Molecular Kinetic Theory Model

The kinetic theory of gases is one of the most elegant parts of A-Level Physics. It derives the macroscopic gas laws from first principles by considering the microscopic behaviour of individual molecules. The central result is the equation pV = ⅓Nmc̄², which connects pressure and volume to the motion of molecules. AQA requires you to understand and be able to reproduce the key steps of this derivation.

Spec mapping (AQA 7408 §3.6.2.3 — Molecular kinetic theory model): This lesson addresses the derivation of pV = (1/3)Nm〈c²〉 from first principles, the model's underlying assumptions (point particles, elastic collisions, no intermolecular forces, random motion, large N), the meaning of root-mean-square speed c_rms = √〈c²〉, and the link to macroscopic gas behaviour. Candidates must be able to reproduce the key steps of the derivation in a structured argument and identify which assumption underpins each step. (Refer to the official AQA specification document for exact wording.)

Synoptic links: The kinetic theory is the microscopic counterpart of the empirical gas laws (§3.6.2.2) — both lead to the same macroscopic relationship pV ∝ T. Equating pV = (1/3)Nm〈c²〉 with pV = NkT yields the kinetic-energy/temperature relation (1/2)m〈c²〉 = (3/2)kT (§3.6.2.3), the cornerstone of the next lesson. It connects backwards to mechanics (§3.4) — the derivation invokes Newton's second law, impulse-momentum, and kinematics — and forwards to statistical mechanics at undergraduate level (Maxwell–Boltzmann distribution, equipartition). Brownian motion (§3.6.2.3) is the experimental evidence that validates the underlying particle model assumed here.

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The Model and Its Assumptions

The kinetic theory considers a large number of identical gas molecules moving randomly inside a container. The assumptions (identical to those for an ideal gas) are:

1. The gas contains a very large number N of identical molecules, each of mass m.
2. The molecules are in constant random motion — there is no preferred direction.
3. The volume of the molecules is negligible compared to the volume of the container.
4. There are no intermolecular forces between the molecules (except during collisions).
5. All collisions (between molecules and with the container walls) are perfectly elastic.
6. The duration of a collision is negligible compared to the time between collisions.
7. The molecules obey Newton's laws of motion.

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The Derivation of pV = ⅓Nmc̄²

This derivation is one of the most important in A-Level Physics. It proceeds in several clear steps.

Step 1: Consider a Single Molecule in a Cubic Box

Consider a cubic container of side length L. A single molecule of mass m moves with velocity c. The molecule has velocity components cₓ, c_y, and c_z in the three perpendicular directions.

The speed is related to the components by:

c² = cₓ² + c_y² + c_z²

Step 2: Calculate the Momentum Change at One Wall

Consider the molecule hitting the wall perpendicular to the x-direction. The molecule approaches with x-component of velocity +cₓ and bounces back with −cₓ (elastic collision, wall is much more massive).

Change in momentum per collision:

Δp = mcₓ − (−mcₓ) = 2mcₓ

Step 3: Calculate the Time Between Successive Collisions with the Same Wall

After bouncing off the wall, the molecule must travel across the box and back — a total distance of 2L in the x-direction.

Time between collisions with the same wall:

Δt = 2L/cₓ

Step 4: Calculate the Force from One Molecule on One Wall

The rate of change of momentum gives the force (Newton's second law):

F = Δp/Δt = 2mcₓ / (2L/cₓ) = mcₓ²/L

Step 5: Sum Over All N Molecules

The total force on the wall from all N molecules is:

F_total = (m/L)(c₁ₓ² + c₂ₓ² + c₃ₓ² + ... + cₙₓ²)

F_total = (m/L) × N⟨cₓ²⟩

where ⟨cₓ²⟩ is the mean of the squared x-components of velocity for all molecules.

Step 6: Use the Isotropy of Random Motion

Because the motion is random, there is no preferred direction. Therefore the average behaviour is the same in all three directions:

⟨cₓ²⟩ = ⟨c_y²⟩ = ⟨c_z²⟩

Since c² = cₓ² + c_y² + c_z², taking the mean:

⟨c²⟩ = ⟨cₓ²⟩ + ⟨c_y²⟩ + ⟨c_z²⟩ = 3⟨cₓ²⟩

Therefore:

⟨cₓ²⟩ = ⟨c²⟩/3

Step 7: Calculate the Pressure

The force on one wall is:

F = Nm⟨cₓ²⟩/L = Nm⟨c²⟩/(3L)

Pressure = Force / Area. The area of one wall of the cube is L².

p = F/L² = Nm⟨c²⟩/(3L³)

Since L³ = V (the volume of the cube):

pV = ⅓Nm⟨c²⟩

This is the fundamental result of kinetic theory.

Important Notes on the Derivation

• The result applies to any shape of container, not just a cube. The cube is used for simplicity.
• ⟨c²⟩ is the mean square speed — the average of the squared speeds of all the molecules. It is NOT the square of the mean speed.
• The notation c̄² is sometimes used interchangeably with ⟨c²⟩.

Exam Tip: When reproducing this derivation, the most common mistakes are: (1) forgetting the factor of 2 in the momentum change (it is 2mcₓ, not mcₓ, because the molecule reverses direction); (2) using ⟨c²⟩ instead of ⟨cₓ²⟩ too early; (3) forgetting to explain why ⟨cₓ²⟩ = ⟨c²⟩/3 (you must state that the motion is random/isotropic).

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Root Mean Square (r.m.s.) Speed

The root mean square speed is defined as:

c_rms = √⟨c²⟩

It is the square root of the mean of the squared speeds. The r.m.s. speed is different from (and always greater than) the mean speed.

From pV = ⅓Nm⟨c²⟩, we can find c_rms if we know p, V, N, and m:

⟨c²⟩ = 3pV/(Nm) = 3p/ρ

where ρ = Nm/V is the density of the gas.

Therefore:

c_rms = √(3p/ρ)

Worked Example 1 — Calculating r.m.s. Speed from Density

Question: Calculate the r.m.s. speed of nitrogen molecules at atmospheric pressure (1.01 × 10⁵ Pa) and room temperature, given that the density of nitrogen at these conditions is 1.25 kg m⁻³.

Solution:

c_rms = √(3p/ρ) = √(3 × 1.01 × 10⁵ / 1.25)

c_rms = √(3.03 × 10⁵ / 1.25)

c_rms = √(2.424 × 10⁵)

c_rms = 492 m s⁻¹

This is about 1770 km h⁻¹ — remarkably fast!

Worked Example 2 — Calculating r.m.s. Speed from Temperature

Using ½m⟨c²⟩ = ³⁄₂kT (derived in the next lesson), we get:

c_rms = √(3kT/m)

Question: Calculate the r.m.s. speed of helium atoms at 300 K. (Mass of helium atom = 4.0 × 10⁻³ / 6.022 × 10²³ = 6.64 × 10⁻²⁷ kg)

Solution:

c_rms = √(3kT/m) = √(3 × 1.38 × 10⁻²³ × 300 / 6.64 × 10⁻²⁷)

c_rms = √(1.242 × 10⁻²⁰ / 6.64 × 10⁻²⁷)

c_rms = √(1.870 × 10⁶)

c_rms = 1370 m s⁻¹

Helium atoms move about 2.8 times faster than nitrogen molecules at the same temperature, because they are about 7 times lighter. Since c_rms ∝ 1/√m (at constant T), lighter molecules move faster.

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Pressure in Terms of Density

From pV = ⅓Nm⟨c²⟩, and noting that the total mass of gas is Nm and the density is ρ = Nm/V:

p = ⅓ρ⟨c²⟩

This form is useful when you know the density of the gas rather than the number of molecules.

Worked Example 3 — Finding Pressure from Molecular Data

Question: A container of volume 0.030 m³ holds 2.0 × 10²⁴ molecules of an ideal gas. Each molecule has a mass of 5.3 × 10⁻²⁶ kg. The r.m.s. speed of the molecules is 400 m s⁻¹. Calculate the pressure of the gas.

Solution:

Using pV = ⅓Nm⟨c²⟩:

p = ⅓Nm⟨c²⟩ / V

p = (2.0 × 10²⁴ × 5.3 × 10⁻²⁶ × 400²) / (3 × 0.030)

p = (2.0 × 10²⁴ × 5.3 × 10⁻²⁶ × 160 000) / 0.090

p = (0.106 × 160 000) / 0.090

p = 16 960 / 0.090

p = 1.88 × 10⁵ Pa

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Why the Derivation Matters

The kinetic theory derivation is powerful because it:

1. Explains the macroscopic gas laws (Boyle's, Charles's, pressure law) from microscopic molecular behaviour.
2. Shows that pressure is caused by molecular collisions with the walls.
3. Provides a definition of temperature in terms of molecular kinetic energy (see next lesson).
4. Predicts relationships that can be experimentally verified.

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Common Misconceptions

1. "⟨c²⟩ is the same as ⟨c⟩²." These are different. The mean of the squares is NOT equal to the square of the mean. For a distribution of speeds, ⟨c²⟩ > ⟨c⟩².

2. "All molecules in a gas move at the same speed." They do not — there is a distribution of speeds (the Maxwell-Boltzmann distribution). The r.m.s. speed is a useful average, but individual molecules have a wide range of speeds.

3. "The derivation only works for a cube." The cube is used for mathematical convenience, but the result applies to any shape of container.

4. "Pressure depends on the shape of the container." Pressure depends on p = ⅓ρ⟨c²⟩ — it is the same throughout the gas regardless of container shape (for an ideal gas in equilibrium).

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Additional Worked Example — Applying pV = (1/3)Nm〈c²〉 to a Real Gas

A cylinder of internal volume 5.00 × 10⁻³ m³ contains 0.200 mol of oxygen (O₂, molar mass M = 32.0 × 10⁻³ kg mol⁻¹) at a pressure of 1.50 × 10⁵ Pa and temperature T (to be determined). Use the kinetic-theory equation to: (a) calculate 〈c²〉 and c_rms for the oxygen molecules, (b) determine the gas temperature, and (c) calculate the total translational kinetic energy of the gas.

Step 1 — number of molecules and molecular mass. N = n × N_A = 0.200 × 6.02 × 10²³ = 1.204 × 10²³ molecules. m_molecule = M / N_A = 32.0 × 10⁻³ / 6.02 × 10²³ = 5.32 × 10⁻²⁶ kg per molecule. Total gas mass = N × m = nM = 0.200 × 0.0320 = 6.40 × 10⁻³ kg = 6.40 g.

Step 2 — apply pV = (1/3)Nm〈c²〉 to find 〈c²〉. 〈c²〉 = 3pV / (Nm) = 3pV / (nM). 〈c²〉 = (3 × 1.50 × 10⁵ × 5.00 × 10⁻³) / (0.200 × 0.0320) 〈c²〉 = (2250) / (6.40 × 10⁻³) 〈c²〉 = 3.516 × 10⁵ m² s⁻².

c_rms = √〈c²〉 = √(3.516 × 10⁵) = 593 m s⁻¹.

Step 3 — find temperature via pV = nRT. T = pV/(nR) = (1.50 × 10⁵ × 5.00 × 10⁻³)/(0.200 × 8.31) T = 750 / 1.662 = 451 K = 178 °C.

Cross-check using (1/2)m〈c²〉 = (3/2)kT: LHS: (1/2)(5.32 × 10⁻²⁶)(3.516 × 10⁵) = 9.353 × 10⁻²¹ J. RHS: (3/2)(1.38 × 10⁻²³)(451) = 9.337 × 10⁻²¹ J. ✓ (Rounding agreement to 3 s.f.)

Step 4 — total translational kinetic energy. Total KE = N × (1/2)m〈c²〉 = (1.204 × 10²³)(9.353 × 10⁻²¹) = 1126 J ≈ 1.13 kJ.

Equivalently using internal energy of an ideal monatomic gas U = (3/2)nRT: U = (3/2)(0.200)(8.31)(451) = 1125 J ✓.

(Note: oxygen is diatomic, so its total internal energy at room temperature includes rotational contributions giving U = (5/2)nRT. But the translational component is still (3/2)nRT and is what enters pV = (1/3)Nm〈c²〉.)

Commentary. This worked example demonstrates the central insight of kinetic theory: the macroscopic state (p, V, n, T) is equivalent to the microscopic state (〈c²〉, N, m). Either set determines the other. The kinetic-theory equation pV = (1/3)Nm〈c²〉 is the explicit bridge between the two descriptions, and the consistency check at the end is a routine verification step that examiners reward as evidence of analytical rigour.

Exam Tip: When a question gives n (moles) but asks about molecular quantities, use m = M/N_A to convert. When it gives N (molecules) but asks about macroscopic quantities, use n = N/N_A. The two pictures must always agree numerically.

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Specimen Exam Question

Specimen question modelled on the AQA paper format (9 marks).