The first law of thermodynamics
The first law of thermodynamics is conservation of energy for heating and work. In the form used in AQA A-level Physics, Q = ΔU + W: the energy Q transferred to a system by heating equals the increase in its internal energy, ΔU, plus the work W done by the system. Heat a gas with 500 J while it does 200 J of work pushing a piston out, and its internal energy rises by 300 J.
The first law of thermodynamics equation
| Symbol | Meaning | Unit |
|---|---|---|
| Q | Energy transferred to the system by heating (negative if the system loses heat) | J |
| ΔU | Increase in internal energy (negative if it decreases) | J |
| W | Work done by the system (negative if work is done on it) | J |
AQA's A-level specification defines the symbols exactly this way, in its engineering physics option, and its data booklet gives Q = ΔU + W.
Sign conventions
Books write the first law in two ways, depending on how they define W:
| Convention | Equation | W means |
|---|---|---|
| Physics (AQA) | Q = ΔU + W | Work done by the gas |
| Chemistry and many university texts | ΔU = Q + W | Work done on the gas |
They say the same thing. A gas that expands does work on its surroundings: positive W in the AQA form, negative W in the other. Check which convention a question uses before substituting.
Internal energy
The internal energy of a system is the total kinetic and potential energy of its particles. For an ideal gas the particles have no potential energy between them, so its internal energy depends only on its temperature: ΔU = 0 whenever the temperature does not change. Heating a gas or doing work on it can both raise its internal energy.
Work done by a gas: W = pΔV
When a gas expands against a constant pressure, the work it does is:
AQA's specification gives this for the constant pressure case, and says that, more generally, the work done can be estimated from the area below the graph on a pV diagram.
Example 1: heating and work together
A gas is heated with 500 J of energy and does 200 J of work as it expands. Find the change in internal energy.
Example 2: expansion at constant pressure
A gas at a constant pressure of 1.0 × 10⁵ Pa expands from 2.0 × 10⁻³ m³ to 3.0 × 10⁻³ m³ while 350 J of energy is supplied by heating. Find the work done and the change in internal energy.
The first law in four special processes
AQA's specification applies the first law to isothermal, adiabatic, constant pressure and constant volume changes.
| Process | What is fixed | First law becomes |
|---|---|---|
| Isothermal | Temperature, so ΔU = 0 | Q = W |
| Adiabatic | No heat transfer, Q = 0 | ΔU = −W |
| Constant volume | Volume, so W = 0 | Q = ΔU |
| Constant pressure | Pressure | Q = ΔU + pΔV |
For an ideal gas, an isothermal change obeys pV = constant, and an adiabatic change obeys pVγ = constant, as given in AQA's booklet.
Example 3: adiabatic expansion
A gas expands quickly in an insulated cylinder, doing 120 J of work. Find the change in internal energy, and say what happens to its temperature.
No heat is transferred, so Q = 0 and ΔU = −W = −120 J. The internal energy falls, so the temperature of the gas falls.
Example 4: isothermal compression
A gas is compressed slowly at constant temperature. 80 J of work is done on it. How much energy is transferred by heating, and in which direction?
Constant temperature means ΔU = 0. Work done by the gas is W = −80 J, so Q = ΔU + W = −80 J: 80 J flows out of the gas to the surroundings as heat.
pV diagrams and heat engines
Here x is the volume in units of 10⁻³ m³ and y is the pressure in kPa, for a gas expanding at constant temperature with pV = 300 J. The work done by the gas is the area under the curve between the start and end volumes. For a cyclic process, such as an engine cycle, AQA's specification says the work done per cycle is the area of the loop on the pV diagram.
Example 5: efficiency of a heat engine
A heat engine takes in 3000 J from a hot source each cycle and does 1200 J of useful work. Find its efficiency and the energy passed to the cold sink.
The efficiency is 40%. By the first law, over a whole cycle the gas returns to its starting state, so ΔU = 0 and the energy in must equal the energy out: 3000 − 1200 = 1800 J goes to the cold sink. AQA's booklet gives efficiency = W ÷ QH = (QH − QC) ÷ QH.
The Gas Laws Calculator works with pV = nRT and the other gas laws, which you need alongside the first law for pressure, volume and temperature changes.
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Common mistakes
- Mixing sign conventions. In Q = ΔU + W, W is work done by the gas. Work done on the gas is a negative W.
- Forgetting units in W = pΔV. Use pascals and cubic metres to get joules; 1 dm³ = 10⁻³ m³ and 1 kPa = 10³ Pa.
- Thinking isothermal means no heat transfer. Isothermal means constant temperature; heat can flow. Adiabatic means no heat transfer.
- Thinking heat and temperature are the same. Q is energy transferred; temperature measures the particles' average kinetic energy.
- Assuming ΔU = 0 for real gases or changes of state. That only holds for an ideal gas at constant temperature.
First law of thermodynamics questions with answers
1. A gas is given 900 J by heating and does 350 J of work. Find ΔU.
2. A gas at 2.0 × 10⁵ Pa expands by 0.50 × 10⁻³ m³ at constant pressure. Find the work done.
3. A gas is heated at constant volume with 250 J. Find ΔU.
4. In an isothermal expansion, an ideal gas does 60 J of work. How much heat does it absorb?
5. What does the area of the loop on a pV diagram for a cycle represent?
Practise thermal physics in a short timed paper
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Common questions
What is the first law of thermodynamics?
What is the equation for the first law of thermodynamics?
What is internal energy?
What is the difference between isothermal and adiabatic?
How do you calculate work done by a gas?
Is the first law of thermodynamics the same as conservation of energy?
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