Thermodynamic Systems: Complete A Level H2 Physics Cheatsheet
A Level Physics (H2), Chapter 13 · 9478 Topic 13 · Read time: ~10 minutes
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Key Formulas
Internal energy and temperature
Internal energy (U) is the sum of a random distribution of microscopic kinetic and potential energies of the particles in a system. It is determined by the system's macroscopic state (such as p, V and T).
The thermodynamic temperature is proportional to the mean microscopic kinetic energy of the particles.
For an ideal gas there are no intermolecular forces, so no potential energy: U is purely kinetic and depends only on T.
Thermal equilibrium and the zeroth law
Two systems in thermal contact exchange energy by heating, from the higher temperature to the lower, until they reach the same temperature: thermal equilibrium, with no net energy transfer.
Zeroth law: if two systems are each in thermal equilibrium with a third system, they are in thermal equilibrium with each other. This is why a thermometer works.
Work done and the first law
A gas expanding against a constant external pressure does work on the surroundings: W(by gas) = pΔV. A gas being compressed has work done on it.
First law (9478 form): ΔU = Q + W, where Q is the energy supplied to the system by heating and W is the work done on the system.
Constant volume: W = 0, so ΔU = Q. Expansion: W is negative. Compression: W is positive.
Trick: On a p-V graph, the work done by the gas is the area under the curve. For a full cycle, the net work is the area enclosed by the loop.
Trap: Sign errors cost the most marks here. A gas that expands has work done ON it equal to −pΔV. Write each sign out explicitly before substituting into ΔU = Q + W.
Specific heat capacity and specific latent heat
Specific heat capacity (c): energy per unit mass per unit temperature change, Q = mcΔθ.
Specific latent heat (l): energy per unit mass to change state at constant temperature, Q = ml, with separate values for fusion and vaporisation.
During a change of state, internal energy rises as particle potential energy increases, while mean kinetic energy, and so temperature, stays constant.
Worked Example
A gas in a cylinder is heated and expands from 2.0 × 10⁻³ m³ to 5.0 × 10⁻³ m³ against a constant pressure of 1.0 × 10⁵ Pa. 800 J of energy is supplied to the gas by heating.
(a) Calculate the work done by the gas. [2]
W(by gas) = pΔV = 1.0 × 10⁵ × (5.0 − 2.0) × 10⁻³
= 300 J
(b) Calculate the change in internal energy of the gas. [2]
Work done ON the gas, W = −300 J; Q = +800 J
ΔU = Q + W = 800 − 300 = +500 J
(c) State what happens to the temperature of the gas. Explain. [2]
The temperature rises.
Internal energy has increased, and for an ideal gas U depends only on T.
Why this question is a good test of the topic: the work done BY the gas is +300 J, but the first law needs work done ON it. Getting that sign right is the whole question.
Frequently Asked Questions
What is internal energy?
The sum of the random microscopic kinetic and potential energies of all the particles in a system.
What is the zeroth law of thermodynamics?
If two systems are each in thermal equilibrium with a third, they are in thermal equilibrium with each other.
What sign convention does 9478 use for the first law?
ΔU = Q + W, with Q the energy supplied to the system by heating and W the work done on the system.
How do I calculate work done by a gas?
At constant pressure, W = pΔV. In general, it is the area under the p-V graph.
Why does internal energy increase during melting if temperature doesn't change?
The energy increases the potential energy of the particles, not their mean kinetic energy.
Still mixing up signs in the first law?
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