A Level Physics: Thermodynamics — Revision Notes

Condensed for the final weeks. For the full explanation, use the Thermodynamics study guide.

Internal energy

The sum of the random distribution of kinetic and potential energies of the molecules in a system.

Both halves matter:

• Kinetic component depends on temperature.
• Potential component depends on separation, so it changes during a change of state.

That is why temperature stays constant while a substance melts or boils: the energy supplied increases the potential component only.

The first law of thermodynamics

delta-U = q + W

delta-U = increase in internal energy
q       = heat supplied TO the system
W       = work done ON the system

Sign convention is where marks are lost. Using Cambridge’s convention:

Situation	Sign
Heat supplied to the gas	q positive
Heat lost by the gas	q negative
Gas compressed (work done on it)	W positive
Gas expands (work done by it)	W negative

The magnitude of the work done at constant pressure is:

W = p delta-V     (delta-V is the change in volume; this gives the work done BY the gas during an expansion)

Since W in delta-U = q + W is defined as work done ON the system (line 46 above), its sign must be fixed up to match: for an expansion (ΔV positive), the gas does work on its surroundings, so work done on the system is negative, W = −pΔV; for a compression (ΔV negative), the surroundings do work on the gas, so work done on the system is positive, W = +p|ΔV|. This matches the sign convention table above: expansion → W negative, compression → W positive.

Worked example. A gas absorbs 500 J of thermal energy and does 200 J of work by expanding against a constant external pressure. Since the gas does work on its surroundings, W (work done on the system) = −200 J.

delta-U = q + W = 500 + (-200) = 300 J

The gas’s internal energy increases by 300 J.

The four processes

Process	Condition	Consequence
Isothermal	Constant T	ΔU = 0, so q = −W
Adiabatic	No heat transfer	q = 0, so ΔU = W
Isobaric	Constant p	W = pΔV
Isovolumetric	Constant V	W = 0, so ΔU = q

Two of these follow directly from ΔU depending only on temperature for an ideal gas.

Internal energy is a state function — it depends only on the current state of the system (its temperature, for an ideal gas), never on the process or path used to reach that state. Two different routes between the same start and end states give the same ΔU, even if q and W individually differ along each route.

Heating equations

temperature change:   E = m c delta-theta
change of state:      E = m L

• Specific heat capacity c — energy to raise 1 kg by 1 K, J kg⁻¹ K⁻¹.
• Specific latent heat L — energy to change the state of 1 kg with no temperature change, J kg⁻¹.

Specific heat capacity is defined per kilogram; heat capacity (without “specific”) is a property of a particular object and is defined per kelvin only, without dividing by mass — the two are easily confused in an exam answer.

Latent heat of vaporisation > latent heat of fusion because boiling must overcome the intermolecular forces completely and also do work pushing back the atmosphere, whereas melting only loosens them.

Absolute zero and the ideal gas

Absolute zero (0 K, −273.15 °C, often rounded to −273 °C) is the temperature at which molecules have minimum internal energy — not zero energy.

T(K) = theta(C) + 273.15   (often rounded to +273 for working)
pV = nRT

Exam traps

• Defining internal energy as kinetic energy only.
• Getting the sign of W wrong for expansion versus compression.
• Using E = mcΔθ during a change of state, where temperature is constant.
• Saying molecules “stop moving” at absolute zero.
• Forgetting to convert °C to K in gas equations.
• Confusing specific heat capacity (per kg per K) with heat capacity (per K).
• Assuming internal energy depends on the process used to reach a state — it is a state function, determined only by the current state.
• Applying W = pΔV when pressure is not constant during the process.

Self-test

1. Define internal energy, covering both components.
2. State the first law and the sign convention for a gas being compressed.
3. Why does temperature stay constant during boiling?
4. An ideal gas expands isothermally. What are ΔU and q?
5. Why is latent heat of vaporisation larger than latent heat of fusion?
6. A gas absorbs 500 J of thermal energy and does 200 J of work by expanding. Find the change in internal energy.
7. Why does internal energy not depend on the process used to reach a given state?

Answers: 1. The sum of the random distribution of kinetic and potential energies of the molecules; the kinetic part depends on temperature, the potential part on molecular separation. 2. ΔU = q + W; compressing the gas means work is done on it, so W is positive. 3. The energy supplied increases the potential component of internal energy by separating the molecules, not the kinetic component, so temperature is unchanged. 4. ΔU = 0 for an ideal gas (internal energy depends only on temperature, which is constant here), so q = −W — this does not hold for a real gas, where internal energy also depends on the (changing) separation between molecules. 5. Vaporisation must completely overcome the intermolecular forces and also do work against atmospheric pressure as the vapour expands; melting only loosens the forces. 6. W = −200 J (the gas does work on its surroundings); ΔU = q + W = 500 + (−200) = 300 J. 7. Internal energy is a state function — it depends only on the system’s current state (its temperature, for an ideal gas), never on the route taken to reach it, even though q and W individually can differ between different routes.