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B.4

Thermodynamics

Theme B · The particulate nature of matter · HL only

Two laws, and they do very different jobs. The first says energy is conserved when a gas is heated or compressed — it tells you what is possible. The second says that some of those possible things never happen anyway, no matter how carefully you engineer them — it tells you what is allowed. Everything on this page, including the reason no engine can ever be perfectly efficient, comes from holding those two apart.

🎯What you need to be able to do

  • State and apply the first law, \( Q = \Delta U + W \), with the sign conventions correct.
  • Calculate the work done by an expanding gas, \( W = p\Delta V \), and recognise it as the area under a \(p\)–\(V\) curve.
  • Describe isobaric, isovolumetric, isothermal and adiabatic changes, and sketch each on \(p\)–\(V\) axes.
  • Use \( pV^{5/3} = \) constant for an adiabatic change in a monatomic ideal gas.
  • Find the net work done in a cycle from the area it encloses.
  • Define entropy and use \( \Delta S = \Delta Q / T \).
  • State the second law in words, and explain what it forbids.
  • Calculate the efficiency of a heat engine and compare it with the Carnot limit \( \eta_{\text{C}} = 1 - T_{\text{C}}/T_{\text{H}} \).

⚖️The first law: conservation of energy, written for a gas

Thermal energy can enter a gas, the gas can do work on its surroundings, and its internal energy can change. The first law says those three are one accounting statement:

\[ Q = \Delta U + W \]
A gas in a cylinder shown as an energy account. Thermal energy Q flows in from a heat source at the bottom; work W is done by the gas as it pushes a piston outwards at the top; and what is left over changes the internal energy delta U of the gas itself. A sign table beside it gives the conventions: Q is positive when thermal energy is added to the gas and negative when it leaves; W is positive when the gas expands and does work on the surroundings and negative when the surroundings compress the gas; and delta U is positive when the temperature rises.
Energy in equals energy stored plus energy out. The whole difficulty of the first law is not the physics, it is keeping three signs straight.
\(Q\)thermal energy added to the gas
\(W\)work done by the gas
\( \Delta U \)change in the gas’s internal energy

For an ideal monatomic gas, internal energy is entirely kinetic, so from B.3:

\[ U = \tfrac{3}{2}nRT \qquad\text{so}\qquad \Delta U = \tfrac{3}{2}nR\,\Delta T \]

which gives a rule worth memorising in this exact form: a change in internal energy is a change in temperature, and nothing else. If the temperature is unchanged, \( \Delta U = 0 \), whatever else is happening.

The sign of \(W\) is the one that catches people. In the IB convention \(W\) is the work done by the gas, so an expanding gas has \( W > 0 \) and a compressed gas has \( W < 0 \). Some textbooks define \(W\) as the work done on the gas and write the law as \( \Delta U = Q + W \). Both are correct; mixing them is not. Check which convention a question is using by asking whether the gas is expanding, and whether that should help or hinder its internal energy.

🔧Work done by an expanding gas

When a gas expands it pushes the piston out, and pushing something through a distance is work. If the pressure stays constant while the volume changes by \( \Delta V \):

\[ W = p\,\Delta V \]
On the left, gas in a cylinder pushes a piston of area A outwards through a distance d. The force on the piston is pressure times area, so the work done is p A d, and since A d is the increase in volume the work is p times delta V. On the right, the same result shown as a graph of pressure against volume: for a constant-pressure expansion the work is the rectangular area under the line, and for a changing pressure it is the area under the curve, which is why the area under a p-V graph always represents work.
\( W = p\Delta V \) is just \( \text{force} \times \text{distance} \) rearranged. The general statement is the one on the right: the area under a \(p\)–\(V\) curve is the work done.

That second statement is the more useful one, because it still works when the pressure is changing — and it is what makes \(p\)–\(V\) diagrams worth drawing at all.

📊The four processes on a \(p\)–\(V\) diagram

Four special changes appear over and over. Each is defined by one thing being held fixed, and each has a characteristic shape.

Four processes drawn on pressure against volume axes. Isovolumetric, at constant volume, is a vertical line, and no work is done because the volume does not change. Isobaric, at constant pressure, is a horizontal line, and the work is the rectangular area beneath it. Isothermal, at constant temperature, is a hyperbola along which p V is constant, so the internal energy does not change and all the heat supplied becomes work. Adiabatic, with no thermal energy transferred at all, is a steeper curve along which p V to the power five thirds is constant, so any work done comes entirely out of the internal energy and the gas cools as it expands.
Each column is the first law with one term deliberately set to zero. Learn which term dies in each case and the four processes stop needing to be memorised separately.
isovolumetric\( \Delta V = 0 \Rightarrow W = 0 \), so \( Q = \Delta U \)
isobaric\( p \) constant, \( W = p\Delta V \)
isothermal\( \Delta T = 0 \Rightarrow \Delta U = 0 \), so \( Q = W \)
adiabatic\( Q = 0 \), so \( \Delta U = -W \)

The last two are worth reading again, because they are opposites. In an isothermal expansion every joule of heat supplied comes straight out again as work, and the gas neither warms nor cools — which requires the change to be slow enough for the gas to stay in thermal contact with its surroundings. In an adiabatic expansion no heat enters at all, so the work has to be paid for out of the internal energy, and the gas cools as it expands. That requires the change to be fast, or the gas to be well insulated.

An isotherm and an adiabatic curve drawn from the same starting point on pressure against volume axes. The adiabatic curve, along which p V to the five thirds is constant, falls more steeply than the isotherm, along which p V is constant. Expanding along the adiabatic ends at a lower pressure and on a lower isotherm, which means a lower temperature, and the shaded area under it is smaller, so less work is done.
From the same start, the adiabatic always falls more steeply: the gas is losing pressure both because the volume is growing and because it is cooling.
\[ pV^{5/3} = \text{constant} \qquad \text{(adiabatic, monatomic ideal gas)} \]

✏️Worked example 1 — the first law, twice

(a) A gas absorbs 500 J of thermal energy and does 200 J of work on its surroundings. Find the change in its internal energy. (b) An ideal gas at a constant pressure of \( 2.0\times10^{5} \) Pa expands from \( 1.0\times10^{-3} \) m\(^{3}\) to \( 3.0\times10^{-3} \) m\(^{3}\) while 800 J of thermal energy is supplied. Find the work done and the change in internal energy.

(a) Straight substitution, with both signs positive because energy went in and the gas did work:

\[ \Delta U = Q - W = 500 - 200 = +300\ \text{J} \]

(b) Constant pressure, so the work is the rectangle:

\[ W = p\Delta V = 2.0\times10^{5} \times (3.0 - 1.0)\times10^{-3} = 400\ \text{J} \]
\[ \Delta U = Q - W = 800 - 400 = +400\ \text{J} \]
Both answers are positive, and both should be. In each case more energy was supplied than the gas spent on work, so the surplus went into internal energy — and since \( \Delta U = \frac{3}{2}nR\Delta T \), that means the gas got hotter. Had the gas been compressed instead, \(W\) would have been negative and \( \Delta U \) larger than \(Q\).

🔄Cycles: where the useful work comes from

A single expansion is not an engine. An engine has to return to its starting state so it can do it again, which on a \(p\)–\(V\) diagram means going round a closed loop.

A closed cycle on pressure against volume axes, traversed clockwise. On the upper path the gas expands at high pressure, and the area under that path is the work done by the gas. On the lower return path the gas is compressed at low pressure, and the area under that path is the work done on the gas. The second is smaller than the first, so the difference, which is the area enclosed by the loop, is the net work output per cycle. Because the gas returns to its starting point, the change in internal energy over a whole cycle is zero, so the net work equals the net thermal energy absorbed.
Expand at high pressure, return at low pressure, and the difference is yours. The enclosed area is the net work — go round clockwise and the engine does work, anticlockwise and it consumes it (a refrigerator).

Because the gas ends where it began, its internal energy is unchanged over one complete cycle: \( \Delta U = 0 \), so \( Q_{\text{net}} = W_{\text{net}} \). Every joule of useful work came from a joule of thermal energy that was absorbed and not given back.

🚫The second law: what conservation of energy allows but nature forbids

Drop a hot block into cold water and the two reach a common temperature. Nothing in the first law forbids the reverse — the water spontaneously cooling and the block spontaneously heating conserves energy perfectly. It simply never happens. The second law is the statement of what direction things go.

Two equivalent statements of the second law, each drawn as a forbidden process. The Clausius statement: thermal energy cannot flow spontaneously from a colder body to a hotter one, drawn as an arrow from cold to hot with a cross through it. The Kelvin statement: no cyclic engine can take thermal energy from a single reservoir and convert all of it into work, drawn as an engine with a single input and a work output and no waste, also crossed out. Beneath, the entropy statement: in an isolated system the total entropy never decreases.
Three statements, one law. Each forbids something that conserves energy perfectly well — which is exactly why a second law was needed at all.

Entropy is the quantity that makes this precise. For thermal energy \( \Delta Q \) transferred at absolute temperature \(T\):

\[ \Delta S = \frac{\Delta Q}{T} \qquad \text{J K}^{-1} \]
Thermal energy of 1000 joules flowing from a hot reservoir at 600 kelvin to a cold one at 300 kelvin. The hot reservoir loses entropy of 1000 over 600, which is 1.67 joules per kelvin; the cold one gains 1000 over 300, which is 3.33 joules per kelvin; and the total change is a gain of 1.67 joules per kelvin, which is positive, so the process happens. Running it backwards would give a total entropy change of minus 1.67 joules per kelvin, which the second law forbids. Beside it, entropy is illustrated as the number of ways of arranging the molecules: an ordered arrangement in one corner has few, a spread-out one has overwhelmingly more.
The same 1000 J is worth more entropy at the cold end than it cost at the hot end, because entropy is energy divided by temperature. That asymmetry is the whole of the second law in one line of arithmetic.

Notice what the division by \(T\) does: the same energy carries a bigger entropy change at a low temperature than at a high one. So energy moving from hot to cold always increases the total, and energy moving from cold to hot always decreases it — which is precisely the process nobody has ever seen.

✏️Worked example 2 — which way will it go?

1000 J of thermal energy passes from a reservoir at 600 K to one at 300 K. Find the total entropy change, and say what it tells you.
\[ \Delta S_{\text{hot}} = \frac{-1000}{600} = -1.67\ \text{J K}^{-1} \qquad \Delta S_{\text{cold}} = \frac{+1000}{300} = +3.33\ \text{J K}^{-1} \]
\[ \Delta S_{\text{total}} = -1.67 + 3.33 = +1.67\ \text{J K}^{-1} \]
Positive, so it happens. Reverse the transfer and every sign flips: \( \Delta S_{\text{total}} = -1.67 \) J K\(^{-1}\), which the second law forbids in an isolated system. Note that energy was conserved in both directions — the first law had nothing to say about which one nature picks. That is what the second law is for.

⚙️Heat engines and the efficiency limit

A heat engine takes thermal energy from a hot reservoir, converts part of it to work, and dumps the rest into a cold reservoir. The dumping is not sloppy engineering — the second law requires it.

A heat engine drawn as a block between two reservoirs. Thermal energy Q sub H flows in from the hot reservoir at temperature T sub H, useful work W comes out of the side, and waste thermal energy Q sub C is rejected to the cold reservoir at temperature T sub C. Energy conservation over a cycle gives W equal to Q sub H minus Q sub C, and the efficiency is W over Q sub H, which is one minus Q sub C over Q sub H. The Carnot limit, one minus T sub C over T sub H, is shown as the best any engine between those two temperatures could ever achieve.
The cold reservoir is not optional. An engine with no \( Q_{\text{C}} \) would be exactly what the Kelvin statement forbids, and its efficiency would be 1.
\( W = Q_{\text{H}} - Q_{\text{C}} \)
\( \eta = \dfrac{W}{Q_{\text{H}}} = 1 - \dfrac{Q_{\text{C}}}{Q_{\text{H}}} \)
\( \eta_{\text{Carnot}} = 1 - \dfrac{T_{\text{C}}}{T_{\text{H}}} \)
The Carnot cycle on pressure against volume axes: an isothermal expansion at the hot temperature, then an adiabatic expansion during which the gas cools to the cold temperature, then an isothermal compression at the cold temperature, then an adiabatic compression back to the start. The two isotherms are labelled with the temperatures of the reservoirs and the two adiabatics are steeper. The enclosed area is the net work per cycle. A note states that the Carnot efficiency, one minus T cold over T hot, is the maximum possible for any engine working between those two temperatures, and that it depends only on the temperatures and not on the working substance.
Two isotherms and two adiabatics. No engine working between the same two temperatures can beat it, and the reason is the second law rather than any detail of the machinery.

The Carnot efficiency depends only on the two absolute temperatures. That has a blunt practical consequence: to make an engine more efficient you must either raise the hot temperature or lower the cold one, and the cold one is usually the surroundings, which you cannot change. It also shows why \( \eta = 1 \) requires \( T_{\text{C}} = 0 \) — a cold reservoir at absolute zero, which is unattainable.

✏️Worked example 3 — a real engine against its limit

An engine takes 1000 J per cycle from a reservoir at 600 K and rejects 700 J to a reservoir at 300 K. Find its work output, its efficiency, and the maximum efficiency possible between those temperatures.
\[ W = Q_{\text{H}} - Q_{\text{C}} = 1000 - 700 = 300\ \text{J} \qquad \eta = \frac{300}{1000} = 0.30 \]
\[ \eta_{\text{Carnot}} = 1 - \frac{T_{\text{C}}}{T_{\text{H}}} = 1 - \frac{300}{600} = 0.50 \]
0.30 against a ceiling of 0.50 — the engine achieves 60% of what is thermodynamically possible. That is a realistic figure for a good engine. If a question ever gives you an efficiency above the Carnot value, the numbers are wrong, and saying so is worth a mark. Note also that both temperatures went in as kelvin: with 327 °C and 27 °C the answer would have been 0.92, which is impossible.

📉Degradation: why energy runs downhill

Energy is always conserved, so the familiar phrase “wasting energy” needs unpacking. Nothing is lost. What happens is that energy becomes degraded: it ends up spread thinly, at low temperature, among the random motions of a great many molecules, and in that form there is no way to gather it back up and do work with it.

That is the second law expressed as a practical statement. A joule at 1000 K can drive an engine; the same joule at 300 K, dispersed into the surroundings, cannot. Every real process increases total entropy, and every increase in entropy is a loss of the usefulness of energy, never of the energy itself.

🔭See it happen

PhET, Reversible Reactions and the gas simulations from B.3 both show entropy as a counting statement: start every molecule in one corner and it never returns there, not because it is forbidden, but because there are overwhelmingly more ways to be spread out than to be tidy. That is what \( \Delta S \geq 0 \) means at the molecular level.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. 250 J of thermal energy is supplied to a gas, which expands and does 100 J of work. Find the change in internal energy.
\( \Delta U = Q - W = 250 - 100 = +150 \) J. Positive, so the internal energy has risen and, since \( \Delta U = \frac{3}{2}nR\Delta T \) for an ideal gas, so has the temperature.
2. A gas is compressed, 300 J of work being done on it, while 500 J of thermal energy leaves it. Find \( \Delta U \).
Both signs are negative in the IB convention: \( Q = -500 \) J (energy left the gas) and \( W = -300 \) J (the surroundings did the work, not the gas). So \( \Delta U = Q - W = -500 - (-300) = -200 \) J. The gas cooled.
3. An ideal gas expands at a constant pressure of \( 1.5\times10^{5} \) Pa from \( 2.0\times10^{-3} \) m\(^{3}\) to \( 5.0\times10^{-3} \) m\(^{3}\). Find the work done by the gas.
\( W = p\Delta V = 1.5\times10^{5} \times 3.0\times10^{-3} = 4.5\times10^{2} \) J. This is the rectangular area under the horizontal line on a \(p\)–\(V\) diagram.
4. Explain why the temperature of an ideal gas falls when it expands adiabatically.
Adiabatic means \( Q = 0 \), so the first law gives \( \Delta U = -W \). The expanding gas does positive work on its surroundings, so \(W\) is positive and \( \Delta U \) is negative. Since \( U = \frac{3}{2}nRT \) for an ideal gas, a fall in internal energy is a fall in temperature. The energy to do the work had to come from somewhere, and with no heat supplied the only source was the gas’s own internal energy.
5. State what is true of \(Q\), \(W\) and \( \Delta U \) in (a) an isothermal change and (b) an isovolumetric change.
(a) Isothermal: \( \Delta T = 0 \), so \( \Delta U = 0 \) and therefore \( Q = W \) — every joule supplied leaves again as work. (b) Isovolumetric: \( \Delta V = 0 \), so \( W = 0 \) and therefore \( Q = \Delta U \) — every joule supplied goes into internal energy and raises the temperature.
6. A cycle on a \(p\)–\(V\) diagram encloses an area of 250 J and is traversed clockwise. State the net work done by the gas per cycle and the net thermal energy absorbed.
Clockwise means the gas does net work on its surroundings, so \( W_{\text{net}} = +250 \) J. The gas returns to its starting state, so \( \Delta U = 0 \) over the cycle, and the first law gives \( Q_{\text{net}} = W_{\text{net}} = 250 \) J absorbed.
7. 600 J of thermal energy flows from a body at 400 K to one at 250 K. Find the total entropy change and say whether the process is allowed.
\( \Delta S = -600/400 + 600/250 = -1.50 + 2.40 = +0.90 \) J K\(^{-1}\). It is positive, so the second law permits it — and indeed this is the direction thermal energy always flows on its own.
8. An engine absorbs 2000 J per cycle and rejects 1400 J. Find its efficiency. If its reservoirs are at 500 K and 300 K, is this engine possible?
\( W = 2000 - 1400 = 600 \) J, so \( \eta = 600/2000 = 0.30 \). The Carnot limit is \( 1 - 300/500 = 0.40 \). Since \( 0.30 < 0.40 \), the engine is possible — it reaches 75% of the theoretical maximum.
9. Explain why no heat engine can have an efficiency of 100%.
The Kelvin statement of the second law says no cyclic engine can take thermal energy from a single reservoir and convert all of it into work; some must always be rejected to a colder reservoir. Quantitatively, \( \eta = 1 - T_{\text{C}}/T_{\text{H}} \) reaches 1 only if \( T_{\text{C}} = 0 \) K, and absolute zero is unattainable. Equivalently: an engine with no waste heat would leave the total entropy unchanged or decreasing, which is forbidden.
10. A car engine and its exhaust gases together conserve energy exactly. Explain what is meant by saying that energy has nevertheless been “degraded”.
None of the energy has been destroyed — it has been transferred to the exhaust gases, the engine block and eventually the atmosphere. But it is now spread among an enormous number of molecules at a temperature close to that of the surroundings, and an engine can only extract work from energy that flows between a hot and a cold reservoir. With no temperature difference left to exploit, that energy can no longer do useful work. Degradation is a loss of usefulness, measured by the increase in entropy, not a loss of energy.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • HyperPhysics — the laws of thermodynamics and the Carnot cycle
  • The Physics Hypertextbook — entropy
  • MIT OpenCourseWare — thermodynamics lectures