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

This study guide follows the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017. It covers section 5.1.4 (Ideal gases), statements (a) to (i), all part of the full A level. Temperature scales, the solid-liquid-gas model, specific heat capacity and latent heat are taught in a separate guide. After this guide use the revision notes and the practice set. The course hub is OCR A Level Physics, the printable list is the checklist, and the free 10-minute diagnostics show which topics need work. The practical skills behind the gas experiments are in Practical skills.

Values used here: NA = 6.02 × 10²³ mol⁻¹ (the specification gives this value), R = 8.31 J mol⁻¹ K⁻¹ and k = R/NA = 1.38 × 10⁻²³ J K⁻¹. Where a question needs other constants, they are given in the question.

What this unit covers

Statement	What you must be able to do
5.1.4(a)	Use amount of substance in moles and the Avogadro constant
5.1.4(b)	State the assumptions of the kinetic theory model of a gas
5.1.4(c)	Explain gas pressure in terms of the model
5.1.4(d)	Use pV = nRT; describe experiments for Boyle’s law and for p/T; estimate absolute zero
5.1.4(e)	Use pV = ⅓Nm<c²> (derivation not required)
5.1.4(f)	Define r.m.s. speed and mean square speed; know the general features of the Maxwell-Boltzmann distribution
5.1.4(g)	Use the Boltzmann constant, k = R/NA
5.1.4(h)	Use pV = NkT; derive ½m<c²> = 3/2 kT from the two equations
5.1.4(i)	Give the internal energy of an ideal gas

5.1.4(a) The mole and the Avogadro constant

One mole is the amount of substance containing 6.02 × 10²³ particles, the Avogadro constant NA. The number of particles is N = n NA, where n is the number of moles. For a sample of mass m and molar mass M, n = m/M.

Worked example: 10.0 g of argon (molar mass 40 g mol⁻¹)
n = 10.0 / 40 = 0.25 mol
N = 0.25 × 6.02 × 10²³ = 1.5 × 10²³ atoms

Argon is monatomic, so atoms and molecules are the same thing here. For oxygen gas, O₂, each molecule holds two atoms, so check which particle a question is counting.

5.1.4(b) The kinetic theory model

The model describes an ideal gas. Its assumptions are:

• A large number of molecules in random, rapid motion.
• The molecules (atoms or molecules) occupy negligible volume compared with the volume of the gas.
• All collisions are perfectly elastic, and each collision lasts a negligible time compared with the gaps between collisions.
• There are negligible forces between the molecules except during collisions.

A real gas behaves most like this at low pressure and high temperature, because the molecules are then far apart and move fast.

5.1.4(c) Pressure in terms of the model

Use Newton’s laws.

1. A molecule travelling towards a wall rebounds elastically. Its velocity component perpendicular to the wall reverses, so its momentum changes.
2. By Newton’s second law, the wall exerts an impulse on the molecule, and by Newton’s third law the molecule exerts an equal and opposite force on the wall.
3. Enormous numbers of molecules hit the wall every second at random. The average force is steady, and pressure is this force divided by the area of the wall.

The same argument explains the gas laws. Squeezing the same gas into half the volume doubles the number of molecules per unit volume, so collisions with the wall are twice as frequent and the pressure doubles. Heating the gas makes molecules faster, so each collision has a larger momentum change and collisions are more frequent. The pressure rises.

5.1.4(d) The equation of state

For an ideal gas, pV = nRT, with p in pascals, V in m³, T in kelvin and n in moles. Convert cm³ to m³ by multiplying by 10⁻⁶.

Worked example 1: 0.250 mol at 310 K in 4.0 × 10⁻³ m³
p = nRT / V = (0.250 × 8.31 × 310) / (4.0 × 10⁻³)
p = 1.6 × 10⁵ Pa

When n is constant, pV/T is constant, so p₁V₁/T₁ = p₂V₂/T₂.

Worked example 2: gas at 1.0 × 10⁵ Pa, 2.4 × 10⁻³ m³ and 290 K
is heated to 350 K and compressed to 1.8 × 10⁻³ m³.
p₂ = p₁V₁T₂ / (T₁V₂)
   = (1.0 × 10⁵ × 2.4 × 10⁻³ × 350) / (290 × 1.8 × 10⁻³)
   = 1.6 × 10⁵ Pa

Boyle’s law experiment (d)(ii)

Boyle’s law: for a fixed mass of gas at constant temperature, pV is constant.

• Trap a fixed mass of air in a sealed tube or syringe connected to a pressure gauge or sensor.
• Change the volume in steps, reading volume from the scale and pressure from the gauge.
• Wait after each change so the gas returns to room temperature, and change the volume slowly so it does not heat up.
• Plot p against 1/V. A straight line through the origin confirms Boyle’s law, and its gradient is nRT.

Sample results at constant temperature:

p / 10⁵ Pa	1.00	1.20	1.50	2.00	2.50	3.00
V / cm³	60	50	40	30	24	20
pV / J	6.0	6.0	6.0	6.0	6.0	6.0

pV is constant, for example 1.00 × 10⁵ × 60 × 10⁻⁶ = 6.0 J. At 293 K this gives n = 6.0 / (8.31 × 293) = 2.5 × 10⁻³ mol.

Pressure and temperature, and absolute zero (d)(ii), (d)(iii)

For fixed mass and volume, p/T is constant, so p is proportional to the thermodynamic temperature.

• Seal a flask of air to a pressure gauge by a short, narrow tube. Immerse the whole flask in a stirred water bath.
• Heat the bath in steps. At each step wait and stir until the thermometer reading is steady, then record θ and p.
• Plot p against θ in °C and draw the best-fit straight line. Extend it back until p = 0. The temperature there is the estimate of absolute zero.

Worked example: p = 100 kPa at 20 °C and 128 kPa at 100 °C (on the best-fit line)
gradient = (128 − 100) / 80 = 0.35 kPa per °C
p = 0 at θ = 20 − 100 / 0.35 = −266 °C

The accepted value of absolute zero is about −273 °C. The estimate is a long way from the data, so a small error in gradient gives a large error in the intercept. Reduce error by using a wide temperature range, keeping the flask fully immersed, and keeping the gas in the connecting tube small.

5.1.4(e) pV = ⅓Nm<c²>

For N molecules each of mass m in volume V:

pV = ⅓ N m <c²>

Here <c²> is the mean square speed, the mean of the squares of the speeds. You do not have to derive this equation. Dividing by V and using density ρ = Nm/V gives p = ⅓ρ<c²>.

Worked example: nitrogen at p = 1.00 × 10⁵ Pa and density 1.15 kg m⁻³
<c²> = 3p / ρ = (3 × 1.00 × 10⁵) / 1.15 = 2.6 × 10⁵ m² s⁻²

5.1.4(f) R.m.s. speed and the Maxwell-Boltzmann distribution

The root mean square speed is c_rms = √<c²>. The order of operations is in the name: square each speed, take the mean, then square-root.

Five molecules with speeds 120, 150, 180, 210 and 240 m s⁻¹
mean speed = 900 / 5 = 180 m s⁻¹
<c²> = (14 400 + 22 500 + 32 400 + 44 100 + 57 600) / 5 = 34 200 m² s⁻²
c_rms = √34 200 = 185 m s⁻¹

The r.m.s. speed is slightly larger than the mean speed, because squaring weights fast molecules more.

The speeds of molecules in a gas at one temperature are spread out. The Maxwell-Boltzmann distribution shows the number of molecules at each speed. Its general features are:

• The curve starts at the origin: almost no molecules are stationary.
• It rises to a peak and then falls with a long tail towards high speeds. It is not symmetrical.
• The area under the curve is proportional to the total number of molecules.
• At a higher temperature the peak moves to a higher speed and becomes lower and broader, with a larger fraction of fast molecules. The area is unchanged if the number of molecules is the same.

5.1.4(g), (h) The Boltzmann constant and molecular kinetic energy

Writing the amount of gas as a number of molecules, n = N/NA, turns nRT into (N/NA)RT. The Boltzmann constant is k = R/NA:

k = 8.31 / 6.02 × 10²³ = 1.38 × 10⁻²³ J K⁻¹

so pV = NkT.

Derivation of ½m<c²> = 3/2 kT

pV = ⅓ N m <c²>      and      pV = N k T
so   ⅓ N m <c²> = N k T
     m <c²> = 3 k T
     ½ m <c²> = 3/2 k T

The left-hand side is the mean translational kinetic energy of a molecule. It depends only on the thermodynamic temperature, not on the type of gas.

Worked example: helium atoms (m = 6.64 × 10⁻²⁷ kg) at 300 K
mean kinetic energy = 3/2 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J
<c²> = 3kT / m = (3 × 1.38 × 10⁻²³ × 300) / 6.64 × 10⁻²⁷ = 1.87 × 10⁶ m² s⁻²
c_rms = 1.37 × 10³ m s⁻¹

Because c_rms is proportional to √(T/m), quadrupling T doubles c_rms, and at the same temperature a lighter molecule is faster. Hydrogen molecules (molar mass 2) have an r.m.s. speed four times that of oxygen molecules (molar mass 32), since √(32/2) = 4.

5.1.4(i) Internal energy of an ideal gas

Internal energy is the sum of the random kinetic and potential energies of the molecules. In an ideal gas there are negligible forces between molecules except in collisions, so there is no potential energy. The internal energy is the total kinetic energy of the molecules:

U = N × 3/2 kT = 3/2 nRT

Worked example: 2.0 mol of ideal gas at 300 K, then heated to 360 K
U (300 K) = 3/2 × 2.0 × 8.31 × 300 = 7.5 × 10³ J
U (360 K) = 3/2 × 2.0 × 8.31 × 360 = 9.0 × 10³ J
change = 1.5 × 10³ J

The internal energy of a fixed amount of ideal gas depends only on its thermodynamic temperature. Doubling the kelvin temperature doubles U.

Common errors

• Using °C in pV = nRT. Always convert to kelvin.
• Leaving volume in cm³ or dm³ when p is in Pa.
• Taking the square root of the mean speed instead of the mean of the squares.
• Writing ½mc² for the kinetic energy of the whole gas rather than one molecule.
• Stating that molecules have no volume, instead of negligible volume compared with the gas.
• Saying pressure is caused by molecules “hitting each other” instead of hitting the walls.

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).