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Heat in reversible isothermal expansions

  1. Sep 28, 2012 #1
    Heat is defined as "the transfer of energy owing to a temperature difference between two bodies".

    Consider a gas expanding against a massless frictionless piston. Assume that the process is both isothermal and reversible, and that the gas is ideal, so that its internal energy does not change.

    Since its internal energy does not change, the work done by the gas must be compensated by heat added to the gas.

    However, we initially assumed that the gas was isothermal, i.e. in thermal equilibrium with its surroundings. So how can heat, by its definition, be added to the gas if its already in thermal equilibrium with its surroundings?

    Obviously, the first law of thermo mandates the heat transfer, but the definition of heat contradicts this viewpoint.

    Perhaps the definition of heat is only valid for isovolumetric processes?

  2. jcsd
  3. Sep 28, 2012 #2

    Simon Bridge

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    That is not correct - isothermal just means that the temperature does not change ... in order for this to happen in the expansion, energy (heat) must flow between the gas and something external.

    It follows that
    is also incorrect.
    Heat will flow spontaneously when there is a temperature difference. It can also flow when we do work.

    An example would be when you first turn a refrigerator on ... the inside is the same temperature as the outside yet the fridge moves heat from the inside to the outside.
  4. Sep 28, 2012 #3
    OK but my textbook uses the "isothermal" fact to prove the second law of thermodynamics by saying that the temperature of the system must be in thermal equilibrium with the surroundings during a reversible process.

    Perhaps during a reversible isothermal process the system must be in thermal equilibrium with the surroundings?

  5. Sep 28, 2012 #4
    In an isothermal reversible gas expansion, the temperature of the surroundings is only infinitescimally higher than that of the system, and the pressure of the surroundings is only infinitescimally lower than the pressure of the gas in the system. This is sufficient to transfer the required heat to the system at a very slow rate, and to allow the expansion to occur at a very slow rate. What we are talking about here is the limiting case of infinitescimally slow expansion.

  6. Sep 28, 2012 #5
    Thank you Chet, that explains it precisely. Thus, would it be correct to say that for an isothermal reversible expansion,

    [tex] P_{gas} = P_{ext} + dP [/tex]

    [tex] T_{sys} + dT = T_{sur} [/tex]

  7. Sep 28, 2012 #6
    Yes. That's how to envision it. Now I have a question for you.

    Suppose you wanted to determine the work done (and heat added) for a prescribed isothermal reversible expansion experimentally, without having to wait an infinite amount of time to get the results. The gas is in a cylinder, and there is an insulated piston at the end of the cylinder. You can preform a sequence of different expreiments controlling the temperature at the outer surface of the cylinder and controlling either the transient motion of the piston or the transient pressure on the outer surface of the piston. How would you set up appropriate pressure- and temperature measurement devices, and what finite set of experiments would you perform so that you could extrapolate the results to the case of isothermal reversible expansion (without having to actually use infinitescimal temperature- and pressure driving forces to get the desired results)?
  8. Sep 28, 2012 #7
    Hey Chet I don't have any experience in this, but I will take a shot:

    The piston should be in a heat reservoir of the desired temperature so that the temperature remains constant?

    And to control the external pressure, perhaps we could keep a tank of water on the piston (so that the gas works against the water above the piston). To remove the water, we can have a valve somewhere so that the water can be allowed to steadily flow out some auxiliary pipe.

    Although, this raises the question, what is the rate at which the pressure needs to be reduced? It surely depends on the rate at which the gas expands, which in turn will probably involve a lot of differential equations.

    I can't fathom how work would be calculated. Perhaps if we had the water being pumped by the gas being used in some type of hydraulic lift, we could check the displacement of that lift and use its weight to measure the transfer of energy (work done).

    That's just my guess!

  9. Sep 28, 2012 #8

    Simon Bridge

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    To be reversable. Yes.
    In practise we put the system in thermal contact with a heat bath (so large that heat drawn off does not affect it's statistics) and the plunger has some weight on it ... otherwise it is in a large room so plunger movements don't affect the pressure in the room. Get the idea? You can figure work from mgh for eg.

    A lot of the processes you are learning refer to idealisations and won't stand up to close scrutiny. They also tend to hide the details of how something happens. You end up approximating real systems by some small tweak on an ideal one. The ideals, though, let you uncover fundamental rules of Nature.
  10. Sep 28, 2012 #9

    Andrew Mason

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    Just to avoid possible confusion here, the only way for heat flow to occur is when it occurs spontaneously due to a temperature difference. Work done on or by the gas may be used to create the temperature differences.

  11. Sep 29, 2012 #10

    Simon Bridge

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    Hmmm... I'm trying to indicate that heat may flow during an isothermal process.
    But you are right - we usually think of creating a small temperature difference and then letting the system return to thermal equilibrium, then another small difference - and so on... and we "call" it isothermal.

    OTOH: don't we usually think of work done by heat engines as moving the heat rather than creating the temperature difference ... like when heat is pumped from a cold place to a hot one?
  12. Sep 29, 2012 #11
    If you are a householder, yes you think of refrigeration as in your earlier example.

    If you are a refrigeration engineer you break it down into component stages, because your system is actually more complicated than simply a blob of heat in a bottle that is moved into the surroundings.

    The machine that does the work actually has hot and cold reservoirs along with a working fluid and heat is exchanged at both reservoirs by temp diference as Andrew said.
    Last edited: Sep 29, 2012
  13. Sep 29, 2012 #12
    I like the methodology you are alluding to here better than the experimental methodology that I had initially thought of, and it also touches on the finite incremental approach that Simon Bridge referred to in reply #11. Let me see if I can paraphrase what you are driving at:

    Use a constant temperature bath to control the temperature in the cylinder, and employ a sequence of small finite irreversible isobaric expansions with small finite pressure increments (of equal magnitude) to impose the surroundings pressure. In each incremental expansion, allow the system to equilibrate before imposing the next pressure increment. Continue the small finite pressure increments until the final desired pressure and volume are attained. Then calculate the amount of work that was done. Repeat this same process with different values of the small (equal) finite pressure increments, and than make a plot of the work done as a function of the pressure increment size used in each experiment. Extrapolate the graph (linearly) to zero pressure increment. This process should give you the ideal work done for the reversible isothermal expansion.

    See if you can do some modeling calculations to predict theoretically how this methodology would work.
  14. Sep 30, 2012 #13

    Simon Bridge

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    I meant in terms of what we actualy tell students in freshman undergraduate courses.

    ... I notice that the cooling coils in the fridge are usually warmer than the inside of the fridge.
    (What I like about this example is that it does the break down as well... courses often leave that out.)

    In terms of the thermodynamics models that get taught at the same level - don't the equations balance that way too ... that heat can flow at constant temperature by applying work? Isn't what we are doing just abstracting out the details of how the work gets done?

    I'll agree that the fridge is not a constant temp machine.
    Last edited: Sep 30, 2012
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