Carnot Engine

The Carnot Engine: The Ultimate Heat Engine

The central question of practical heat-engine design is this: given a hot reservoir at temperature $T_1$ and a cold reservoir at $T_2$, what is the maximum possible efficiency, and what cycle of processes achieves it? In 1824, the French engineer Sadi Carnot answered both questions correctly, remarkably, before the First Law of Thermodynamics was even fully established.

The answer begins with a crucial insight. Any real engine has irreversible steps — friction, turbulence, heat flow across a finite temperature difference — and these dissipate useful work, lowering efficiency. The ideal engine, therefore, must be a reversible engine. A process is reversible only if it is both quasi-static and non-dissipative.

Now, a process cannot be quasi-static if there is a finite temperature difference between the system and the reservoir. This forces a specific structure on the reversible engine. Heat must be absorbed from the hot reservoir isothermally (at $T_1$) and rejected to the cold reservoir isothermally (at $T_2$). That gives us two steps. But to complete a cycle, the working substance must be taken from $T_1$ down to $T_2$, and then back up from $T_2$ to $T_1$. What reversible processes can do this without involving any heat reservoirs? Only reversible adiabatic processes — processes with no heat exchange at all. Any other path (like an isochoric one) would require a continuous series of reservoirs between $T_2$ and $T_1$ to keep the process quasi-static, which violates the condition that the engine operates between only two temperatures.

Thus, a reversible heat engine operating between two fixed temperatures must consist of exactly four steps: two isothermal processes and two adiabatic processes. This sequence is called the Carnot cycle, and the engine itself is a Carnot engine.

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The Carnot Cycle with an Ideal Gas

We take an ideal gas as the working substance. The cycle, shown in Fig. 11.9 of the textbook, consists of four reversible steps. Let's go through each one, tracking the state variables $(P, V, T)$.

Step 1 → 2: Isothermal Expansion at $T_1$

The gas is in contact with the hot reservoir at $T_1$. It expands isothermally from $(P_1, V_1, T_1)$ to $(P_2, V_2, T_1)$. Since the internal energy of an ideal gas depends only on temperature, $\Delta U = 0$ for an isothermal process. From the First Law, $Q = \Delta U + W$, so the heat absorbed $Q_1$ equals the work done by the gas $W_{1 \to 2}$.

For an isothermal reversible expansion of $\mu$ moles of an ideal gas:

Step 2 → 3: Adiabatic Expansion from $T_1$ to $T_2$

The gas is now thermally insulated. It expands adiabatically from $(P_2, V_2, T_1)$ to $(P_3, V_3, T_2)$. No heat is exchanged ($Q = 0$). The gas does work at the expense of its internal energy, so its temperature drops from $T_1$ to $T_2$.

The work done by the gas in a reversible adiabatic process is given by:

where $\gamma = C_P / C_V$.

Step 3 → 4: Isothermal Compression at $T_2$

The gas is placed in contact with the cold reservoir at $T_2$. It is compressed isothermally from $(P_3, V_3, T_2)$ to $(P_4, V_4, T_2)$. Again, $\Delta U = 0$. The work done on the gas is $W_{3 \to 4}$, and the heat $Q_2$ is released by the gas to the cold reservoir. Since the gas is being compressed ($V_4 < V_3$), the work done by the gas is negative. The textbook defines $W_{3 \to 4}$ as the work done on the gas by the environment, but in the efficiency calculation, it is subtracted as a negative contribution. The magnitude is:

Step 4 → 1: Adiabatic Compression from $T_2$ to $T_1$

The gas is insulated again. It is compressed adiabatically from $(P_4, V_4, T_2)$ back to the initial state $(P_1, V_1, T_1)$. Work is done on the gas, raising its temperature from $T_2$ to $T_1$. The work done on the gas is:

Notice that $W_{4 \to 1}$ has the same magnitude as $W_{2 \to 3}$, but it is work done on the system, so it contributes negatively to the net work done by the system.

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Net Work and Efficiency of the Carnot Engine

The total work done by the gas in one complete cycle is the sum of the work in each step, taking signs into account:

Substituting the expressions:

The efficiency $\eta$ of any heat engine is defined as the ratio of net work output to heat input:

Substituting $Q_1$ and $Q_2$ from Eqs. (11.18) and (11.20):

This expression still contains the volumes. To simplify it, we need a relation between the volume ratios. This comes from the adiabatic steps.

Relating the Volume Ratios

For the adiabatic expansion (step 2→3), the relation $TV^{\gamma-1} = \text{constant}$ holds:

For the adiabatic compression (step 4→1):

From Eqs. (11.24) and (11.25), we get the crucial result:

Substituting Eq. (11.26) into Eq. (11.23), the logarithmic terms cancel, giving the famous result:

This is the maximum possible efficiency for any heat engine operating between two reservoirs at temperatures $T_1$ and $T_2$. It depends only on the two temperatures, not on the working substance.

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Carnot's Theorem: The Proof

The result above leads to two profound statements, collectively known as Carnot's theorem:

1. No engine operating between two given temperatures can have an efficiency greater than that of a Carnot engine operating between the same two temperatures.
2. The efficiency of a Carnot engine is independent of the nature of the working substance.

The textbook provides a proof of the first statement using a contradiction argument based on the Second Law of Thermodynamics.

›Proof

Proof of Carnot's Theorem (Part a)

Imagine a reversible Carnot engine $R$ and an irreversible engine $I$ operating between the same hot reservoir (source at $T_1$) and cold reservoir (sink at $T_2$). Let $I$ act as a heat engine and $R$ act as a refrigerator. The arrangement is shown in Fig. 11.10.

• Engine $I$ absorbs heat $Q_1$ from the source, does work $W'$, and rejects heat $Q_1 - W'$ to the sink.
• Refrigerator $R$ takes heat $Q_2$ from the sink, requires work $W = Q_1 - Q_2$ to be done on it, and returns heat $Q_1$ to the source.

Now, suppose for contradiction that the irreversible engine $I$ is more efficient than the reversible engine $R$. That is, $\eta_I > \eta_R$. For the same heat input $Q_1$, this means $W' > W$.

Consider the combined system of $I$ and $R$ as a single device. What is the net effect after one cycle?

• Net heat extracted from the cold sink: The sink gives $Q_2$ to $R$ and receives $Q_1 - W'$ from $I$. The net heat taken from the sink is $Q_2 - (Q_1 - W')$.
• Net heat delivered to the hot source: The source gives $Q_1$ to $I$ and receives $Q_1$ from $R$. The net heat delivered to the source is zero.
• Net work output: The combined system does work $W'$ (from $I$) and has work $W$ done on it (by $R$). The net work output is $W' - W$. …