The Ideal Gas Equation (A-Level Physics)

What the ideal gas equation tells you

The ideal gas equation links the pressure, volume and temperature of a fixed amount of gas. If you know any three of p, V, n (or N) and T, you can find the fourth. It is one of the most useful equations in A-level Physics because it connects what you can measure (pressure, volume, temperature) to what you cannot see (the number of molecules and how fast they move).

An ideal gas is a model. Real gases behave very much like an ideal gas at low pressure and at temperatures well above their boiling point, because then the molecules are far apart and rarely interact. Helium, nitrogen and air at room conditions are all close to ideal.

Use pV = nRT when the question gives you moles. Use pV = NkT when it gives you a number of molecules or atoms, or when you are linking to kinetic energy.

The three gas laws and their experiments

Before the ideal gas equation, there were three separate laws, each found by holding two quantities constant. They all follow from pV = nRT.

• Boyle's law: for a fixed mass of gas at constant temperature, pressure is inversely proportional to volume, so pV = constant.
• The pressure law: for a fixed mass of gas at constant volume, pressure is directly proportional to the thermodynamic temperature, so p/T = constant.
• Charles's law: for a fixed mass of gas at constant pressure, volume is directly proportional to the thermodynamic temperature, so V/T = constant.

These experiments sit within the practical endorsement (the PAGs), so be ready to describe the method, identify the control variables, and explain how a graph is used to test the relationship. They also link to the work on specific heat capacity and latent heat, where absolute zero and the kelvin scale are introduced.

Graphs you must be able to draw and interpret

Three graphs come up repeatedly. Make sure you can sketch each one and say what the shape tells you.

The gradient of a pV against T graph is nR, so you can find the number of moles from a measured gradient, or the number of molecules from the gradient divided by k.

Assumptions of the kinetic theory of gases

The kinetic model explains gas pressure as the result of molecules bumping into the walls of the container. For the maths to work, you need to make these assumptions about an ideal gas:

• There is a very large number of identical molecules, and they move randomly in all directions.
• The molecules have negligible volume compared with the volume of the container.
• All collisions (between molecules and with the walls) are perfectly elastic, so there is no loss of kinetic energy.
• There are no intermolecular forces except during collisions, so the molecules move in straight lines at constant speed between collisions.
• The time of a collision is negligible compared with the time between collisions.
• Newton's laws of motion apply to the molecules.

The random motion is supported by Brownian motion: smoke particles seen through a microscope jitter about at random because they are hit unevenly by air molecules that are too small to see.

Deriving pV = ⅓Nm⟨c²⟩

You need to be able to reproduce this derivation. Take a cubic box of side L containing N molecules of mass m each.

1. A molecule has velocity component cₓ towards a wall. It rebounds elastically, so its velocity component reverses and its change in momentum is 2mcₓ.
2. The molecule crosses the box and returns to the same wall, a distance 2L, in time 2L/cₓ.
3. The force on the wall from one molecule is the rate of change of momentum: F = 2mcₓ ÷ (2L/cₓ) = mcₓ²/L.
4. For N molecules the total force is F = (Nm/L)⟨cₓ²⟩, where ⟨cₓ²⟩ is the mean square of the x components.
5. Each speed has components so c² = cₓ² + cᵧ² + cz². The motion is random, so the mean values of the three squares are equal: ⟨cₓ²⟩ = ⅓⟨c²⟩.
6. Pressure = force ÷ area = F/L² = Nm⟨c²⟩/(3L³). Since L³ = V, this gives pV = ⅓Nm⟨c²⟩.

Because the density is ρ = Nm/V, the result can also be written p = ⅓ρ⟨c²⟩. The symbol ⟨c²⟩ is the mean square speed: you square every speed and then take the average.

Mean kinetic energy and root mean square speed

Now put pV = ⅓Nm⟨c²⟩ together with pV = NkT. The pV cancels, N cancels, and you are left with ⅓m⟨c²⟩ = kT. Multiplying both sides by 3/2 gives the key result.

For an ideal gas all the internal energy is kinetic (there are no intermolecular forces), so the internal energy of N molecules is U = (3/2)NkT = (3/2)nRT. Mass of a molecule in kilograms comes from the relative molecular mass: for nitrogen, N₂ has mass 28 u = 28 × 1.661 × 10⁻²⁷ kg.

Worked examples

Common mistakes

How OCR A examines the ideal gas equation

Typical questions include:

• Calculations using pV = nRT or pV = NkT, often with a unit conversion step built in.
• Stating the assumptions of kinetic theory, or deriving pV = ⅓Nm⟨c²⟩ (several marks).
• Calculating the mean kinetic energy or root mean square speed of molecules.
• Describing and evaluating the Boyle's law or pressure law experiments, including graph plotting and the value of absolute zero.
• Interpreting p-V, p-T or pV-T graphs, including changes between two temperatures.

The topic is synoptic with mechanics: the derivation uses momentum and impulse for each collision, and the conversion between energy and temperature is part of the thermal physics covered in specific heat capacity and latent heat. See the Module 5 hub for all the topics in this module.

Practice questions

1. A gas at a pressure of 2.0 × 10⁵ Pa occupies 1.5 × 10⁻³ m³ at a temperature of 37 °C. Calculate the number of molecules in the gas.3 marks

   Show the answer

   Answer: 7.0 × 10²² molecules

   T = 37 + 273 = 310 K. N = pV/kT = (2.0 × 10⁵ × 1.5 × 10⁻³) / (1.38 × 10⁻²³ × 310) = 300 / 4.278 × 10⁻²¹ = 7.01 × 10²².

2. Calculate the mean kinetic energy of a gas molecule at 400 K.2 marks

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   Answer: 8.3 × 10⁻²¹ J

   mean KE = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 400 = 8.28 × 10⁻²¹ J.

3. Calculate the internal energy of 2.0 mol of an ideal monatomic gas at 300 K.2 marks

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   Answer: 7.5 × 10³ J

   U = (3/2)nRT = 1.5 × 2.0 × 8.31 × 300 = 7479 J.

4. Air has a density of 1.25 kg m⁻³ at a pressure of 1.01 × 10⁵ Pa. Use p = ⅓ρ⟨c²⟩ to estimate the root mean square speed of the molecules.3 marks

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   Answer: 4.9 × 10² m s⁻¹

   ⟨c²⟩ = 3p/ρ = 3 × 1.01 × 10⁵ / 1.25 = 2.424 × 10⁵ m² s⁻². c(rms) = √(2.424 × 10⁵) = 492 m s⁻¹.

5. A fixed volume of gas is heated. Use the kinetic model to explain why the pressure rises.4 marks

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   Answer: The molecules gain kinetic energy and move faster on average. They collide with the walls more often, and each collision involves a larger change in momentum. The force on the walls (rate of change of momentum) increases, so pressure (force per unit area) increases.

   Four ideas: faster molecules, more frequent collisions, bigger momentum change, so a larger force and pressure.

Next steps

Now that you can use the ideal gas equation, look at how motion at a very different scale is described in circular motion, or see where the physics of hot gases matters on a cosmic scale in the Hertzsprung-Russell diagram guide, where gas pressure holds stars up against gravity. You can also practise this topic with step-by-step marking, which is free to start.

Related guides

Module 5 · Thermal physicsSpecific Heat Capacity and Specific Latent Heat

Module 3 · Newton's laws of motion and momentumMomentum and Impulse: Collisions, Explosions and Force-Time Graphs

Module 5 · Circular motionCentripetal Force and Acceleration: Circular Motion Explained

Module 5 · Astrophysics and cosmologyThe Hertzsprung-Russell Diagram