Class XI · Chapter 12
Kinetic Theory — Exam Practice
Practice questions in CBSE, NEET, and JEE Main style covering gas laws, kinetic theory pressure, rms speed, and specific heats.
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
1CBSE Board Exam Style
Q1. Why does the pressure of a gas increase when it is heated at constant volume?
Answer
Increased temperature raises the average kinetic energy (and hence speed) of gas molecules, causing more frequent and more forceful collisions with the container walls — this increased rate of momentum transfer is exactly what pressure measures.
Increased temperature raises the average kinetic energy (and hence speed) of gas molecules, causing more frequent and more forceful collisions with the container walls — this increased rate of momentum transfer is exactly what pressure measures.
Derive the expression relating gas pressure to the mean square speed of its molecules, based on the kinetic theory.
Answer
Considering molecular collisions with a container wall and the resulting momentum transfer, kinetic theory analysis (summed over all molecules and averaged over all directions) gives:P = (1/3) ρ (v²)avg, where ρ is gas density and (v²)avg is the mean square speed of the molecules — showing pressure emerges directly from molecular motion, not as a separate assumption.
Considering molecular collisions with a container wall and the resulting momentum transfer, kinetic theory analysis (summed over all molecules and averaged over all directions) gives:P = (1/3) ρ (v²)avg, where ρ is gas density and (v²)avg is the mean square speed of the molecules — showing pressure emerges directly from molecular motion, not as a separate assumption.
Q3. (a) State the assumptions of the kinetic theory of gases. (b) Find the rms speed of nitrogen molecules at 300 K. (Molar mass of N₂ = 28×10⁻³ kg/mol, R = 8.314 J/mol·K)
Answer
(a) Molecules are in continuous random motion; collisions are perfectly elastic; molecules exert no force on each other except during collision; the total volume of the molecules is negligible compared to the container; molecules obey Newton’s laws.(b) vrms = √(3RT/M) = √[(3×8.314×300)/0.028] = √267,236 ≈ 517 m/s.
(a) Molecules are in continuous random motion; collisions are perfectly elastic; molecules exert no force on each other except during collision; the total volume of the molecules is negligible compared to the container; molecules obey Newton’s laws.(b) vrms = √(3RT/M) = √[(3×8.314×300)/0.028] = √267,236 ≈ 517 m/s.
2NEET Style (Single-Correct MCQ)
Q4. The average kinetic energy of a gas molecule depends only on:
Solution
KEavg = (3/2)kT — this depends only on temperature, not on the gas’s identity, pressure, or volume.
KEavg = (3/2)kT — this depends only on temperature, not on the gas’s identity, pressure, or volume.
Q5. At the same temperature, lighter gas molecules have:
Solution
vrms = √(3RT/M) — for the same average kinetic energy at a given temperature, lighter molecules (smaller M) must move faster.
vrms = √(3RT/M) — for the same average kinetic energy at a given temperature, lighter molecules (smaller M) must move faster.
Q6. The number of degrees of freedom for a monatomic gas molecule is:
Solution
A monatomic molecule can only translate in three independent directions (x, y, z) — no rotational or vibrational modes are relevant at ordinary temperatures.
A monatomic molecule can only translate in three independent directions (x, y, z) — no rotational or vibrational modes are relevant at ordinary temperatures.
Q7. The mean free path of gas molecules decreases with an increase in:
Solution
λ = 1/(√2πd²n) — a higher number density n (more molecules per unit volume, from higher density/pressure) means more frequent collisions and a shorter mean free path.
λ = 1/(√2πd²n) — a higher number density n (more molecules per unit volume, from higher density/pressure) means more frequent collisions and a shorter mean free path.
3JEE Main Style
Q8. Find the average translational kinetic energy of one mole of an ideal gas at 300 K. (R = 8.314 J/mol·K)
Solution
KE (per mole) = (3/2)RT = (3/2) × 8.314 × 300 ≈ 3741 J.
KE (per mole) = (3/2)RT = (3/2) × 8.314 × 300 ≈ 3741 J.
Q9. The ratio of specific heats (Cp/Cv) for a diatomic gas is:
Solution
For a diatomic gas, Cv = (5/2)R, Cp = (7/2)R, giving γ = Cp/Cv = 7/5 = 1.4.
For a diatomic gas, Cv = (5/2)R, Cp = (7/2)R, giving γ = Cp/Cv = 7/5 = 1.4.
Q10. At the same temperature, find the ratio of rms speeds of hydrogen (M=2 g/mol) to oxygen (M=32 g/mol).
Solution
vrms ∝ 1/√M, so ratio (H₂/O₂) = √(MO₂/MH₂) = √(32/2) = √16 = 4 — hydrogen molecules move 4× faster on average.
vrms ∝ 1/√M, so ratio (H₂/O₂) = √(MO₂/MH₂) = √(32/2) = √16 = 4 — hydrogen molecules move 4× faster on average.
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