Kinetic Theory — Class 11 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Physics chapter "Kinetic Theory" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Kinetic Theory" — 8 important questions with detailed answers for CBSE board exam prep…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 11 Physics — Kinetic Theory. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Kinetic Theory — All 8 Questions Solved
Q1. Calculate the RMS (root mean square) speed of nitrogen gas molecules at 300 K. Given: M = 28 g/mol = 0.028 kg/mol, R = 8.314 J/mol·K.
Step 1: RMS speed formula:
v_rms = √(3RT/M) = √(3kT/m)
where R = gas constant, T = temperature, M = molar mass
Step 2: Substitute values:
T = 300 K
M = 0.028 kg/mol
R = 8.314 J/mol·K
v_rms = √(3 × 8.314 × 300 / 0.028)
v_rms = √(7482.6 / 0.028)
v_rms = √(267,235.7)
v_rms = 516.9 m/s ≈ 517 m/s
Final Answer: RMS speed of N₂ = 517 m/s
Q2. At what temperature will the RMS speed of oxygen molecules be equal to 400 m/s? (M_O₂ = 32 g/mol, R = 8.314 J/mol·K)
Step 1: RMS speed formula:
v_rms = √(3RT/M)
Step 2: Square both sides:
v_rms² = 3RT/M
Step 3: Rearrange for T:
T = (v_rms² × M)/(3R)
Step 4: Substitute values:
v_rms = 400 m/s
M = 32 g/mol = 0.032 kg/mol
R = 8.314 J/mol·K
T = (400² × 0.032)/(3 × 8.314)
T = (160,000 × 0.032)/(24.942)
T = 5120/24.942
T = 205.2 K ≈ 205 K
Step 5: Convert to Celsius:
T = 205 - 273 = -68°C
Final Answer: Temperature = 205.2 K or -68°C
Q3. Calculate the average kinetic energy of a gas molecule at 27°C. Given: Boltzmann constant k = 1.38 × 10⁻²³ J/K.
Step 1: Average kinetic energy per molecule:
KE_avg = (3/2)kT
where k = Boltzmann constant, T = absolute temperature
Step 2: Convert temperature to Kelvin:
T = 27°C = 27 + 273 = 300 K
Step 3: Calculate KE_avg:
KE_avg = (3/2) × 1.38 × 10⁻²³ × 300
KE_avg = 1.5 × 1.38 × 10⁻²³ × 300
KE_avg = 1.5 × 414 × 10⁻²³
KE_avg = 621 × 10⁻²³
KE_avg = 6.21 × 10⁻²¹ J
Final Answer: Average kinetic energy = 6.21 × 10⁻²¹ J
Q4. Using kinetic theory, derive the relation between pressure and mean kinetic energy of gas molecules.
Step 1: Consider a gas container with N molecules, each of mass m.
Step 2: Pressure from molecular collisions:
Each collision transfers momentum 2mv to the wall (for elastic collision).
Number of collisions per unit area per unit time = (nv/4)
where n = number density, v = mean speed
Step 3: Pressure P = (force)/(area) = (momentum change × number of collisions)/(area × time)
P = (N/V) × (m × v²)/3
P = (n × m × v²)/3
Step 4: Relate to kinetic energy:
Mean KE per molecule = (1/2)m(v_rms)² = (3/2)kT
But for mean square speed: (1/2)m<v²> = (3/2)kT
m<v²> = 3kT
Step 5: Therefore:
P = (n × m × v²)/3 = (n × 3kT)/3 = nkT = (N/V)kT
or: PV = NkT which gives P = (2/3)(N/V) × (1/2)m<v²> = (2/3) × (N/V) × KE_avg
Final Answer: P = (2/3) × (N/V) × KE_avg or P × V = (2/3) × N × KE_avg
Q5. A gas has mean free path λ = 10⁻⁷ m at pressure 1 atm. Calculate the mean free path at pressure 0.1 atm (same temperature).
Step 1: Mean free path formula:
λ = 1/(√2 × π × d² × n)
where d = molecular diameter, n = number density
Step 2: At constant temperature:
n ∝ P (from ideal gas law: n = P/kT)
Therefore: λ ∝ 1/n ∝ 1/P
Step 3: Ratio of mean free paths:
λ₁/λ₂ = P₂/P₁
Step 4: Substitute values:
λ₁ = 10⁻⁷ m at P₁ = 1 atm
P₂ = 0.1 atm
λ₂ = λ₁ × (P₁/P₂) = 10⁻⁷ × (1/0.1) = 10⁻⁷ × 10 = 10⁻⁶ m
Final Answer: Mean free path at 0.1 atm = 10⁻⁶ m or 1 μm
Q6. At 0°C, the density of a gas is 1.29 kg/m³. Calculate its density at 100°C (at same pressure).
Step 1: Ideal gas law: P = ρRT/M or ρ = PM/RT
At constant pressure and molar mass:
ρ ∝ 1/T
Step 2: Ratio of densities:
ρ₁/ρ₂ = T₂/T₁
Step 3: Substitute values:
T₁ = 0°C = 273 K
T₂ = 100°C = 373 K
ρ₁ = 1.29 kg/m³
ρ₂ = ρ₁ × (T₁/T₂) = 1.29 × (273/373)
ρ₂ = 1.29 × 0.732
ρ₂ = 0.944 kg/m³
Final Answer: Density at 100°C = 0.944 kg/m³
Q7. Calculate the number of molecules in 1 cm³ of a gas at STP. Given: k = 1.38 × 10⁻²³ J/K, at STP: P = 1.01 × 10⁵ Pa, T = 273 K.
Step 1: From ideal gas law: PV = NkT
where N = number of molecules
Step 2: Rearrange for N:
N = PV/(kT)
Step 3: Substitute values:
P = 1.01 × 10⁵ Pa
V = 1 cm³ = 1 × 10⁻⁶ m³
k = 1.38 × 10⁻²³ J/K
T = 273 K
N = (1.01 × 10⁵ × 1 × 10⁻⁶)/(1.38 × 10⁻²³ × 273)
N = (101)/(3.766 × 10⁻²¹)
N = 2.68 × 10¹⁹
Alternatively: At STP, 1 mole = 22.4 L
1 cm³ = 1 mL = 0.001 L
N = (6.022 × 10²³ × 0.001)/22.4 = 6.022 × 10²⁰/22.4 = 2.69 × 10¹⁹
Final Answer: Number of molecules = 2.68 × 10¹⁹ or approximately 2.7 × 10¹⁹
Q8. The speeds of five gas molecules are 100, 200, 300, 400, and 500 m/s. Calculate (a) the mean speed, (b) the RMS speed, (c) the most probable speed.
Step 1: Mean speed (arithmetic average):
v_mean = (v₁ + v₂ + v₃ + v₄ + v₅)/5
v_mean = (100 + 200 + 300 + 400 + 500)/5
v_mean = 1500/5 = 300 m/s
Step 2: RMS speed:
v_rms = √[(v₁² + v₂² + v₃² + v₄² + v₅²)/5]
v_rms = √[(100² + 200² + 300² + 400² + 500²)/5]
v_rms = √[(10,000 + 40,000 + 90,000 + 160,000 + 250,000)/5]
v_rms = √[550,000/5]
v_rms = √110,000
v_rms = 331.7 m/s
Step 3: Most probable speed:
For a set of given speeds, most probable speed is the one appearing most frequently.
Since each speed appears once, any could be considered.
But in this discrete case: v_prob = 300 m/s (middle value, or any of them)
Note: In Maxwell distribution, v_prob = √(2RT/M) ≈ 0.816 × v_rms
Final Answer: (a) Mean speed = 300 m/s; (b) RMS speed = 331.7 m/s or 332 m/s; (c) Most probable speed = 300 m/s (or any single speed in discrete case)
Kinetic Theory — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following is NOT a postulate of the kinetic theory of gases?
- A. Gas molecules are in random, ceaseless motion.
- B. The volume occupied by gas molecules is negligible compared to the total volume of the gas.
- C. Collisions between gas molecules and container walls are perfectly inelastic.
- D. There are no intermolecular forces between gas molecules.
Answer: C. Collisions between gas molecules and container walls are perfectly inelastic. — According to the kinetic theory of gases, collisions between gas molecules and the container walls (and among molecules themselves) are considered perfectly elastic, meaning kinetic energy is conserved during collisions.
MCQ 2. The average translational kinetic energy of a gas molecule is directly proportional to which of the following?
- A. Volume of the gas
- B. Pressure of the gas
- C. Absolute temperature of the gas
- D. Density of the gas
Answer: C. Absolute temperature of the gas — The average translational kinetic energy of a gas molecule is given by (3/2)kT, where k is the Boltzmann constant and T is the absolute temperature. Thus, it is directly proportional to the absolute temperature.
MCQ 3. The root mean square (RMS) speed of gas molecules is given by the formula:
- A. √(3RT/M)
- B. √(2RT/M)
- C. √(RT/M)
- D. √(8RT/πM)
Answer: A. √(3RT/M) — The root mean square speed of gas molecules is given by v_rms = √(3RT/M), where R is the universal gas constant, T is the absolute temperature, and M is the molar mass.
MCQ 4. How many degrees of freedom does a monoatomic gas molecule possess at room temperature?
- A. 1
- B. 2
- C. 3
- D. 5
Answer: C. 3 — A monoatomic gas molecule can only undergo translational motion along three independent axes (x, y, z), hence it has 3 translational degrees of freedom and no rotational or vibrational degrees of freedom at room temperature.
MCQ 5. According to the law of equipartition of energy, the energy associated with each degree of freedom for a gas molecule in thermal equilibrium is:
- A. kT
- B. (1/2)kT
- C. (3/2)kT
- D. RT
Answer: B. (1/2)kT — The law of equipartition of energy states that for a system in thermal equilibrium, the total energy is equally distributed among its independent degrees of freedom, with each degree of freedom having an average energy of (1/2)kT.
MCQ 6. For an ideal gas, the relation between molar specific heat at constant pressure (Cp) and molar specific heat at constant volume (Cv) is given by Meyer's formula. What is this relation?
- A. Cp + Cv = R
- B. Cp - Cv = R
- C. Cp / Cv = R
- D. Cv - Cp = R
Answer: B. Cp - Cv = R — Meyer's formula states that for an ideal gas, the difference between the molar specific heat at constant pressure and molar specific heat at constant volume is equal to the universal gas constant R (Cp - Cv = R).
MCQ 7. What is the theoretical value of the ratio of specific heats (γ = Cp/Cv) for an ideal diatomic gas at room temperature?
- A. 1.33
- B. 1.40
- C. 1.67
- D. 1.00
Answer: B. 1.40 — A diatomic gas at room temperature has 3 translational and 2 rotational degrees of freedom, totaling 5 degrees of freedom (f=5). Then Cv = (f/2)R = (5/2)R and Cp = (f/2 + 1)R = (7/2)R. Thus, γ = Cp/Cv = (7/2)R / (5/2)R = 7/5 = 1.40.
MCQ 8. If the absolute temperature of an ideal gas is doubled, its RMS speed will change by a factor of:
- A. 2
- B. √2
- C. 1/2
- D. 4
Answer: B. √2 — The RMS speed is directly proportional to the square root of the absolute temperature (v_rms ∝ √T). If T is doubled, v_rms will increase by a factor of √2.
MCQ 9. The mean free path of gas molecules at constant temperature is inversely proportional to:
- A. Volume of the gas
- B. Pressure of the gas
- C. Density of the gas
- D. Average kinetic energy
Answer: B. Pressure of the gas — The mean free path (λ) is inversely proportional to the number density of molecules (n). Since pressure P = nkT, at constant temperature, P is proportional to n, so λ is inversely proportional to the pressure.
MCQ 10. Which of the following equations represents the ideal gas law for n moles of gas?
- A. PV = nRT
- B. P/V = nRT
- C. PV = nR/T
- D. P + V = nRT
Answer: A. PV = nRT — The ideal gas law, also known as the general gas equation, relates the pressure (P), volume (V), temperature (T), and amount (n moles) of an ideal gas through the equation PV = nRT.
MCQ 11. An ideal gas is heated from 27°C to 127°C. The ratio of the average translational kinetic energies of its molecules (KE₁:KE₂) will be:
- A. 1:2
- B. 2:1
- C. 3:4
- D. 4:3
Answer: C. 3:4 — The average translational kinetic energy is directly proportional to the absolute temperature. Convert temperatures to Kelvin: T₁ = 27 + 273 = 300 K, T₂ = 127 + 273 = 400 K. The ratio of kinetic energies KE₁:KE₂ will be T₁:T₂ = 300:400 = 3:4.
MCQ 12. If the temperature of 1 mole of a monoatomic ideal gas is increased by 1°C at constant volume, the change in its internal energy will be:
- A. (1/2)R
- B. (3/2)R
- C. R
- D. (5/2)R
Answer: B. (3/2)R — For a monoatomic gas, the number of translational degrees of freedom (f) is 3. The change in internal energy (ΔU) for n=1 mole is (f/2)RΔT. With f=3 and ΔT=1°C (which is 1K), ΔU = (3/2)R(1) = (3/2)R.
MCQ 13. For a given gas at a specific temperature, arrange the most probable speed (v_mp), average speed (v_avg), and root mean square speed (v_rms) in increasing order:
- A. v_mp < v_avg < v_rms
- B. v_rms < v_avg < v_mp
- C. v_avg < v_mp < v_rms
- D. v_mp = v_avg = v_rms
Answer: A. v_mp < v_avg < v_rms — The characteristic speeds are related as v_mp = √(2RT/M), v_avg = √(8RT/πM), and v_rms = √(3RT/M). Numerically, √2 ≈ 1.414, √(8/π) ≈ 1.596, √3 ≈ 1.732. Thus, v_mp < v_avg < v_rms.
MCQ 14. An ideal gas behaves as such primarily because:
- A. Its molecules are infinitely small.
- B. There are no forces of attraction or repulsion between its molecules.
- C. Its molecules occupy a negligible volume.
- D. All of the above.
Answer: D. All of the above. — The kinetic theory of ideal gases makes several key assumptions, including negligible molecular volume, no intermolecular forces, and random elastic collisions, which contribute to its ideal behavior.
MCQ 15. If the pressure of an ideal gas is doubled while its temperature is kept constant, what happens to the RMS speed of its molecules?
- A. It doubles.
- B. It becomes half.
- C. It remains unchanged.
- D. It increases by a factor of √2.
Answer: C. It remains unchanged. — The RMS speed (v_rms = √(3RT/M)) depends only on the absolute temperature (T) and the molar mass (M) of the gas. Since the temperature is constant, the RMS speed of the molecules remains unchanged, irrespective of pressure changes.
MCQ 16. The total translational kinetic energy of N molecules of an ideal gas at temperature T is given by:
- A. (1/2)NkT
- B. (3/2)NkT
- C. NkT
- D. (3/2)RT
Answer: B. (3/2)NkT — The average translational kinetic energy per molecule is (3/2)kT. For N molecules, the total translational kinetic energy is N times this value, i.e., (3/2)NkT.
MCQ 17. The SI unit of molar specific heat capacity is:
- A. J kg⁻¹ K⁻¹
- B. J mol⁻¹ K⁻¹
- C. J K⁻¹
- D. J m⁻³ K⁻¹
Answer: B. J mol⁻¹ K⁻¹ — Molar specific heat capacity is defined as the heat required to raise the temperature of one mole of a substance by one Kelvin. Its SI unit is Joule per mole per Kelvin (J mol⁻¹ K⁻¹).
MCQ 18. A non-linear polyatomic gas molecule has how many degrees of freedom at room temperature?
- A. 3
- B. 5
- C. 6
- D. 7
Answer: C. 6 — A non-linear polyatomic molecule has 3 translational degrees of freedom and 3 rotational degrees of freedom (rotation about three perpendicular axes). Vibrational degrees of freedom are typically not active at room temperature, making a total of 3+3=6 degrees of freedom.
MCQ 19. For an ideal gas, if its temperature is kept constant, which of the following quantities remains constant?
- A. Pressure
- B. Volume
- C. Internal energy
- D. Number of molecules
Answer: C. Internal energy — For an ideal gas, the internal energy is solely a function of its absolute temperature. Therefore, if the temperature is constant (isothermal process), the internal energy of the gas remains constant.
MCQ 20. Two different ideal gases, A and B, are at the same temperature. If the mass of a molecule of gas A is twice that of gas B, what is the ratio of their average translational kinetic energies per molecule (KE_A : KE_B)?
- A. 2:1
- B. 1:2
- C. 1:1
- D. √2:1
Answer: C. 1:1 — The average translational kinetic energy of gas molecules (3/2 kT) depends only on the absolute temperature, not on the mass of the individual molecules. Since both gases are at the same temperature, their average translational kinetic energies per molecule are equal (1:1).
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.