CBSE Class 11 Physics Chapter 13 Notes: Kinetic Theory

Chapter Introduction

How can the invisible motion of trillions of molecules explain the pressure of a gas, its temperature, and the laws it obeys? Kinetic theory answers this by building a microscopic model of an ideal gas. Chapter 13 of NCERT Physics for Class 11 derives Boyle's law, Charles's law, and the ideal gas equation from the molecular picture, introduces the equipartition theorem to explain why different gases have different specific heats, and connects the microscopic (molecular speeds, collisions) to the macroscopic (pressure, temperature, heat capacity). This chapter is the bridge between thermodynamics (Ch 12) and statistical mechanics.

Key Definitions

Detailed Concepts

Molecular Nature of Matter and Behaviour of Gases

The kinetic theory makes the following assumptions for an ideal gas: 1. Gas consists of a large number of identical point molecules (negligible volume) 2. Molecules are in constant random motion 3. All collisions (molecule-molecule and molecule-wall) are perfectly elastic 4. No intermolecular forces except during collisions 5. The effect of gravity on molecular motion is negligible

Gas Laws from Kinetic Theory

Boyle's law: At constant temperature, the pressure of a given mass of gas is inversely proportional to its volume.

P ∝ 1/V (at constant T) → PV = constant

From kinetic theory: P = (1/3)nmv²_rms, and at constant T, v_rms is constant. If N is fixed, P ∝ 1/V.

Charles's law: At constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature.

V ∝ T (at constant P) → V/T = constant

From kinetic theory: P = (1/3)(N/V)mv²_rms and v_rms ∝ √T. At constant P, V must increase proportionally with T.

Ideal gas equation: Combining Boyle's and Charles's laws:

PV = nRT

where n = number of moles, R = 8.314 J mol⁻¹ K⁻¹ (universal gas constant)

Kinetic Theory of an Ideal Gas — Pressure and RMS Speed

Consider N molecules of mass m in a cubical container of side L. The pressure on a wall is derived from the momentum transfer of molecules colliding with it:

P = (1/3) (N/V) m v²_rms

where v²_rms = mean of the squares of molecular speeds = (v₁² + v₂² + ... + v_N²)/N

Using the ideal gas equation PV = NkT (where N = number of molecules, k = Boltzmann constant):

PV = NkT = (1/3) Nmv²_rms

Therefore:

• v_rms = √(3kT/m) = √(3RT/M) where M = molar mass
• v_avg = √(8kT/πm) = √(8RT/πM)
• v_mp = √(2kT/m) = √(2RT/M)

Ratio: v_mp : v_avg : v_rms = √2 : √(8/π) : √3 ≈ 1 : 1.128 : 1.224

Kinetic interpretation of temperature:

From (1/2)mv²_rms = (3/2)kT, the average translational kinetic energy per molecule is:

E = (3/2)kT

Temperature is a measure of the average kinetic energy of molecules. At T = 0 K, all molecular motion ceases (classical picture).

Law of Equipartition of Energy

The law states: Each degree of freedom contributes ½kT of energy per molecule (or ½RT per mole).

Why diatomic has 2 rotational DOF: A diatomic molecule can rotate about two axes perpendicular to its bond axis. Rotation about the bond axis has negligible moment of inertia (point masses), so it is not counted.

Why polyatomic has 3 rotational DOF: A polyatomic molecule (non-linear) can rotate about all three mutually perpendicular axes.

Specific Heat Capacity of Gases

For an ideal gas, the specific heats are determined entirely by the degrees of freedom:

C_v = (f/2)R (per mole), where f = total DOF

C_p = C_v + R (Mayer's relation)

γ = C_p/C_v = (f + 2)/f

Note: At moderate temperatures, vibrational DOF are "frozen" (not excited). At very high temperatures, they contribute. This explains why measured γ values deviate from the simple model at extreme temperatures.

Mean Free Path

The mean free path (λ) is the average distance a molecule travels between successive collisions:

λ = 1 / (√2 π n d²)

where n = number density of molecules (N/V), d = molecular diameter.

• λ increases when pressure decreases (n decreases) — fewer molecules per unit volume means fewer collisions.
• λ increases when temperature increases at constant pressure (n decreases as V increases).
• At STP, λ for air ≈ 68 nm.

Important Derivations

1. Pressure of an Ideal Gas

Consider a molecule of mass m and velocity v hitting a wall of the container (area A = L²) at the x-face.

Change in momentum per collision: Δp = 2mv_x

Time between collisions with the same wall: Δt = 2L/v_x

Force by one molecule: F₁ = 2mv_x / (2L/v_x) = mv_x²/L

Total force from N molecules: F = Σ(mv_x²/L) = (Nm/L) × v²_x_mean = (N/V) × m × v²_x_rms

Since v²_rms = v²_x + v²_y + v²_z and by symmetry v²_x = v²_y = v²_z = v²_rms/3:

P = F/A = (1/3)(N/V)mv²_rms

2. Mayer's Relation (C_p – C_v = R)

For 1 mole of ideal gas at constant volume: Q_v = C_v ΔT = ΔU (since W = 0)

For 1 mole of ideal gas at constant pressure: Q_p = C_p ΔT = ΔU + W = C_v ΔT + PΔV = C_v ΔT + RΔT

3. Degrees of Freedom and γ

For f degrees of freedom: C_v = fR/2, C_p = C_v + R = (f + 2)R/2

γ = C_p/C_v = (f + 2)/f

• Monoatomic (f = 3): γ = 5/3
• Diatomic (f = 5): γ = 7/5
• Polyatomic (f = 6): γ = 8/6 = 4/3

Common Mistakes

1. Using R in molecular formula instead of k: PV = NkT uses k (Boltzmann constant) for N molecules; PV = nRT uses R (gas constant) for n moles. Nk = nR. Do not substitute R where k belongs. 2. Confusing m (molecular mass) with M (molar mass): v_rms = √(3kT/m) = √(3RT/M). Using molar mass M in the k formula (without dividing by N_A) gives wrong results. 3. Forgetting that equipartition gives energy per molecule, not per mole: Each DOF contributes ½kT per molecule (= ½RT per mole). When calculating total energy, decide whether you need per-molecule or per-mole quantities. 4. Assuming all molecules have the same speed: The Maxwell-Boltzmann distribution shows a spread. v_rms, v_avg, and v_mp are different statistical measures of this distribution. 5. Counting vibrational DOF at room temperature: At room temperature, vibrational modes of diatomic molecules are "frozen" (energy gap too large for kT to excite). Only translational (3) and rotational (2) DOF are active, giving f = 5, not 7. 6. Mixing up C_p and C_v: C_p is always greater than C_v by R. C_p is the heat at constant pressure (gas expands and does work); C_v is at constant volume (no work done).

Numerical Practice

Example 1: RMS Speed

Calculate the RMS speed of nitrogen molecules (M = 28 g/mol) at 300 K.

v_rms = √(3RT/M) = √(3 × 8.314 × 300 / 0.028) = √(7482.6 / 0.028) = √(267235) ≈ 517 m/s

Example 2: Degrees of Freedom and Specific Heats

Find C_v, C_p, and γ for oxygen (O₂, diatomic).

f = 5 (3 translational + 2 rotational)

C_v = (5/2)R = 2.5 × 8.314 = 20.8 J mol⁻¹ K⁻¹

C_p = C_v + R = 20.8 + 8.314 = 29.1 J mol⁻¹ K⁻¹

γ = C_p/C_v = 29.1/20.8 = 1.40

Example 3: Mean Free Path

If the molecular diameter of nitrogen is 0.3 nm and the number density at STP is 2.7 × 10²⁵ m⁻³, find the mean free path.

λ = 1/(√2 π n d²) = 1/(√2 × π × 2.7 × 10²⁵ × (0.3 × 10⁻⁹)²)

= 1/(1.414 × 3.14 × 2.7 × 10²⁵ × 9 × 10⁻²⁰)

= 1/(1.414 × 3.14 × 2.43 × 10⁶)

= 1/(1.08 × 10⁷) ≈ 9.3 × 10⁻⁸ m = 93 nm

Exam Tips

1. Pressure derivation: The derivation of P = (1/3)(N/V)mv²_rms from molecular collisions is a 5-mark question favourite. Practise it step by step with a diagram of the cubical container. 2. Gas laws from kinetic theory: Be ready to show how Boyle's and Charles's laws follow from P = (1/3)(N/V)mv²_rms and v_rms ∝ √T. 3. Equipartition theorem: Know the DOF table for monoatomic, diatomic, and polyatomic gases. This is the most commonly tested concept. 4. γ values: Memorise γ = 5/3 (monoatomic), 7/5 (diatomic), 4/3 (polyatomic). Quick recall saves time in MCQs. 5. Mayer's relation derivation: A 2–3 mark standard derivation using the first law of thermodynamics. 6. RMS speed numericals: Always convert molar mass to kg/mol before using v_rms = √(3RT/M). 7. Mean free path: Usually a 1-mark formula recall or a short numerical.

NCERT Exercise Focus Areas

The table below highlights the most frequently tested NCERT exercise question types for this chapter, along with typical mark allocation and difficulty level as observed in CBSE board exams.

Board exam pattern: This chapter typically carries 3–5 marks in the theory paper. Derivations (pressure, Mayer's relation) are the high-value questions. Numerical problems on RMS speed and mean free path appear regularly in the 2-mark segment. MCQs frequently test γ values and DOF counts for quick recall.

Conceptual Questions for Practice

Test your understanding of kinetic theory with these conceptual questions. Try answering them before checking your notes.

1. Gases have two principal specific heats (C_p and C_v), but solids generally have only one. Why?

2. On a humid day, sound travels faster than on a dry day. How does kinetic theory explain this?

3. The RMS speed of hydrogen molecules at room temperature is very high (nearly 2000 m/s), yet hydrogen does not escape from an open bottle instantly. Why?

4. Two gases — helium and carbon dioxide — are at the same temperature. Which has the higher RMS speed, and why?

5. According to the equipartition theorem, each degree of freedom contributes ½kT of energy. Why then do vibrational degrees of freedom of diatomic molecules not contribute at room temperature?

Quick Revision Notes

DOF summary: Monoatomic = 3, Diatomic = 5, Polyatomic = 6

γ values: 5/3 (mono), 7/5 (di), 4/3 (poly)

Speed ratio: v_mp : v_avg : v_rms = 1 : 1.128 : 1.224