Kinetic Theory of Gases Class 11 Physics Quick Recall (Short Notes 2026-27)

[!TIP] 🚀 2-Minute Quick Recall Summary (Save for Exam Day)

• Ideal Gas Pressure: P = ⅓ ρ vᵣₘₛ².
• RMS Speed: v_rms = √(3RT/M) = √(3kT/m).
• Average K.E.: E = (3/2) kT (per molecule).
• Degrees of Freedom (f): Monoatomic = 3; Diatomic = 5; Triatomic = 6.
• Mayer's Relation: Cp - Cv = R. γ = Cp/Cv = 1 + 2/f. 📥 Download 1-Page Short Notes PDF (Zero-Friction)

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Introduction

The Kinetic Theory of Gases (KTG) provides a bridge between the macroscopic properties of gases (Pressure, Volume, Temperature) and the microscopic behavior of individual molecules. It treats a gas as a collection of billions of tiny, rapidly moving particles in constant random motion. Understanding these molecular dynamics is essential for explaining heat, thermodynamics, and the very nature of matter. In this "Comprehensive" guide, we provide exhaustive derivations for the Pressure of an Ideal Gas, the Kinetic Interpretation of Temperature, and the Law of Equipartition of Energy—providing the technical depth required for top-tier performance in JEE and NEET.

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1. Postulates of the Kinetic Theory

To model an "Ideal Gas," we assume:

1. Gases consist of large numbers of identical, tiny, spherical, and perfectly elastic particles (molecules).
2. Molecules are in continuous, random, straight-line motion.
3. The volume of actual molecules is negligible compared to the volume of the container.
4. There are no attractive or repulsive forces between molecules except during collisions.
5. Collisions are perfectly elastic and instantaneous.

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2. Derivation Master-Sheet: Pressure of an Ideal Gas

Theorem: The pressure exerted by an ideal gas is P = 1/3 ρ v_rms².

Derivation:

1. Consider a cubic container of side L containing N molecules each of mass m.
2. A molecule moving with velocity v_x strikes the wall perpendicular to the X-axis.
3. Change in momentum (Δp): Final - Initial = (-mv_x) - (mv_x) = -2mv_x.
4. Force (F): Rate of change of momentum. F = Δp / Δt.
5. Time between collisions with the same wall: Δt = 2L / v_x.
6. Force of one molecule (f): 2mv_x / (2L/v_x) = mv_x² / L.
7. Total Force (F): Σ (mv_xi²/L) = (m/L) Σ v_xi².
8. Pressure (P): Force / Area = (m/L³) Σ v_xi² = (m/V) Σ v_xi².
9. By symmetry, Σ v_x² = Σ v_y² = Σ v_z² = 1/3 Σ v².
10. P = (m N / 3V) v_rms² = 1/3 ρ v_rms². (Proven)

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3. Kinetic Interpretation of Temperature

Derivation:

1. From P = 1/3 ρ v_rms² => PV = 1/3 M v_rms².
2. From Ideal Gas Law: PV = nRT.
3. 1/3 M v_rms² = nRT.
4. Average Kinetic Energy of one mole: E = 1/2 M v_rms² = 3/2 nRT.
5. Average Kinetic Energy per molecule: E_avg = 3/2 (R/N_A) T = 3/2 kT. (where k = Boltzmann constant). Conclusion: Temperature is a direct measure of the average kinetic energy per molecule.

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4. Degrees of Freedom and Equipartition

Degree of Freedom (f): The number of independent ways in which a molecule can possess energy.

• Monoatomic (He, Ar): 3 Translational. f = 3.
• Diatomic (H2, O2): 3 Translational + 2 Rotational. f = 5.
• Polyatomic: 3 Translational + 3 Rotational + Vibrational. f > 6.

Law of Equipartition of Energy: In thermal equilibrium, the total energy is equally distributed among all degrees of freedom, and the energy associated with each degree is 1/2 kT.

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5. Specific Heat Capacities & Mayer's Relation

I. Derivation: Relation between Cv and f

1. Internal Energy (U) = f × (1/2 RT) per mole.
2. Cv = dU/dT = f/2 R.
3. Cp = Cv + R = (f/2 + 1) R.
4. γ = Cp / Cv = 1 + 2/f. (Proven)

II. Derivation: Mayer's Relation (Cp - Cv = R)

1. For 1 mole of gas: dQ = dU + dW.
2. At constant volume: Cv dT = dU (since dW = 0).
3. At constant pressure: Cp dT = dU + P dV.
4. Substitute dU: Cp dT = Cv dT + P dV.
5. From PV = RT (at constant P): P dV = R dT.
6. Cp dT = Cv dT + R dT.
7. Cp - Cv = R. (Proven)

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Comprehensive Exam Strategy (Q&A)

Q1: Why does the root-mean-square (rms) velocity increase with temperature? Answer: According to the Kinetic Theory, v_rms = √(3RT/M). As the temperature T increases, the kinetic energy supplied to the molecules increases their speed. Thus, v_rms is directly proportional to the square root of the absolute temperature.

Q2: Does an ideal gas have potential energy? Answer: No. By postulate 4 of KTG, there are no intermolecular forces in an ideal gas. Since potential energy is defined by mutual attraction/repulsion between particles, an ideal gas consists only of kinetic energy.

Q3: Compare the specific heat ratio (γ) for Helium and Oxygen. Answer:

• Helium (Monoatomic, f=3): γ = 1 + 2/3 = 1.67.
• Oxygen (Diatomic, f=5): γ = 1 + 2/5 = 1.40. Helium has a higher ratio of specific heats than Oxygen.

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Related Revision Notes

• Chapter 11: Thermodynamics (Internal Energy Deep-Dive)
• Chapter 10: Thermal Properties (Specific Heat Basics)
• KTG and Gas Laws: Advanced Numerical Vault

Conclusion

The Kinetic Theory of Gases transforms our view of matter from static substances to a dynamic dance of particles. By mastering the molecular derivations of pressure and energy, you gain the ability to predict the macroscopic behavior of any gas from its microscopic components. Master the derivation of the Pressure of an Ideal Gas and the Law of Equipartition—these are the tools that allow us to understand the atmosphere, chemical reactions, and the physics of the stars themselves. Stay fast, stay elastic, and keep your degrees of freedom wide!

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Reference: Encyclopaedia Britannica: Kinetic Theory of Gases