Carnot Cycle: Derivation, Stages & Properties

Although physics is used to describe complex, real-world systems, many of the problems you'll encounter in real-life were first solved using approximations and simplifications. This is one of the greatest skills you'll learn as a physicist: The ability to drill down to the most crucial components of a problem and leave all of the messy detail for afterward, when you already have a good grasp of how a system works.

So while you might think of a physicist trying to understand a thermodynamic process as going through a long struggle over some even longer equations, in reality, the real-life physicist is more likely to be looking at the problem using an idealization like the ​Carnot cycle​.

The Carnot cycle is a special heat engine cycle that ignores the complexities that come from the second law of thermodynamics – the tendency of all closed systems to increase in entropy over time – and simply assumes maximum efficiency for the system. This allows physicists to treat the thermodynamic process as a ​reversible cycle​, making things much easier to calculate and understand conceptually, before taking the step up to real systems, and the usually irreversible processes that govern them.

Learning how to work with the Carnot cycle involves learning about the nature of reversible processes like adiabatic and isothermal processes and about the stages of the Carnot cycle.

A heat engine is a type of thermodynamic system that turns heat energy into mechanical energy, and most engines in real life, including car engines, are some type of heat engine.

Since the ​first law​ of thermodynamics tells you that energy isn't created, just converted from one form into another (since it states the conservation of energy), the heat engine is one way of extracting usable energy from a form of energy that's easier to generate, in this case, heat. In simple terms, the heating of a substance causes it to expand, and the energy from this expansion is harnessed into some form of mechanical energy that can go on to do other work.

The basic theoretical parts of a heat engine include a heat bath or high-temperature heat source, a low-temperature cold reservoir and the engine itself, which contains a gas. The heat bath or heat source transfers heat energy to the gas, which leads to expansion that drives a piston. This expansion is the engine doing ​work​ on the environment, and in the process, it releases heat energy into the cold reservoir, which returns the system to its initial state.

There can be many different thermodynamic processes in a heat engine cycle, but the idealized Carnot cycle – named after the "father of thermodynamics" Nicolas Leonard Sadi Carnot – involves ​reversible processes​. Real-world processes generally aren't reversible because any change in a system tends to increase entropy, but if the processes are theoretically assumed to be perfect, then this complication can be ignored.

A reversible process is one that can essentially be run "backward in time" to return the system to its initial state without violating the second law of thermodynamics (or any other law of physics).

An isothermal process is an example of a reversible process that happens at a constant temperature. This isn't possible in real life because in order to maintain thermal equilibrium with the environment, it would take an infinite amount of time to complete the process. In practice, you could approximate an isothermal process by having it occur very, very slowly, but as a theoretical construct, it works well enough to serve as a tool for understanding real-world thermodynamic processes.

An adiabatic process is one which occurs without heat transfer between the system and the environment. Again, this isn't really possible because there will always be ​some​ heat transfer in a real system, and for it to truly occur it would have to happen instantaneously. But, as with an isothermal process, it can be a useful approximation for a real-world thermodynamic process.

The Carnot cycle is an idealized, maximally efficient heat engine cycle composed of adiabatic and isothermal processes. It's a simple way to describe a real-world heat engine (and a similar engine is sometimes called a Carnot engine), with the idealizations simply ensuring that it's a completely reversible cycle. This also makes it easier to describe using the first law of thermodynamics and the ideal gas law.

In general, a Carnot engine is built about a central reservoir of gas, with a piston attached to the top that moves when the gas expands and contracts.

In the first stage of the Carnot cycle, the temperature of the system remains constant (it is an isothermal process) as the system expands, drawing heat energy from the hot reservoir and converting it into work. In a heat engine, work is only done when the volume of the gas changes, so in this stage the engine does work on the environment as it expands.

However, the internal energy of an ideal gas only depends on its temperature, and so in an isothermal process, the internal energy of the system remains constant. Noting that the first law of thermodynamics states that:

\(∆U = Q – W\)

Where ​U​ is the change in internal energy, ​Q​ is the heat added and ​W​ is the work done, for ∆​U​ = 0 this gives:

\(Q = W\)

Or in words, the heat transfer to the system is equal to the work done by the system on the environment. If you don't want to use the heat directly (or the problem doesn't give you enough information to calculate it), you can calculate the work done by the system on the environment using the expression:

\(W = nRT_{high} \ln \bigg(\frac{V_2}{ V_1}\bigg)\)

Where ​T​_high refers to the temperature at this stage of the cycle (the temperature reduces to ​T​_low later in the process, so you call this one the "high temperature"), ​n​ is the number of moles of gas in the engine, ​R​ is the universal gas constant, ​V​₂ is the final volume and ​V​₁ is the starting volume.

In this stage, the word "isentropic" or "adiabatic" tells you that no heat is exchanged between the system and its surroundings, so by the first law, the entire change in internal energy is given by the work the system does.

The system expands adiabatically, so the increase in volume (and therefore the work done) leads to a decrease in temperature within the system. You can also think about the temperature difference from the beginning to the end of the process as explaining the reduction in internal energy of the system, according to the expression:

\(∆U = \frac{3}{2}nR∆T\)

Where ∆​T​ is the change in temperature. These two facts imply that the work done by the system (​W​) can be related to the change in temperature, and the expression for this is:

\(W = nC_v∆T\)

Where ​C​_v is the heat capacity for the substance at constant volume. Remember that the work done is taken as negative because it is done ​by​ the system rather than ​on​ it, which is given automatically here by the fact that the temperature reduces.

This is also called "isentropic" because the entropy of the system remains the same during this process, which means it's completely reversible.

Isothermal compression is a reduction in volume while the system is kept at a constant temperature. However, when you increase the pressure of a gas, this is usually accompanied by an increase in temperature, and so the extra heat energy has to go somewhere. In this stage of the Carnot cycle, the additional heat is transferred to the cold reservoir, and in terms of the first law, it's worth noting that in order to compress the gas, the environment must be doing work on the system.

As an isothermal part of the cycle, the internal energy of the system remains constant throughout. As before, this means that the work done by the system is exactly balanced by the heat lost to the system, by the first law of thermodynamics. There is an analogous expression to the one in stage 1 for this part of the process:

\(W = nRT_{low} \ln \bigg(\frac{V_4}{ V_3}\bigg)\)

In this case, ​T​_low is the lower temperature, ​V​₃ is the starting volume and ​V​₄ is the final volume. Note that this time, the natural logarithm term will come out with a negative result, which reflects the fact that in this case, the work is done on the system by the environment, and the heat transfers from the system to the environment.

The final stage involves adiabatic compression, or in other words, the system being compressed due to work done on it by its surroundings but with ​no​ heat transfer between the two. This means the temperature of the gas increases, and so there is a change in the internal energy of the system. Because there is no heat exchange in this part of the process, the change in internal energy comes entirely from the work done on the system.

In an analogous way to stage 2, you can relate the change in temperature to the work done on the system, and in fact the expression is exactly the same:

\(W = nC_v∆T\)

However, this time, you have to remember that the change in temperature is positive, and so the change in internal energy is also positive, by the equation:

\(∆U = \frac{3}{2}nR∆T\)

At this point, the system has returned to its initial state, and so it's initial internal energy, volume and pressure. The Carnot cycle forms a closed loop on a ​PV​-diagram (a plot of pressure vs. volume) or indeed on a T-S diagram of temperature vs. entropy.

In a full Carnot cycle, the total change in internal energy is zero because the final state and the initial state are the same. Adding the work done from all four stages, and remembering that in stage 1 and 3 the work is equal to the heat transferred, the total work done is given by:

\(\begin{aligned}
W &= Q_h + nC_v∆T – Q_c – nC_v∆T \
&= Q_h- Q_c
\end{aligned}\)

Where ​Q​_h is the heat added to the system in stage 1 and ​Q​_c is the heat lost from the system in stage 3, and the expressions for the work in stages 2 and 4 cancel out (because the size of the temperature changes are the same). Since the engine is designed to turn heat energy into work, you calculate the efficiency of a Carnot engine using: efficiency = work / heat added, so:

\(\begin{aligned}
\text{Efficiency }&= \frac{W}{Q_h} \
\
&= \frac{Q_h – Q_c}{Q_h} \
\
&= 1 – \frac{T_c}{T_h}
\end{aligned}\)

Here, ​T​_c is the temperature of the cold reservoir and ​T​_h is the temperature of the hot reservoir. This gives the limit of maximum efficiency for heat engines, and the expression shows that the Carnot efficiency is greater when the difference between the temperatures of the hot and cold reservoirs is bigger.