Syllabus coverage
- Understand that the internal energy of a system is determined by the state of the system
- Understand that internal energy 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 the temperature of an object to an increase in its internal energy
- Recall and use W = p Delta V for the work done when the volume of a gas changes at constant pressure
- 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 Delta U = 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
Part of the AS & A Level Physics syllabus 9702.
Internal energy: the energy stored inside a system
The internal energy of a system is the sum of the random distribution of kinetic and potential energies associated with the molecules of the system. Every molecule is moving, so it has kinetic energy, and the molecules also attract and repel one another, so the forces between them store potential energy. The word random matters: the molecules are not marching in one direction, they move in every direction at a range of speeds, and the internal energy is the total of all those energies added together.
Internal energy is determined by the state of the system, not by the route the system took to reach that state. The state of a fixed mass of gas is fixed by its pressure, its volume and its temperature. Two samples in the same state therefore have the same internal energy, whatever happened to them beforehand, and any change of internal energy depends only on the starting and finishing states.
For an ideal gas there are no forces between the molecules except during the brief collisions, so there is no potential energy term and the internal energy is entirely the random kinetic energy of the molecules. That is why the internal energy of an ideal gas depends only on its temperature. In a real solid, liquid or gas the potential energy term matters as well, and that is why a substance can take in a great deal of energy without its temperature rising at all.
- Internal energy is the sum of the random kinetic and potential energies associated with the molecules of a system.
- It is determined by the state of the system: the pressure, the volume and the temperature of a fixed mass of material.
- For an ideal gas there are no intermolecular forces to store potential energy, so the internal energy depends only on the temperature.
- Internal energy is measured in joules, J.
Temperature, internal energy and changes of state
A rise in the temperature of an object is an increase in its internal energy. Temperature is a measure of the average random kinetic energy of the molecules, so as an object gets hotter the molecules move faster on average, the kinetic energy part of the internal energy grows and the internal energy is larger. Cool the object and the molecules slow down, so the internal energy falls.
The two quantities are not the same thing, and the distinction is worth marks. Temperature is an average per molecule, while internal energy is a total for the whole object. A mug of water at 80 degrees C has a far smaller internal energy than a bath of water at 40 degrees C, because the bath contains vastly more molecules even though each molecule in the mug has more kinetic energy on average.
Internal energy can also change with no change of temperature at all. While ice melts at 0 degrees C or water boils at 100 degrees C, energy is transferred to the substance and the internal energy rises, but the thermometer does not move, because the energy goes into the potential energy of the molecules as the forces between them are pulled apart. The energy transferred to do that is the latent heat of the substance, and work done on a system raises its internal energy in the same way, with no heating needed.
Work done when a gas changes volume at constant pressure
A gas trapped in a cylinder behind a movable piston does work when it expands. The force the gas exerts on the piston is F = pA, where p is the pressure of the gas and A is the area of the piston. If the piston moves a distance d, the work done by the gas is F x d = pA x d, and the area multiplied by the distance moved is the change in volume, so the work done by the gas at constant pressure is W = p Delta V.
The units confirm the equation. Pressure in N m-2 multiplied by a volume change in m3 gives N m, which is the joule, so the work done comes out in joules (J). A gas at a constant pressure of 1.5 x 10^5 Pa expanding by a volume of 3.0 x 10^-3 m3 does 1.5 x 10^5 x 3.0 x 10^-3 = 450 J of work on the piston.
W = p Delta V holds only while the pressure stays constant. If the pressure changes as the volume changes, the work done is the area under the p-V graph between the two volumes, found by counting squares or by evaluating the area. Two cases need no calculation at all: a gas expanding into a vacuum does no work, because there is nothing for it to push against, and a gas in a rigid sealed container does no work, because its volume does not change.
- Work done by the gas at constant pressure: W = p Delta V, with p in N m-2 and Delta V in m3.
- The unit check: N m-2 x m3 = N m = J.
- If the pressure is not constant, the work done is the area under the p-V graph between the two volumes.
- Expansion into a vacuum and heating a gas in a rigid container both give zero work done.
Work done by the gas and work done on the gas
Most mistakes in this topic are mistakes of direction, so be precise about the wording. Work done by the gas is work done by the gas on its surroundings, and it happens when the gas expands and pushes the piston outwards. Work done on the gas is work done by the surroundings on the gas, and it happens when the gas is compressed.
Cambridge writes the first law with W as the work done on the system, so the two quantities are equal in size but opposite in sign. If a gas expands and does 450 J of work on the piston, then the work done on the gas is -450 J, and that is the number you substitute into the equation. If a piston is pushed in and does 200 J of work on the gas, then W = +200 J. A sign error flips the sign of your answer for the change in internal energy, so write the convention you are using before you substitute anything.
The table below collects the cases you are expected to recognise. Note that the third and fourth rows both give zero: a confined gas cannot do work against a rigid wall, and a gas expanding freely into a vacuum meets no resistance, so no energy is transferred as work.
| Change to the gas | Work done by the gas | Value of W, the work done on the gas |
|---|---|---|
| Expands at constant pressure | p Delta V, positive | -p Delta V, negative |
| Compressed at constant pressure | -p Delta V, negative | +p Delta V, positive |
| Heated at constant volume in a rigid container | Zero | Zero |
| Expands freely into a vacuum | Zero | Zero |
The first law of thermodynamics
The first law of thermodynamics is the law of conservation of energy applied to a system that can be heated and can do work: Delta U = q + W. In words, the increase in the internal energy of a system equals the energy transferred to it by heating plus the work done on it. Every symbol carries a sign, and the signs are what the examiner is testing.
Here is the convention to learn. q is positive when energy is transferred to the system by heating and negative when the system transfers energy away by heating. W is positive when work is done on the system, as in a compression, and negative when the system itself does work, as in an expansion. Delta U is positive when the internal energy increases. A gas compressed by 350 J of work while 120 J leaves it by heating has q = -120 J and W = +350 J, so Delta U = -120 + 350 = +230 J and the gas becomes hotter.
The same equation answers every process question you will meet. Heat a gas in a rigid container: W = 0, so all the energy transferred by heating becomes internal energy and the temperature rises. Let a gas expand into a vacuum with no heating: q = 0 and W = 0, so Delta U = 0 and nothing changes. Supply energy by heating at exactly the rate at which an expanding gas does work on its surroundings: q = -W, so Delta U = 0 and the temperature stays constant. Over a complete cycle the gas returns to its original state, so the internal energy returns to its original value and Delta U = 0, which means the net work done by the gas equals the net energy transferred to it by heating.
| Quantity | Positive when | Negative when |
|---|---|---|
| Delta U | The internal energy of the system increases | The internal energy of the system decreases |
| q | Energy is transferred to the system by heating | Energy is transferred from the system by heating |
| W | Work is done on the system, as in a compression | The system does work, as in an expansion |
The first law at work in engines and machines
The diesel engine, developed by Rudolf Diesel, is built on the first law. Air in the cylinder is compressed very rapidly by the piston, so W is large and positive, and there is almost no time for energy to escape by heating, so q is close to zero. Delta U = q + W is therefore positive and the air reaches several hundred degrees C, hot enough to ignite the fuel injected into the cylinder with no spark plug at all. Part of the reason a diesel engine is more efficient than a petrol engine is that it compresses its charge much further.
A power station works through the same bookkeeping. Fuel is burned to transfer energy by heating to water in a boiler, producing high-pressure steam. The steam expands through a turbine, doing work on the blades that drive the generator, and its internal energy falls as it hands over that work. The steam is then condensed, transferring energy by heating to cooling water so that it returns to liquid water and can be pumped back to the boiler. Over one complete cycle the water is back in the same state, so the change in internal energy is zero, and the work taken out of the turbine is paid for by the energy transferred by heating in the boiler.
A bicycle pump warms up for exactly the same reason as the diesel cylinder. Pushing the piston in does work on the air inside, the push is fast so little energy escapes by heating, and Delta U = q + W is positive, so the air and the pump barrel both get hotter. Release a compressed gas quickly and the reverse happens: the gas does work on its surroundings with almost no energy transferred to it by heating, so its internal energy and its temperature fall. That is why the jet from an aerosol can feels cold and why expanding gas is used to cool things down.
Working through first law calculations
Nearly every calculation in this topic follows four steps. Write down the sign convention you are using. Identify q and W from the wording of the question, watching for the words 'by the gas' and 'on the gas'. Substitute into Delta U = q + W, converting cm3 to m3 and degrees C to kelvin where they are needed. Then give the answer with its unit and say whether the internal energy has increased or decreased, because a bare number is not a complete answer.
Worked example 1. In an engine cylinder a gas is compressed, 350 J of work being done on it, while 120 J of energy is transferred from the gas to the surroundings by heating. The heating is away from the gas, so q = -120 J; the compression means work is done on the gas, so W = +350 J. Then Delta U = -120 + 350 = +230 J, so the internal energy of the gas has increased by 230 J and the gas is hotter than it was.
Worked example 2. A fixed mass of gas expands at a constant pressure of 1.5 x 10^5 Pa from a volume of 2.0 x 10^-3 m3 to 5.0 x 10^-3 m3, and 600 J of energy is transferred to the gas by heating during the expansion. The volume change is 5.0 x 10^-3 - 2.0 x 10^-3 = 3.0 x 10^-3 m3. The work done by the gas is p Delta V = 1.5 x 10^5 x 3.0 x 10^-3 = 450 J, so the work done on the gas is W = -450 J, and the heating into the gas gives q = +600 J. Therefore Delta U = 600 - 450 = +150 J: the internal energy rises by 150 J and the temperature of the gas rises, even though the gas was doing work on its surroundings.
Keep the units disciplined. To change a volume from cm3 to m3 multiply by 1.0 x 10^-6, and remember that a temperature change is the same number in kelvin as in degrees C. In first law questions the temperature is usually needed only to say what happens to the gas, not to work out a number, so the arithmetic stays short and the marks come from the signs.
Key terms you must be able to define
- Internal energy
- The sum of the random distribution of kinetic and potential energies associated with the molecules of a system, measured in joules; it is determined by the state of the system.
- State of a system
- The set of properties that fixes the condition of a fixed mass of material, in particular its pressure, its volume and its temperature; the internal energy depends on the state and not on how the state was reached.
- Temperature
- A measure of the average random kinetic energy of the molecules of a substance; a rise in temperature of an object is accompanied by an increase in its internal energy.
- Heating, q
- The energy transferred to a system by heating; q is positive when energy is transferred to the system and negative when the system transfers energy to its surroundings by heating.
- Work done on a system, W
- The energy transferred to a system by a force moving through a distance; W is positive when work is done on the system, as in a compression, and negative when the system does work, as in an expansion.
- First law of thermodynamics
- The increase in internal energy of a system equals the energy transferred to it by heating plus the work done on it: Delta U = q + W.
- Work done by a gas at constant pressure
- W = p Delta V, where p is the constant pressure in N m-2 and Delta V is the change in volume in m3, giving the work done by the gas in joules; the work done on the gas is -p Delta V for an expansion and +p Delta V for a compression.
- Adiabatic change
- A change in which no energy is transferred to or from the system by heating, so q = 0 and the first law reduces to Delta U = W; rapid compression or expansion of a gas is approximately adiabatic.
Applied to the real world
Cambridge sets 9702 in unfamiliar contexts - a fairground ride, a power station, a hospital scanner, a satellite, an astronomical observatory. Use these as models for how to apply the physics, then write your own.
James Joule's paddle-wheel experiment showed that work and heating are both ways of raising internal energy. A falling weight turned a paddle inside a barrel of water, and the temperature of the water rose by an amount that matched the work done by the weight. No energy had been transferred by heating, so q was zero and Delta U = W: the work done on the water appeared as internal energy, and the rise in temperature measured it.
Pushing a bicycle pump warms the barrel. The piston does work on the air inside, and because the push is quick very little energy escapes by heating, so q is nearly zero and Delta U = q + W is positive. The same reasoning explains why a diesel engine needs no spark plug: compressing its air charges it with internal energy and takes it well above the ignition temperature of the fuel.
In a four-stroke petrol engine the piston compresses the fuel and air mixture, a spark ignites it, and the hot gas expands and does p Delta V of work on the piston during the power stroke. Over the complete cycle the gas returns to its starting state, so Delta U is zero for the cycle as a whole and the net work delivered by the piston equals the net energy taken in by heating from the burning fuel.
In a coal or gas fired power station the boiler transfers energy by heating to water, making high-pressure steam. The steam expands through a turbine and does work on the rotor, which drives the generator, so the internal energy of the steam falls. A condenser then takes energy out of the steam by heating cooling water, returning it to liquid so the feed pump can send it back to the boiler and the cycle can begin again.
A laboratory refrigerator moves energy from a cold interior to a warmer room, and the first law keeps the accounts straight. A compressor does work W on the refrigerant vapour, which then transfers energy to the room by heating, so q is negative there, and over one complete cycle Delta U is zero because the refrigerant returns to its starting state. Hospital fridges holding vaccines and blood products rely on this cycle running continuously.
Flashcards
Click a card to flip it. Mark honestly — the tracker uses this.
Exam-style questions
8 questions, 27 marks. Write your answer, then reveal the mark scheme and tick each point you actually made. Answering honestly here is what makes the rest of the site useful.
Mark scheme — tick each point you actually made
Mark scheme — tick each point you actually made
Mark scheme — tick each point you actually made
Mark scheme — tick each point you actually made
Mark scheme — tick each point you actually made
Mark scheme — tick each point you actually made
Mark scheme — tick each point you actually made
Mark scheme — tick each point you actually made
Exam technique for this topic
- Write down the sign convention before you substitute anything. Cambridge defines W as the work done on the system, so an expanding gas has negative W and a compressed gas has positive W. Nearly every lost mark in this topic is a sign error.
- Use W = p Delta V only for a constant-pressure change. If the pressure varies, the work done is the area under the p-V graph, and if the volume does not change, or the gas expands into a vacuum, the work done is zero.
- Convert volumes before you multiply. cm3 to m3 means multiplying by 1.0 x 10^-6, and forgetting it makes the answer a million times too large. The pressure must be in N m-2, and 1 kPa is 1.0 x 10^3 Pa.
- Finish an internal energy answer with a unit and a direction. 'Delta U = +230 J, so the internal energy increases by 230 J' scores the reasoning mark that '-230' or a bare number does not.
- Remember that internal energy is fixed by the state. Over a complete cycle Delta U is zero, and for an ideal gas internal energy depends only on temperature, so a change at constant temperature changes neither the kinetic energy of the molecules nor the internal energy.