Where Does the Carnot Cycle Come From?

We want to build a device that turns heat into work. How much can be converted? Sadi Carnot (1824), at a time when steam engines were still only about 3 % efficient, derived the upper limit without any reference to what is inside the device. Here we take an ideal gas as the working substance, write down the heat and work of all four stages, and arrive at η = 1 − T_C/T_H. Along the way, a quantity called entropy appears of its own accord.

0. What Does a Heat Engine Do?

A heat engine is a device that returns to its original state after one cycle while delivering net work to the outside. “Returning to the original state” is the crucial point; without it, we would have just a one-time expansion. After one cycle all the state functions return to their original values, so the change in internal energy is zero.

\oint d U = 0 \Rightarrow W_{net} = Q_{net} = Q_{H} + Q_{C}

Here Q_H > 0 is the heat received from the hot side and Q_C < 0 is the heat rejected to the cold side. Work is taken as positive when it is “work done by the gas” (first law of thermodynamics: ΔU = Q − W). The efficiency is “the fraction of the heat put in that could be turned into work,” so

η \equiv \frac{W}{Q_{H}} = \frac{Q_{H} + Q_{C}}{Q_{H}} = 1 - \frac{| Q_{C} |}{Q_{H}}

Start here If Q_C = 0 can be achieved, then η = 1. In other words, an “engine that converts all of the heat into work” is an engine that rejects nothing to the cold side. What we see below is that, as long as the engine has to run in a cycle, this cannot be done; that there is a lower limit on the amount rejected; and that this limit is determined by the temperatures alone.

1. Preparation — Only Two Properties of the Ideal Gas Are Needed

We use an ideal gas as the working substance. Only the following two properties are used.

\begin{array}{l} (a) P V = n R T \\ (b) U = U (T) — independent of volume \end{array}

(b) is another way of saying that “the molecules exert no forces on each other.” Pulling the molecules apart does not change the potential energy, so changing the volume does not move U. On the ideal gas derivation page, this is derived from the partition function. Here we do not assume that C_V is constant — we allow it to vary with temperature, and write it as $C_{V} (T) \equiv \frac{d U_{m}}{d T}$ throughout.

2. The Isothermal Process — Turning Heat Entirely into Work

Keeping the temperature T fixed, expand the gas quasistatically from V_a to V_b. By (b), ΔU = 0, so the first law gives Q = W. For a quasistatic process W = ∫PdV, which on a P–V diagram is the area under the curve (the net work summed over one full cycle is the area enclosed by the cycle, ∮PdV). Since T is constant here,

Q = W = \int_{V_{a}}^{V_{b}} \frac{n R T}{V} d V = n R T \ln \frac{V_{b}}{V_{a}}

(1)

Seen on its own, the efficiency here is 100 % The isothermal expansion turns the heat it receives into work without wasting even 1 J. So why does the efficiency of an engine not reach 1? Because this process does not go around a cycle. The gas has simply kept expanding and has not returned to where it started. What it costs to bring it back is the next part of the story.

3. The Quasistatic Adiabatic Process — Work in Exchange for Cooling

Cut off the exchange of heat (Q = 0) and expand the gas quasistatically. The first law becomes dU = −PdV, that is,

n C_{V} (T) d T = - \frac{n R T}{V} d V \Rightarrow C_{V} \frac{d T}{T} = - R \frac{d V}{V}

(2)

Integrate both sides. If C_V is constant, this takes a familiar form, but in general it is more accurate to write it as follows.

\int_{T_{2}}^{T_{1}} C_{V} (T) \frac{d T}{T} = R \ln \frac{V_{2}}{V_{1}} (T_{1} \to T_{2}, V_{1} \to V_{2})

(3)

If we set C_V constant, the left-hand side becomes C_V ln(T₁/T₂), and using γ = C_P/C_V = 1 + R/C_V to tidy up, we get

T V^{γ - 1} = const. \Leftrightarrow P V^{γ} = const.

That is the textbook form. However, γ is constant only if, over the temperature range in question, the degrees of freedom that are “awake” do not change. Cool H₂ to 80 K and rotation begins to freeze out; heat CO₂ to 1000 K and vibration wakes up. This page and the simulator use Eq. (3) as it stands, numerically, so they can handle even the temperature dependence of C_V (although at very low temperatures the H₂ result is an approximation that ignores the nuclear-spin statistics of ortho- and para-hydrogen).

The left-hand side of Eq. (3) should look familiar When the temperature is raised from T₂ to T₁ at constant volume, the entropy change is ∫nC_VdT/T, and when the volume is expanded from V₁ to V₂ at constant temperature it is nR ln(V₂/V₁). Eq. (3) says that these two are equal. In an actual adiabatic expansion the temperature falls from T₁ to T₂, so the temperature part changes sign and exactly cancels the volume part. In other words, a quasistatic adiabatic process is a process in which the entropy does not change. — We have not yet defined “entropy,” so for now just remember that the forms look similar. §8 will prove that this quantity is a state function.

What you can see in the simulator In the cylinder of tab ① of the simulator, the molecules are colored by speed (blue = slow, red = fast). Even though the exchange of heat is cut off, the color shifts toward blue during the adiabatic expansion. This means the gas loses internal energy by the amount it uses to push the piston, so dU = −PdV, written at the start of this section, is something you can watch happening.

4. Lining Up the Four Stages

The Carnot cycle uses the two kinds of processes we have just prepared, alternately, twice each. Let the temperature of the hot reservoir be T_H and that of the cold reservoir T_C, and start from state ① (T_H, V₁).

Stage	What happens	Q	W = work done by the gas	ΔU
① → ②	Isothermal expansion in contact with the T_H reservoir	nRT_H ln(V₂/V₁)	same as left	0
② → ③	Removed from the reservoir; adiabatic expansion (T_H → T_C)	0	n[U_m(T_H) − U_m(T_C)]	−(same as left)
③ → ④	Isothermal compression in contact with the T_C reservoir	nRT_C ln(V₄/V₃)	same as left	0
④ → ①	Removed from the reservoir; adiabatic compression (T_C → T_H)	0	−n[U_m(T_H) − U_m(T_C)]	−(same as left)

The work in the two adiabatic stages exactly cancels. This is because the gas goes back and forth between the same two temperatures, so the change in internal energy has simply changed sign. The work taken out in the adiabatic stage is spent entirely on the next adiabatic compression and comes back. What remains is only the heat of the two isothermal stages:

\begin{array}{rcl} Q_{H} & = & n R T_{H} \ln \frac{V_{2}}{V_{1}} > 0 \\ Q_{C} & = & - n R T_{C} \ln \frac{V_{3}}{V_{4}} < 0 \end{array}

(4)

5. The Condition for the Cycle to Close

To find the efficiency we need to compare the two ln terms in Eq. (4). Here we use the fact that ②→③ and ④→① are both adiabatic processes between T_H and T_C. Applying Eq. (3) to both (since the temperature runs the opposite way in ④→①, we flip the signs of both sides), the left-hand sides are the same integral, so

R \ln \frac{V_{3}}{V_{2}} = \int_{T_{C}}^{T_{H}} C_{V} \frac{d T}{T} = R \ln \frac{V_{4}}{V_{1}} \Rightarrow \frac{V_{2}}{V_{1}} = \frac{V_{3}}{V_{4}}

(5)

The volume ratios of the two isothermal stages are equal. This is the crux of the Carnot cycle. Whatever the form of C_V(T) and whatever the gas, it holds as long as the adiabatic processes connect the same two temperatures.

The figure below plots the cycle in the coordinates in which this situation is easiest to see — ln V on the horizontal axis and ln T on the vertical axis. The isothermal stages are horizontal segments, and from Eq. (2) the adiabatic stages are lines with slope d lnT/d lnV = −R/C_V. And the two adiabats are the same curve, just shifted sideways (the left-hand side of Eq. (3) is the same integral for both, so on any adiabat ln V is the same function of temperature plus a constant, and a translation in the x direction makes them coincide exactly). So the width at the top and the width at the bottom are necessarily equal — that is Eq. (5).

Fig. 1: The Carnot cycle in ln V – ln T coordinates. Because the two adiabats are congruent (they coincide under a translation), the widths at the top and bottom are always equal. Changing the gas changes the slopes of the adiabats, and on the low-temperature side of H₂ or the high-temperature side of CO₂ the lines curve because C_V varies — but they remain congruent.

Gas Hot reservoir T_H = 600 K Cold reservoir T_C = 300 K Expansion ratio V₂/V₁ = 2.0

V₃/V₂ (stretch factor of the adiabatic expansion)—

ln(V₃/V₂)—

∫C_VdT/T ÷ R (numerical integration)—

C_V/R (at T_H)—

C_V/R (at T_C)—

Efficiency η = 1 − T_C/T_H—

6. The Efficiency — Everything Cancels Out

Put Eq. (5) into Eq. (4). Replacing V₃/V₄ by V₂/V₁, the two ln terms become the same thing.

\begin{array}{rcl} η & = & 1 - \frac{n R T_{C} \ln (V_{3} / V_{4})}{n R T_{H} \ln (V_{2} / V_{1})} \\ = & 1 - \frac{T_{C}}{T_{H}} \end{array}

(6)

Count what has disappeared The amount of substance n, the gas constant R, the expansion ratio V₂/V₁, the pressure and volume at the starting point, and C_V — that is, the kind of gas. The value is exactly the same for monatomic He and for CO₂, where vibration also matters. All that is left is the temperatures of the two reservoirs. In the table at the bottom of tab ② of the simulator, C_V, γ and V₃/V₂ all differ from gas to gas, yet you can check that only Q_H, W and η come out the same.

Only the efficiency drops out; the dimensions of the cylinder do not For T_H = 600 K and T_C = 300 K, the stretch factor of the adiabatic expansion V₃/V₂ is about 2.8 for monatomic He, and about 17 for CO₂, where vibration also matters (you can check this in the output fields of Fig. 1). This is because a gas with a larger C_V has to be expanded more to lower the temperature by the same amount. Even so, η stays as in Eq. (6) — what changes from gas to gas is not the efficiency but the length of cylinder that is needed.

7. The T–S Diagram — Seeing It as a Rectangle

Rewriting Eq. (4), the heat exchanged in the two isothermal stages is

\frac{Q_{H}}{T_{H}} = n R \ln \frac{V_{2}}{V_{1}} \equiv Δ S \frac{Q_{C}}{T_{C}} = - n R \ln \frac{V_{3}}{V_{4}} = - Δ S

(7)

Thanks to Eq. (5), this gives quantities that are exactly equal in magnitude and differ only in sign. If we draw the cycle with entropy S on the horizontal axis and temperature T on the vertical axis, so that this ΔS is the horizontal width, the Carnot cycle becomes a rectangle (isothermal = moving horizontally, adiabatic = moving vertically). For a quasistatic process Q = ∫TdS, so the area is directly the heat.

Part of the figure	Area	Meaning
The whole column (height T_H)	T_H ΔS	Heat Q_H absorbed from the hot reservoir
The lower band (height T_C)	T_C ΔS	Heat \|Q_C\| rejected to the cold reservoir
The rectangle (height T_H − T_C)	(T_H − T_C) ΔS	Work W that can be extracted

The efficiency is the ratio of “the upper rectangle (height T_H − T_C)” to “the whole column,” that is, the ratio of the vertical lengths. η = (T_H − T_C)/T_H can be read off at a glance in the T–S diagram of tab ① of the simulator. If T_C could be made zero, the lower band would vanish and η = 1, but that would require a reservoir at absolute zero, which the third law forbids.

8. A Quantity That Returns to 0 After One Cycle — The Discovery of Entropy

Adding the two relations in Eq. (7) gives

\frac{Q_{H}}{T_{H}} + \frac{Q_{C}}{T_{C}} = 0 that is, \oint \frac{d Q_{rev}}{T} = 0

(8)

This is the path Clausius took. Q itself does not return to 0 after one cycle (if it did, no work would come out). But if we divide Q by T before adding, it returns neatly to 0. A “quantity that returns to 0 after a trip around any path” is a quantity fixed by the state alone, independent of the path — that is, an infinitesimal change of a state function. So we can define

d S \equiv \frac{d Q_{rev}}{T}

This is the entropy. Any reversible cycle can be divided into small Carnot cycles (the adjacent adiabats cancel each other), so Eq. (8) is not a property of the Carnot cycle alone but holds for every reversible cycle.

If the process is irreversible, the equality breaks In real engines, friction and expansion or compression at a finite speed make ∮dQ/T < 0 (the Clausius inequality). Here T is the temperature of the reservoir that exchanges the heat — in an irreversible process the temperature of the working substance is not uniquely defined, so we measure it by the temperature of the surroundings. Restated for an isolated system, this becomes ΔS ≥ 0, that is, the “law of increase of entropy.” The various statements of the second law — Kelvin (an engine that turns all of the heat into work cannot be built), Clausius (heat does not flow by itself from low to high temperature), and the increase of entropy — are all equivalent, and Eq. (6) can be derived from any one of them.

9. Carnot’s Theorem — Why Can We Say It Is an “Upper Limit”?

So far this has been a calculation for one particular engine that uses an ideal gas. To say that “no engine can exceed it,” one more step is needed. Carnot’s theorem asserts the following.

Carnot’s theorem For heat engines operating between the same two reservoirs,
(i) the efficiency of a reversible engine is at least as great as that of an irreversible engine.
(ii) the efficiency of a reversible engine is the same whatever the working substance.

The proof is by contradiction. Suppose there were an engine X with efficiency η_X > η_rev. Use the work obtained by running X to drive a reversible engine in reverse (= as a refrigerator). Run in reverse, a reversible engine pumps heat up from the cold side and returns it to the hot side. Looking at the combined device of the two engines, we can adjust the scales so that the net work is exactly zero, and what remains is the result that heat has simply been pumped up from the cold reservoir and handed to the hot reservoir. This is a “device that transfers heat from low to high temperature with no other change left behind,” which contradicts Clausius’s statement. Hence η_X ≤ η_rev. As for (ii), it follows by running the same argument in both directions with two reversible engines.

It is important to see where (ii) comes into play. Since the efficiency of a reversible engine does not depend on the working substance, when computing it, we may choose the substance that is easiest to calculate with. Choosing the ideal gas gave Eq. (6). So Eq. (6) is not a formula for the ideal gas alone but a universal upper limit determined by the two temperatures alone.

10. The Definition of Temperature Itself

From (ii), |Q_C|/Q_H of a reversible engine is a function of the two reservoirs alone. Kelvin therefore adopted this ratio itself as the definition of temperature.

\frac{| Q_{C} |}{Q_{H}} \equiv \frac{T_{C}}{T_{H}}

This scale does not depend at all on what is inside the thermometer. It is fixed neither by how mercury expands nor by how a gas expands, but only by the ratio of quantities of heat. This is why it is called the “thermodynamic temperature.” That the ideal-gas temperature agrees with it is shown by the derivation of Eq. (6) itself. The kelvin today was fixed in the 2019 SI revision by assigning the defined value k_B = 1.380649×10⁻²³ J/K, but what lies behind the scale is this Carnot relation.

11. So Why Do Real Engines Fall Short?

Eq. (6) is a value with the condition “if reversible.” Real engines always fall below it, for the following reasons.

They run at a finite speed. “Quasistatic” means infinitely slowly; if an engine is actually run, nonuniformities of pressure and temperature arise in the gas, and their evening-out is an irreversible process.
A temperature difference is needed to make heat flow. If the reservoir and the working substance are at the same temperature, no heat flows. The moment a difference is introduced to make it flow, that heat transfer itself generates entropy.
Friction and heat leaks. Friction of the piston, heat loss from the piping.
The Carnot cycle is hard to build as a machine in the first place. If you increase T_H/T_C for CO₂ in the simulator, the adiabatic expansion alone makes the volume tens of times larger, and an absurdly long cylinder is needed. Real machines use other cycles — Otto, Diesel, Rankine, Brayton — because they trade some efficiency for a manageable device size and power density.

Heat engine	Approximate T_H / T_C	Carnot limit	Actual
Steam locomotive	473 / 293 K	38 %	about 8 %
Steam turbine power generation (supercritical)	873 / 303 K	65 %	about 45 %
Gasoline engine	2500 / 800 K	68 %	about 30 %
Combined-cycle power generation	1600 / 300 K	81 %	about 62 %

If you want a higher efficiency, the only way is to raise T_H — and what sets T_H is usually not thermodynamics but how high a temperature the material can withstand. This is why the competition for power-plant efficiency has become a competition in heat-resistant alloys and cooling technology for turbine blades.

12. Summary

Question	Answer
Why four stages?	To expand the gas and get work, and then push it back at a lower pressure. That is why it is first cooled by an adiabatic expansion
Why is a cold reservoir needed?	In the compression, the gas cannot return to its original state unless it dumps the heat it has received somewhere
Why does the efficiency not depend on the gas?	Because the two adiabats are congruent, V₂/V₁ = V₃/V₄ and the ln factors cancel (Eq. (5))
Why can we call it an “upper limit”?	Because if an engine could exceed it, a device violating Clausius’s statement could be built (§9)
Where did entropy come from?	From the fact that Q/T returns to 0 after one cycle (Eq. (8))

In three lines ① A heat engine is a device that passes heat from high temperature to low temperature and receives the difference as work.
② The upper limit on the fraction that can be received is η = 1 − T_C/T_H, independent of both the working substance and the construction of the device.
③ In the course of that calculation, Q/T appears as a state function — and that is entropy.

The physical constants are the 2019 SI defined values (k_B = 1.380649×10⁻²³ J/K, h = 6.62607015×10⁻³⁴ J·s, N_A = 6.02214076×10²³ mol⁻¹); the spectroscopic constants are the values for the ground vibrational state (B₀ and the fundamental frequency ν₀). The adiabats do not assume a constant C_V; the condition of constant entropy is solved as it stands. The efficiencies of real engines are representative published values, rounded off.