IB Physics B.4 Thermodynamics — Notes (HL)
Written by Evan Kotronis for the IB Diploma Physics course (first assessment 2025). B.4 is entirely Additional Higher Level content — there is no SL material in this sub-topic, and it is examined only on HL papers. It builds directly on B.1 (internal energy and temperature) and B.3 (the ideal gas and its P–V diagrams), and in quantitative problems the system is always a monatomic ideal gas.
What this page covers
- The first law of thermodynamics, \(Q=\Delta U+W\), as energy conservation for a closed system, with the Clausius sign convention
- Work done by or on a gas when its volume changes, \(W=P\Delta V\), and as the area under a P–V graph
- The change in internal energy of a monatomic ideal gas, \(\Delta U=\tfrac32Nk_B\Delta T=\tfrac32nR\Delta T\)
- Entropy as a measure of the disorder of the particles of a system
- Entropy from macroscopic quantities, \(\Delta S=\dfrac{\Delta Q}{T}\), and from microstates, \(S=k_B\ln\Omega\)
- The second law: the entropy of an isolated system, and the limits it places on what processes can happen
- Why real processes are irreversible, so the entropy of a real isolated system increases
- How entropy can fall locally in a non-isolated system while the surroundings gain at least as much
- Isovolumetric, isobaric, isothermal and adiabatic processes, and \(PV^{5/3}=\) constant for adiabatic changes
- Cyclic processes, heat engines and their efficiency \(\eta=\dfrac{\text{useful work}}{\text{input energy}}\)
- The Carnot cycle and the maximum efficiency \(\eta_{\text{Carnot}}=1-\dfrac{T_c}{T_h}\)
In one minute
- First law: \(Q=\Delta U+W\). \(Q\) is thermal energy supplied to the gas; \(W\) is work done by the gas. Expansion: \(W\gt0\). Compression: \(W\lt0\).
- For a monatomic ideal gas, \(U\) depends only on temperature: \(\Delta U=\tfrac32nR\Delta T\). No temperature change, no change in internal energy.
- Work is the area under a P–V graph; for a cycle, the net work is the area enclosed.
- Four processes, each with one thing fixed: isovolumetric (\(W=0\)), isobaric (\(W=P\Delta V\)), isothermal (\(\Delta U=0\)), adiabatic (\(Q=0\), \(PV^{5/3}=\) constant).
- Entropy measures disorder: \(\Delta S=\Delta Q/T\) and \(S=k_B\ln\Omega\). The entropy of an isolated system never decreases; real (irreversible) processes increase it.
- A heat engine takes \(Q_h\) from a hot reservoir, does work \(W\) and must reject \(Q_c\) to a cold one. No engine can beat the Carnot efficiency \(1-T_c/T_h\).
What is the first law of thermodynamics?
Thermodynamics describes a system — here, a sample of gas — and the energy that crosses its boundary. Two kinds of system matter in B.4:
- A closed system: no mass can enter or leave, but energy can be transferred in either direction, as thermal energy (heat) or as work. Gas trapped in a cylinder by a piston is a closed system.
- An isolated system: neither mass nor energy can enter or leave. Its total energy is fixed.
- \(Q\) = resultant thermal energy supplied to the system (J)
- \(\Delta U\) = change in internal energy of the system (J)
- \(W\) = resultant work done by the system (J)
This equation is in the data booklet. IB problems use the Clausius sign convention, and every sign error in B.4 comes from forgetting it:
| Quantity | Positive (+) when… | Negative (−) when… |
|---|---|---|
| \(Q\) | thermal energy flows into the gas | thermal energy flows out of the gas |
| \(W\) | the gas does work on the surroundings (it expands) | work is done on the gas (it is compressed) |
| \(\Delta U\) | internal energy increases (temperature rises) | internal energy decreases (temperature falls) |
Rearranged, \(\Delta U=Q-W\): the internal energy goes up by whatever thermal energy comes in, minus whatever the gas spends doing work. Recall from B.1 that internal energy is the total kinetic and potential energy of the particles; for an ideal gas there are no intermolecular forces, so \(U\) is entirely the random kinetic energy of the particles.
0.20 mol of a monatomic ideal gas is held in a cylinder. During a process, 350 J of thermal energy is supplied to the gas while the piston does 120 J of work compressing it.
(a) State the values of \(Q\) and \(W\) in the first law. (b) Calculate the change in internal energy of the gas. (c) Determine the change in temperature of the gas.
(a) Thermal energy is supplied to the gas: \(Q=+350\ \text{J}\). Work is done on the gas, so the work done by the gas is negative: \(Q=+350\ \text{J}\), \(W=-120\ \text{J}\)
(b) \(\Delta U=Q-W=350-(-120)=\) +470 J
(c) \(\Delta U=\tfrac32nR\Delta T\Rightarrow\Delta T=\dfrac{2\Delta U}{3nR}=\dfrac{2\times470}{3\times0.20\times8.31}=\) 190 K (188.5 K) — a rise, since \(\Delta U\gt0\).
Both energy inputs raise the internal energy. The most common wrong answer, 230 J, comes from writing \(W=+120\ \text{J}\).
How do you calculate the work done by a gas?
A gas at pressure \(P\) pushes on a piston of area \(A\) with a force \(F=PA\). If the piston moves out through a small distance \(\Delta x\) while the pressure stays constant, the work done by the gas is \(W=F\Delta x=PA\Delta x\). Since \(A\Delta x\) is the change in volume:
- \(W\) = work done by the gas (J)
- \(P\) = pressure of the gas (Pa), constant during the change
- \(\Delta V\) = change in volume, final − initial (m3)
- This equation is in the data booklet. The sign comes out automatically: expansion (\(\Delta V\gt0\)) gives \(W\gt0\); compression gives \(W\lt0\).
- Convert volumes to m3: 1 cm3 = 10−6 m3, 1 dm3 = 1 litre = 10−3 m3. Then Pa × m3 = N m−2 × m3 = N m = J.
- When the pressure is not constant, the work done is the area under the P–V graph between the initial and final volumes. Count squares, or split the area into rectangles and triangles. (You are not expected to integrate.)
- Work depends on the path between two states, not just on the start and end points — different paths on a P–V diagram enclose different areas.
How is internal energy related to temperature?
In B.3 you met \(U=\tfrac32Nk_BT=\tfrac32nRT\) for a monatomic ideal gas. Its particles have no potential energy, and the mean kinetic energy of a particle is proportional to the absolute temperature, so the internal energy changes only when the temperature changes.
- \(N\) = number of particles; \(k_B\) = Boltzmann constant, 1.38 × 10−23 J K−1
- \(n\) = amount of substance (mol); \(R\) = gas constant, 8.31 J K−1 mol−1
- \(\Delta T\) = change in temperature (K; a change of 1 K equals a change of 1 °C)
- This equation is in the data booklet. It holds for any process of a monatomic ideal gas, whatever the pressure or volume does.
- Combining with \(PV=nRT\) gives a useful shortcut (not in the data booklet): \(\Delta U=\tfrac32\Delta(PV)\). You can find \(\Delta U\) straight from a P–V graph without knowing \(n\) or \(T\).
- \(U\) is a function of state: it depends only on where the gas is on the P–V diagram, not on how it got there. \(Q\) and \(W\) are not — they depend on the path. Over a complete cycle, \(\Delta U=0\).
- Any two states on the same isotherm (curve of constant \(T\), where \(PV\) is constant) have the same internal energy.
0.048 mol of a monatomic ideal gas is taken from state A to state B at a constant pressure of 1.2 × 105 Pa, and then at constant volume from B to state C, where the pressure is 0.60 × 105 Pa, as shown in the P–V diagram above.
(a) Calculate the temperatures at A and B. (b) For the change A → B, determine the work done by the gas, the change in internal energy and the thermal energy supplied. (c) For B → C, determine \(Q\), \(W\) and \(\Delta U\). (d) Comment on the change in internal energy from A to C.
(a) \(T_A=\dfrac{P_AV_A}{nR}=\dfrac{1.2\times10^5\times1.0\times10^{-3}}{0.048\times8.31}=\) 300 K (301 K). The volume doubles at constant pressure, so \(T_B=2T_A=\) 600 K (602 K).
(b) \(W=P\Delta V=1.2\times10^5\times(2.0-1.0)\times10^{-3}=\) 120 J (the shaded area).
\(\Delta U=\tfrac32nR\Delta T=\tfrac32\times0.048\times8.31\times301=\) 180 J (check: \(\tfrac32P\Delta V=\tfrac32\times120=180\ \text{J}\)).
\(Q=\Delta U+W=180+120=\) 300 J
(c) No volume change, so \(W=0\). \(\Delta U=\tfrac32V\Delta P=\tfrac32\times2.0\times10^{-3}\times(0.60-1.2)\times10^5=\) −180 J, so \(Q=\Delta U+W=\) −180 J: 180 J of thermal energy leaves the gas.
(d) \(\Delta U_{AC}=+180-180=0\). \(P_AV_A=P_CV_C=120\ \text{J}\), so A and C are at the same temperature (300 K). Yet the gas absorbed a net 120 J of thermal energy, all of which was used to do the 120 J of work.
At constant pressure only \(\tfrac35\) of the thermal energy supplied raises the internal energy; the other \(\tfrac25\) is spent pushing the piston out.
What are the four thermodynamic processes?
Each standard process keeps one quantity fixed. Apply the first law to each one by asking which term is zero.
| Process | Fixed | Work done by gas \(W\) | \(\Delta U\) | First law becomes | P–V graph |
|---|---|---|---|---|---|
| Isovolumetric (isochoric) | \(V\) | 0 | \(\tfrac32nR\Delta T\) | \(Q=\Delta U\) | vertical line |
| Isobaric | \(P\) | \(P\Delta V\) | \(\tfrac32nR\Delta T=\tfrac32P\Delta V\) | \(Q=\Delta U+P\Delta V\) | horizontal line |
| Isothermal | \(T\) | area under curve | 0 | \(Q=W\) | \(PV=\) constant (hyperbola) |
| Adiabatic | no thermal energy transfer, \(Q=0\) | area under curve \(=-\Delta U\) | \(\tfrac32nR\Delta T\) | \(\Delta U=-W\) | \(PV^{5/3}=\) constant (steeper) |
- Isothermal: all the thermal energy absorbed is turned into work (expansion), or all the work done on the gas leaves as thermal energy (compression). The change must be slow, with the gas in good thermal contact with its surroundings, so that its temperature never moves away from that of the surroundings.
- Adiabatic: no thermal energy crosses the boundary — the gas is insulated, or the change is so fast that there is no time for heat transfer. An adiabatic expansion cools the gas (it does work using its own internal energy); an adiabatic compression warms it.
- “Adiabatic” does not mean “constant temperature”, and “isothermal” does not mean “no heat”. They are opposite cases.
The adiabatic equation
- This is in the data booklet. It applies only to monatomic ideal gases and only to adiabatic changes. \(P\) and \(V\) can be in any units, as long as you are consistent on both sides.
- Combined with \(PV=nRT\) (which still holds at every instant) it gives \(TV^{2/3}=\) constant — not in the data booklet, but a quick route to the new temperature. Alternatively find \(P_2\) first and then use \(\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\).
- From any starting point, the adiabatic curve is steeper than the isotherm through that point (by a factor of \(\tfrac53\)). On an adiabatic expansion, the pressure falls for two reasons: the volume increases and the temperature falls.
A cylinder contains a monatomic ideal gas at 1.0 × 105 Pa, 300 K and volume 6.0 × 10−4 m3. The gas is compressed very rapidly to a volume of 1.5 × 10−4 m3, so that the change may be treated as adiabatic.
(a) Show that the final pressure is about 1.0 × 106 Pa. (b) Calculate the final temperature. (c) Determine the work done by the gas. (d) Explain, in terms of the first law, why the temperature of the gas rises.
(a) \(P_2=P_1\left(\dfrac{V_1}{V_2}\right)^{5/3}=1.0\times10^5\times4.0^{5/3}=1.0\times10^5\times10.08=\) 1.01 × 106 Pa ≈ 1.0 × 106 Pa
(b) \(T_2=T_1\dfrac{P_2V_2}{P_1V_1}=300\times\dfrac{1.008\times10^6\times1.5\times10^{-4}}{1.0\times10^5\times6.0\times10^{-4}}=300\times2.52=\) 760 K (756 K)
(c) \(Q=0\), so \(W=-\Delta U\). Using \(\Delta U=\tfrac32\Delta(PV)\):
\(\Delta U=\tfrac32\left(1.008\times10^6\times1.5\times10^{-4}-1.0\times10^5\times6.0\times10^{-4}\right)=\tfrac32(151.2-60.0)=137\ \text{J}\)
Work done by the gas \(W=\) −137 J (137 J of work is done on the gas).
Check with \(\Delta U=\tfrac32nR\Delta T\): \(n=\dfrac{P_1V_1}{RT_1}=0.0241\ \text{mol}\), and \(\tfrac32\times0.0241\times8.31\times456=137\ \text{J}\) ✓
(d) Work is done on the gas (\(W\lt0\)) and no thermal energy leaves it (\(Q=0\)), so \(\Delta U=Q-W=-W\gt0\). The internal energy — the total random kinetic energy of the particles — increases, so the mean kinetic energy of the particles and hence the temperature increase.
What is entropy?
Entropy can be found in two equivalent ways — from measurable (macroscopic) quantities, or by counting the arrangements of the particles (microscopic).
Macroscopic: thermal energy and temperature
\[ \Delta S=\frac{\Delta Q}{T} \]- \(\Delta S\) = change in entropy of the system (J K−1)
- \(\Delta Q\) = thermal energy transferred to the system (J); negative if thermal energy leaves
- \(T\) = absolute temperature at which the transfer happens (K)
- This is in the data booklet. Use it when the temperature stays constant during the transfer: a phase change (melting, boiling), a large reservoir whose temperature barely changes, or an isothermal process of a gas.
- The same \(\Delta Q\) produces a larger entropy change at a lower temperature. This is why thermal energy flowing from hot to cold always raises the total entropy.
Microscopic: counting microstates
A macrostate is described by bulk quantities (for coins: how many heads; for a gas: \(P\), \(V\), \(T\)). A microstate is one particular arrangement of every individual particle consistent with that macrostate. All microstates are equally probable.
\[ S=k_B\ln\Omega \]- \(S\) = entropy of a macrostate (J K−1)
- \(k_B\) = Boltzmann constant (J K−1)
- \(\Omega\) = number of microstates that give that macrostate (no unit)
- This is in the data booklet. \(\ln\) is the natural logarithm. A macrostate with only one microstate (perfect order) has \(S=k_B\ln1=0\).
- Because every microstate is equally likely, the macrostate with the most microstates is the most probable. Disordered macrostates have vastly more microstates than ordered ones, so systems drift towards disorder simply because it is overwhelmingly likely.
- The IB model uses coins (or particles in two halves of a box): with \(N\) two-sided objects there are \(2^N\) microstates in total.
(a) Five coins are tossed. Calculate the entropy of the macrostate “three heads and two tails”, and determine the probability of obtaining it.
(b) An ice cube of mass 0.250 kg at 0 °C (273 K) melts completely in a room whose temperature stays at 295 K. The specific latent heat of fusion of ice is 3.34 × 105 J kg−1. Determine the entropy change of the ice, of the room, and of the ice and room together.
(a) \(\Omega=10\): the number of ways of choosing which 3 of the 5 coins show heads (HHHTT, HHTHT, HHTTH, HTHHT, HTHTH, HTTHH, THHHT, THHTH, THTHH, TTHHH).
\(S=k_B\ln\Omega=1.38\times10^{-23}\times\ln10=\) 3.2 × 10−23 J K−1 (3.18 × 10−23)
Probability \(=\dfrac{10}{2^5}=\dfrac{10}{32}=\) 0.31 (0.3125)
(b) Thermal energy absorbed by the ice: \(Q=mL=0.250\times3.34\times10^5=8.35\times10^4\ \text{J}\).
Ice (at constant 273 K): \(\Delta S=\dfrac{+8.35\times10^4}{273}=\) +306 J K−1
Room (loses the same energy at 295 K): \(\Delta S=\dfrac{-8.35\times10^4}{295}=\) −283 J K−1
Together: \(306-283=\) +23 J K−1 (22.8 J K−1)
The room's entropy falls — allowed, because the room is not isolated — but the ice gains more, so the total for the isolated ice-plus-room system rises, as the second law requires.
What does the second law of thermodynamics say?
The first law says energy is conserved, but it does not forbid a cold drink from spontaneously cooling further while warming the room. The second law says which processes actually happen. You must be able to state it in three equivalent forms.
- The heat pump in (c) obeys both laws. First law: 600 J + 200 J = 800 J delivered. Second law: entropy change \(=-\dfrac{600}{278}+\dfrac{800}{293}=-2.16+2.73=+0.57\ \text{J K}^{-1}\gt0\).
- The three forms are equivalent: if you could build a device that broke one, you could combine it with an ordinary engine or heat pump to break the others.
Reversible and irreversible processes
- A reversible process is an idealisation: it happens infinitely slowly, with no friction or turbulence, and with thermal energy only ever flowing across a vanishingly small temperature difference. It could be run backwards, restoring the system and its surroundings exactly. For a reversible process, the total entropy of system + surroundings is unchanged.
- Real processes are irreversible: friction, thermal energy flowing across a finite temperature difference, mixing, a gas expanding freely into a vacuum. Real processes in isolated systems are almost always irreversible, so the entropy of a real isolated system increases.
Can entropy ever decrease?
- Water freezing in a freezer: the water's entropy falls (ordered crystal), but the freezer releases even more thermal energy into the kitchen, whose entropy rises by more.
- A living organism builds ordered molecules, lowering its own entropy, while releasing thermal energy and disordered waste that raise the entropy of its surroundings by more.
- The total entropy (system + surroundings, treated together as an isolated system) never decreases.
How do heat engines use cycles?
A single expansion can do work only once — the piston runs out of cylinder. A heat engine repeats a cycle of processes that returns the gas (the working substance) to its starting state, so it can run continuously.
- Over one complete cycle the gas is back where it started, so \(\Delta U=0\) and the first law gives net \(Q\) = net \(W\).
- On a P–V diagram, the net work done by the gas per cycle is the area enclosed by the loop. A loop traced clockwise gives positive net work (an engine); anticlockwise gives net work done on the gas (a heat pump or refrigerator).
- For a positive net work, the gas must be compressed when it is cooler (low pressure, little work needed) and expand when it is hotter (high pressure, more work done).
- Every heat engine absorbs thermal energy \(Q_h\) from a hot reservoir at \(T_h\), does useful work \(W\), and — by the Kelvin form of the second law — must reject thermal energy \(Q_c\) to a cold reservoir at \(T_c\): \(Q_h=W+Q_c\).
- An engine can be designed around different cycles (a petrol engine, for example, is modelled by a cycle of two adiabatic and two isovolumetric processes). Each cycle gives a different efficiency between the same temperatures.
- The first form is in the data booklet; the other two follow from \(Q_h=W+Q_c\) and are not. Efficiency has no unit and may be given as a decimal or a percentage.
- The “input energy” is the thermal energy absorbed in all stages where \(Q\gt0\) — not the net thermal energy, which equals the work.
- Power = work per cycle × number of cycles per second.
A monatomic ideal gas is taken round the cycle A → B → C → D → A shown above. The temperature at A is 300 K.
(a) Show that the net work done by the gas per cycle is 100 J. (b) Determine the temperatures at B, C and D. (c) Determine the thermal energy absorbed by the gas per cycle, and hence the efficiency of the cycle. (d) Calculate the Carnot efficiency for reservoirs at the highest and lowest temperatures of the cycle, and suggest why the cycle's efficiency is much lower.
(a) Net work = enclosed area \(=(2.0-1.0)\times10^5\times(2.0-1.0)\times10^{-3}=\) 100 J
(b) For a fixed amount of gas \(T\propto PV\). \(PV\) at A is 100 J. At B (200 J) and D (200 J): \(T_B=T_D=600\ \text{K}\). At C (400 J): \(T_C=1200\ \text{K}\)
(c) Use \(\Delta U=\tfrac32\Delta(PV)\) and \(Q=\Delta U+W\) for each stage:
| Stage | \(W\) / J | \(\Delta U\) / J | \(Q\) / J |
|---|---|---|---|
| A → B (isovolumetric) | 0 | \(\tfrac32(200-100)=150\) | +150 |
| B → C (isobaric) | \(2.0\times10^5\times1.0\times10^{-3}=200\) | \(\tfrac32(400-200)=300\) | +500 |
| C → D (isovolumetric) | 0 | −300 | −300 |
| D → A (isobaric) | −100 | −150 | −250 |
| Cycle | 100 | 0 | net +100 |
Thermal energy absorbed \(=150+500=\) 650 J. \(\eta=\dfrac{W}{Q_h}=\dfrac{100}{650}=\) 0.15 (15.4 %)
(d) \(\eta_{\text{Carnot}}=1-\dfrac{300}{1200}=\) 0.75. Much of the thermal energy is absorbed while the gas is far below 1200 K (A → B) and rejected while it is far above 300 K (C → D). Thermal energy flowing across large temperature differences between reservoirs and gas is irreversible, and generates entropy, so less of the input can become work.
What is the Carnot cycle?
The Carnot cycle is a theoretical, perfectly reversible cycle operating between two reservoirs. It consists of two isothermal and two adiabatic processes:
- A → B, isothermal expansion at \(T_h\): the gas absorbs \(Q_h\) from the hot reservoir and does an equal amount of work (\(\Delta U=0\)).
- B → C, adiabatic expansion: the gas does work using its internal energy and cools from \(T_h\) to \(T_c\).
- C → D, isothermal compression at \(T_c\): work is done on the gas and it rejects \(Q_c\) to the cold reservoir.
- D → A, adiabatic compression: work done on the gas raises its temperature from \(T_c\) back to \(T_h\).
- \(T_h\) = temperature of the hot reservoir (K); \(T_c\) = temperature of the cold reservoir (K)
- This is in the data booklet. Temperatures must be in kelvin.
- Why it is the limit: the cycle is reversible, so the total entropy does not change. The hot reservoir loses \(Q_h/T_h\), the cold one gains \(Q_c/T_c\), and these are equal, so \(\dfrac{Q_c}{Q_h}=\dfrac{T_c}{T_h}\). Any irreversibility makes \(Q_c\) larger and the efficiency smaller.
- Even a perfect engine cannot reach \(\eta=1\) unless \(T_c=0\) K. To raise the maximum efficiency, raise \(T_h\) or lower \(T_c\).
- Real engines fall well short of \(\eta_{\text{Carnot}}\): they run fast (not quasi-statically), have friction, and exchange heat across finite temperature differences.
B.4 Heat Engines & Carnot simulation: run a gas round a cycle on a P–V diagram, watch the enclosed area (the work) change, and compare the engine's efficiency with the Carnot limit as you change the reservoir temperatures.
An engine uses 0.40 mol of a monatomic ideal gas in a Carnot cycle between reservoirs at 500 K and 320 K (see the P–V diagram above). In each cycle the gas absorbs 1200 J from the hot reservoir.
(a) Calculate the efficiency of the engine. (b) Determine the work done by the gas during the adiabatic expansion B → C, and explain why the two adiabatic stages make no net contribution to the work per cycle. (c) Determine the work done per cycle and the thermal energy rejected per cycle. (d) Show that the total entropy change of the two reservoirs per cycle is zero. (e) A real engine working between the same reservoirs also absorbs 1200 J per cycle but has an efficiency of 0.25. Determine the entropy change of the reservoirs per cycle and comment on your answer.
(a) \(\eta=1-\dfrac{T_c}{T_h}=1-\dfrac{320}{500}=\) 0.36
(b) \(Q=0\), so \(W=-\Delta U=-\tfrac32nR\Delta T=-\tfrac32\times0.40\times8.31\times(320-500)=\) +897 J ≈ 900 J.
In D → A the temperature changes by the same amount in the opposite direction (320 K → 500 K), so the work done by the gas is −897 J. The two cancel.
(c) \(W=\eta Q_h=0.36\times1200=\) 432 J (the enclosed area). \(Q_c=Q_h-W=1200-432=\) 768 J
(d) Hot reservoir: \(\Delta S=-\dfrac{1200}{500}=-2.40\ \text{J K}^{-1}\). Cold reservoir: \(\Delta S=+\dfrac{768}{320}=+2.40\ \text{J K}^{-1}\). Total = 0 (the gas itself returns to its initial state, so its entropy change over the cycle is also zero).
(e) \(W=0.25\times1200=300\ \text{J}\), so \(Q_c=900\ \text{J}\).
\(\Delta S=-\dfrac{1200}{500}+\dfrac{900}{320}=-2.40+2.81=\) +0.41 J K−1 per cycle
The total entropy increases, as the second law requires for a real (irreversible) engine. The Carnot engine, being reversible, leaves the total entropy unchanged; an engine with \(\eta\gt0.36\) would need \(Q_c\lt768\ \text{J}\), giving a total entropy decrease — which is impossible.
Exam tips for B.4
Start every first-law calculation by writing \(Q=\) …, \(W=\) …, with a sign on each. \(W\) is the work done by the gas: negative whenever the gas is compressed. Students who carry over the convention used in some other courses (where \(W\) is the work done on the gas) while still writing \(Q=\Delta U+W\) get the wrong value of \(\Delta U\) and lose the marks.
For any process, say which quantity is zero and why: “isothermal, so \(\Delta U=0\)”, “adiabatic, so \(Q=0\)”, “constant volume, so \(W=0\)”. That sentence often earns the first mark, and it tells you which form of the first law to use. In a cycle, also state “\(\Delta U=0\) over a cycle”.
When asked where on a cycle the gas has the most internal energy, find where \(PV\) (and so \(T\)) is largest. In a Carnot cycle the whole hot isothermal stage is at the highest temperature, so the answer is a process (the whole stage), not a single corner of the diagram.
\(\eta_{\text{Carnot}}\), \(PV=nRT\) and \(\Delta S=\Delta Q/T\) all need kelvin. Convert volumes to m3 before using \(W=P\Delta V\) (a common factor-of-1000 error). For a cycle's efficiency, divide by the thermal energy absorbed, not the net heat.
When a system's entropy falls (water freezing, a refrigerator's contents cooling), full marks need both halves: “the entropy of the system decreases, but the entropy of the surroundings increases by more, so the total entropy increases”. For coin problems, \(\Omega\) is the number of microstates of the macrostate asked about, not the total \(2^N\).
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| In an isothermal process no heat is transferred. | The temperature stays constant because thermal energy flows in or out. | Isothermal: \(\Delta U=0\) and \(Q=W\). It is the adiabatic process that has \(Q=0\). |
| In an adiabatic process the temperature is constant. | With \(Q=0\), any work changes the internal energy. | Adiabatic expansion cools the gas; adiabatic compression warms it. |
| Heating a gas always raises its temperature. | If the gas does more work than the heat supplied, \(\Delta U=Q-W\lt0\). | Temperature follows \(\Delta U\), not \(Q\). In an isothermal expansion, heat is supplied but \(T\) is constant. |
| Entropy can never decrease. | That applies only to an isolated system. | A non-isolated system's entropy can fall if its surroundings gain at least as much. |
| A better-engineered engine could reach 100 % efficiency. | The Kelvin form forbids converting all the input heat into work in a cycle. | The upper limit is \(1-T_c/T_h\), even for a perfect, reversible engine. |
| The net work of a cycle is the area under the top curve. | The work done on the gas on the return path must be subtracted. | Net work = area enclosed by the loop; clockwise = positive. |
How this topic connects
The second law and entropy: why no engine can reach 100 % efficiency
Any device that turns thermal energy into work runs in a cycle, and the second law (Kelvin form) forbids converting all the absorbed thermal energy into work: some must be rejected to a cooler reservoir. The minimum that must be rejected is set by entropy. The hot reservoir loses entropy \(Q_h/T_h\), so the cold reservoir must gain at least as much, \(Q_c/T_c\ge Q_h/T_h\), which gives \(\eta\le1-T_c/T_h\). Power stations, car engines and jet engines are all limited this way; this is why they run their hot side as hot as materials allow, and why roughly half or more of the input energy in a thermal power station leaves as waste heat.
Energy losses in motors and generators, and why Carnot does not cap them
Motors (D.3) and generators (D.4) convert between electrical and mechanical energy without passing through a heat cycle, so the Carnot limit does not apply to them directly, and well-designed machines can exceed 90 % efficiency. Their losses are dissipative: resistive heating of the coils and friction in the bearings. Both are irreversible processes that turn ordered energy into internal energy and raise the entropy of the surroundings. However, a generator in a fossil-fuel or nuclear power station is driven by a heat engine (a turbine), so the overall efficiency of the station is limited by the Carnot efficiency of its steam cycle.
Entropy on a cosmic scale: the second law and the long-term fate of the universe
The universe can be treated as an isolated system: there is nothing outside it to exchange energy with. Because real processes are irreversible, its total entropy keeps increasing. Energy is conserved, but it is steadily spread out into less useful forms — thermal energy at ever more uniform temperatures, from which less work can be extracted. Local order (stars, planets, living things) can still form, but always at the cost of a larger entropy increase elsewhere. The one-way growth of entropy also gives time a direction: it is how we distinguish past from future.
Tools & inquiry — investigating compression of a gas
- Connect a gas syringe to a pressure sensor and data logger. Compress the trapped air slowly, recording \(P\) and \(V\); then repeat quickly. Slow compression is close to isothermal; rapid compression is closer to adiabatic and gives a steeper P–V curve.
- Plot \(\ln P\) against \(\ln V\). The gradient is −1 for an isothermal change. For an adiabatic change of a monatomic gas it would be −5/3; air is mainly diatomic, so its adiabatic gradient is smaller in magnitude (about −1.4). A real fast compression lies between the two because some heat leaks through the walls.
- Systematic errors: the dead volume in the tube connecting the syringe to the sensor (add it to every reading), and leaks past the plunger, which reduce \(n\) during the experiment.
- NOS — paradigm shifts and models: the steam engine drove the Industrial Revolution before anyone understood why it worked. Sadi Carnot found the maximum efficiency in 1824 using the caloric theory — the idea that heat is a fluid — which was later shown to be wrong. His result survived the change of model, and the attempt to explain it led Clausius and Kelvin to the concept of entropy and the laws of thermodynamics.
Key equations
| Equation | Use | In data booklet? |
|---|---|---|
| \(Q=\Delta U+W\) | first law (Clausius convention: \(W\) done by the gas) | yes |
| \(W=P\Delta V\) | work done by a gas at constant pressure; otherwise area under P–V graph | yes |
| \(\Delta U=\tfrac32nR\Delta T=\tfrac32Nk_B\Delta T\) | change in internal energy, monatomic ideal gas | yes |
| \(\Delta U=\tfrac32\Delta(PV)\) | internal energy change read from a P–V graph | no — combine with \(PV=nRT\) |
| \(PV^{5/3}=\) constant | adiabatic change, monatomic ideal gas | yes |
| \(TV^{2/3}=\) constant | temperature after an adiabatic change | no — derive from the two above |
| \(\Delta S=\dfrac{\Delta Q}{T}\) | entropy change for transfer at constant temperature | yes |
| \(S=k_B\ln\Omega\) | entropy from the number of microstates | yes |
| \(\eta=\dfrac{\text{useful work}}{\text{input energy}}\) | efficiency of an engine | yes |
| \(Q_h=W+Q_c\), \(\eta=1-\dfrac{Q_c}{Q_h}\) | energy per cycle of an engine | no — follow from the first law |
| \(\eta_{\text{Carnot}}=1-\dfrac{T_c}{T_h}\) | maximum efficiency between two reservoirs | yes |
Check your understanding
1. A monatomic ideal gas is compressed at a constant pressure of 1.5 × 105 Pa from 4.0 × 10−3 m3 to 2.5 × 10−3 m3. Determine \(W\), \(\Delta U\) and \(Q\).
\(W=P\Delta V=1.5\times10^5\times(2.5-4.0)\times10^{-3}=-225\ \text{J}\) (work done on the gas). \(\Delta U=\tfrac32P\Delta V=-338\ \text{J}\) (337.5 J; the gas cools). \(Q=\Delta U+W=-338+(-225)=-563\ \text{J}\): about 560 J of thermal energy leaves the gas.
2. 2400 J of thermal energy is conducted from a large body at 480 K to a large body at 320 K. Calculate the total entropy change and state what it shows.
\(\Delta S=-\dfrac{2400}{480}+\dfrac{2400}{320}=-5.0+7.5=+2.5\ \text{J K}^{-1}\). The total entropy increases, so the process is irreversible — the energy will not flow back from 320 K to 480 K by itself (Clausius form).
3. Calculate the maximum possible efficiency of an engine working between 650 K and 290 K. Give two reasons why a real engine has a lower efficiency.
\(\eta_{\text{Carnot}}=1-\dfrac{290}{650}=0.55\) (0.554). A real engine runs quickly, so its processes are not reversible; there is friction between moving parts; and thermal energy is exchanged across finite temperature differences. All of these generate entropy and increase the energy rejected.
4. Six coins are tossed. Calculate the entropy of the macrostate “three heads, three tails”, and explain why this macrostate is the most likely.
\(\Omega=20\) (the number of ways of choosing which 3 of the 6 coins are heads), so \(S=k_B\ln20=1.38\times10^{-23}\times3.00=4.1\times10^{-23}\ \text{J K}^{-1}\). All \(2^6=64\) microstates are equally probable, and this macrostate contains more of them (20) than any other, so it is the most probable (20/64 ≈ 0.31).
5. A bicycle pump is pushed in rapidly with the outlet blocked, and the air inside gets warmer. Explain this using the first law of thermodynamics.
The compression is rapid, so there is no time for thermal energy to leave: the process is approximately adiabatic, \(Q=0\). Work is done on the air, so \(W\lt0\). Then \(\Delta U=Q-W=-W\gt0\): the internal energy of the air increases, so the mean kinetic energy of its molecules, and hence its temperature, rises.
Practise B.4 with exam-style questions
Original Paper 1A, 1B and 2 questions on thermodynamics, with full mark schemes. B.4 is HL only, so there is no SL bank for this sub-topic. Tick the questions you want and print them as your own test.
Revise and explore
- Flashcards — choose IB Physics and filter to B.4 Thermodynamics. Free, no sign-up.
- B.4 Heat Engines & Carnot simulation
- Full IB Physics question bank (all sub-topics, SL & HL)