# Why only isothermal and adiabatic for Carnot engine?

What is the reason behind choosing adiabatic and isothermal process for carnot engine?

My book states the following reason. I am not able to understand it. It would be great if you could put it in simpler words.

• A process is not quasi-static if it involves finite temperature difference between the system and reservoir. Hence, in a reversible heat engine operating between two temperatures, heat should be absorbed and released isothermally.

• Employing any other processes other than adiabatic, say isochoric, to take the system from one temperature to another, we shall need series of reservoirs in the temperature range $$T_2$$ and $$T_1$$ to ensure that the process is quasi-static.

I feel there are much better ways to explain the need for reversible isothermal and adiabatic processes in the Carnot cycle than the way it is explained in your book. But since you have to deal with the material you have, I will attempt to clarify what they are saying.

Regarding the first bullet, you start with the requirement that a Carnot cycle only exchanges heat with two fixed temperature thermal reservoirs. The second requirement is that the heat transfers must be reversible. That means the temperature of the system has to be essentially the same as the reservoir during the process, i.e., the system temperature has to be constant. The only process where heat transfer occurs reversibly with the system temperature constant is a reversible isothermal process.

Regarding the second bullet, to complete the cycle you need processes that link together the high and low temperature isothermal processes. In order to do this you need to lower and raise the temperature of the system. The only way to lower and raise the system temperature without transferring heat with additional heat reservoirs are with processes that don't require heat transfer. Those are reversible adiabatic processes.

Hope this helps.

• I think you should add a little more detail on the 2nd bullet. Jan 22 '20 at 14:02
• @ChetMiller Are you thinking details of the adiabatic processes for ideal gas? Jan 22 '20 at 14:39
• No; something like to get the working fluid between the two temperatures, the fluid has to pass through intermediate temperatures. If the path is not adiabatic, heat transfer will be occurring at these intermediate temperatures using reservoirs at these intermediate temperatures, which is contrary to the use of only two reservoirs. Jan 22 '20 at 14:56
• @ChetMiller When I said "the only way to lower and raise the system temperature without transferring heat with additional reservoirs..." with the emphasis on "additional" I was alluding to the need for intermediate temperatures. But I can elaborate. Jan 22 '20 at 15:50

The beauty of the Carnot cycle is that reversible heat transfer (isothermal) is separated from pure reversible work (adiabatic) and they can be analyzed individually. Almost all of thermostatics can be derived by using Carnot cycles, reversible isothermal ($$Q_1, T_1$$)-adiabatic-isothermal ($$Q_2, T_2$$)-adiabatic.

Since the processes are well separated one knows exactly how the entropy changes in the system; for example, in the isothermal stage the entropy changes exactly by $$Q_1/T_1$$ or $$Q_2/T_2$$, while in the adiabatic stage there is no entropy change.

It is also possible to derive the existence of the entropy function from postulating the efficiency of the Carnot cycle, ie., $$1-T_2/T_1$$.

The first paragraph is because you want the simplest possible engine , one that extracts heat from only two sources of temperature. Having more sources will make the analysis more complicated. Then you need two other processes that connect the two isothermal ones into a cycle. You do not want to have a heat exchange durig this process because then it will violate the two temperatures assumption, the process moves between two isothermals so it will have variable temperature along its path. Only an adiabatic process prevents heat exchange with these variable sources.