Class 11 Physics: Kinetic Theory | Physics Q&A

CBSE Class 11 › Unit IX: Behaviour of Perfect Gases

Kinetic Theory

NCERT Chapter 12 • Ideal Gas, Pressure, Equipartition of Energy & Mean Free Path

NCERT 2025–26 Unit IX • ~6 Marks JEE Main • 1-2 Questions

Visualization of random molecular motion in an ideal gas. — Kinetic Theory: Explaining macroscopic properties like pressure and temperature through microscopic molecular motion.

1. Introduction & Historical Context

In 1661, Robert Boyle discovered the law named after him. Boyle, Newton and several others tried to explain the behaviour of gases by considering that gases are made up of tiny atomic particles. The actual atomic theory got established more than 150 years later. Kinetic theory explains the behaviour of gases based on the idea that the gas consists of rapidly moving atoms or molecules. This is possible as the inter-atomic forces, which are short range forces that are important for solids and liquids, can be neglected for gases. The kinetic theory was developed in the nineteenth century by Maxwell, Boltzmann and others. It has been remarkably successful. It gives a molecular interpretation of pressure and temperature of a gas, and is consistent with gas laws and Avogadro’s hypothesis. It correctly explains specific heat capacities of many gases. It also relates measurable properties of gases such as viscosity, conduction and diffusion with molecular parameters, yielding estimates of molecular sizes and masses.

2. Molecular Nature of Matter

Richard Feynman, one of the great physicists of 20th century considers the discovery that “Matter is made up of atoms” to be a very significant one…

Atomic Hypothesis: All things are made of atoms—little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another.

Ancient Ideas: Kanada (India, 6th century BCE) proposed Paramanu. Democritus (Greece, 4th century BCE) coined “atom” (indivisible).

3. Behaviour of Gases & Ideal Gas Equation

Gases at low pressures and high temperatures much above that at which they liquefy approximately satisfy:

PV = µRT = N k_BT

Where R = 8.314 J/mol·K, k_B = 1.38 × 10⁻²³ J/K, N_A = 6.02 × 10²³ mol⁻¹.

At STP, 1 mole occupies 22.4 litres.

Dalton’s Law of Partial Pressures: P = P₁ + P₂ + …

3A. Real Gases vs. Ideal Gases

Real gases obey the Ideal Gas Law only at low pressure and high temperature. Deviations occur because:

Intermolecular forces are not negligible at high pressure.
Volume of molecules is not negligible compared to container volume.

Deviation of real gases from ideal behavior at high pressure. — Real gases deviate from Boyle’s law (PV=constant) at high pressures.

Van der Waals Equation (for real gases):

(P + a n²/V²)(V − n b) = n R T

where a accounts for intermolecular attraction and b for excluded volume.

4. Pressure of an Ideal Gas

Pressure arises from momentum transfer during molecular collisions with walls.

Diagram showing the origin of gas pressure from molecular collisions. — Pressure is the result of billions of molecular collisions with the container walls.

Derivation: Pressure Formula 3 Marks

Step 1: Momentum Change
A molecule (mass m, x-velocity v_x) hits wall elastically.
Δp = –mv_x – (mv_x) = –2mv_x.
Momentum imparted to wall = +2mv_x.

Step 2: Collision Rate
In time Δt, molecules within distance v_xΔt hit wall.
Volume = A v_x Δt → Number hitting = ½ n A v_x Δt.

Step 3: Total Force
Total momentum transferred = (½ n A v_x Δt)(2m v_x) = n m A v_x² Δt.
Force F = Δp/Δt = n m A v_x².

Step 4: Average & Isotropy
P = F/A = n m v_x².
Due to randomness: ⟨v_x²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩ = ⅓ ⟨v²⟩.
P = ⅓ n m ⟨v²⟩ = ⅓ ρ ⟨v²⟩

4A. Deduction of Gas Laws from Kinetic Theory

We know from kinetic theory: PV = ⅔ E, where E is total translational kinetic energy.

Boyle’s Law: If T is constant, average KE is constant ⇒ E constant ⇒ PV = constant.
Charles’ Law: Since E ∝ T, at constant P: V ∝ T.
Avogadro’s Law: For two gases at same P, V, T:
⅓ (N₁/V) m₁⟨v₁²⟩ = ⅓ (N₂/V) m₂⟨v₂²⟩.
But ½ m₁⟨v₁²⟩ = ½ m₂⟨v₂²⟩ (same T) ⇒ N₁ = N₂.

5. Kinetic Interpretation of Temperature

From PV = ⅓ N m ⟨v²⟩ and PV = N k_B T:

½ m ⟨v²⟩ = (3/2) k_B T

Conclusion: The average translational kinetic energy of a molecule is proportional to absolute temperature and independent of the gas type.

Root Mean Square (RMS) Speed

v_rms = √⟨v²⟩ = √(3k_BT/m) = √(3RT/M)

Problem 1: Calculate the RMS speed of oxygen molecules at 27°C. (Molar mass of O₂ = 32 g/mol)

Show Answer

T = 300 K, M = 32 × 10⁻³ kg/mol.
v_rms = √(3 × 8.314 × 300 / 0.032) = √(233831) ≈ 484 m/s.

5A. Maxwell-Boltzmann Speed Distribution

Molecules in a gas do not all move with the same speed. Maxwell and Boltzmann derived a distribution curve showing how many molecules have a specific speed at a given temperature.

Maxwell-Boltzmann speed distribution graph showing most probable, average, and rms speeds. — The peak represents the Most Probable Speed. Higher T flattens the curve and shifts the peak to the right.

Different Types of Speeds

Most Probable Speed (v_mp): The speed possessed by the maximum number of molecules.
v_mp = √(2RT/M)
Average Speed (v̄): Arithmetic mean of speeds.
v̄ = √(8RT/πM) ≈ 1.59 √(RT/M)
RMS Speed (v_rms): Root mean square speed.
v_rms = √(3RT/M) ≈ 1.73 √(RT/M)

Ratio v_mp : v̄ : v_rms :: 1 : 1.128 : 1.224

6. Law of Equipartition of Energy

In thermal equilibrium, energy is equally distributed among all quadratic degrees of freedom. Each degree contributes ½ k_BT per molecule.

Type	f (Total)	U (per mole)	C_v	γ
Monatomic	3	3/2 RT	3/2 R	5/3 ≈ 1.67
Diatomic (rigid)	5	5/2 RT	5/2 R	7/5 = 1.40
Diatomic (vibrating)	7	7/2 RT	7/2 R	9/7 ≈ 1.29
Polyatomic (non-linear)	6+	3RT+	3R+	<1.33

7. Specific Heat Capacity

7.1 Gases

From equipartition: C_v = f/2 R, C_p = C_v + R, γ = 1 + 2/f.

7.2 Mixture of Non-Reacting Gases

For a mixture of $n_1$ moles of Gas 1 and $n_2$ moles of Gas 2:

C_v,mix = (n₁ C_v1 + n₂ C_v2) / (n₁ + n₂)

The specific heat ratio for the mixture is $\gamma_{mix} = \frac{C_{p,mix}}{C_{v,mix}}$ .

Problem (JEE): 1 mole of Helium (Monatomic) is mixed with 1 mole of Hydrogen (Diatomic). Find $\gamma$ for the mixture.

Show Answer

He: $C_{v1} = 1.5R$ , H $_2$ : $C_{v2} = 2.5R$ .
$C_{v,mix} = \frac{1(1.5R) + 1(2.5R)}{2} = 2R$ .
$C_{p,mix} = C_{v,mix} + R = 3R$ .
$\gamma_{mix} = \frac{3R}{2R} = 1.5$ .

7.3 Solids (Dulong-Petit Law)

Each atom vibrates in 3D → 6 degrees → U = 3RT per mole.

C = dQ/dT ≈ dU/dT = 3R ≈ 25 J/mol·K

Problem 2: A 44.8 L cylinder of He at STP is heated by 15°C. Find heat required.

Show Answer

Volume = 44.8 L = 2 moles.
He is monatomic → C_v = 3/2 R.
Q = n C_v ΔT = 2 × (3/2 × 8.314) × 15 ≈ 374 J.

8. Mean Free Path (λ)

The average distance a molecule travels between two successive collisions.

Illustration of the zig-zag path of a gas molecule between collisions. — Mean Free Path (λ) is the average distance a molecule travels between successive collisions.

λ = 1 / (√2 π n d²)

Application:

In uranium enrichment, lighter ²³⁵UF₆ molecules have higher v_rms and effuse faster through porous barriers.

Problem 3: Estimate λ for N₂ at STP (d = 3 Å, n = 2.7×10²⁵ m⁻³).

Show Answer

λ = 1 / (√2 π × 2.7×10²⁵ × (3×10⁻¹⁰)²) ≈ 1.9 × 10⁻⁷ m.

← Previous Chapter Next Chapter: Oscillations →