OxfordAQA International AS and A-level Physics: Ideal gases and kinetic theory (9630) – Revision Notes

These revision notes are aligned to the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, sections 3.11.3 (ideal gases) and 3.11.4 (kinetic theory of gases). Section 3.11 is International A-level only. For full explanations and the derivation, read the study guide first, then test yourself with the practice questions.

Other links: the course page, the checklist, the free diagnostics and the neighbouring unit on the first law and conduction.

Key definitions

• Absolute zero: 0 K, about −273 °C. The temperature at which an ideal gas would, by extrapolation, have zero volume at constant pressure or zero pressure at constant volume.
• Ideal gas: a gas that obeys pV = nRT at all pressures and temperatures.
• Avogadro constant N_A: particles per mole, 6.02 × 10²³ mol⁻¹.
• Molar mass M: the mass of one mole, in kg mol⁻¹. Molecular mass m: the mass of one molecule, m = M / N_A.
• Empirical: based on observation and measurement. The gas laws are empirical. Kinetic theory is a model built from theory.
• Root mean square speed c_rms: the square root of the mean of the squared speeds.
• Internal energy of an ideal gas: the total kinetic energy of its atoms, with no potential energy.

Formulas

Quantity	Equation	Notes
Boyle’s law	pV = constant	fixed mass, constant T
Charles’s law	V/T = constant	fixed mass, constant p, T in K
Pressure law	p/T = constant	fixed mass, constant V, T in K
Combined	p₁V₁/T₁ = p₂V₂/T₂	fixed mass of gas
Ideal gas, moles	pV = nRT	R = 8.31 J mol⁻¹ K⁻¹
Ideal gas, molecules	pV = NkT	k = 1.38 × 10⁻²³ J K⁻¹
Links	R = N_A k, n = N / N_A, n = mass / M
Temperature	T / K = θ / °C + 273
Work done	W = pΔV	constant pressure; area under a p–V graph
Kinetic theory	pV = ⅓Nm(c_rms)²	m is the mass of one molecule
Density form	p = ⅓ρ(c_rms)²	because ρ = Nm / V
Average kinetic energy	½m(c_rms)² = 3kT/2 = 3RT/2N_A	per molecule
Internal energy	U = 3nRT/2 = 3NkT/2	ideal gas, total kinetic energy of the atoms

Methods in steps

Gas-law calculation (fixed mass)

1. List p, V and T before and after.
2. Convert every temperature to kelvin.
3. Write p₁V₁/T₁ = p₂V₂/T₂, cancelling any quantity that stays constant.
4. Rearrange, substitute and check the sign of the change makes physical sense.

Using pV = nRT or pV = NkT

1. Convert p to Pa, V to m³, T to K and mass to kg.
2. Use n = mass / M if the question gives a mass, or N = nN_A if it asks for a count.
3. Rearrange symbolically before substituting.
4. Quote the result with its unit and an appropriate number of significant figures.

Kinetic-theory calculation

1. Find the mass of one molecule: m = M / N_A.
2. Use pV = ⅓Nm(c_rms)² to link pressure and speed, or ½m(c_rms)² = 3kT/2 to link speed and temperature.
3. Take the square root last. The result is c_rms, not the mean speed.

Required practical 9

• Boyle’s law: trap air in a syringe on a pressure gauge. Change V slowly, wait for thermal equilibrium, record p. Plot p against 1/V. Pressure and 1/V give a straight line through the origin if pV is constant.
• Charles’s law: trap dry air in a capillary tube with a liquid index, held in a stirred water bath. Plot the air column length against θ and extrapolate to zero length. The intercept estimates absolute zero.
• Improve accuracy by waiting at each step, keeping the whole gas sample in the bath and measuring the temperature next to the tube.

Must-know distinctions

Idea	Do not confuse with
Molar mass M (kg mol⁻¹)	Molecular mass m (kg)
R, used with moles	k, used with molecules
c_rms, from squaring first	Mean speed, a plain average
Empirical gas laws	Kinetic theory, derived from assumptions
Work done by the gas (expansion)	Work done on the gas (compression)
Kelvin temperature	Celsius temperature
Average kinetic energy per molecule	Internal energy of the whole sample

Assumptions of the kinetic model

1. A very large number of molecules in random motion.
2. Molecular volume negligible compared with the container.
3. No intermolecular forces except during collisions, so no potential energy.
4. Collisions with each other and with walls are elastic.
5. Collision time negligible compared with time between collisions.
6. Newton’s laws apply.

Brownian motion in two lines

Visible smoke or pollen particles jitter randomly because unequal numbers of fast air or water molecules hit them from different sides. That is evidence that the molecules exist and move at random.

Worked reminders

Boyle’s law. Gas at 1.2 × 10⁵ Pa in 90 cm³ is compressed slowly to 36 cm³. p₂ = 1.2 × 10⁵ × 90 / 36 = 3.0 × 10⁵ Pa.

Pressure law. A rigid tyre holds air at 2.2 × 10⁵ Pa and 280 K. At 310 K, p₂ = 2.2 × 10⁵ × 310 / 280 = 2.4 × 10⁵ Pa (2.44 × 10⁵).

Moles and molecules. 0.220 kg of carbon dioxide, M = 0.0440 kg mol⁻¹. n = 0.220 / 0.0440 = 5.0 mol. N = 5.0 × 6.02 × 10²³ = 3.0 × 10²⁴. Mass of one molecule = 0.0440 / 6.02 × 10²³ = 7.31 × 10⁻²⁶ kg.

Density form. A gas of density 1.2 kg m⁻³ has c_rms = 500 m s⁻¹. p = ⅓ × 1.2 × 500² = 1.0 × 10⁵ Pa.

Area under a p–V graph. Pressure falls in a straight line from 5.0 × 10⁵ Pa to 3.0 × 10⁵ Pa while V rises from 0.8 × 10⁻³ m³ to 2.3 × 10⁻³ m³. The area is a trapezium: W = ½(5.0 + 3.0) × 10⁵ × 1.5 × 10⁻³ = 600 J, done by the gas.

Comparing gases. At one temperature, helium and argon atoms have the same average kinetic energy. Since c_rms is proportional to 1/√m, c_rms(He) / c_rms(Ar) = √(40 / 4) = 3.2.

Explaining a law in words. For the pressure law say: higher temperature, so higher average kinetic energy and speed; molecules hit the walls harder and more often; the force per unit area rises, so p rises while V is fixed.

Quick self-test

1. Convert 37 °C to kelvin.
2. Convert −196 °C to kelvin.
3. A fixed mass of gas at constant temperature is compressed to half its volume. What happens to its pressure?
4. How many molecules are in 0.50 mol?
5. Find the mass of one oxygen molecule if M = 0.0320 kg mol⁻¹.
6. A gas expands by 4.0 × 10⁻³ m³ at a constant 150 kPa. Find the work it does.
7. Find the average kinetic energy of a molecule at 600 K.
8. By what temperature must a gas at 300 K be heated for c_rms to double?
9. Find the internal energy of 0.30 mol of ideal gas at 400 K.
10. Find c_rms for hydrogen molecules (M = 0.0020 kg mol⁻¹) at 273 K.
11. Explain why smoke particles seen through a microscope move randomly.
12. State one reason the gas laws are called empirical.

Answers

1. 37 + 273 = 310 K.
2. −196 + 273 = 77 K.
3. The pressure doubles, since pV is constant.
4. N = 0.50 × 6.02 × 10²³ = 3.0 × 10²³.
5. m = 0.0320 / 6.02 × 10²³ = 5.32 × 10⁻²⁶ kg.
6. W = pΔV = 1.5 × 10⁵ × 4.0 × 10⁻³ = 600 J.
7. 3kT/2 = 1.5 × 1.38 × 10⁻²³ × 600 = 1.24 × 10⁻²⁰ J.
8. Doubling c_rms quadruples T, so T = 1200 K and the rise is 900 K.
9. U = 1.5 × 0.30 × 8.31 × 400 = 1.5 × 10³ J (1496 J).
10. m = 0.0020 / 6.02 × 10²³ = 3.32 × 10⁻²⁷ kg. c_rms = √(3kT/m) = √(3 × 1.38 × 10⁻²³ × 273 / 3.32 × 10⁻²⁷) = 1.84 × 10³ m s⁻¹.
11. Air molecules, too small to see, strike a smoke particle at random and fast. The numbers hitting each side are unequal at any moment, so the net force changes in size and direction.
12. They were found by experiment and describe how p, V and T behave. They do not explain why.

Where marks are usually lost

• Using Celsius temperatures in pV = nRT or in a gas-law ratio.
• Leaving molar mass in g mol⁻¹ while other quantities are in SI units.
• Using the molar mass in place of the mass of one molecule in pV = ⅓Nm(c_rms)².
• Forgetting the square root at the end, so (c_rms)² is quoted as a speed.
• Writing “molecules move faster” when explaining Boyle’s law, where temperature is constant.
• Describing Brownian motion as the movement of the molecules themselves, rather than the visible particles they bump.
• Writing that molecules occupy zero space, instead of negligible volume compared with the container.
• Dropping the “random” in the definition of internal energy, or adding potential energy to an ideal gas.
• Describing the Charles’s law method without saying that the gas must reach the bath temperature.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams from May/June 2020, published by OxfordAQA. Sections 3.11.3 and 3.11.4.