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Kinetic Theory – Class 11 Physics notes, important questions and MCQs

NCERT Class 11 Physics, chapter 12. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Kinetic theory explains gas behaviour from the motion of molecules. It links pressure and temperature to molecular speeds and energy.

Gas laws and pressure

Ideal gas equation PV = nRT = Nk_BT; R = 8.314 J/(mol K).
Pressure P = ⅓ρv²_rms = ⅓(N/V)mv²_rms.
Average KE per molecule = (3/2)k_BT; temperature measures average KE.

Speeds and energy

v_rms = √(3RT/M); v_avg = √(8RT/πM); v_mp = √(2RT/M); v_rms > v_avg > v_mp.
Law of equipartition: each degree of freedom gets ½k_BT.
Cᵥ for monatomic gas (3/2)R, γ = 5/3; diatomic (5/2)R, γ = 7/5; mean free path λ = 1/(√2 πd²n).

Key terms

Degrees of freedom: independent ways a molecule can store energy
Mean free path: average distance between successive collisions
Avogadro's hypothesis: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules

Formulas

PV = nRT
v_rms = √(3RT/M)
KE_avg = (3/2)k_BT

Important questions with answers

Why does the pressure of a gas increase when it is heated in a closed container?2 marks

The molecules move faster, hitting the walls more often and with greater momentum change, so the force per unit area (pressure) rises.

State the law of equipartition of energy and find Cᵥ and γ for a diatomic gas.3 marks

In equilibrium, each degree of freedom gets ½k_BT of energy per molecule. A diatomic molecule (at room temperature) has 3 translational + 2 rotational = 5 degrees of freedom: U = (5/2)RT per mole, Cᵥ = (5/2)R, Cₚ = (7/2)R, γ = 7/5 = 1.4.

Derive the expression for the pressure exerted by an ideal gas, P = ⅓ρv²_rms, stating the assumptions, and show that average KE per molecule is (3/2)k_BT.5 marks

Assumptions: molecules are point masses; collisions are elastic; no forces except during collisions; random motion. A molecule of mass m with velocity component vₓ hits a wall and its momentum changes by 2mvₓ. In a cube of side l, it hits the same wall every 2l/vₓ, so force = mvₓ²/l. For N molecules, F = (m/l)Σvₓ², and by symmetry Σvₓ² = ⅓Σv² = ⅓Nv²_rms. P = F/l² = (1/3)(Nm/V)v²_rms = ⅓ρv²_rms. Then PV = ⅓Nmv²_rms; comparing with PV = Nk_BT: ½mv²_rms = (3/2)k_BT.

Quick revision

Temperature ∝ average KE.
Monatomic γ = 5/3.
Cₚ − Cᵥ = R.

Exam tips

v_rms ∝ √T: doubling T multiplies v_rms by √2.
Use T in kelvin in all kinetic theory formulas, and M in kg/mol for v_rms = √(3RT/M).

Practice MCQs

Value of R is about
1. 6.02 × 10²³
2. 1.38 × 10⁻²³ J/K
3. 8.314 J/(mol K)
4. 9.8 J/K
Show answer
Answer: (c) 8.314 J/(mol K). Gas constant.
Average KE of a gas molecule depends on
1. pressure only
2. volume only
3. temperature
4. shape
Show answer
Answer: (c) temperature. (3/2)k_BT.
For a monatomic gas, γ is
1. 7/5
2. 4/3
3. 5/3
4. 1
Show answer
Answer: (c) 5/3. Cₚ/Cᵥ = (5/2)/(3/2).
If absolute temperature becomes 4 times, v_rms becomes
1. ½
2. 4 times
3. 16 times
4. 2 times
Show answer
Answer: (d) 2 times. v ∝ √T.
v_rms of oxygen (M = 0.032 kg/mol) at 300 K is about (R = 8.31)
1. 48 m/s
2. 4830 m/s
3. 483 m/s
4. 1500 m/s
Show answer
Answer: (c) 483 m/s. √(3 × 8.31 × 300/0.032).
Correct order of molecular speeds is
1. v_rms > v_avg > v_mp
2. v_mp > v_avg > v_rms
3. v_avg > v_rms > v_mp
4. all equal
Show answer
Answer: (a) v_rms > v_avg > v_mp. Standard order.
Cₚ − Cᵥ for an ideal gas equals
1. 0
2. 2R
3. R/2
4. R
Show answer
Answer: (d) R. Mayer's relation.
Ratio of v_rms of hydrogen to oxygen at the same temperature is
1. 1 : 4
2. 16 : 1
3. 4 : 1
4. 1 : 16
Show answer
Answer: (c) 4 : 1. v ∝ 1/√M.

All chapters of Class 11 Physics

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.