AQA 7408 · Section 3.6.2 · Paper 2
Thermal physics, gas laws and kinetic theory
Connect internal energy and changes of state to ideal gases, work and the molecular model. Diagnose the misconception, test a prediction in the existing Thermal Laws simulation, then complete the exam assessment.
Content diagnostic
Commit before you calculate
Active revision
Use the same three-pass structure
Name the system
Decide what is inside the boundary and whether energy crosses it by heating or mechanical work.
Choose the model
Separate macroscopic gas laws from the microscopic collision model, then identify fixed quantities.
Check units and meaning
Use kelvin, pascals, cubic metres and kilograms. Finish by interpreting the sign or molecular change.
Thermal energy transfer
Track kinetic energy, potential energy and transfers
Internal energy
Internal energy is the sum of the randomly distributed kinetic and potential energies of the particles in a body. Heating a system or doing work on it can increase its internal energy.
The equivalent convention is ΔU = Q − Wby. State which work convention you use. Expansion gives positive work by the gas and negative work on the gas.
Temperature change
Use mass in kilograms, specific heat capacity in J kg⁻¹ K⁻¹ and a temperature change in K or °C. A temperature interval has the same numerical size in kelvin and degrees Celsius.
Change of state
Specific latent heat is energy per unit mass for a state change at constant temperature.
Continuous flow
Mass flow rate ṁ is measured in kg s⁻¹. This follows by dividing Q = mcΔθ by time.
Ideal gases
Move between experimental laws and the ideal-gas equation
| Fixed quantities | Relationship | Useful plot | Molecular interpretation |
|---|---|---|---|
| Fixed mass and temperature | pV = constant | p against 1/V is linear through the origin | Smaller volume means wall collisions occur more frequently. |
| Fixed mass and pressure | V/T = constant | V against T in kelvin is linear through the origin | Faster particles require more volume to keep collision rate per area unchanged. |
| Fixed mass and volume | p/T = constant | p against T in kelvin is linear through the origin | Faster impacts produce a greater rate of change of momentum at the walls. |
Use n in mol and R = 8.31 J mol⁻¹ K⁻¹.
Use N as the number of molecules and k = 1.38 × 10⁻²³ J K⁻¹.
NA = 6.02 × 10²³ mol⁻¹ and k = R/NA.
This expression applies directly for expansion or compression at constant pressure.
Useful simulation
Reuse the PhysicsUK Thermal Laws model
The existing model connects the macroscopic state to molecular speed, collision density and four equivalent equations. It reports every result numerically, so the animation is never the only source of information.
Molecular kinetic theory
Explain pressure from momentum changes
Simple model assumptions
- Molecules are identical point masses with negligible volume compared with the container.
- They move randomly and obey Newton's laws.
- There are no intermolecular forces except during collisions.
- Collisions with each other and the walls are perfectly elastic and very brief.
Brownian motion
A visible smoke particle follows an irregular path because many unseen air molecules strike it unevenly. It is the visible particle—not an individual air molecule—that is observed moving.
Temperature and kinetic energy
For an ideal gas of atoms, internal energy is entirely random translational kinetic energy. Therefore U = 3NkT/2 = 3nRT/2 and depends only on thermodynamic temperature.
AQA derivation route
- For one molecule of mass m with x-component cx, an elastic collision with a wall changes its momentum by 2mcx.
- The time between successive collisions with the same wall is 2l/cx, so its mean force on that wall is mcx²/l.
- Divide by wall area A. Since V = Al, the pressure contribution is mcx²/V.
- Sum over N molecules: pV = Nm⟨cx²⟩. Random motion gives ⟨cx²⟩ = ⟨c²⟩/3 = crms²/3.
- Therefore pV = ⅓Nmcrms².
Experimental relationships
Collect evidence for a gas law
Boyle's law
- Trap a fixed mass of gas in a sealed syringe connected to a pressure sensor.
- Compress slowly so the gas can return to room temperature after each change.
- Record pressure and volume over a wide range without exceeding apparatus limits.
- Plot p against 1/V, or calculate pV. A straight line through the origin for p against 1/V supports pV = constant.
Temperature relationships
- Keep mass fixed and control either pressure or volume.
- Place the gas container in stirred water baths and wait for thermal equilibrium.
- Measure thermodynamic temperature and the corresponding V or p.
- Plot V or p against T in kelvin. Extrapolation towards zero supports the concept of absolute zero, but do not claim an experiment reaches 0 K.
Misconception repair
Six ideas that commonly cost marks
- “Heating always raises temperature.” False during a state change: potential energy can increase at constant temperature.
- “Use degrees Celsius in pV = nRT.” False: thermodynamic temperature must be in kelvin.
- “Brownian motion is a molecule being watched.” False: the visible particle is buffeted by unseen molecules.
- “Pressure is caused by particles pushing continuously.” The model uses discrete wall collisions and rates of momentum change.
- “crms is the ordinary mean speed.” It is the square root of the mean of c².
- “Work done by and on a gas have the same sign.” They are opposites; state the convention before using the first law.
Original exam-style assessment
33 marks with worked feedback
Attempt every question before opening its answer. Show substitutions with SI units, use kelvin for gas equations and explain changes using energy or molecular momentum—not slogans.
1. Define internal energy. A block of ice melts at constant temperature. Explain how its internal energy can increase while its temperature remains constant.[4 marks]
Internal energy is the sum of the randomly distributed kinetic and potential energies of the particles. During melting, energy transferred to the ice increases particle potential energy as intermolecular separation changes. Mean kinetic energy, and therefore temperature, stays constant.
2. Calculate the energy needed to raise the temperature of 0.45 kg of water by 18 K. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.[2 marks]
Q = mcΔθ = 0.45 × 4200 × 18 = 3.40 × 10⁴ J, or 34 kJ.
3. A 1.5 kW heater warms water continuously through 12 K. Assuming no energy loss, calculate the mass flow rate. Use c = 4200 J kg⁻¹ K⁻¹.[3 marks]
For continuous flow, P = ṁcΔθ. Therefore ṁ = 1500/(4200 × 12) = 2.98 × 10⁻² kg s⁻¹.
4. A gas at 320 K occupies 0.024 m³ at a pressure of 1.20 × 10⁵ Pa. Calculate the amount of gas in moles. Use R = 8.31 J mol⁻¹ K⁻¹.[3 marks]
n = pV/(RT) = (1.20 × 10⁵ × 0.024)/(8.31 × 320) = 1.08 mol.
5. A container holds an ideal gas at 9.5 × 10⁴ Pa, 3.0 × 10⁻³ m³ and 290 K. Calculate the number of molecules. Use k = 1.38 × 10⁻²³ J K⁻¹.[3 marks]
N = pV/(kT) = (9.5 × 10⁴ × 3.0 × 10⁻³)/(1.38 × 10⁻²³ × 290) = 7.1 × 10²² molecules.
6. A gas expands at constant pressure 1.8 × 10⁵ Pa from 4.0 × 10⁻³ m³ to 7.5 × 10⁻³ m³. It receives 1000 J by heating. Calculate the work done by the gas and the increase in internal energy.[4 marks]
Wby = pΔV = 1.8 × 10⁵ × 3.5 × 10⁻³ = 630 J. Using ΔU = Q − Wby, ΔU = 1000 − 630 = 370 J. Equivalently, ΔU = Q + Won with Won = −630 J.
7. Describe Brownian motion observed through a microscope and explain why it is evidence for atoms or molecules.[3 marks]
Small visible particles move in an irregular, random zig-zag path. They are struck unevenly by many much smaller, unseen fluid molecules. The changing resultant impulses provide evidence that the molecules exist and move randomly.
8. The kelvin temperature of a fixed amount of gas in a rigid sealed container doubles. State what happens to its pressure and explain this using molecules.[3 marks]
The pressure doubles. Higher temperature gives molecules greater mean kinetic energy and speed. They collide with the walls more often and with a greater rate of change of momentum, so the force per unit area increases.
9. Nitrogen gas has molar mass 0.028 kg mol⁻¹ at 350 K. Calculate its root-mean-square molecular speed and the mean kinetic energy per molecule. Use R = 8.31 J mol⁻¹ K⁻¹ and k = 1.38 × 10⁻²³ J K⁻¹.[4 marks]
cᵣₘₛ = √(3RT/M) = √(3 × 8.31 × 350/0.028) = 5.58 × 10² m s⁻¹. Mean kinetic energy = 3kT/2 = 1.5 × 1.38 × 10⁻²³ × 350 = 7.25 × 10⁻²¹ J.
10. State two assumptions of the simple kinetic-theory model. Then explain the difference between the empirical gas laws and kinetic theory, including how scientific understanding changed.[4 marks]
Any two valid assumptions include negligible molecular volume, random motion, no intermolecular forces except during collisions, perfectly elastic collisions or negligible collision duration. Gas laws are empirical relationships established from measurements. Kinetic theory is a theoretical molecular model that explains those relationships; later evidence such as Brownian motion strengthened the particle interpretation and allowed the model to be refined.
Specification coverage
AQA 7408 section 3.6.2
- 3.6.2.1(a) Define internal energy as the sum of randomly distributed kinetic and potential energies of particles in a body.
- 3.6.2.1(b) Explain that internal energy increases when energy is transferred by heating or when work is done on a system.
- 3.6.2.1(c) Use a qualitative treatment of the first law of thermodynamics.
- 3.6.2.1(d) Explain that during change of state particle potential energy changes but kinetic energy does not.
- 3.6.2.1(e) Use Q = mcΔθ for energy transfer during temperature change.
- 3.6.2.1(f) Solve calculations involving continuous flow and specific heat capacity.
- 3.6.2.1(g) Use Q = ml for energy transfer during a change of state.
- 3.6.2.2(a) Use gas laws as experimental relationships between pressure, volume, temperature and mass of gas.
- 3.6.2.2(b) Use the concept of absolute zero of temperature.
- 3.6.2.2(c) Use ideal gas equation pV = nRT for n moles.
- 3.6.2.2(d) Use ideal gas equation pV = NkT for N molecules.
- 3.6.2.2(e) Use work done = pΔV for a gas at constant pressure.
- 3.6.2.2(f) Use Avogadro constant, molar gas constant and Boltzmann constant.
- 3.6.2.2(g) Use molar mass and molecular mass.
- 3.6.2.3(a) Describe Brownian motion as evidence for the existence of atoms.
- 3.6.2.3(b) Explain relationships between pressure, volume and temperature using a simple molecular model.
- 3.6.2.3(c) Recognise that gas laws are empirical whereas kinetic theory is a theoretical model.
- 3.6.2.3(d) State and use assumptions leading to pV = ⅓Nmc_rms².
- 3.6.2.3(e) Derive pV = ⅓Nmc_rms² using a simple algebraic approach with conservation of momentum.
- 3.6.2.3(f) Explain that for an ideal gas internal energy is kinetic energy of atoms.
- 3.6.2.3(g) Use average molecular kinetic energy = ½mc_rms² = 3kT/2 = 3RT/(2NA).
- 3.6.2.3(h) Appreciate how knowledge of gas behaviour has changed over time.
Continue through AQA Paper 2
Return to the AQA 3.6 module hub, connect force and work ideas through circular motion and mechanics and materials, or revisit measurements, units and uncertainties before the practical section. Then use the AQA Paper 2 revision hub, problem-solving practice and MCQ practice.
Written against all 22 AQA Physics 7408 points in sections 3.6.2.1–3.6.2.3. Questions are original. The static figure is an original generated PNG with no assessed text; the linked simulation is the existing PhysicsUK Thermal Laws model.