# Graphs of thermodynamic transformations

I received a question from one of my students of a high school (17 years old) with reference to a graph of a Clayperon-plane. Suppose that it is the one shown in the figure.

From state $$A \to B$$ is evidently an isobar (with volume expansion); and from $$B\to C$$ an isochore.

Are the transformations from $$C\to D$$ and from $$D \to A$$ hypothetical or not real transformations referring to ideal gases or to real gases or are they approximations of particular thermodynamic transformations that I am not known to except for adiabatic, isothermal, isochore and isobaric? What is the best justification that I can give my students of a high school about transformations such as those going from $$C$$ to $$D$$ and from $$D$$ to $$A$$?

• Why do they have to have names? – Chet Miller Apr 23 at 21:01
• @ChetMiller Hi. The question is from my students and I have formulated it. To me they don't have names but in my humble opinion they are not thermodynamic transformations that exist in nature. – Sebastiano Apr 23 at 21:04
• You're right. They are contrived for the problem being considered. That's done all the time for pedagogical purposes. – Chet Miller Apr 23 at 21:17
• @ChetMiller I certainly agree with you although I like questions asked by students who also reflect and reason about the graphs. For me they are to be rewarded. I have never asked myself the problem of which transformation is involved, but to be unprepared to give an exhaustive answer is not a good sign. :-) Anyway thanks for your collaboration as always. – Sebastiano Apr 23 at 21:20

What is the best justification that I can give my students of a high school about transformations such as those going from $$C$$ to $$D$$ and from $$D$$ to $$A$$?

For one, you can say that not all processes necessarily have a name and need not be either isobaric, isochoric, isothermal, adiabatic. It so happens that in the case of an ideal gas these processes are specific cases for what is called a polytropic process, which is given by

$$pV^{n}=C$$

Where $$n$$ is called the polytropic index (not to be confused with $$n$$ being the number of moles of a gas) and $$C$$ is a constant. Each of the four processes mentioned can be derived from the polytropic process with the selection of an appropriate value of $$n$$. (See https://en.wikipedia.org/wiki/Polytropic_process).

This is not to say that C to D and D to A are even possible in the form of a single "process". But perhaps by including multiple equilibrium states along each path it may be possible to break each path down to a combination of multiple polytropic processes which applicable values of $$n$$ connecting the equilibrium states. Might be an interesting student exercise.

Hope this helps.

• Hi, :-)...surely your answer help me...I not know that exist a "Polytropic process" never studied or treated at the university. I am a mathematician :-(....And I try to give my students the best. So any point on the curve that is not a known transformation could be a state of a particular polytropic? That is, the union of many polytropic points generating a semicircle or a segment not parallel to the Cartesian axes? – Sebastiano Apr 23 at 21:56
• @Sebastiano I am not a mathematician but to a thermodynamicist a point on the pV graph is not a process or a transformation. A point on the graph is an equilibrium state that typically initiates or ends a process. The process connects the points. Since $n$ can theoretically take on any value between + or - infinity it may be possible, given enough points between C and D , you might (but not necessarily) identify a polytropic process between each pair of points. If it were me, I would start with D-A since it is linear and see if there is a value of $n$ corresponding to that path. – Bob D Apr 23 at 22:21
• Always kind Bob D. Thank you again for your kindness, courtesy and your subsequent comment. I had my students read your response this morning and I elegantly admitted my ignorance of these types of polytropic transformations. We have also seen on Wikipedia the same site but in Italian language but I don't give information on topics that I have not studied. I think it is right that first I study in depth the topic to be able to make it understood by my students. – Sebastiano Apr 24 at 20:33
• In any case, you (a mathematician) would be better at coming up with formulas or mathematical steps to fit the paths C-D and D-A than I and perhaps then I could comment on whether the process is feasible from a thermodynamics point of view. Sorry I can’t help you further at this point. – Bob D Apr 25 at 18:19
• @Sebastiano Same to you, Ciao! – Bob D Apr 25 at 19:48