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Class 11 · CBSE

Kinetic Theory — NCERT Class 11 Physics

Complete NCERT Solutions for CBSE Class 11 Physics Chapter 13 — all intext and chapter-end exercise questions on kinetic theory, RMS speed, equipartition of energy, specific heats, degrees of freedom answered clearly.

Updated 11 June 20267 min read

How to Use This Page

Try each NCERT question yourself before checking the solution. For numericals, every step is shown: formula, substitution, arithmetic, and final answer with correct units. Watch out for unit conversions between kJ, J, and calories, and remember that molar mass M must be in kg mol⁻¹ when using R = 8.314 J mol⁻¹ K⁻¹.

Key Concepts (Quick Recap)

  • Ideal gas equation: PV = nRT; also PV = (1/3)Nm v²_rms
  • RMS speed: v_rms = √(3RT/M) = √(3kT/m) = √(3P/ρ)
  • Mean (average) speed: v_avg = √(8RT/πM)
  • Most probable speed: v_mp = √(2RT/M)
  • Kinetic energy per molecule: E = (3/2)kT (for monatomic)
  • Kinetic energy per mole: E = (3/2)RT
  • Equipartition: ½kT per molecule per degree of freedom
  • Monatomic: f = 3, C_v = (3/2)R, C_p = (5/2)R, γ = 5/3
  • Diatomic: f = 5, C_v = (5/2)R, C_p = (7/2)R, γ = 7/5
  • Mayer's relation: C_p − C_v = R
  • Mean free path: λ = 1/(√2 π d² n)
  • Boltzmann constant: k = R/N_A = 1.38 × 10⁻²³ J K⁻¹

Intext Questions

Page 329

Q1. Estimate the average thermal energy of a helium atom at (a) room temperature (300 K) (b) at 6000 K.
E = (3/2)kT for a monatomic gas (a) E = (3/2) × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J (b) E = (3/2) × 1.38 × 10⁻²³ × 6000 = 1.242 × 10⁻¹⁹ J
Q2. At what temperature is the RMS speed of nitrogen equal to that of hydrogen at 300 K? M_N₂ = 28, M_H₂ = 2.
v_rms(N₂) = v_rms(H₂) √(3RT/M_N₂) = √(3RT'/M_H₂) T'/M_H₂ = T/M_N₂ T' = T × M_H₂/M_N₂ = 300 × 28/2 = 4200 K
Answer

T(N₂) = 4200 K

Page 335

Q3. Calculate the mean free path of a nitrogen molecule at 1 atm and 300 K. Diameter of N₂ = 0.3 nm.
n = P/(kT) = 1.013 × 10⁵ / (1.38 × 10⁻²³ × 300) = 2.45 × 10²⁵ m⁻³ d = 0.3 × 10⁻⁹ m λ = 1/(√2 × π × d² × n) = 1/(1.414 × 3.14 × (0.3 × 10⁻⁹)² × 2.45 × 10²⁵) = 1/(1.414 × 3.14 × 9 × 10⁻²⁰ × 2.45 × 10²⁵) = 1/(1.414 × 3.14 × 2.205 × 10⁶) = 1/(9.78 × 10⁶) ≈ 1.02 × 10⁻⁷ m
Answer

Mean free path ≈ 1.0 × 10⁻⁷ m = 100 nm

Chapter-End Exercise

Q1. Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take diameter of O₂ = 3 Å.
Molecular volume per molecule = (4/3)π(d/2)³ = (4/3)π(1.5 × 10⁻¹⁰)³ = 1.414 × 10⁻²⁹ m³ At STP, 1 mol occupies 22.4 L = 22.4 × 10⁻³ m³ Number per m³ = N_A / (22.4 × 10⁻³) = 6.022 × 10²³ / 0.0224 = 2.69 × 10²⁵ Volume of molecules = 2.69 × 10²⁵ × 1.414 × 10⁻²⁹ = 3.8 × 10⁻⁴ m³ Fraction = 3.8 × 10⁻⁴ / (22.4 × 10⁻³) ≈ 0.017 = 1.7 × 10⁻²
Answer

Fraction ≈ 0.017 (about 1.7%)

Q2. Molar volume is the volume occupied by 1 mol of any (ideal) gas at STP (22.4 L). Show that it is consistent with the ideal gas equation.
PV = nRT → V = nRT/P For n = 1, T = 273.15 K, P = 1.013 × 10⁵ Pa V = 1 × 8.314 × 273.15 / (1.013 × 10⁵) = 2271.3 / 101300 = 0.02242 m³ = 22.42 L
Answer

This matches the standard molar volume of 22.4 L.

Q3. Figure shows plots of PV/T vs P for 1.00 × 10⁻³ kg of three gases. Interpret the plots.
Answer

For an ideal gas, PV/T = nR = constant, so the plot should be a horizontal line. Deviations at high pressure are due to intermolecular forces and finite molecular volume. The gas that deviates least is closest to ideal behaviour (likely hydrogen or helium, which have weak intermolecular forces).

Q4. An oxygen molecule has mass 5.3 × 10⁻²⁶ kg and a nitrogen molecule 4.7 × 10⁻²⁶ kg. At what temperature will the RMS speed of O₂ equal that of N₂ at 300 K?
v_rms(O₂) = v_rms(N₂) √(3kT'/m_O) = √(3kT/m_N) T'/m_O = T/m_N T' = T × m_O/m_N = 300 × 5.3/4.7 = 338.3 K
Answer

Temperature ≈ 338 K

Q5. Two molecules of a gas have speeds 1 km/s and 2 km/s. What is the RMS speed?
v_rms = √((v₁² + v₂²)/2) = √((1² + 2²)/2) = √(5/2) = √2.5 = 1.581 km/s
Answer

RMS speed = 1.58 km/s

Q6. From a certain apparatus, the diffusion rate of hydrogen has an RMS speed that is 4 times that of oxygen. Find the mass of the hydrogen molecule.
v_H = 4 v_O √(3kT/m_H) = 4 √(3kT/m_O) 1/m_H = 16/m_O m_H = m_O/16 m_O = 32/N_A = 32/(6.022 × 10²³) = 5.31 × 10⁻²⁶ kg m_H = 5.31 × 10⁻²⁶/16 = 3.32 × 10⁻²⁷ kg
Q7. A gas in equilibrium has uniform density and pressure throughout. Is this also true for molecular speed and molecular kinetic energy?
Answer
  • Molecular speed: Not uniform — there is a Maxwell-Boltzmann distribution of speeds. Different molecules have different speeds at any instant.
  • Molecular kinetic energy: The average kinetic energy per molecule is uniform (½kT per DOF), but individual molecules have different kinetic energies at any instant.
Q8. At what temperature is the RMS speed of an atom in an argon gas cylinder equal to the RMS speed of a helium atom at −20°C? M_Ar = 39.9, M_He = 4.
√(3RT_Ar/M_Ar) = √(3RT_He/M_He) T_Ar/M_Ar = T_He/M_He T_He = −20 + 273 = 253 K T_Ar = T_He × M_Ar/M_He = 253 × 39.9/4 = 2523 K
Answer

Temperature of argon = 2523 K

Q9. Estimate the total number of air molecules in a room of volume 30 m³ at 300 K and 1 atm.
n = PV/(RT) = (1.013 × 10⁵ × 30) / (8.314 × 300) = 3.039 × 10⁶/2494.2 = 1218.2 mol N = n × N_A = 1218.2 × 6.022 × 10²³ = 7.33 × 10²⁶ molecules
Answer

Total molecules ≈ 7.3 × 10²⁶

Q10. Estimate the average energy and RMS speed of a nitrogen molecule at 300 K. Given M_N₂ = 28 g mol⁻¹.
Average KE per molecule = (5/2)kT (diatomic, f = 5) = (5/2) × 1.38 × 10⁻²³ × 300 = 1.035 × 10⁻²⁰ J v_rms = √(3RT/M) = √(3 × 8.314 × 300 / 0.028) = √(7482.6/0.028) = √(267235) = 517 m/s
Answer

Average energy ≈ 1.04 × 10⁻²⁰ J, v_rms ≈ 517 m/s

Q11. Calculate the RMS speed of a helium atom at 300 K. M_He = 4 × 10⁻³ kg mol⁻¹.
v_rms = √(3RT/M) = √(3 × 8.314 × 300 / 4 × 10⁻³) = √(7482.6 / 4 × 10⁻³) = √(1.87 × 10⁶) = 1368 m/s
Answer

v_rms ≈ 1368 m/s

Q12. At what temperature is the RMS speed of helium twice that of nitrogen at 300 K?
v_He = 2 v_N₂ √(3RT_He/M_He) = 2√(3RT_N₂/M_N₂) T_He/M_He = 4 × T_N₂/M_N₂ T_He = 4 × 300 × (4 × 10⁻³)/(28 × 10⁻³) = 1200 × 4/28 = 171.4 K
Answer

Temperature of helium = 171.4 K

Q13. What is the ratio of C_p/C_v for a monatomic gas? For a diatomic gas?
Monatomic: f = 3, C_v = (3/2)R, C_p = (5/2)R, γ = C_p/C_v = 5/3 ≈ 1.67 Diatomic: f = 5, C_v = (5/2)R, C_p = (7/2)R, γ = C_p/C_v = 7/5 = 1.4
Q14. Explain why the Maxwell-Boltzmann distribution is asymmetric.
Answer

The speed distribution f(v) ∝ v² exp(−mv²/2kT). The v² factor (from the volume element in velocity space) pushes the distribution toward higher speeds, while the exponential factor suppresses high speeds. The competition between these two opposing effects creates an asymmetric curve with a tail extending toward high speeds. The most probable speed, mean speed, and RMS speed are all different (v_mp < v_avg < v_rms).

Internal Links

  • [Chapter 13 Guide — Kinetic Theory](/school/cbse/class/11/physics/cbse-kinetic-theory)
  • [All Class 11 Physics Solutions](/school/cbse/class/11/physics/solutions)
  • [Class 11 Physics Subject Guide](/school/cbse/class/11/physics)

Frequently Asked Questions

v_rms = √(3kT/m) = √(3RT/M), where k is Boltzmann's constant, T is absolute temperature, m is mass of one molecule, R is the gas constant, and M is the molar mass.
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