Contents: 5 sections
Both subtopics here are printed under "A Level subject content" in the 9702 syllabus, and every objective carries the tier "A Level". None of it is AS. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper, and the whole 9702 bank on this site comes from Paper 1, so no practice is tagged to this topic.
This is a short topic with only four objectives, but the sign convention in it accounts for a large share of the marks lost across the whole A Level paper, so it is worth more time than its length suggests.
Syllabus points
16.1 Internal energy
- Understand that internal energy is determined by the state of the system and that it can be expressed as the sum of a random distribution of kinetic and potential energies associated with the molecules of a system.
- Relate a rise in temperature of an object to an increase in its internal energy.
16.2 The first law of thermodynamics
- Recall and use W = pΔV for the work done when the volume of a gas changes at constant pressure and understand the difference between the work done by the gas and the work done on the gas.
- Recall and use the first law of thermodynamics ΔU = q + W expressed in terms of the increase in internal energy, the heating of the system (energy transferred to the system by heating) and the work done on the system.
Internal energy
The internal energy of a system is the sum of a random distribution of the kinetic and potential energies associated with its molecules.
Every word in that definition is doing work:
- Random. The energy of ordered motion does not count. A brick flying through the air has kinetic energy, but the internal energy of the brick is unchanged by throwing it.
- Kinetic. Molecules move, vibrate and rotate. This part depends on temperature.
- Potential. Molecules attract one another, so separating them stores energy. This part depends on the separation of the molecules, which is to say on the state and the volume.
- Distribution. Molecules do not all have the same energy; there is a spread, and internal energy is the total across it.
Internal energy is determined by the state of the system, meaning its temperature, pressure and volume, and not by how the system arrived at that state. Two identical samples at the same temperature, pressure and volume have the same internal energy, whether one was heated and the other compressed. This is why ΔU depends only on the start and end points.
A rise in temperature means an increase in the mean kinetic energy of the molecules, and therefore an increase in internal energy.
Two cases worth separating:
- For an ideal gas, intermolecular forces are assumed to be zero, so there is no potential energy component. The internal energy of an ideal gas is therefore purely kinetic and depends only on its temperature. If the temperature does not change, ΔU is zero, however much the gas is compressed or expanded.
- During a change of state, the temperature does not change, so the kinetic component is unchanged, yet internal energy still rises because the potential component increases as the molecules are pulled apart. That is what latent heat is.
Work done by a gas at constant pressure
When a gas expands, it pushes the surroundings back, so it does work on them.
W = pΔV
The derivation is one line. If a gas in a cylinder pushes a piston of area A out through a distance d against a constant pressure p, the force on the piston is pA, so the work done is pAd, and Ad is the increase in volume ΔV.
Units matter here. p must be in pascals and ΔV in cubic metres for W to come out in joules. A volume in cm³ must be divided by 10⁶ first.
Worked example. A gas at a constant pressure of 2.0 × 10⁵ Pa expands from 300 cm³ to 800 cm³.
Convert the volume change to cubic metres:
ΔV = (800 - 300) / 1000000 = 0.0005
W = 2.0 × 10⁵ × 0.0005 = 100
so 100 J of work is done by the gas on the surroundings.
The equation only applies at constant pressure. If the pressure varies, the work done is the area under a graph of pressure against volume, which is beyond what is required here.
The first law of thermodynamics
ΔU = q + W
where
- ΔU is the increase in internal energy of the system,
- q is the energy transferred to the system by heating, and
- W is the work done on the system.
This is simply conservation of energy applied to a system that can exchange energy in two ways. There is no third way.
The sign convention, which is where the marks go
Get these four statements straight and the topic is done. The convention in this syllabus is written from the point of view of the system, so anything that increases the system's energy is positive.
| Quantity | Positive when | Negative when |
|---|---|---|
| ΔU | Internal energy increases, so temperature rises | Internal energy decreases, temperature falls |
| q | Energy is supplied to the system by heating | The system loses energy by heating the surroundings |
| W | Work is done on the system, so the gas is compressed | Work is done by the system, so the gas expands |
The trap sits in the last row. The formula W = pΔV gives the work done by the gas, and the first law wants the work done on the gas. When a gas expands, ΔV is positive and W in the first law is negative. Writing W = +pΔV straight into ΔU = q + W after an expansion is the commonest single error in this topic.
A useful way to hold it: a gas that expands has spent energy pushing the surroundings out of the way, so unless it was heated, its internal energy and therefore its temperature must fall. That is why a can of aerosol goes cold when it is emptied and why air cools as it rises and expands.
Worked examples on the four standard cases
A gas is compressed and 150 J of work is done on it, while 60 J of energy escapes by heating.
q = -60, W = +150
ΔU = -60 + 150 = 90
so the internal energy rises by 90 J and the gas gets hotter.
A gas expands at constant pressure, doing 400 J of work, while 700 J of energy is supplied by heating.
q = +700, W = -400
ΔU = 700 - 400 = 300
so the internal energy rises by 300 J.
Constant volume heating. No volume change means no work, so W = 0 and
ΔU = q
All the energy supplied goes into internal energy, which is why a gas heated at constant volume shows a larger temperature rise than the same gas heated at constant pressure with the same energy input.
Isothermal expansion of an ideal gas. The temperature is constant, so for an ideal gas ΔU = 0, and
q = -W
meaning the energy the gas spends doing work on the surroundings is exactly replaced by heating from them.
Melting ice. Energy is supplied, so q is positive. The volume barely changes, so W is almost zero. Therefore ΔU is positive even though the temperature is constant, and the whole increase is in the potential energy of the molecules. This example is worth remembering because it is the clearest case of internal energy rising with no temperature change at all.
Common mistakes
- Substituting W = +pΔV into the first law after an expansion, instead of W = -pΔV.
- Saying internal energy is the total kinetic energy of the molecules, and leaving out the potential term.
- Including the kinetic energy of the whole object moving, or its gravitational potential energy, in the internal energy.
- Saying that internal energy depends on how the system got to its state. It depends only on the state.
- Saying an ideal gas has potential energy between its molecules. The model assumes there are no intermolecular forces except during collisions.
- Saying the temperature must rise whenever internal energy rises. During melting or boiling it does not.
- Leaving the volume in cm³ in W = pΔV, so the answer is out by a factor of a million.
- Applying W = pΔV where the pressure is not constant.
- Confusing q with temperature. Energy transferred by heating and temperature are different quantities with different units.
- Saying a gas cools when it expands "because it does work", without adding the condition that no energy is supplied by heating.