The Second Law, Carnot Cycles, and Entropy

The first law permits any process that conserves energy, including many that are never observed: heat does not flow from cold to hot on its own, and a gas does not spontaneously compress into a corner of its container. The second law is the principle that selects, from among the energy-conserving processes, the ones that actually occur. Historically it grew from the engineering question of how efficiently heat can be turned into work; the answer, Carnot's bound, turns out to encode a new state function, the entropy, whose behavior is the deepest of the thermodynamic laws.

The Kelvin and Clausius statements

The second law admits two classic phrasings, each a denial that a certain too-good device can exist.

Both forbid a sole result: the impossibility is of a cyclic device that returns to its initial state and leaves nothing else changed. An engine can of course turn heat into work in one stroke — an isothermal expansion does exactly that — but not cyclically, because restoring the piston undoes the conversion.

The two statements are equivalent. Each can be built into a violation of the other by coupling the hypothetical device to a Carnot engine.

Proof. Suppose a Kelvin-violating engine $K$ exists, drawing heat $Q$ from a hot reservoir at $T_{h}$ and delivering work $W = Q$ with no other effect. Use that work to drive a refrigerator that moves heat $Q^{'}$ from a cold reservoir at $T_{c}$ to the hot one, depositing $Q^{'} + W$ . The net effect of the pair, per cycle, is heat $Q^{'}$ moved from cold to hot with no work input from outside — a Clausius violation. Conversely, suppose a Clausius-violating device moves heat $Q$ from cold to hot for free. Run an ordinary engine between the same reservoirs that draws $Q$ from the hot one, does work $W$ , and rejects $Q - W$ to the cold one. The hot reservoir is left unchanged (it received $Q$ and gave up $Q$ ), so the pair extracts $Q - (Q - W) = W$ from the cold reservoir alone and turns it entirely into work — a Kelvin violation. Each device implies the other, so the two prohibitions stand or fall together.

A heat engine draws

Q_{h}

from the hot reservoir, delivers work

W

, and must reject

Q_{c} = Q_{h} - W

to the cold reservoir; the second law forbids

Q_{c} = 0

, so some heat is always dumped.

The Carnot cycle

The Carnot cycle is the reversible engine that operates between two reservoirs using only isothermal and adiabatic strokes. For an ideal-gas working substance the four strokes are:

1 → 2, isothermal expansion at $T_{h}$ . The gas absorbs $Q_{h} = N k_{B} T_{h} ln (V_{2} / V_{1})$ from the hot reservoir and does work.
2 → 3, adiabatic expansion. No heat exchanged; the gas cools from $T_{h}$ to $T_{c}$ along $T V^{γ - 1} = const$ .
3 → 4, isothermal compression at $T_{c}$ . The gas rejects $Q_{c} = N k_{B} T_{c} ln (V_{3} / V_{4})$ to the cold reservoir.
4 → 1, adiabatic compression. The gas warms from $T_{c}$ back to $T_{h}$ .

The two adiabats link the volumes: $T_{h} V_{2}^{γ - 1} = T_{c} V_{3}^{γ - 1}$ and $T_{h} V_{1}^{γ - 1} = T_{c} V_{4}^{γ - 1}$ . Dividing gives $V_{2} / V_{1} = V_{3} / V_{4}$ , so the two logarithms are equal and the heats reduce to a clean ratio,

\frac{Q _{c}}{Q _{h}} = \frac{T _{c}}{T _{h}} .

The efficiency is the work delivered per unit heat drawn from the hot source. With $W = Q_{h} - Q_{c}$ by the first law over the cycle,

η = \frac{W}{Q _{h}} = 1 - \frac{Q _{c}}{Q _{h}} = 1 - \frac{T _{c}}{T _{h}} .

The Carnot cycle on the

P

–

V

plane: two isotherms at

T_{h}

and

T_{c}

joined by two adiabats; heat

Q_{h}

enters along the top isotherm and

Q_{c}

leaves along the bottom, and the enclosed area is the net work.

On a temperature–entropy diagram the cycle is a rectangle. The isothermal strokes are horizontal at $T_{h}$ and $T_{c}$ ; the adiabatic strokes are vertical because a reversible adiabat holds entropy fixed (shown below). Heat absorbed is $T d S$ integrated along a stroke, so $Q_{h} = T_{h} Δ S$ and $Q_{c} = T_{c} Δ S$ with the same entropy width $Δ S$ , and the net heat — equal to the net work — is the enclosed area $(T_{h} - T_{c}) Δ S$ .

On the

T

–

S

plane the Carnot cycle is a rectangle of height

T_{h} - T_{c}

and width

Δ S

; the enclosed area equals the net work, and the efficiency

1 - T_{c} / T_{h}

is the height ratio.

Carnot's theorem and thermodynamic temperature

The efficiency $1 - T_{c} / T_{h}$ was computed for an ideal gas, but its significance is that no engine can beat it and every reversible engine matches it, whatever the working substance.

Because the reversible efficiency is a universal function of the two temperatures alone, it can be used to define temperature. Setting $Q_{c} / Q_{h} = f (T_{c}, T_{h})$ and demanding consistency for engines in series forces $f (T_{c}, T_{h}) = Θ (T_{c}) /Θ (T_{h})$ for some universal function $Θ$ . The choice $Θ (T) = T$ defines the thermodynamic (Kelvin) temperature, and it coincides with the ideal-gas absolute temperature of the zeroth-law lesson because the ideal-gas Carnot calculation gave exactly $Q_{c} / Q_{h} = T_{c} / T_{h}$ . Thermodynamic temperature is thus substance-independent by construction.

The Clausius inequality and entropy

Carnot's ratio $Q_{c} / Q_{h} = T_{c} / T_{h}$ can be rewritten, restoring signs so that heat absorbed is positive, as $Q_{h} / T_{h} + Q_{c} / T_{c} = 0$ for a reversible cycle across two reservoirs. Any reversible cycle is a limit of many infinitesimal Carnot cycles, and summing the relation over them gives $\oint δ Q_{rev} / T = 0$ . For an irreversible cycle Carnot's theorem makes the engine less efficient, which turns the equality into an inequality.

For reversible cycles the loop integral vanishes; this is what it means for $δ Q_{rev} / T$ to be the exact differential of a state function.

Entropy is defined by a reversible path even when the actual process is irreversible: to find $Δ S$ for an irreversible change, connect the same endpoints by any reversible path and integrate there. Because $S$ is a state function, the answer is the true entropy change regardless of how the system actually got between the states.

A cycle made of an irreversible leg

a \to b

closed by a reversible return

b \to a

; the Clausius inequality forces

\int_{a}^{b} δ Q / T

on the irreversible leg to fall below the entropy change

S_{b} - S_{a}

Entropy production and the arrow of time

Applying the Clausius inequality to a cycle whose first leg is an irreversible process $a \to b$ and whose return $b \to a$ is reversible gives, since the reversible return contributes $S_{a} - S_{b}$ ,

a \int b \frac{δ Q}{T} + (S_{a} - S_{b}) \leq 0 ⟹ d S \geq \frac{δ Q}{T} .

The entropy change of a system meets or exceeds the heat it absorbs divided by temperature, with equality only for reversible heating. For a thermally isolated system $δ Q = 0$ , so

d S \geq 0,

the entropy principle: the entropy of an isolated system never decreases, and increases in any irreversible process. Equilibrium is the state of maximum entropy subject to the constraints, the fact the microcanonical ensemble later builds on.

The entropy created within the system during an irreversible process is the entropy production $δ S_{gen} = d S - δ Q / T \geq 0$ . It measures irreversibility directly: reversible processes generate no entropy, dissipative ones generate positive entropy. Because macroscopic processes generate entropy and their reverses would destroy it, the sign of $δ S_{gen}$ distinguishes past from future — the thermodynamic arrow of time. The first law is symmetric under time reversal; the second law is not, and it is the second law that a film run backward violates. The statistical origin of this asymmetry, the overwhelming multiplicity of high-entropy configurations, is the subject of the entropy lessons to come.

Summary

The Kelvin and Clausius statements forbid, respectively, a cyclic engine that fully converts single-reservoir heat to work and a cyclic device that moves heat cold-to-hot for free; the two are equivalent.
The Carnot cycle between $T_{h}$ and $T_{c}$ has efficiency $η = 1 - T_{c} / T_{h}$ , a rectangle of area $(T_{h} - T_{c}) Δ S$ on the $T$ – $S$ plane. Carnot's theorem makes this the maximum for any engine and defines thermodynamic temperature.
The Clausius inequality $\oint δ Q / T \leq 0$ constructs entropy as the state function with $d S = δ Q_{rev} / T$ ; entropy changes are computed along a reversible path even for irreversible processes.
For an isolated system $d S \geq 0$ : entropy is non-decreasing, equilibrium maximizes it, and the positive entropy production of irreversible processes sets the arrow of time.