The Carnot Cycle

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In the second quarter of the Nineteenth Century Sadi Carnot and Émile Clapeyron established the foundations upon which other physicists built the modern science of thermodynamics in the form that we call classical thermodynamics to distinguish it from the statistical thermodynamics of Maxwell and Boltzmann. Carnot wanted to devise a way to calculate the efficiency of a steam engine and to that end he conceived the idea of an ideal heat engine, an abstraction that he could use as an aid to his imagination in working out the mathematical description of what we now call the Carnot Cycle, the cycle that started modern thermodynamics. Nicolas Léonard Sadi Carnot (1796 Jun 01 B 1832 Aug 24) originally described the cycle in his book AReflections on the Motive Power of Fire@ (1824). In 1834 Benôit Paul Émile Clapeyron (1799 Feb 26 B 1864 Jan 28) updated Carnot=s theory, in his book ADriving Force of Heat@, by replacing the caloric theory that Carnot used with the modern concept of heat and he devised the diagram that we now use.

The Carnot Cycle consists of four steps, two isothermal processes and two adiabatic (isentropic) processes. The cycle gives us an abstract view of an engine that converts heat into work and enables us to calculate the maximum efficiency with which a heat engine can carry out such a conversion. The engine consists of an expandable volume (such as a cylinder fitted with a piston) filled with an ideal gas. We can put the engine into or out of contact with two heat reservoirs, a hot reservoir (kept at absolute temperature Th) and a cold reservoir (kept at absolute temperature Tc).

At the first step the engine touches the hot reservoir and the gas in it expands. As the gas does work upon the engine it draws heat from the reservoir to keep the its temperature unchanged, the heat compensating the gas for the energy lost in doing work on the engine. We call this step isothermal because it occurs while the temperature of the gas remains unchanged.

At the second step the engine breaks contact with the hot reservoir while the gas continues to expand and do work upon it. In this step the temperature of the gas goes down as the expansion of the gas continues to convert heat into work. We call this step adiabatic because no heat flows into or out of the engine during its occurrence. We also call it isentropic because the entropy of the system does not change.

At the third step the engine touches the cold reservoir and the engine does work upon it to compress it. As the engine does work on the gas heat flows into the cold reservoir to remove the energy that the work puts into the gas. Again we have an isothermal process.

At the fourth and final step the engine breaks contact with the cold reservoir while it continues to compress the gas to its original volume. Now adiabatic compression heats the gas back to its original temperature and the engine stands ready to repeat the cycle.

We can calculate the efficiency of the Carnot Cycle with ease. We need only invoke the First Law of Thermodynamics and infer that the net work done by the engine upon whatever application we have attached to it must equal the difference between the amount of heat the engine absorbs from the hot reservoir and the amount of heat it puts into the cold reservoir B W=Qh-Qc. We define the efficiency of the engine as equaling the ratio of the work harnessed to the amount of energy passing into the engine; that is,

(Eq=n 1)

But of course we can=t measure the amounts of heat entering and leaving the engine. We want to calculate the efficiency as a function of some quantity that we can measure. We start by noting that for an ideal gas consisting of N particles occupying a volume V we have as true to Reality the statement that

(Eq=n 2)

which we call the ideal gas law. In an isothermal expansion or compression, that equation tells us, we can calculate the pressure as

(Eq=n 3)

which means that the work done in taking the gas from volume V1 to volume V2 equals

(Eq=n 4)

That work equals the amount of heat that the engine absorbs from the hot reservoir or discharges into the cold reservoir, which equality is necessitated by Equation A-2 telling us that the total energy in the gas does not change during an isothermal process. If we take the change from V1 to V2 as the expansion and the change from V3 to V4 as the compression in the Carnot Cycle, then we have the efficiency of the cycle as

(Eq=n 5)

What can we say about the two volume ratios? We know that the adiabatic expansion and compression steps both involve the same amount of work because they involve the same change in the temperature of the gas: if that statement did not stand true to Reality, the gas would have to do more work on the engine or absorb more heat in one of those steps to make up the difference.

One of the assumptions that we bring into our premises when we conceive the idea of an ideal gas tells us that the particles in the gas interact weakly, if at all. That means that we can take a gas at pressure p and volume V and separate the particles into M identical sets with volume V and partial pressure pM=p/M. If we put those sets adjacent to each other, they occupy a volume VM=MV, so the total energy pV does not change (that is, we have as true to Reality the statement that pV=pMVM). Assume that we expand or compress the original gas and the separated gas by the same ratio R. Because in essence nothing has changed within the gas (the particles comprising it have no way of Aknowing@ whether they are in the original gas or the separated gas), the energy relations within the gases must change in the same way. In an adiabatic expansion or compression both gases do the same amount of work on their respective mechanical systems. Necessarily the temperatures of the gases changes in the same way; if both gases start with temperature T1, then both must end up with temperature T2. We can reverse that statement to infer that the change in the volume of one or the other gases conforms to the same function, f(T1,T2), a function independent of the pressure or the volume of the gas. That fact tells us that our volumes V1 and V2 must expand in the same ratio as their temperatures go from Th to Tc and that, in turn, necessitates that

(Eq=n 6)

We know that statement stands true to Reality because we know that we could have simply taken our gas at volume V1 and used our separation process to convert it into the gas at volume V2 without making it do work or absorb heat. Substituting Equation 6 into Equation 5 then gives us

(Eq=n 7)

the standard equation describing the efficiency of an ideal heat engine in terms of its working temperatures.

If we now compare Equation 1 with Equation 7, we find that in the Carnot Cycle we have

(Eq=n 8)

We can also write that statement, more suggestively, as

(Eq=n 9)

That statement tells us that the act of absorbing heat Qh at temperature Th confers some property Sh on the system and that the act of discarding heat Qc at temperature Tc removes some of that property Sc from the system such that Sh-Sc=0. Following Rudolf Clausius, we call that property entropy. Mathematically we describe entropy through the differential

(Eq=n 10)

in which the indexed dee represents an inexact differential (usually represented by a bar-dee), and we express Equation 9 as

(Eq=n 11)

That latter equation expresses Clausius Theorem, the statement that the net change of entropy in a system taken through a reversible thermodynamic cycle equals zero. Note that the definition of entropy does not include work done: any process that does mechanical work is perfectly reversible and, thus, does not change the entropy of the system doing the work.

In general the efficiency of a real heat engine will not match that of a perfect Carnot engine. For a given amount of work done by the engine we must discard a greater proportion of the heat drawn from the hot reservoir into the cold reservoir. The engine thus draws an extra ΔQ of heat from the hot reservoir and discards all of it into the cold reservoir, having converted none of it into work. This leads to a net change in the system= s entropy of

(Eq=n 12)

Thus we must rewrite Equation 11 as Clausius= Inequality,

(Eq=n 13)

In that equation the equality applies only to a perfect Carnot engine. So now we know that a heat engine, of any kind, draws heat with its associated entropy from a hot reservoir and discards part of that heat with its associated entropy into a cold reservoir, thereby imposing a net increase on the entropy of its environment. Thus we get the basic statement of the second law of thermodynamics, the law of entropy.

We can also make our mechanism run the Carnot Cycle backwards, thereby giving us the ideal heat pump (or refrigerator). In this case we must do work upon the gas to raise the heat it absorbs from the cold reservoir to the temperature of the hot reservoir. Of course, in a real heat pump operating between two given temperatures we must do more net work upon the gas than we calculate for the ideal heat pump, so again we put more entropy into the hot reservoir than we take out of the cold reservoir. Whichever way we run the thermodynamic cycle, we will always increase the entropy of our environment.

For the Carnot Cycle itself, we can calculate the entropy change wrought by each phase. Bearing in mind the fact that work done during the isothermal expansion or compression of the gas equals the amount of heat drawn into or pushed out of the gas, we can compare Equation 4 to Equation 10 and write for the expansion phase, for example,

(Eq=n 14)

as the amount of entropy drawn into the gas with the heat coming from the hot reservoir. The appearance of the natural logarithm demonstrates the relation to Boltzmann=s theorem,

(Eq=n 15)

which relates the entropy of a gas to the number of microstates of the particles in the gas that correspond to a given macrostate. So the Carnot Cycle gives us a touchstone for comparing statistical thermodynamics with classical thermodynamics, thereby giving us the means to ensure that the theories will give us a consistent description of thermal Reality.

Note also that in the ideal refrigerator we must do some work upon the gas to move heat from the cold reservoir to the hot reservoir. If we have any temperature difference at all between the hot and cold reservoirs, we must do work to move heat from the cold reservoir to the hot reservoir. In this fact we see a reflection of Clausius= statement of the law of entropy: AHeat will not, of itself, flow from a cold body to a hot body.@

The Carnot Cycle thus gives us something like the Platonic Ideal of a thermodynamic cycle. As a perfectly reversible process it gives us a connection between mechanics and heat. As such it gives us a touchstone against which we can compare other thermodynamic processes.

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