# Kinetic theory of gases

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u19-kinetic-theory-of-gases/

This subtopic explains macroscopic properties of ideal gases by modelling the motion of individual molecules. You will learn core assumptions, derive the key kinetic theory equation, and calculate molecular speeds from gas properties.

**Prerequisites:** [Ideal gas equation $pV = nRT$](https://www.owlsprep.com/study/cie-9702-u19-ideal-gas-law/)

## Learning objectives

- State the core assumptions of the kinetic theory of ideal gases
- Derive the kinetic theory equation $pV = \frac{1}{3}Nm\langle c^2 \rangle$ from first principles
- Calculate root-mean-square speed of gas molecules from temperature and molar mass
- Relate mean molecular kinetic energy to absolute temperature

## Assumptions of the Kinetic Theory Model

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.

**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

## Derivation of the Kinetic Theory Equation

**Derivation:** 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 $-2mv_x$, so the change in momentum of the wall is $+2mv_x$.
2. Time between collisions with the same wall is $\Delta t = \frac{2L}{v_x}$, 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 $F = \frac{Nm \langle v_x^2 \rangle}{L}$, where $\langle v_x^2 \rangle$ is the mean square x-speed. For random motion, $\langle c^2 \rangle = 3\langle v_x^2 \rangle$ where $\langle c^2 \rangle$ is the overall mean square speed.
5. Pressure $p = \frac{F}{A} = \frac{F}{L^2} = \frac{Nm \langle v_x^2 \rangle}{L^3} = \frac{Nm \langle v_x^2 \rangle}{V}$. Substituting $\langle v_x^2 \rangle = \frac{\langle c^2 \rangle}{3}$ gives:

*Conclusion:* 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 $2 \times 10^{23}$ molecules of mass $5 \times 10^{-26}$ kg. If the gas pressure is $1.0 \times 10^5$ 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. 2. Rearrange for $\langle c^2 \rangle$:
4. $$\langle c^2 \rangle = \frac{3pV}{Nm}$$
5. 3. 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

## Root-Mean-Square Speed

**Root-mean-square (r.m.s) speed** — The square root of the mean square speed of gas molecules, a measure of the typical molecular speed: $c_{\text{rms}} = \sqrt{\langle c^2 \rangle}$

*Notation:* $c_{\text{rms}}$

Equate the kinetic theory equation to the ideal gas equation $pV = NkT$ (where $k$ is the Boltzmann constant) to get a relation for $c_{\text{rms}}$ 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. 2. Substitute into the formula:
4. $$c_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 300}{0.028}} \approx 517 \, \text{m} \, \text{s}^{-1}$$

## Mean Kinetic Energy of Gas Molecules

Rearranging the relation $3kT = mc_{\text{rms}}^2$ 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 $\frac{3}{2}kT$. For 1 mole, the number of molecules is Avogadro's constant $N_A$:
2. 2. Total kinetic energy = $N_A \times \frac{3}{2}kT = \frac{3}{2}(N_A k) T = \frac{3}{2}RT$, since $R = N_A k$
3. 3. Substitute values:
4. $$\text{Total } E_k = 1.5 \times 8.31 \times 300 = 3740 \, \text{J} = 3.74 \, \text{kJ}$$

## Common pitfalls

- **Wrong:** Stating "molecules have negligible volume" without context
  - Why it fails: CIE examiners require you to specify what the molecular volume is negligible compared to
  - Correct: Always write that the volume of individual molecules is negligible compared to the total volume occupied by the gas
- **Wrong:** Confusing $\langle c^2 \rangle$ (mean of squares) with $(\langle c \rangle)^2$ (square of mean)
  - Why it fails: 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: Remember r.m.s speed is defined as $\sqrt{\langle c^2 \rangle}$, not the square of the average speed
- **Wrong:** Using molar mass in g mol⁻¹ instead of kg mol⁻¹ when calculating $c_{\text{rms}}$
  - Why it fails: 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: Always convert molar mass from g mol⁻¹ to kg mol⁻¹ before substituting into the formula
- **Wrong:** Using Celsius temperature instead of Kelvin in kinetic energy or r.m.s speed calculations
  - Why it fails: The relation $\langle E_k \rangle = \frac{3}{2}kT$ uses absolute (Kelvin) temperature, so using Celsius gives wrong results
  - Correct: Always add 273 to a Celsius temperature to get the absolute temperature $T$
- **Wrong:** Stating all molecules have the same kinetic energy at a given temperature
  - Why it fails: Temperature only determines the mean kinetic energy, individual molecules have a wide range of speeds and energies
  - Correct: Always specify that the *mean* kinetic energy of molecules is fixed for a given absolute temperature

## Cheatsheet

| Quantity | Formula | Key Notes |
| --- | --- | --- |
| Kinetic theory equation | $pV = \frac{1}{3}Nm\langle c^2 \rangle$ | N = number of molecules |
| r.m.s speed definition | $c_{\text{rms}} = \sqrt{\langle c^2 \rangle}$ | Typical molecular speed |
| r.m.s speed vs temperature | $c_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3kT}{m}}$ | M = molar mass in kg mol⁻¹ |
| Mean KE per molecule | $\langle E_k \rangle = \frac{3}{2}kT$ | Only depends on T |
| Total KE for n moles | $E_{\text{total}} = \frac{3}{2}nRT$ | For monatomic ideal gas |

## 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.

- [Root mean square speed](https://www.owlsprep.com/study/cie-9702-u19-root-mean-square-speed/)
- [Internal energy of ideal gas](https://www.owlsprep.com/study/cie-9702-u19-internal-energy-of-ideal-gas/)
- [Thermodynamics](https://www.owlsprep.com/study/cie-9702-u20-overview/)

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