A-Level Physics
Thermal physics
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At a glance
Subtopics and learning goals
- Apply pV = nRT and thermal-energy equations.
- Relate microscopic particle motion to macroscopic pressure and temperature.
- Apply the ideal-gas equation pV = nRT.
- Explain gas pressure and temperature using a kinetic model.
- Use specific heat capacity in thermal-energy calculations.
Revision notes
Temperature is linked to the microscopic energy distribution of particles. In an ideal gas, particles are treated as point-like, moving randomly with negligible intermolecular forces except during elastic collisions. For a fixed amount of ideal gas, pV = nRT. Gas pressure arises from momentum changes when particles collide with container walls.
Temperature, internal energy and thermal equilibrium
Temperature is associated with the distribution of microscopic particle energies, while internal energy is the total microscopic kinetic and potential energy of a system. Two objects in thermal equilibrium have the same temperature and no net thermal-energy transfer between them.
Absolute temperature is measured in kelvin. Temperature differences in kelvin and degrees Celsius have the same numerical size, but equations involving ratios or ideal gases require absolute temperature.
Specific heat capacity
For heating without a change of state, Q = mcΔT. Specific heat capacity is the energy required per kilogram per kelvin of temperature rise. In calorimetry, energy losses to the surroundings or the apparatus can cause systematic differences between the ideal model and measured value.
The kinetic model
An ideal gas is modelled as a large number of particles moving randomly with negligible volume and negligible intermolecular forces except during elastic collisions. Pressure arises from momentum changes when particles collide with container walls.
Increasing temperature increases mean kinetic energy. At fixed volume this increases the rate and magnitude of momentum transfer to the walls, increasing pressure.
Ideal-gas equation
For n moles of ideal gas, pV = nRT. For a fixed mass of gas, relationships between pressure, volume and absolute temperature follow from this equation. Always convert temperature to kelvin and volume to cubic metres when using SI units.
The kinetic theory equation pV = ⅓Nm(\(c_{\text{rms}}\))² connects the two models. Comparing it with pV = NkT gives the mean kinetic energy of a molecule, ½m(\(c_{\text{rms}}\))² = \(\dfrac{3}{2}kT\).
Internal energy of gases
For an ideal monatomic gas, internal energy depends only on temperature in the school-level model because intermolecular potential energy is neglected. Work done on or by a gas and heating can therefore change its internal energy.
Exam technique
Use kelvin in gas-law calculations. Distinguish temperature from internal energy. For state changes, explain where transferred energy goes when temperature is constant. In practical calorimetry, identify heat-loss mechanisms and their effect on the calculated quantity.
Kinetic theory and the Boltzmann constant
The ideal gas equation can be written for molecules as pV = NkT, where N is the number of molecules and k = \(\dfrac{R}{N_{A}}\) = 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant. One mole contains \(N_{A}\) = 6.02 × 10²³ particles, and the number of moles is n = mass/molar mass.
Kinetic theory models pressure as the result of molecular collisions and gives pV = ⅓Nm(\(c_{\text{rms}}\))², where \(c_{\text{rms}}\) is the root-mean-square speed. Combining this with pV = NkT gives the mean kinetic energy of one molecule: ½m(\(c_{\text{rms}}\))² = \(\dfrac{3}{2}kT\). Mean molecular kinetic energy is therefore proportional to absolute temperature.
Gas laws and absolute zero
For a fixed mass of gas: at constant temperature pV is constant (Boyle's law); at constant pressure V/T is constant (Charles's law); at constant volume \(\dfrac{p}{T}\) is constant. Extrapolating a graph of pressure or volume against temperature to zero gives absolute zero, 0 K = −273.15 °C.
Worked example
Worked example 1: For 1.00 mol of ideal gas at 300 K occupying 0.0240 m³, p = \(\dfrac{nRT}{V}\) = \(\dfrac{1.00 \times 8.314 \times 300}{0.0240}\) ≈ 1.04×10⁵ Pa.
Worked example 2: Heating 0.50 kg of water by 30 K with c = 4200 J kg⁻¹ K⁻¹ requires Q = 0.50 × 4200 × 30 = 63000 J.
For 1.00 mol of ideal gas at 300 K in 0.0240 m³, p = \(\dfrac{nRT}{V}\) = \(\dfrac{1.00 \times 8.314 \times 300}{0.0240}\) ≈ 1.04×10⁵ Pa.
Worked example 3: For 0.40 mol of ideal gas at 320 K in 0.010 m³, p = \(\dfrac{nRT}{V}\) = \(\dfrac{0.40 \times 8.31 \times 320}{0.010}\) = 1.06 × 10⁵ Pa.
Misconception check
Gas particles do not expand when a gas is heated in the ideal-particle model. The average kinetic energy and typical separations can change; the particles themselves are not modelled as swelling.
Assessment coverage
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Retrieval cards
State the ideal-gas equation in molar form.
pV = nRT.
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Course specification reference
This topic has been matched to the official course specification.
- CCEA — CCEA GCE Physics (1210): Thermal physics — 1210 specification.. Checked 6 September 2026. View official course specification