Kinetic theory of gases

CIE A-Level Physics· 45 min read

1. Assumptions of the Kinetic Theory Model★★☆☆☆⏱ 10 min

Kinetic theory uses a microscopic model to explain the macroscopic behaviour of ideal gases, based on a set of simplifying assumptions that match real gas behaviour at low pressure and high temperature.

📘 Definition

Ideal gas (kinetic theory)

A gas that obeys all kinetic theory assumptions and the ideal gas equation of state

Example:

Helium gas at room temperature and standard atmospheric pressure approximates an ideal gas

All molecules are identical, with negligible volume compared to the total volume of the gas
Molecules move randomly and at constant speed between collisions
All collisions (between molecules and with the container walls) are perfectly elastic
There are no intermolecular forces between molecules except during collisions
The time between collisions is much longer than the duration of a collision

Exam tip:

CIE usually asks for 4 out of 5 assumptions, so memorise the first four for full marks

2. Derivation of the Kinetic Theory Equation★★★★☆⏱ 20 min

🔬 Derivation

Goal:

Derive the relation between pressure, volume and molecular speed for an ideal gas

Starting from:

Newton's laws of motion and kinetic theory assumptions

1
Consider a molecule of mass $m$ moving at speed $v_{x}$ along the x-axis in a cubic box of side length $L$ . On elastic collision with the wall, the change in momentum of the molecule is $- 2 m v_{x}$ , so the change in momentum of the wall is $+ 2 m v_{x}$ .
2
Time between collisions with the same wall is $Δ t = \frac{2 L}{v _{x}}$ , so force on the wall is:
3
$F = \frac{\Delta p}{\Delta t} = \frac{mv_x^2}{L}$
4
For $N$ molecules, total force is $F = \frac{N m ⟨ v _{x}^{2} ⟩}{L}$ , where $⟨ v_{x}^{2} ⟩$ is the mean square x-speed. For random motion, $⟨ c^{2} ⟩ = 3 ⟨ v_{x}^{2} ⟩$ where $⟨ c^{2} ⟩$ is the overall mean square speed.
5
Pressure $p = \frac{F}{A} = \frac{F}{L ^{2}} = \frac{N m ⟨ v _{x}^{2} ⟩}{L ^{3}} = \frac{N m ⟨ v _{x}^{2} ⟩}{V}$ . Substituting $⟨ v_{x}^{2} ⟩ = \frac{⟨ c ^{2} ⟩}{3}$ gives:

Result:

The key kinetic theory equation:

pV = \frac{1}{3} N m \langle c^2 \rangle

📐 Worked Example

A cubic box of volume 0.01 m³ contains $2 \times 1 0^{23}$ molecules of mass $5 \times 1 0^{- 26}$ kg. If the gas pressure is $1.0 \times 1 0^{5}$ Pa, calculate the mean square speed of the molecules.

1
1. Recall the kinetic theory equation:
2
$pV = \frac{1}{3} N m \langle c^2 \rangle$
3
1. Rearrange for $⟨ c^{2} ⟩$ :
4
$\langle c^2 \rangle = \frac{3pV}{Nm}$
5
1. Substitute values:
6
$\langle c^2 \rangle = \frac{3 \times 1.0 \times 10^5 \times 0.01}{2 \times 10^{23} \times 5 \times 10^{-26}} = 3 \times 10^5 \, \text{m}^2 \text{s}^{-2}$

Exam tip:

A full derivation from first principles is a common 6-7 mark question in CIE Paper 2

3. Root-Mean-Square Speed★★★☆☆⏱ 15 min

📘 Definition

Root-mean-square (r.m.s) speed

$c_{rms}$

The square root of the mean square speed of gas molecules, a measure of the typical molecular speed: $c_{rms} = ⟨ c^{2} ⟩$

Equate the kinetic theory equation to the ideal gas equation $p V = N k T$ (where $k$ is the Boltzmann constant) to get a relation for $c_{rms}$ in terms of temperature:

NkT = \frac{1}{3}Nm c_{\text{rms}}^2 \implies c_{\text{rms}} = \sqrt{\frac{3kT}{m}}

In terms of molar mass $M$ (mass per mole of gas) and molar gas constant $R$ , this becomes:

c_{\text{rms}} = \sqrt{\frac{3RT}{M}}

📐 Worked Example

Calculate the r.m.s speed of nitrogen molecules at 27°C. Molar mass of nitrogen is 28 g mol⁻¹, $R = 8.31$ J mol⁻¹ K⁻¹.

1
1. Convert temperature to Kelvin and molar mass to kg:
2
$T = 27 + 273 = 300 \, \text{K}, \quad M = 0.028 \, \text{kg} \, \text{mol}^{-1}$
3
1. Substitute into the formula:
4
$c_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 300}{0.028}} \approx 517 \, \text{m} \, \text{s}^{-1}$

4. Mean Kinetic Energy of Gas Molecules★★★☆☆⏱ 10 min

Rearranging the relation $3 k T = m c_{rms}^{2}$ gives the mean kinetic energy of a single gas molecule:

\langle E_k \rangle = \frac{1}{2} m \langle c^2 \rangle = \frac{3}{2} kT

This is a core result: the mean kinetic energy of an ideal gas molecule depends only on absolute temperature, and is directly proportional to $T$ . Absolute temperature is therefore a measure of the average random kinetic energy of gas molecules.

📐 Worked Example

Calculate the total kinetic energy of 1 mole of an ideal gas at 300 K, $R = 8.31$ J mol⁻¹ K⁻¹.

1
1. Mean kinetic energy per molecule is $\frac{3}{2} k T$ . For 1 mole, the number of molecules is Avogadro's constant $N_{A}$ :
2
1. Total kinetic energy = $N_{A} \times \frac{3}{2} k T = \frac{3}{2} (N_{A} k) T = \frac{3}{2} R T$ , since $R = N_{A} k$
3
1. Substitute values:
4
$\text{Total } E_k = 1.5 \times 8.31 \times 300 = 3740 \, \text{J} = 3.74 \, \text{kJ}$

5. Common Pitfalls

Wrong move:

Stating "molecules have negligible volume" without context

Why:

CIE examiners require you to specify what the molecular volume is negligible compared to

Correct move:

Always write that the volume of individual molecules is negligible compared to the total volume occupied by the gas

Wrong move:

Confusing $⟨ c^{2} ⟩$ (mean of squares) with $(⟨ c ⟩)^{2}$ (square of mean)

Why:

The mean of squares is never equal to the square of the mean for a range of values, leading to incorrect r.m.s speed calculations

Correct move:

Remember r.m.s speed is defined as $⟨ c^{2} ⟩$ , not the square of the average speed

Wrong move:

Using molar mass in g mol⁻¹ instead of kg mol⁻¹ when calculating $c_{rms}$

Why:

The gas constant $R$ uses SI units (J = kg m² s⁻²), so mass must be in kilograms to get the correct speed in m s⁻¹

Correct move:

Always convert molar mass from g mol⁻¹ to kg mol⁻¹ before substituting into the formula

Wrong move:

Using Celsius temperature instead of Kelvin in kinetic energy or r.m.s speed calculations

Why:

The relation $⟨ E_{k} ⟩ = \frac{3}{2} k T$ uses absolute (Kelvin) temperature, so using Celsius gives wrong results

Correct move:

Always add 273 to a Celsius temperature to get the absolute temperature $T$

Wrong move:

Stating all molecules have the same kinetic energy at a given temperature

Why:

Temperature only determines the mean kinetic energy, individual molecules have a wide range of speeds and energies

Correct move:

Always specify that the mean kinetic energy of molecules is fixed for a given absolute temperature

6. Quick Reference Cheatsheet

Quantity	Formula	Key Notes
Kinetic theory equation	$p V = \frac{1}{3} N m ⟨ c^{2} ⟩$	N = number of molecules
r.m.s speed definition	$c_{rms} = ⟨ c^{2} ⟩$	Typical molecular speed
r.m.s speed vs temperature	$c_{rms} = \frac{3 R T}{M} = \frac{3 k T}{m}$	M = molar mass in kg mol⁻¹
Mean KE per molecule	$⟨ E_{k} ⟩ = \frac{3}{2} k T$	Only depends on T
Total KE for n moles	$E_{total} = \frac{3}{2} n R T$	For monatomic ideal gas

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
Derive kinetic theory equation
2023 · 1
Calculate r.m.s. speed of molecules
2024 · 2
State kinetic theory assumptions

Going deeper

syllabusCIE 9702 Syllabus Page for Ideal Gases

What's Next

Kinetic theory of gases is the foundation of thermal physics in A-Level, connecting microscopic molecular behaviour to measurable macroscopic properties like pressure and temperature. This model is extended in later topics to explain internal energy of ideal gases, thermal energy transfer, and changes of state. Exam questions often combine kinetic theory with the ideal gas law, so mastering the derivation and key formulas here will help you earn full marks in multi-part questions. The relationship between temperature and mean kinetic energy is also a core concept for understanding entropy and thermal equilibrium in advanced thermal physics.