Carnot cycle — full derivation, efficiency = 1 − T_C - T_H: Visual walkthrough

The pictures we live on: the P–V plane

Before any cycle, we need a stage. Imagine a gas trapped in a cylinder under a piston.

Push the piston down → the gas takes up less room → its volume $V$ gets smaller.
Squeeze a gas into less room → it pushes back harder → its pressure $P$ (force per unit area on the piston) goes up.

So the state of the gas is one dot on a flat map: horizontal axis = ==volume $V$ == (how much room), vertical axis = ==pressure $P$ == (how hard it pushes). As the gas changes, the dot slides along a curve. The area swept under that curve is the mechanical work. Why area? Because work is force times the distance the piston moves, and on this map "force" lives on the $P$ axis while "distance moved" lives on the $V$ axis — multiply them and you get area.

Figure — Carnot cycle — full derivation, efficiency = 1 − T_C - T_H

Step 1 — The ideal-gas law gives us the isothermal curve

WHAT. For one mole of ideal gas we are handed one fact: $P V = R T$ , where $R$ is a fixed number (the gas constant) and $T$ is the absolute temperature in kelvin (see Absolute temperature scale).

WHY. We need to know the shape of an isothermal move on our map so we can measure the area (the work). Hold $T$ fixed and solve for pressure: $P = pressure as room shrinks \frac{R T}{V}$

Read this term by term: $R T$ on top is a constant while $T$ is held fixed; $V$ on the bottom means pressure and volume are inversely linked. Double the room, halve the push. That is a hyperbola — a curve that swoops down and to the right.

PICTURE. A higher temperature gives a hyperbola sitting higher up (bigger $R T$ on top). So the hot curve ( $T_{H}$ ) floats above the cold curve ( $T_{C}$ ).

Step 2 — Isothermal expansion at $T_{H}$ : grabbing heat $Q_{H}$

WHAT. Sit the gas on the hot reservoir at $T_{H}$ and let it slowly expand from volume $V_{1}$ to a larger $V_{2}$ , staying on the hot hyperbola.

WHY. Because temperature is held fixed and (for an ideal gas) internal energy depends only on temperature, the gas's stored energy $U$ does not change: $Δ U = 0$ . By the First law of thermodynamics, $Δ U = Q - W$ , so with $Δ U = 0$ every joule of heat that comes in leaves as work: $Q_{H} = W_{1 \to 2} .$

The work is the area under the hot curve from $V_{1}$ to $V_{2}$ . To get it we add up $P d V$ over every tiny sliver of volume — that is what the integral sign means, "sum of thin strips": $W_{1 \to 2} = \int_{V_{1}}^{V_{2}} P d V = \int_{V_{1}}^{V_{2}} \frac{R T _{H}}{V} d V = R T_{H} ln \frac{V _{2}}{V _{1}} .$

Term by term: $\int \frac{1}{V} d V$ is exactly the operation that produces a logarithm — that is why $ln$ appears (it is the one function whose slope is $1/ V$ ). Inside it, $\frac{V _{2}}{V _{1}} > 1$ because we expanded, so $ln$ of it is positive → heat genuinely flows in.

$Q_{H} = R T_{H} ln \frac{V _{2}}{V _{1}} (> 0)$

Step 3 — Adiabatic expansion: coasting from $T_{H}$ down to $T_{C}$

WHAT. Wrap the cylinder in a perfect blanket ( $Q = 0$ ) and let it keep expanding, $V_{2} \to V_{3}$ . With no heat coming in, the gas pays for its outward push out of its own energy, so it cools from $T_{H}$ to $T_{C}$ .

WHY. We must move the gas to the cold temperature without touching any reservoir in between — touching a mid-temperature body would let heat cross a temperature gap, which is wasteful and irreversible. Blanketing keeps it reversible. Now we must actually derive the shape of this "blanketed" move — we cannot just quote it.

Read the result: as $V$ grows, $T$ must shrink to keep the product fixed — that is exactly the cooling from $T_{H}$ to $T_{C}$ you feel in a blanketed expansion.

PICTURE. The adiabatic curve is steeper than either hyperbola — it dives across from the high hot curve down to the low cold curve.

Applying the law to this stroke ( $T_{H}$ at $V_{2}$ down to $T_{C}$ at $V_{3}$ ) gives our first bridge equation: $T_H\,V_2^{\gamma-1} = T_C\,V_3^{\gamma-1}. \tag{A}$

Step 4 — Isothermal compression at $T_{C}$ : dumping heat $Q_{C}$

WHAT. Sit the gas on the cold reservoir at $T_{C}$ and slowly squeeze it, $V_{3} \to V_{4}$ (with $V_{4} < V_{3}$ ), staying on the cold hyperbola.

WHY. Same isothermal logic as Step 2, but reversed: we do work on the gas, and since $Δ U = 0$ again, that energy must leave as heat into the cold reservoir. We define $Q_{C}$ as the positive amount of heat rejected: $Q_{C} = R T_{C} ln \frac{V _{3}}{V _{4}} .$

Term by term: $\frac{V _{3}}{V _{4}} > 1$ (we compressed, so we flipped the fraction to keep it above 1), hence $ln$ is positive and $Q_{C} > 0$ — heat really does leave. This is where the Second law of thermodynamics bites: you cannot skip this step. Some heat must be dumped.

$Q_{C} = R T_{C} ln \frac{V _{3}}{V _{4}} (> 0)$

Step 5 — Adiabatic compression: climbing back to $T_{H}$

WHAT. Blanket the gas again and squeeze $V_{4} \to V_{1}$ , closing the loop. With no heat exchange, the compression work heats it back up from $T_{C}$ to $T_{H}$ .

WHY. We need to return to the exact starting dot to make it a cycle (repeatable forever). The same adiabatic law $T V^{γ - 1} = const$ (derived in Step 3) applied to this stroke ( $T_{C}$ at $V_{4}$ up to $T_{H}$ at $V_{1}$ ) gives our second bridge equation: $T_H\,V_1^{\gamma-1} = T_C\,V_4^{\gamma-1}. \tag{B}$

PICTURE. The loop is now closed: two hyperbola arcs (hot top, cold bottom) joined by two steep adiabatic arcs. The enclosed area is the net work per cycle.

Step 6 — The magic cancellation: why the volumes disappear

WHAT. Divide bridge equation (A) by bridge equation (B): $\frac{T _{H} V _{2}^{γ - 1}}{T _{H} V _{1}^{γ - 1}} = \frac{T _{C} V _{3}^{γ - 1}}{T _{C} V _{4}^{γ - 1}} .$

WHY. The $T_{H}$ 's cancel on the left, the $T_{C}$ 's cancel on the right, and taking the $(γ - 1)$ -th root of both sides collapses everything to: $\frac{V _{2}}{V _{1}} = \frac{V _{3}}{V _{4}}$

This is the whole secret. The volume ratio hiding inside $Q_{H}$ is exactly equal to the one inside $Q_{C}$ . Picture it: the two adiabats are parallel "slides," so they carry the same stretch factor from the hot curve to the cold curve on both the left and right sides of the loop.

Step 7 — Assemble the efficiency

WHAT. Efficiency $η$ asks: of the heat you paid for, what fraction became work? Over one full cycle the gas returns to start, so $Δ U = 0$ and net work equals net heat: $η = \frac{W _{net}}{Q _{H}} = \frac{Q _{H} - Q _{C}}{Q _{H}} = 1 - \frac{Q _{C}}{Q _{H}} .$

WHY. Now substitute the two boxed heats and use the cancellation from Step 6: $\frac{Q _{C}}{Q _{H}} = \frac{R T _{C} l n ( V _{3} / V _{4} )}{R T _{H} l n ( V _{2} / V _{1} )} = \frac{T _{C}}{T _{H}} \cdot = 1 by Step 6 \frac{ln ( V _{3} / V _{4} )}{ln ( V _{2} / V _{1} )} = \frac{T _{C}}{T _{H}} .$

The $R$ 's cancel; the two logs are equal so their ratio is $1$ ; only the temperature ratio survives. This is why the answer does not care what gas you used.

Step 8 — Where the reversible identity $Q_{H} / T_{H} = Q_{C} / T_{C}$ comes from

WHAT. We never assumed this — it falls out of the two boxed heats we already earned. Divide $Q_{H}$ by $T_{H}$ and $Q_{C}$ by $T_{C}$ :

\qquad \frac{Q_C}{T_C} = \frac{R\,T_C\ln(V_3/V_4)}{T_C} = R\ln\frac{V_3}{V_4}.$$ **WHY.** In each case the temperature on the bottom **cancels** the temperature inside the heat, leaving only the volume-ratio log. But Step 6 proved $V_2/V_1 = V_3/V_4$, so these two right-hand sides are *identical*: $$\boxed{\dfrac{Q_H}{T_H} = \dfrac{Q_C}{T_C}}$$ **WHAT IT MEANS.** Rearranged, the quantity $Q/T$ that you *gain* on the hot stroke is *exactly returned* on the cold stroke — nothing is lost around the loop. Writing this as a loop-sum over the whole reversible cycle, $$\oint \frac{dQ_{\text{rev}}}{T} = 0,$$ which says there is a hidden bookkeeping quantity that comes back to its starting value every cycle. That quantity is ==[[Entropy]]==, and this identity is its birth certificate. --- ## Step 9 — The edge and degenerate cases (never left uncovered) **WHAT / WHY / PICTURE for every corner:** - **$T_C = T_H$ (no temperature difference).** Then $\eta = 1 - 1 = 0$. The hot and cold hyperbolas merge into one line; the loop has **zero enclosed area** — no work at all. You cannot run an engine between two things at the same temperature. - **$T_C \to 0$ (perfectly cold sink).** Then $\eta \to 1 - 0 = 1$. But absolute zero is unreachable, so $\eta = 1$ is a limit you approach, never touch. The cold hyperbola sinks toward the $V$-axis. - **$T_C > T_H$ (mislabelled reservoirs).** Then $\eta < 0$: heat spontaneously flows the wrong way and the "engine" is really a refrigerator being driven — see [[Heat engines and refrigerators]]. The formula honestly warns you with a negative sign. - **Zero-width strokes ($V_1 = V_2$).** No expansion means $\ln 1 = 0$, so $Q_H = 0$: nothing absorbed, nothing done. A degenerate flat loop. ![[deepdives/dd-physics-1.7.21-d2-s08.png]] > [!mistake] The most common slip > Using Celsius. The formula is a **ratio** $T_C/T_H$, and ratios explode if you shift the zero point. $300\,\text{K}/600\,\text{K} = 0.5$ but $27\,°\text{C}/327\,°\text{C} = 0.083$ — totally different. Always kelvin. --- ## The one-picture summary Everything above compressed into a single annotated loop: heat $Q_H$ in along the hot top, heat $Q_C$ out along the cold bottom, the two adiabatic slides that force $V_2/V_1 = V_3/V_4$, and the enclosed area that is your net work. ![[deepdives/dd-physics-1.7.21-d2-s09.png]] > [!recall]- Feynman retelling of the whole walkthrough > We drew a map where left–right is *how much room the gas has* and up–down is *how hard it pushes*. On the hot stove the gas swells and grabs heat — that heat becomes pushing, and the amount is the area under the top curve (a logarithm, because we summed thin strips of $1/V$). Then we blanket it so no heat sneaks in and let it coast, cooling from hot to cold as it keeps pushing. On the ice block we squeeze it slowly and it dumps its leftover heat — a second logarithm. Finally we blanket and squeeze it back to the exact start. Now here's the trick: the two blanketed slides are parallel, so the stretch factor on the left equals the stretch factor on the right — the two logarithms are *identical* and cancel. All that's left of the whole story is the ratio of the two temperatures, which is why $\eta = 1 - T_C/T_H$ and why it doesn't matter what gas you used. And you can never reach 100 %, because that would need an infinitely cold ice block. > [!recall]- Quick self-check > Why do the volume logs cancel? ::: The two adiabats force $V_2/V_1 = V_3/V_4$, so $\ln(V_3/V_4)/\ln(V_2/V_1) = 1$. > What survives the cancellation? ::: Only $T_C/T_H$ — hence the efficiency ignores the working substance. > What does the enclosed loop area represent? ::: The net work done per cycle. > Why must Step 4 (rejecting $Q_C$) exist? ::: The second law forbids turning all heat into work; some must be dumped. > Where does $Q_H/T_H = Q_C/T_C$ come from? ::: Divide each heat by its own temperature; the $T$'s cancel and Step 6 makes the two logs equal. > Why is $\gamma > 1$? ::: Because $\gamma = C_P/C_V$ and $C_P = C_V + R > C_V$ (expanding while heating costs extra work). --- > [!mnemonic] The shape, in one breath > **Top curve = heat IN. Bottom curve = heat OUT. Steep sides = free coasting. Area = work. Answer = one minus cold-over-hot.**