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Contradiction in thermodynamics problem?

  1. Feb 22, 2015 #1
    Consider a gas as your system, confined in the usual frictionless piston-cylinder. The piston is massless, external pressure is constant, Pext. Let the system be at initial state T1 and P1 = Pext. We want to compare the following two processes: in the first process, we reversibly heat the gas against constant external pressure to final volume V2. In the second process, we heat the gas irreversibly from the same initial conditions against the same constant pressure to reach the same final volume V2, and then we wait for the gas to reach internal equilibrium, so that P2= Pext and both the initial and final states of the two process are identical. Assuming the gas has constant Cp over the temperature range, we wish to compare qrev and qirrev
    for the two processes, and also ∆S for the two processes.

    Now here's the contradiction: ∆S should be equal for the two processes, since the initial and final states are the same. So let's calculate ∆S for the reversible process. Since it is reversible, we have ∆S = ∫dqrev/T, where dqrev = CpdT. So ∆S = Cp Ln T2/T1. This is also ∆S for the irreversible process.
    Also qrev must equal qirrev since both are constant pressure processes, with ∆H, ∆U, W and q all being equal for both processes, right?

    But what if we consider the inequality dS>dqirrev/T? If we use dqirrev=dqrev= CpdT, then we have ∆S>Cp Ln T2/T1, which contradicts our previous finding. Where have I gone wrong?
    Last edited: Feb 22, 2015
  2. jcsd
  3. Feb 22, 2015 #2


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    How might you go about heating irreversibly from T1 to T2?
  4. Feb 23, 2015 #3
    What a great question!!!!

    It is not widely discussed, but the Clausius inequality is based on use of the temperature at the location where the heat transfer dq is occurring (i.e., at the interface between the system and the surroundings). During the irreversible heating, the temperature at this interface is higher than the bulk average temperature of the gas, so that, even though the same amount of heat is transferred, the integral of dq/TI is less for the irreversible case than for the reversible case (where TI represents the temperature at the interface).

  5. Feb 23, 2015 #4
    The entire setup is physically possible. We could for instance- without negatively affecting the rest of the problem- have a substance, say a monatomic gas, which has nearly constant Cp, especially if we keep the temperature range from being too big. It is perfectly possible to start from the same initial conditions, we could hold the external pressures constant in both cases, the piston doesn't really have to be massless if the set up is vertical (we'd just add the mass of the piston to the external pressure to calculate the work directly on the gas which we take as our system), we could even assume the irreversible heating is done with say a Bunsen-burner fume and ask to calculate the change in entropy of the surroundings as a bonus brain teaser (now how would you do that?).

    Also we could reach the same final states because in either case when the piston is at rest, the ultimate equilibrium pressure will be P2= Pext.
    So all we have to do is reach the same final volumes for the final states to be identical, and that too is possible in this particular problem; we irreversibly heat the system and find that sweet spot with trial and error ;)

    The specific details of the irreversible heating, are not given in the problem though, for instance the finite temperature difference, and you would be correct to suspect that as a key to resolving the contradiction (I was actually able to solve the problem a few hours after posting the question, but plz do provide your own resolution so we can all learn from it ;) )
    Last edited: Feb 23, 2015
  6. Feb 23, 2015 #5
    Did you not see my post #3?

  7. Feb 23, 2015 #6
    Yes I did Chet, it's just that your answer was deep and got me thinking on a LOOOOOT of things lol. I understand what you said and think that you are correct, however your answer has opened doors to new questions for me, will get back to you.
  8. Feb 23, 2015 #7
    Actually, having thought on Chet's answer, It seems to me that the only way to establish a final state of equilibrium during irreversible heating would be to emerge the system in a heat reservoir (constant temperature bath) with the desired final temperature. This way we ensure that both the final pressure and temperature of our system in the irreversible process eventually equal that of the reversible heating.

    If however, we choose to do the heating with a Bunsen-burner fume, all the while the temperature of the surrounding atmosphere remaining equal to the initial temperature (say 298 K), then we can't hope to establish a final state of equilibrium in the irreversible process (see why?). While calculating the change in entropy of surroundings in this condition is intriguing, I cannot think of any simple way to do so. Would it be necessary to integrate over the surface of the surroundings in contact with system, or is there a way to simplify the problem? Any thoughts Chet?
  9. Feb 23, 2015 #8
    Excellent. While you are thinking about it, I'm going to give you something else to look over. This is a write up I did which at one time was my PF blog (when PF still have blogs). It is a brief description of the First and Second laws, presented in a slightly different way than what you're used to.

    I'm starting this blog on thermodynamics because I have some ways of presenting the material that I think can help people learn the subject more easily.


    Suppose that we have a closed system that at initial time ti is in an initial equilibrium state, with internal energy Ui, and at a later time tf, it is in a new equilibrium state with internal energy Uf. The transition from the initial equilibrium state to the final equilibrium state is brought about by imposing a time-dependent heat flow across the interface between the system and the surroundings, and a time-dependent rate of doing work at the interface between the system and the surroundings. Let [itex]\dot{q}(t)[/itex] represent the rate of heat addition across the interface between the system and the surroundings at time t, and let [itex]\dot{w}(t)[/itex] represent the rate at which the system does work on the surroundings at the interface at time t. According to the first law (basically conservation of energy),
    where Q is the total amount of heat added and W is the total amount of work done by the system on the surroundings at the interface.

    The time variation of [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex] between the initial and final states uniquely characterizes the so-called process path. There are an infinite number of possible process paths that can take the system from the initial to the final equilibrium state. The only constraint is that Q-W must be the same for all of them.

    If a process path is irreversible, then the temperature and pressure within the system are typically inhomogeneous (i.e., non-uniform, varying with spatial position), and one cannot define a unique pressure or temperature for the system (except at the initial and the final equilibrium state). However, the pressure and temperature at the interface can be measured and controlled using the surroundings to impose the temperature and pressure boundary conditions that we desire. Thus, TI(t) and PI(t) can be used to impose the process path that we desire. Alternately, and even more fundamentally, we can directly control, by well established methods, the rate of heat flow and the rate of doing work at the interface [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex]).

    Both for reversible and irreversible process paths, the rate at which the system does work on the surroundings is given by:
    where [itex]\dot{V}(t)[/itex] is the rate of change of system volume at time t. However, if the process path is reversible, the pressure P within the system is uniform, and

    [itex]P_I(t)=P(t)[/itex] (reversible process path)

    Therefore, [itex]\dot{w}(t)=P(t)\dot{V}(t)[/itex] (reversible process path)

    Another feature of reversible process paths is that they are carried out very slowly, so that [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex] are both very close to zero over then entire process path. However, the amount of time between the initial equilibrium state and the final equilibrium state (tf-ti) becomes exceedingly large. In this way, Q-W remains constant and finite.


    In the previous section, we focused on the infinite number of process paths that are capable of taking a closed thermodynamic system from an initial equilibrium state to a final equilibrium state. Each of these process paths is uniquely determined by specifying the heat transfer rate [itex]\dot{q}(t)[/itex] and the rate of doing work [itex]\dot{w}(t)[/itex] as functions of time at the interface between the system and the surroundings. We noted that the cumulative amount of heat transfer and the cumulative amount of work done over an entire process path are given by the two integrals:
    In the present section, we will be introducing a third integral of this type (involving the heat transfer rate [itex]\dot{q}(t)[/itex]) to provide a basis for establishing a precise mathematical statement of the Second Law of Thermodynamics.

    The discovery of the Second Law came about in the 19th century, and involved contributions by many brilliant scientists. There have been many statements of the Second Law over the years, couched in complicated language and multi-word sentences, typically involving heat reservoirs, Carnot engines, and the like. These statements have been a source of unending confusion for students of thermodynamics for over a hundred years. What has been sorely needed is a precise mathematical definition of the Second Law that avoids all the complicated rhetoric. The sad part about all this is that such a precise definition has existed all along. The definition was formulated by Clausius back in the 1800's.

    Clausius wondered what would happen if he evaluated the following integral over each of the possible process paths between the initial and final equilibrium states of a closed system:
    where TI(t) is the temperature at the interface with the surroundings at time t. He carried out extensive calculations on many systems undergoing a variety of both reversible and irreversible paths and discovered something astonishing. He found that, for any closed system, the values calculated for the integral over all the possible reversible and irreversible paths (between the initial and final equilibrium states) was not arbitrary; instead, there was a unique upper bound (maximum) to the value of the integral. Clausius also found that this result was consistent with all the "word definitions" of the Second Law.

    Clearly, if there was an upper bound for this integral, this upper bound had to depend only on the two equilibrium states, and not on the path between them. It must therefore be regarded as a point function of state. Clausius named this point function Entropy.

    But how could the value of this point function be determined without evaluating the integral over every possible process path between the initial and final equilibrium states to find the maximum? Clausius made another discovery. He determined that, out of the infinite number of possible process paths, there existed a well-defined subset, each member of which gave the same maximum value for the integral. This subset consisted of what we call today the reversible process paths. So, to determine the change in entropy between two equilibrium states, one must first conceive of a reversible path between the states and then evaluate the integral. Any other process path will give a value for the integral lower than the entropy change.

    So, mathematically, we can now state the Second Law as follows:

    [tex]I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt} \le ΔS = \int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}[/tex]
    where [itex]\dot{q}_{rev}(t)[/itex] is the heat transfer rate for any of the reversible paths between the initial and final equilibrium states, and T(t) is the system temperature at time t (which, for a reversible path, is equal to the temperature at the interface with the surroundings). This constitutes a precise mathematical statement of the Second Law of Thermodynamics.

  10. Feb 23, 2015 #9
    The way to do the irreversible path is to guarantee that the total amount of heat transferred is the same. This will guarantee that the final temperature and volume will be the same as for the reversible path. How this is accomplished in practice is a separate issue.

    As far as the surroundings are concerned, we don't need to concern ourselves with that if our focus is the system, and we are testing the Clausius inequality with respect to the system. That's what's nice about doing it all with the Clausius inequality. You don't need to address all the possible things that could be happening in the surroundings.

  11. Feb 23, 2015 #10


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    Period, end of sentence, problem solved.
  12. Feb 23, 2015 #11
  13. Feb 23, 2015 #12
    OK I see what you are doing there, trying to get the students to calculate and apply the forces and works and everything correctly. you may want to emphasize a few points though:
    1- absence of macroscopic kinetic energy and potential energies, or else it is E (total energy), not U (internal energy) that should be stated in the first law.
    2- You've omitted the definition of quasi-static processes and replaced it with reversible processes.
    3- Choosing the boundary, defining the boundary and keeping in mind what the boundary of the system is throughout. This has two implications: first, it means that we need to add ALL the works done throughout this boundary. And since work is defined in terms of forces, we must first account for all the working forces in our diagram. Same goes for calculating the remaining microscopic work (heat). Also, this implies that if we choose a different system as our focus, the terms for work and heat could change. So for instance a force that is external to the piston may not necessarily be external to our system, depending on whether our system boundary comes in direct contact with the external surface of the piston or not (which it usually doesn't, usually it's the internal surface of the piston that kisses the system).
    4- The definition of reversible processes has nothing to do with the first law, and it doesn't even make sense to define such a process while analyzing the first law. The motivation for definition of reversible processes arises when one tries to prove (based on the assumption of validity of the second law), that ∫dqrev/T = 0 in any closed path. Here, a reversible process is defined as one where you can restore BOTH the system and surroundings to their original states. This means if you replace w with -w, and q with minus q (the only way to completely restore the surroundings), you would have also restored the system. SO in essence, a reversible process is one where you can reverse or "flip" everything(i.e. the inputs, outputs and the process within the system) in the diagram. We see this definition or property of reversible processes used explicitly in proving things like maximum efficiency of carnot cycles, which ultimately leads to the definition of entropy.
    So this is why they define the reversible process, and they do so in the second law, not the first.

    Defining a quasi-static process, one where the only necessary condition is the process being infinitely slow (whether or not we can flip everything in the diagram), allows us to use dW= -pdV, where p is the bulk pressure of system, instead of dw= -pextdV, since the two pressures are equal in such processes. For instance, we could use dW= -pdV even if the process is not reversible, provided it is quasi-static (infinitely slow). One such process could be an infinitely slow process in the presence of friction (can't flip the diagrams!).
    Also you may want to distinguish between being mechanically, thermally and chemically reversible. All of this requires lots of careful teaching (or self learning).
    Last edited: Feb 23, 2015
  14. Feb 23, 2015 #13
    As you know Chet, all reversible processes are infinitely slow, but not all infinitely slow processes are reversible, so you might want to emphasize that.
    Except I have one question, are really all reversible processes infinitely slow? Is there no reversible process that can be done over finite time? What about converting Ice to water at 1 atm, 0 C?
  15. Feb 23, 2015 #14
    Yes thnx, I got that from your first post by writing dSt= dS + dSsurr >0 and working backwards (forwards?).
  16. Feb 23, 2015 #15
    Hi amin2014. Thanks for your constructive criticism and suggestions. You need to understand my motivation for writing this. My target audience was new thermo students who are struggling to learn the basic concepts, but who have developed many misconceptions because of the poor way that thermo is presented in the typical textbooks that are used. Constantly, on Physics Forums, we encounter these same misconceptions. So I wanted to provide the students with supplementary material that was presented in a little different way, and I wanted to keep it very concise so that they would not lose interest. Some of the misconceptions I hoped to address were:

    1. Isn't the work PdV? Why do we have to use PextdV to get the work when the process is irreversible?

    2. Isn't the change in entropy always equal to the integral of dq/T, irrespective of whether the path is reversible or irreversible?

    3. The entropy is not a material property of the system. It depends on path, and can only be determined for a specific process path.

    4. The temperature T in the Clausius inequality is the temperature of the system.

    I'm going to try to address some of the points that you raised, and then I have a proposal for you.

    Yes. This is, of course, correct, but I was trying to keep the development as simple as possible (and as close to what I deemed they had previoiusly learned) so that I could emphasize the points that I thought were most important.

    Yes. This is very true. We need to carefully specify the system that we are focusing upon. Also, we need to include all the heats that enter, sometimes through different parts of the interface with the surroundings. This also goes for applying the Clausius Inequality.

    This goes back to item 1 on my list of misconceptions. Maybe I should have used "quasistatic" rather than "reversible" in what I said.
    You're preaching to the choir. But, another reason for introducing reversible paths is to provide a basis for determining the change in entropy.

    Yes. As I said, maybe I should have said quasistatic instead of reversible.
    Yes. Up to the point I had shown, I was just trying to keep it as simple as possible.

    I have been toying with the idea of doing additional writing, but just haven't gotten around to it. That would certainly include approaches to achieving mechanical, thermal, and chemical reversibility.

    Here's my proposal. How about we collaborate on improving what I have written and maybe even adding additional sections? We could do this in private conversations. I would be glad to participate, provided I had final literary say. Any interest?

  17. Feb 23, 2015 #16
    This is a situation that is expected to be much closer to being reversible than others. However, even here, you need to have some temperature gradients in the ice water to provide the heat transfer to the ice, and the amount of irreversibility will be greater if you try to carry out the change at a very high heat transfer rate. Irreverisiblity will always be present if you have significant rates of deformation (viscous heating), rates of heat transfer (heat conduction), rates of mass transfer (mixing/diffusion), or rates of chemical reaction.

  18. Feb 24, 2015 #17
    Intriguing suggestion Chet, let me know the details of how you are writing this, what you are writing about.
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