9702 · 16.2The first law of thermodynamics

Internal energy and energy transfers

0:00Pump up a bicycle tyre quickly, then touch the end of the pump. It's warm. You never put it near a flame, but it still got hotter. Now think about a car engine. You burn fuel, the gas inside the cylinder gets very hot, and it pushes a piston out. Somehow, heating a gas turns into a moving car. Both of these are the same idea running in opposite directions.

0:20There are two ways to change the internal energy of a gas: you can heat it, or you can do work on it. And the rule connecting them is the first law of thermodynamics. In this video, we'll work out how much work is done when a gas expands or is compressed, get very clear on the difference between work done by a gas and work done on a gas, and then use the first law to track exactly where energy goes.

0:41A quick reminder from the last video. Internal energy, U, is the sum of the random kinetic and potential energies of the molecules in a system. It depends only on the state of the system. And for an ideal gas, there are no intermolecular forces, so the internal energy is purely random kinetic energy, and it's directly proportional to thermodynamic temperature. So for an ideal gas: if the temperature goes up, U goes up.

1:05If the temperature stays the same, U stays the same. That last point is the most useful shortcut in this entire topic, so keep it in mind. There are exactly two ways to transfer energy into or out of a system like a gas. The first is heating. If the gas is in contact with something hotter, energy flows into it because of the temperature difference. Faster molecules in the hotter wall collide with the gas molecules and pass energy on to them.

1:29The second is work. If a force moves something through a distance, energy is transferred. For a gas, that usually means a piston moving in or out. When you push a piston in, the piston is moving toward the molecules as they hit it. Each molecule bounces off faster than it arrived, a bit like a tennis ball hit by a racket that's moving toward it. So the molecules gain kinetic energy, the internal energy rises, and the gas gets hotter.

1:53That's exactly why the bicycle pump warms up. And when a gas pushes a piston out, the reverse happens. The piston is moving away from the molecules as they hit it, so they bounce back slower. The gas loses internal energy, and that energy goes into the work done moving the piston.

Work done by expanding gases

2:09Let's work out how much work a gas does when it expands. Picture gas trapped in a cylinder by a piston of cross-sectional area A. The gas is at pressure p, and it expands, pushing the piston outward through a small distance delta x. We'll keep the pressure constant while this happens. Pressure is force per unit area, so the force the gas exerts on the piston is F equals p times A.

2:31Work done is force times distance moved in the direction of the force. So the work done by the gas is F times delta x, which is p times A times delta x. But look at A times delta x. That's the cross-sectional area multiplied by the distance the piston moved, which is exactly the extra volume the gas now occupies. That's delta V, the change in volume. So the work done is: W equals p delta V.

2:58Pressure times the change in volume. Two things to notice. First, this only works directly when the pressure is constant while the volume changes. If the pressure were changing, you couldn't just multiply by one value of p. Second, check the units. Pressure in pascals, which is newtons per square metre, times volume in cubic metres, gives newton metres, which are joules. So the units work out as energy, as they should.

3:24There's also a nice way to picture this on a graph. Plot pressure on the vertical axis against volume on the horizontal axis. At constant pressure, the gas follows a horizontal line. The work done is p times delta V, which is the area of the rectangle under that line. In general, the work done is the area under a pressure-volume graph.

Signs and the first law

3:42This is the part of the topic students most often get wrong, so let's slow down. When a gas expands, it pushes the piston out. The gas is doing work on its surroundings. Energy leaves the gas. We call this work done by the gas. When a gas is compressed, something outside, like your hand on the pump, pushes the piston in. The surroundings are doing work on the gas.

4:02Energy enters the gas. We call this work done on the gas. These are the same size but opposite in sign. If a gas expands and does 200 joules of work on its surroundings, then the work done on the gas is minus 200 joules. So in any question, ask yourself two things. Is the volume increasing or decreasing? And is the question asking for the work done by the gas, or on it?

4:26Expansion: the work done by the gas is positive, and the work done on the gas is negative. Compression: the work done on the gas is positive, and the work done by the gas is negative. Now we can put everything together. The first law of thermodynamics is really just conservation of energy, applied to the internal energy of a system. The internal energy of a system can only change if energy is transferred to it by heating, or by work being done on it.

4:51So the increase in internal energy equals the energy transferred to the system by heating, plus the work done on the system. Delta U equals q plus W. Let's define each term carefully, because the signs are everything here. Delta U is the increase in internal energy of the system. If the internal energy goes down, delta U is negative. q is the energy transferred to the system by heating.

5:14If the system loses energy by heating, for example by cooling down in a cold room, q is negative. W is the work done on the system. If the system does work on its surroundings, like a gas expanding, W is negative. The easiest way to remember the sign convention is this: every term is counted as energy going into the system. Energy in is positive. Energy out is negative.

5:36Think of the gas's internal energy as a bank account. Heating it is a deposit. Doing work on it is a deposit. The gas cooling down, or the gas doing work by expanding, are withdrawals. Delta U is just the change in the balance. One warning here. Some textbooks and websites write the first law as delta U equals Q minus W, where W means the work done by the gas.

5:59That's the same physics with a different sign convention. For Cambridge A Level, always use delta U equals q plus W, with W as the work done on the system. If you mix the two conventions, you'll get the sign of your answer wrong.

Worked examples using the first law

6:12Let's use this. A gas is held at a constant pressure of 1.5 times ten to the five pascals. It's heated, and 450 joules of energy is transferred to it by heating. As it warms, it expands from a volume of 2.0 times ten to the minus three cubic metres to 3.2 times ten to the minus three cubic metres. Find the change in internal energy of the gas. Step one: find the change in volume.

6:37Delta V equals 3.2 times ten to the minus three minus 2.0 times ten to the minus three, which is 1.2 times ten to the minus three cubic metres. Step two: find the work done by the gas as it expands. p delta V equals 1.5 times ten to the five, times 1.2 times ten to the minus three, which is 180 joules. Step three: get the sign right. The gas is expanding, so it's doing 180 joules of work on its surroundings.

7:09The work done on the gas is therefore minus 180 joules. So W equals minus 180 joules. Step four: the heating is energy going in, so q equals plus 450 joules. Step five: apply the first law. Delta U equals q plus W, which is 450 plus minus 180, which is 270 joules. So the internal energy increases by 270 joules. Out of the 450 joules supplied by heating, 180 joules left again as work done pushing the piston out, and the remaining 270 joules stayed in the gas as extra internal energy.

7:47Since the internal energy went up, the temperature of the gas went up too. Now the other direction. A gas is compressed at a constant pressure of 1.0 times ten to the five pascals, from 5.0 times ten to the minus four cubic metres down to 3.0 times ten to the minus four cubic metres. During this, the gas loses 50 joules of energy to its surroundings by heating. Find the change in internal energy.

8:12The magnitude of the change in volume is 2.0 times ten to the minus four cubic metres. So p delta V equals 1.0 times ten to the five, times 2.0 times ten to the minus four, which is 20 joules. The gas is being compressed, so work is being done on it. W equals plus 20 joules. The gas is losing energy by heating, so q equals minus 50 joules.

8:36Delta U equals minus 50 plus 20, which is minus 30 joules. The internal energy decreases by 30 joules, so the gas cools down, even though work was done on it. It lost more by heating than it gained from the work.

Special cases of the first law

8:49A lot of exam questions describe a process in words, and you need to spot which term in the first law is zero. There are three situations worth recognising. First: constant volume. If the gas is in a rigid, sealed container, the volume can't change. So delta V is zero, and no work is done. W equals zero. Then delta U equals q. All the energy you supply by heating goes straight into internal energy.

9:15Second: no heat transfer. If a gas is compressed or expanded very quickly, or it's very well insulated, there's no time for energy to flow in or out by heating. So q equals zero, and delta U equals W. This is the bicycle pump from the start. You do work on the air, W is positive, so delta U is positive, and the air gets hotter. Run it the other way, with a gas expanding rapidly, and W is negative, so the gas loses internal energy and cools down.

9:43Third: constant temperature, for an ideal gas. Remember, an ideal gas's internal energy depends only on its temperature. So if the temperature doesn't change, delta U is zero. Then q plus W equals zero. If a gas expands slowly at constant temperature and does 100 joules of work, it must absorb exactly 100 joules by heating to stay at the same temperature. And a fourth idea follows from the previous video.

10:08If a gas goes round a complete cycle and ends up back in exactly its starting state, its internal energy is back where it began. So over the whole cycle, delta U is zero. That's the basic principle behind every heat engine: over one cycle, the net energy put in by heating is equal to the net work the engine does.

Exam mistakes and table question

10:27Let's go through the mistakes that cost marks most often. Mistake one: using the wrong sign for W. In delta U equals q plus W, W is the work done on the gas. When a gas expands, W is negative. The most common error in this whole topic is calculating p delta V for an expansion and then adding it as a positive number. Always ask: is the gas expanding or being compressed?

10:51Mistake two: using the wrong units for volume or pressure. W equals p delta V needs pressure in pascals and volume in cubic metres. Questions often give volumes in cubic centimetres or litres, and pressures in kilopascals. One cubic centimetre is ten to the minus six cubic metres. One litre is ten to the minus three cubic metres. One kilopascal is a thousand pascals. Convert first, then calculate. Mistake three: thinking that heating a gas must raise its temperature.

11:22As worked example two showed, and as the constant-temperature case showed, that isn't always true. The temperature change depends on delta U, which depends on both q and W together. Mistake four: using W equals p delta V when the pressure isn't constant. In that case, the work done is the area under the pressure-volume graph, not a single pressure times the change in volume. Mistake five: saying that a gas contains heat.

11:49Heating is a way of transferring energy. What the gas contains is internal energy. The first law is precisely about how heating and work both change that internal energy. Here's a classic exam-style question. "An ideal gas is compressed very rapidly, so that no energy is transferred by heating. 60 joules of work is done on the gas. The gas is then kept at constant volume and cools until it returns to its original temperature.

12:13Complete a table showing q, W, and delta U for each of the two stages." This kind of table question is very common, and there's a reliable way to fill it in: for each stage, find the two easy boxes first, then use the first law for the third. Stage one, the rapid compression. "No energy transferred by heating" tells you q equals zero. "60 joules of work done on the gas" tells you W equals plus 60 joules.

12:39So delta U equals zero plus 60, which is plus 60 joules. The gas heats up. Stage two, cooling at constant volume. Constant volume means no work is done, so W equals zero. The gas returns to its original temperature, and for an ideal gas, internal energy depends only on temperature. So it must lose exactly the 60 joules it gained. Delta U equals minus 60 joules. Then from the first law, q equals minus 60 joules.

13:05The gas loses 60 joules to its surroundings by heating. Notice the reasoning you used: q equals zero from "no heating", W equals zero from "constant volume", and delta U from "same temperature means same internal energy for an ideal gas". Examiners are testing whether you can translate those phrases into zeros and signs, not just plug numbers into a formula. Let's bring it all together. The internal energy of a system can be changed in two ways: by heating it, or by doing work on it.

13:35When a gas changes volume at constant pressure, the work done is W equals p delta V, which comes from force times distance on a piston. It's also the area under a pressure-volume graph. When a gas expands, it does work on its surroundings, so the work done on the gas is negative. When a gas is compressed, work is done on it, so the work done on the gas is positive.

13:55The first law of thermodynamics is conservation of energy: delta U equals q plus W, where delta U is the increase in internal energy, q is the energy transferred to the system by heating, and W is the work done on the system. Energy in is positive, energy out is negative. Watch for the special cases: constant volume means W is zero; no heat transfer means q is zero; and for an ideal gas at constant temperature, or over a complete cycle, delta U is zero.

14:23Get the signs right, convert your units, and the first law will tell you exactly where every joule goes.