OCR A Level Physics A: Ideal gases and kinetic theory (H556) – Revision Notes

These revision notes condense section 5.1.4 (Ideal gases) of the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, which is part of the full A level. For full explanations and longer worked examples, read the study guide. Then test yourself with the practice set. The course hub is OCR A Level Physics, the printable list is the checklist, and the free 10-minute diagnostics show gaps. Temperature scales, specific heat capacity and latent heat are in the separate revision notes.

Constants used: NA = 6.02 × 10²³ mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹, k = 1.38 × 10⁻²³ J K⁻¹.

Equations at a glance

Statement	Equation	Notes
(a)	N = n NA; n = mass / molar mass	N is the number of particles
(d)	pV = nRT	p in Pa, V in m³, T in K
(d)	p₁V₁/T₁ = p₂V₂/T₂	fixed amount of gas
(d)	Boyle: pV constant	fixed mass, constant temperature
(d)	p/T constant	fixed mass and volume, T in kelvin
(e)	pV = ⅓ N m <c²>	derivation not required
(f)	c_rms = √<c²>	square, mean, root
(g)	k = R/NA	J K⁻¹
(h)	pV = NkT	N molecules
(h)	½ m <c²> = 3/2 kT	mean kinetic energy per molecule
(i)	U = 3/2 NkT = 3/2 nRT	ideal gas: no potential energy

(a) Moles

• One mole contains NA = 6.02 × 10²³ particles.
• Check the particle: oxygen gas O₂ has two atoms per molecule.

(b), (c) Kinetic theory model and pressure

Assumptions.

• A large number of molecules in random, rapid motion.
• Negligible volume compared with the volume of the gas.
• Collisions perfectly elastic, and each one lasts a negligible time compared with the gaps between collisions.
• Negligible forces between molecules except during collisions.

Pressure in steps.

1. A molecule hits the wall and rebounds, so its momentum changes.
2. Force on the molecule equals its rate of change of momentum (Newton’s second law), so the wall feels an equal and opposite force (Newton’s third law).
3. Very many random collisions give a steady average force.
4. Pressure is the force divided by the wall area.

Faster molecules or a smaller volume mean a higher pressure.

(d) Gas law experiments

Boyle’s law (constant temperature).

1. Trap air in a sealed tube linked to a pressure gauge.
2. Change the volume slowly in steps and wait for each reading to settle.
3. Record p and V. Plot p against 1/V.
4. A straight line through the origin shows pV is constant. The gradient equals nRT.

Pressure against temperature (constant volume) and absolute zero.

1. Immerse a sealed flask of air fully in a stirred water bath, with a pressure gauge on a short, narrow tube.
2. Heat in steps. Wait until the thermometer is steady, then record θ and p.
3. Plot p against θ (°C). Draw a best-fit line and extrapolate to p = 0.
4. The temperature at p = 0 is the estimate of absolute zero, about −273 °C.

Worked reminder: a flask of volume 5.0 × 10⁻⁴ m³ holds air at 1.0 × 10⁵ Pa and 300 K
n = pV / RT = (1.0 × 10⁵ × 5.0 × 10⁻⁴) / (8.31 × 300) = 0.020 mol
N = 0.020 × 6.02 × 10²³ = 1.2 × 10²² molecules

(e), (f) pV = ⅓Nm<c²> and speeds

• <c²> is the mean square speed. c_rms is its square root.
• Density form: p = ⅓ρ<c²>, so c_rms = √(3p/ρ).
• R.m.s. speed is a little larger than the mean speed.

Worked reminder: a gas at 1.0 × 10⁵ Pa with density 0.50 kg m⁻³
c_rms = √(3 × 1.0 × 10⁵ / 0.50) = 7.7 × 10² m s⁻¹

Maxwell-Boltzmann distribution: what to remember.

• Starts at the origin; rises to a peak; long tail at high speed; not symmetrical.
• Area under the curve is proportional to the number of molecules.
• Higher temperature: peak moves right, becomes lower and broader. More molecules have high speeds.
• Molecules at the same temperature have a spread of speeds because collisions keep exchanging energy.

(g), (h) Boltzmann constant and kinetic energy

Derivation (learn it)
pV = ⅓ N m <c²>   and   pV = N k T
⅓ N m <c²> = N k T
½ m <c²> = 3/2 k T

• Mean kinetic energy per molecule depends only on T, not on the gas.
• c_rms is proportional to √(T/m): four times the kelvin temperature doubles c_rms.
• At the same T the lighter molecule has the larger r.m.s. speed, but both have the same mean kinetic energy.

(i) Internal energy of an ideal gas

• No forces between molecules except in collisions, so no potential energy.
• Internal energy is the total random kinetic energy: U = 3/2 NkT = 3/2 nRT.
• It depends only on thermodynamic temperature (for fixed n). Doubling T in kelvin doubles U.

0.50 mol at 400 K: U = 3/2 × 0.50 × 8.31 × 400 = 2.5 × 10³ J

Method in steps: a gas calculation

1. List what is given and what is fixed (amount, volume, pressure or temperature).
2. Convert: °C to K by adding 273, cm³ to m³ by multiplying by 10⁻⁶, kPa to Pa by multiplying by 10³.
3. Choose the equation. Fixed amount with two states: p₁V₁/T₁ = p₂V₂/T₂. One state with n or N: pV = nRT or pV = NkT. Speed or kinetic energy: pV = ⅓Nm<c²> or ½m<c²> = 3/2 kT.
4. Rearrange before substituting, then give the answer with a unit and a sensible number of significant figures.
5. Check the size: room-temperature air molecules move at hundreds of metres per second, and a mole of gas at atmospheric pressure and room temperature occupies a few hundredths of a cubic metre.

Must-know distinctions

• Mean speed is not the r.m.s. speed, and <c²> is not (mean speed)².
• Mean kinetic energy per molecule (3/2 kT) is not the internal energy of the sample (3/2 NkT).
• N is a number of molecules; n is a number of moles.
• A temperature rise in °C equals the rise in K, but pV = nRT needs T in kelvin.

Quick self-test

1. How many moles and how many atoms are in 8.0 g of helium (molar mass 4.0 g mol⁻¹)?
2. Calculate the pressure of 0.50 mol of ideal gas at 300 K in a volume of 2.5 × 10⁻³ m³.
3. A gas at 1.0 × 10⁵ Pa and 3.0 × 10⁻³ m³ is compressed at constant temperature to 2.0 × 10⁵ Pa. Find the new volume.
4. Convert −40 °C to kelvin.
5. Find the r.m.s. speed of molecules in a gas of pressure 1.0 × 10⁵ Pa and density 0.50 kg m⁻³.
6. Calculate the mean kinetic energy of a molecule at 400 K.
7. Explain why an ideal gas has no potential energy.
8. Calculate the internal energy of 0.50 mol of ideal gas at 400 K.
9. By what factor does c_rms change when the temperature rises from 300 K to 1200 K?
10. How many moles are there in 3.01 × 10²⁴ molecules?
11. A sealed rigid container holds gas at 1.2 × 10⁵ Pa. The kelvin temperature doubles. State the new pressure.

Answers

1. n = 8.0 / 4.0 = 2.0 mol; N = 2.0 × 6.02 × 10²³ = 1.2 × 10²⁴ atoms.
2. p = nRT / V = (0.50 × 8.31 × 300) / (2.5 × 10⁻³) = 5.0 × 10⁵ Pa.
3. V₂ = p₁V₁ / p₂ = (1.0 × 10⁵ × 3.0 × 10⁻³) / 2.0 × 10⁵ = 1.5 × 10⁻³ m³.
4. −40 + 273 = 233 K.
5. c_rms = √(3p/ρ) = √(3 × 1.0 × 10⁵ / 0.50) = 7.7 × 10² m s⁻¹.
6. 3/2 × 1.38 × 10⁻²³ × 400 = 8.28 × 10⁻²¹ J.
7. There are negligible forces between molecules except during collisions, so there is no potential energy associated with their separation.
8. U = 3/2 × 0.50 × 8.31 × 400 = 2.5 × 10³ J.
9. c_rms is proportional to √T, and √(1200/300) = 2, so it doubles.
10. n = 3.01 × 10²⁴ / 6.02 × 10²³ = 5.0 mol.
11. At constant volume p is proportional to T, so the pressure doubles to 2.4 × 10⁵ Pa.

Where marks are usually lost

• Using °C in pV = nRT or in p/T calculations.
• Forgetting to convert cm³ to m³ before using pascals.
• Writing “molecules have no size” instead of negligible volume compared with the gas.
• Explaining pressure without the link between momentum change, force and area.
• Drawing a Maxwell-Boltzmann curve that touches the axis at high speed or is symmetrical.
• Quoting 3/2 kT as the energy of the whole sample.
• Describing a gas-law experiment without waiting for thermal equilibrium or without a table of readings.
• Extrapolating a p against θ graph from a short range and not commenting on the uncertainty.

Official syllabus

OCR Advanced GCE in Physics A (H556), specification Version 3.0 (March 2026), first assessment 2017, published by OCR: section 5.1.4, statements (a) to (i).