theory: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 12?
From NCERT Class 11 Physics Chapter 12 (Kinetic Theory), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 11 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula (M^a L^b T^c). For Chapter 12 Kinetic Theory problems, write the dimensions of each quantity, substitute, and simplify. If dimensions match the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 12 exercises?
Common mistakes in NCERT Class 11 Physics Chapter 12 Kinetic Theory exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for vectors, (3) forgetting to consider all forces in free-body diagrams, (4) algebra errors during derivations, and (5) misreading the problem statement. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 11 Physics Chapter 12 Kinetic Theory solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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