# B.3 Kinetic theory of gases

> IB Physics Higher Level · IB Physics 2025+ First Assessment
> Source: https://www.owlsprep.com/study/ib-physics-hl-u2-b-3-kinetic-theory-of/

This sub-topic builds on ideal gas laws to develop a microscopic particle model of gas behavior. You will connect molecular motion to macroscopic properties like pressure and temperature, and use kinetic theory to calculate average molecular speeds.

**Prerequisites:** [Ideal gas laws](https://www.owlsprep.com/study/ib-physics-hl-u2-b-2-ideal-gases/); [Basic Newtonian momentum](https://www.owlsprep.com/study/ib-physics-hl-mechanics-momentum/)

## Learning objectives

- State the core assumptions of the kinetic model of an ideal gas
- Derive the kinetic theory equation of state from first principles
- Connect absolute temperature to average molecular kinetic energy
- Calculate root-mean-square speed of gas molecules

## 1. Core Assumptions of the Kinetic Model

The kinetic theory models an ideal gas as a collection of tiny particles moving according to Newtonian mechanics. The model relies on five key simplifying assumptions that hold well for real gases at low pressure and high temperature.

**Ideal Gas (Kinetic Model)** — A gas that obeys all core assumptions of the kinetic model, and follows the ideal gas law $pV = nRT$ under all conditions.

*Example:* Helium at room temperature and atmospheric pressure approximates closely to an ideal gas.

- The number of molecules is large, so statistical averaging is valid.
- The volume of individual molecules is negligible compared to the total volume of the gas.
- Molecules move randomly at constant speeds between collisions.
- All collisions (between molecules, and with container walls) are perfectly elastic.
- There are no intermolecular forces between molecules except during collisions.

> **note**
>
> All five assumptions are regularly tested in 1-2 mark short answer questions on Paper 1 and Paper 2. You must be able to recall all of them.

> **Exam tip:** Deviations from ideal gas behavior in real gases are always linked to failures of these assumptions: high pressure means molecular volume is not negligible, low temperature means intermolecular forces are significant.

## 2. Derivation of the Kinetic Theory Equation

We can derive the equation relating pressure, volume and molecular speed by considering molecules bouncing between the walls of a cubical container. This full derivation is a common 5-6 mark exam question, so you need to memorize every step.

**Derivation:** Derive the kinetic theory equation of state for an ideal gas

*Starting from:* Newton's laws of motion for a colliding molecule

1. Consider one molecule of mass $m_0$ moving along the x-axis with speed $v_x$ towards a container wall of side length $L$. The change in momentum for the molecule after an elastic collision with the wall is:
2. $$\Delta p_{\text{molecule}} = -m_0 v_x - m_0 v_x = -2m_0 v_x$$
3. By Newton's third law, the momentum transferred to the wall is $+2m_0 v_x$. The time between successive collisions with the same wall is:
4. $$\Delta t = \frac{2L}{v_x}$$
5. Force is the rate of change of momentum, so force from the single molecule on the wall is:
6. $$F_1 = \frac{\Delta p}{\Delta t} = \frac{2m_0 v_x}{2L / v_x} = \frac{m_0 v_x^2}{L}$$
7. Sum over all $N$ molecules, and use symmetry: $\langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle = \frac{1}{3}\langle v^2 \rangle$.
8. Pressure $p = F/A$, where $A = L^2$ and total container volume $V = L^3$. Substitute to get:

*Conclusion:* pV = \frac{1}{3} N m_0 \langle v^2 \rangle

**Exam command terms**

Common command terms for this topic have specific exam expectations:

- **Derive** — You must show all steps from first principles *(You cannot skip the symmetry argument or momentum steps to get full marks)*

- **State** — Just write the final equation, no working needed

## 3. Temperature and Average Molecular Kinetic Energy

We can connect the kinetic theory equation to the ideal gas law to find a fundamental relation between temperature and the kinetic energy of gas molecules.

**Worked example:** Show that the average kinetic energy of an ideal gas molecule is $\frac{3}{2}k_B T$.

1. Start by writing both the kinetic theory equation and the ideal gas law in terms of Boltzmann constant $k_B$ (where $R = N_A k_B$ and $N = n N_A$):
2. $$pV = \frac{1}{3} N m_0 \langle v^2 \rangle \quad \text{and} \quad pV = N k_B T$$
3. Equate the two expressions for $pV$ and cancel the common term $N$:
4. $$k_B T = \frac{1}{3} m_0 \langle v^2 \rangle$$
5. Rearrange to get average kinetic energy $\langle E_k \rangle = \frac{1}{2} m_0 \langle v^2 \rangle$:
6. $$\langle E_k \rangle = \frac{1}{2} m_0 \langle v^2 \rangle = \frac{3}{2} k_B T$$

This is one of the most important results of kinetic theory: it proves that absolute temperature is a direct measure of the average random kinetic energy of the molecules of an ideal gas.

**Check your understanding**

Test your understanding:

1. Two different ideal gases are at the same temperature. What must be true?

   - Molecules of both gases have the same average speed
   - Molecules of both gases have the same average kinetic energy
   - Molecules of both gases have the same rms speed
   - Molecules of both gases have the same mass

   *Answer:* Molecules of both gases have the same average kinetic energy

   *Why:* Correct! Equal absolute temperature means equal average kinetic energy per molecule, regardless of gas type. Heavier molecules have lower average speed at the same temperature.

## 4. Calculating Root-Mean-Square Speed

**Root-mean-square (rms) speed** — A convenient statistical measure of the average speed of molecules in a gas, calculated from the mean of the squares of individual molecular speeds.

*Notation:* v_{\text{rms}} = \sqrt{\langle v^2 \rangle}

**Worked example:** Calculate the rms speed of nitrogen molecules at 20°C, given molar mass $M = 0.028 \text{ kg mol}^{-1}$ and $R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$.

1. First convert temperature to Kelvin:
2. $$T = 20 + 273 = 293 \text{ K}$$
3. Rearrange the kinetic theory equation to get $v_{\text{rms}}$ in terms of molar mass $M = N_A m_0$:
4. $$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$
5. Substitute values and calculate:
6. $$v_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 293}{0.028}} \approx 510 \text{ m s}^{-1}$$

rms speed is proportional to the square root of temperature, and inversely proportional to the square root of molar mass. So lighter gases have much higher average molecular speeds at the same temperature.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Claiming all molecules in a gas have the same speed at a given temperature
  - Why it fails: Kinetic theory only fixes the average kinetic energy, individual molecules have a wide range of speeds
  - Correct: State that average kinetic energy (and thus average rms speed for a given gas) is fixed by temperature
- **Wrong:** Using Celsius temperature instead of Kelvin in calculations
  - Why it fails: All kinetic theory equations rely on absolute (Kelvin) temperature; Celsius gives a drastically wrong result
  - Correct: Always add 273 to Celsius temperatures before substituting into any kinetic theory formula
- **Wrong:** Using molar mass in $\text{g mol}^{-1}$ instead of $\text{kg mol}^{-1}$
  - Why it fails: SI units for energy and mass require kilograms, so this gives a speed ~30 times smaller than the correct value
  - Correct: Convert molar mass to $\text{kg mol}^{-1}$ by dividing by 1000 before calculation
- **Wrong:** Claiming temperature is proportional to total kinetic energy of a gas
  - Why it fails: Temperature is proportional to average kinetic energy per molecule, not total. A larger volume of gas has more total energy but the same temperature
  - Correct: Remember temperature is proportional to average kinetic energy per molecule, not total kinetic energy of the whole sample

## Cheatsheet

| Quantity | Formula | Key Notes |
| --- | --- | --- |
| Kinetic theory equation | $pV = \frac{1}{3}Nm_0 \langle v^2 \rangle$ | N = number of molecules |
| Average KE per molecule | $\langle E_k \rangle = \frac{3}{2}k_B T$ | Same for all gases at same T |
| rms speed | $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ | M = molar mass in kg mol⁻¹ |
| Core assumptions | 5 key assumptions | Common 1-2 mark exam question |

## What's next

Kinetic theory of gases forms the microscopic foundation for all thermal physics topics in IB HL. It connects the random motion of individual particles to the macroscopic properties of pressure and temperature that you measure experimentally. The concepts you learned here will be extended when you study the Maxwell-Boltzmann molecular speed distribution, and are foundational for understanding thermodynamics, entropy, and phase changes later in the course. Mastery of the derivation of the kinetic theory equation and rms speed calculations is frequently tested in long answer questions on Paper 2, so it is important to practice these skills.

- [B.4 Mass, energy and matter structure](https://www.owlsprep.com/study/ib-physics-hl-u2-b-4-mass-energy-and/)
- [B.5 Current and electric circuits (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u2-b-5-current-and-electric/)
- [B.6 Heating effect of current and electric cells (AHL)](https://www.owlsprep.com/study/ib-physics-hl-u2-b-6-heating-effect-of/)

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