I was reading the second edition of "THERMODYNAMICS AND AN INTRODUCTION TO THERMOSTATICS" by (late) Herbert B. Callen which states

(Furthermore) the delivery of work (and of heat) is identical for every reversible process.

Could this be interpreted as work and heat being state variable (or functions) for reversible processes or approximately so for quasi-static processes?


In "Physical Chemistry: A Molecular Approach," (late) Donald A. McQuarrie clearly stated that reversible work and heat are not state functions. Consider the following diagram showing three reversible processes between two given states:

three different reversible processes between the same two states with different work and heat

Since the work for a simple system (as defined by Callen, H. B.) corresponds to the area under the $P$$V$ curve, the reversible work and heat for the 3 different processes will be different. However, I am a little confused about whether we can simply draw some curves on a $P$$V$ graph and expect it to be a reversible process.


In a course of thermodynamics, the prof said that work (as defined in according to $U = W + Q$ by considering $P_{external}$ or that $W = -\int_{V_1}^{V_2}P_{ext}dV$) is dependent on whether the process is reversible or irreversible, which makes sense given that work is a path variable (or function). He further said that work during an irreversible and reversible processes, $W_{rev}$ and $W_{irr}$, respectively, are related for expansion and compression processes as follows:

wrong interpretation of reversible and irreversible work

He reasoned it by saying that the area under the $P$$V$ curve seems to follow the same relation. When I pointed out that by considering the proper sign conventions for work the difference of inequality disappears, he tried to argue by saying that it is the modulus of work that matters and not the sign. Yes! let's all start assigning random positive and negative signs to any variables in our analysis and finally say the value is what matters. Anyways, this is what he meant:

wrong interpretation of reversible and irreversible work

Again, I was not fully convinced and decided to solve my doubts on my own. The good old Callen was surely to come in handy. I jumped over to Section 4-5, THE MAXIMUM WORK THEOREM, where it is stated as paraphrased below.

(Then) the maximum work theorem states that for all processes leading from a specified initial state to a specified final state of a primary system, the delivery of work is maximum (and the delivery of heat is minimum) for a reversible process. Furthermore, the delivery of work (and heat) is identical for every reversible process.

Then, the discussion goes into reversible work and heat sources, which I think are only important considerations for the mathematical proof of the theorem that follows, and I think the maximum work theorem, as stated, disproves the arguments made by the prof.

Anyways, what really got my attention was the statement where it is stated that for all reversible paths from a specified initial state to a specified final state, work (and heat) are the same. In general, under fewer constraints, there are numerous (sometimes infinite) reversible paths between two given states, and work (and heat) being the same for all such reversible paths must be implying that the otherwise path variables (or functions), under the assumption the process is reversible (or quasi-static) must somehow be acting like state variables (or functions).

  • $\begingroup$ Could you provide some more context for that quote? $\endgroup$
    – J. Murray
    Commented Apr 11, 2023 at 4:54
  • $\begingroup$ @J.Murray thank you for requesting some much requried context for the question. I have added context and more comprehensive information from the reference. $\endgroup$
    – ananta
    Commented Apr 11, 2023 at 6:14

3 Answers 3


First, what you understand about signs and $W$ is correct. There are two conventions to write the work: $\Delta U=Q+W$ or $\Delta U=Q-W$. In the first convention, it is the work given to the system, in the second convention, it is the work taken from the system. Usually, it is better to stick to a single convention in a course but your teacher does not seem to do this. With the convention $\Delta U=Q+W$, you always have $W_{irr}>W_{rev}$. I always use this convention.

Then you should force yourself to be very clear about the terms you are using. First there are state variables (or state functions, they are synonyms). A state variable is a property of the system at a given time, when you're looking at it. Imagine a box on the table with a gas in it. You can ask: what are the values of the state variables of the gas NOW. They are pressure, temperature, volume... and then energy, entropy...

Then you have a process. A process is a physical action on a gas. This is for example pushing the piston a certain way, heating the gas a certain way... In a process you have an initial state and a final state : that is the state of the system before you do the action and after. In both, the system has certain values for the state variables. The difference of one variable between the initial and final states is called a variation. You can write for example: $$\Delta U=U_{final}-U_{initial}$$

For a process, you have properties of the process. They are the quantities of energy given to the system in the form of work and heat, written $W$ and $Q$. They depend on the process. They cannot be state variables, because it is about what happens to a system during an action. You cannot look at a gas in a box and ask "what is the heat or work of this gas now". You can ask "I pushed the piston for 5 minutes, what is the work given to the gas during these five minutes".

You can ask however, is the work a variation of "something". Is there a mysterious state variable, call it $X$, such that during any process, we would have: $$W=\Delta X$$

This answer is no. It was believed in the eighteenth century that it was the case for $Q$. People believed that giving heat to a system was increasing it's "heat", and people called it the "calorific fluid". People believed that during a process, you had: $$Q=\Delta C$$

where $C$ would be a state variable meaning "the amount of heat contained in the system : its calorific content". This is false.

Then we realized that "smooth" processes were special in some sense and we called them reversible processes. For a reversible process, the amount of work given to a system in a infinitesimal step of the process is always: $$\delta W=-PdV$$

where $\delta$ just means it is an infinitesimal quantity. $d$ means infinitesimal variation and is the equivalent of $\Delta$ in the infinitesimal case.

Let us call $\delta W_{rev}=-PdV$. The non infinitesimal equivalent is:

$$W_{rev}=\int_{path} -PdV$$

We know that $Q$ and $W$ are not variations of state functions. Could $W_{rev}$ be ? Is there a state variable $X$ such as for all processes:

$$W_{rev}=\Delta X$$.

If it was the case, the reversible work would only depend on the initial and final states of a process. Because we would have:


But $\int_{path} -PdV$ is the area under the path when the path is represented in the $P,V$ diagram. It is clear that the area depends on the path and not only the initial and final states. Thus, $W_{rev}$ cannot be the variation a state variable $X$, whatever this variable is.

As a conclusion, neither work, heat, reversible work or reversible heat are variations of state variables.

Then what does this mysterious sentence means:

(Furthermore) the delivery of work (and of heat) is identical for every reversible process.

I haven't read Callen's book, I don't know the context, I'm only guessing. This is my own interpretation of the sentence without the context.

That's where vocabulary matters. A process is not a path. A path is the sequence of states during a process. It is the curve, for example in the $P,V$ diagram. The path does not tell you what you are doing exactly with the system. A path may show that the temperature of your gas progressively increases during your process a certain way. But it does not tell you what you are doing with the gas: are you heating it with a heat bath? Are you spinning a powerful fan inside it? These are completely different actions. In one case, you give it heat. In the other case, you give it work.

But some quantities do not depend on the process, only on the path. Clearly it is the case for:


It means that whatever process you are doing, as long as the path is the same and the process is reversible, the work will be the same. Same for heat:


I would rephrase:

The delivery of work (and of heat) is identical for every reversible process having the same path.

A good video introductory course (clear and easy) is available here: https://www.khanacademy.org/science/physics/thermodynamics/laws-of-thermodynamics/v/quasistatic-and-reversible-processes

It is not so easy to find a text explaining this, since it is mostly considered as implicit from the start. A good elementary course is available here: https://jfoadi.me.uk/stat_therm.html. The text uses exactly the same denotations as me. (Denotations like $W_{rev}$, $Q_{rev}$ seem to be only used with students, I tried to bridge the gap between these ideas and a more classical text like the one in the link)

  • $\begingroup$ Some references would be helpful. $\endgroup$
    – ananta
    Commented Apr 12, 2023 at 13:37
  • 1
    $\begingroup$ Added at the end of the answer. $\endgroup$
    – Benoit
    Commented Apr 12, 2023 at 14:24
  • $\begingroup$ I added a good video introductory course that is very easy to understand. $\endgroup$
    – Benoit
    Commented Apr 21, 2023 at 11:24

Callen is talking about a reversible process of a system connected to a single thermal reservoir and a single work reservoir. Since there is only one temperature at which the reversible thermal exchange can take place, the temperature of the system is the same as that of the reservoir throughout and then the change $\Delta F$ of the system's Helmholtz free energy F=U-TS is exactly equal to the change of the delivered work $\Delta F = W_{rev}$. Since $F$ is a characteristic function (state variable) and does not depend on the process path, then so is the delivered work in an isothermal reversible process. But then of course the same will also hold for $\Delta U - W = \Delta Q$.


For a reversible process, you can always add a few Carnot cycles between the initial and final states without affecting the two end state, but which would affect the heat and work done.


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