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

This page builds every symbol the parent note uses, starting from absolute zero of assumed knowledge. Read it top to bottom: each idea is the ground the next one stands on.

1. Pressure $P$ , Volume $V$ , Temperature $T$ — the three dials of a gas

The picture. Imagine a syringe with the nozzle sealed. Push the plunger in: $V$ goes down, the trapped air pushes back harder ( $P$ goes up). That pushback is pressure — a real force you feel in your thumb.

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

Why the topic needs all three. The Carnot cycle is a journey of the gas through different $(P, V, T)$ states. To draw and reason about that journey we must be able to name where the gas is — and $P$ , $V$ , $T$ are its coordinates. When the gas visits four corner states in the cycle, we label their volumes $V_{1}, V_{2}, V_{3}, V_{4}$ — just names for "the volume at corner 1", and so on.

2. Absolute temperature $T$ — why kelvin, not Celsius

The picture. Think of a thermometer where the bottom of the tube is nailed to absolute zero, not to "ice melts". Water freezes at $273.15 K$ , boils at $373.15 K$ .

Why the topic needs it — this is not optional. The efficiency formula uses a ratio $T_{C} / T_{H}$ . A ratio only makes sense if zero means nothing. In Celsius, $0° C$ is an arbitrary point (melting ice), so $\frac{20° C}{40° C}$ is meaningless nonsense. In kelvin, $\frac{300 K}{600 K} = \frac{1}{2}$ genuinely means "half as much jiggle-energy". See Absolute temperature scale.

3. The ideal gas law $P V = R T$ — the rule linking the three dials

The picture. If you hold $T$ fixed and let $V$ grow, then $P = R T / V$ must shrink — the curve $P = const / V$ is a smooth downward swoop called a hyperbola. That swoop is an Isothermal process line.

Why the topic needs it. Every integral in the derivation replaces $P$ with $R T / V$ using this law. It is the bridge that lets us compute work purely from temperatures and volumes.

4. Two sizes of change: $Δ$ (finite) vs $d$ (infinitesimal)

5. Internal energy $U$ — and exactly how it depends on $T$

The picture. Picture $U$ as the total kinetic energy of a swarm of bouncing balls. Warm them up (raise $T$ ) and they bounce faster ( $U$ rises). Move the walls without heating (change $V$ alone) and the swarm's total energy is untouched.

Why the topic needs it. Because $U = C_{V} T$ , any process that returns to the same $T$ has $Δ U = 0$ . Around one full cycle the gas returns to its start, so $Δ U = 0$ over the cycle — that single zero collapses the first law into "net work = net heat" (§7).

6. Work $W$ and heat $Q$ — two ways energy moves

The picture — why area = work. When the gas expands by one sliver $d V$ (§4) at pressure $P$ , it does a sliver of work $P d V$ (force × distance, packaged neatly). Stack up all those thin slivers with the integral $\int P d V$ and you get the area under the $P$ – $V$ curve.

See First law of thermodynamics for the accounting, and Heat engines and refrigerators for what "engine" means.

7. The first law — energy bookkeeping

The picture. Think of a bank account. $Q$ is money deposited, $W$ is money withdrawn, $Δ U$ is the change in your balance. Nothing appears or vanishes.

Why the topic needs it. Two special cases do all the heavy lifting:

Isothermal ( $Δ U = 0$ ): all heat becomes work, $Q = W$ .
Adiabatic ( $Q = 0$ ): work comes purely from internal energy, $W = - Δ U$ .

8. $γ$ and $C_{V}$ — and why $T V^{γ - 1}$ is constant on an adiabat

The picture. Heating gas in a sealed box (constant $V$ ): every joule becomes jiggle. Heating it in a box with a free lid (constant $P$ ): some joules lift the lid instead — so you need more heat for the same temperature rise. Hence $C_{P} > C_{V}$ and $γ > 1$ .

Why the topic needs it. Applied to both adiabatic steps, this relation forces $V_{2} / V_{1} = V_{3} / V_{4}$ — the magic cancellation that leaves efficiency depending on temperature alone.

See Isothermal process and Adiabatic process for the two move-types in full.

9. Reversibility, $d Q_{rev}$ , and the loop integral $\oint$

Why the topic needs it. Reversibility is why Carnot is the best possible — it's the theme of the Second law of thermodynamics. And $\oint d Q_{rev} / T = 0$ is the discovery that a new quantity, Entropy, is conserved around the loop.

Prerequisite map

Equipment checklist

Test yourself — you are ready for the parent derivation when you can answer each without peeking.

What is the difference between

Δ V

and

d V

Δ V

is a finite, measurable change you can put a ruler on;

d V

is an infinitely small sliver you add up with an integral.

Why must temperatures be in kelvin, not Celsius, for

T_{C} / T_{H}

Because it is a ratio, and ratios only mean something when zero means "no jiggle at all" — only kelvin has that true zero.

What does the area under a

P

–

V

curve represent?

The work done by the gas,

\int P d V

Why does

\int d V / V

give a logarithm?

ln

is the function whose slope is

1/ V

, so it is the exact antiderivative of

1/ V

For an ideal gas, what is

U

and why does it depend only on

T

U = C_{V} T

; particles don't attract, so changing

V

alone costs no energy — only

T

moves

U

. Hence same

T

⇒

Δ U = 0

What do

V_{1}, V_{2}

and

Q_{H}

mean?

V_{1}, V_{2}

are the volumes at cycle corners 1 and 2;

Q_{H}

is the heat absorbed on the hot isotherm as volume grows from

V_{1}

V_{2}

State the first law in words.

Change in internal energy = heat in minus work out,

Δ U = Q - W

(energy conservation).

Where does

T V^{γ - 1} = const

come from?

From

d U = - P d V

on an adiabat with

U = C_{V} T

and

P = R T / V

, integrated using

R = C_{P} - C_{V}

Why is

γ = C_{P} / C_{V} > 1

At constant pressure some heat lifts the piston, so

C_{P} > C_{V}

, making the ratio exceed 1.

What is

d Q_{rev}

and what does

\oint d Q_{rev} / T = 0

hint at?

A sliver of heat exchanged reversibly (no temperature gap); the loop sum of

d Q_{rev} / T

vanishing hints at a conserved quantity — entropy.

What does "reversible" require physically?

Infinitely slow changes and no finite temperature gaps, so nothing is wasted.

Recall One-breath summary

$P, V, T$ locate the gas; kelvin makes ratios meaningful; $P V = R T$ links the three; $Δ$ is a finite change while $d$ is an infinitesimal sliver we integrate; area under $P$ – $V$ is work (giving logs on isotherms); $U = C_{V} T$ depends on $T$ alone; the first law is energy bookkeeping; $γ$ shapes the adiabat via $T V^{γ - 1} = const$ ; reversibility makes Carnot the champion and births entropy.