If gas volume remains constant, it can do work?

  1. jcsd
  2. Simon Bridge

    Simon Bridge 15,471
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    Let me guess - the amount of gas in the volume is not constant?
  3. My English is not good, but I think I can describe it clear.
    The boxes in figure are rigid. There is a membrane with mass dividing the box into two parts. The left part is filled with high pressure gas, and the right part is filled with low pressure gas (or vacuum). The only difference between the upper fig and the down fig is: there is a small hole in the membrane in the down fig. There is no friction between membrane and the box. Because there is a hole in the membrane, the gas in left part and the right part can be seen a whole gas. And the whole gas volume is constant. The high pressure gas in left part must push the membrane with mass to move. The membrane’s energy is from the gas’s work, but the volume is constant.

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  4. sophiecentaur

    sophiecentaur 14,696
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    The total volume may be constant but the volumes of the two regions are not and I think they need to be considered separately. If you want to find where the energy comes from, to move the membrane, you could look at any temperature changes involved.
  5. Simon Bridge

    Simon Bridge 15,471
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    What he said :)

    What you actually have there is two chambers - and it looks like gas can flow from one to the other. The volume occupied by the gas changes. Another way of looking at it is that the system is not in thermal equilibrium with itself - since it does not have the same pressure or temperature everywhere.

    Whichever way you look at it - this is not an isochoric process so of course it is possible for it to do work.
  6. It looks like the OP is describing a free expansion of a gas.

    It doesn't matter if the gas expands by the ( massless ) membrane moving, or through the pinhole, it ends up as being the same thing. The chosen system consists of both volumes on either side of the membrane to simplify the calculation.

    There is nothing on the other side, ie the vacuum, to provide a force against expansion, and so it happens quickly, and uncontrolled.

    Membranes are usually considered massless. So thus no work is being done.
    It happens quickly, with no time for heat to flow in or out.

    Thus, no energy crosses the system boundary, the internal energy of the gas is constant, and thus so is temperature constant for the gas.

    That is for an ideal gas.

    For a real gas, we see a Joule Thompson effect.
    a real gas has a slight change in temperature during free uncontrolled expansion.
  7. the process is not quasi-static, the equation dw=pdV is not correct.
    the high energy state relaxes to low energy state, the process usually is non-equilibrium. If the volume remains constant and it is no heat transfer, the difference of two energy states must do work to others.
  8. Of course its not quasi-static. An expansion into a vacuum is not quasi-static.

    The temperature of an ideal gas in this scenario does not change from state 1 to state 2, and therefor neither does the internal energy.

    From the first law
    ΔE = Q+W
    In this case,
    ΔU = Q+W, as no other forms of energy are crossing the boundary.
    ΔU = 0, --> the temperatue did not change.
    Q = 0
    It follows that W = 0
    To what work are you referring?
  9. If you take as the system the overall box, then no work is done on the system. And, if the overall box is insulated, there is no heat entering or leaving. Therefore, from the first law (as 256bits points out), there is no change in total internal energy of the system U in going from the initial equilibrium state to the final equilibrium state.

  10. Simon Bridge

    Simon Bridge 15,471
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    Or you can take the system as one side or the other - in which case the volume does change since the "membrane with mass" moves.

    Some notes:
    The mass on the membrane only slightly modifies the statements the others have been making.
    To show work in your diagram, you want to explicitly show the mass (draw a square with "m" in it where the mass is), and you want the movement to be "upwards" (against gravity) and not sideways: this is so the final state of the system is clearly different from the initial.

    The small hole in the membrane is a minor technicality.

    You could have started with a chamber divided into two volumes by a removable wall.
    One side is very low pressure and the other has a much higher pressure.
    On the low-pressure side there is a small cart with a sail attached.
    Quickly remove the dividing wall - now the gas on the two sides can be considered part of a single system much better than the case where there is a small hole. As gas rushes from one side to the other - the cart gets pushed along: accelerated.
    Thus the gas is doing work on the cart.

    Is this how you are thinking?

    Are you also thinking that this violates some rule in thermodynamics about isochoric processes doing no work?
  11. As Simon has suggested, whether or not work is involved depends on the choice of the system (or subsystems). All choices must lead to the same result for the final solution to the problem, but some choices (such as selecting the constant volume box as the system) make solving the problem much easier than others.

  12. An isochoric process, also called a constant-volume process, an isovolumetric process, or an isometric process, is a thermodynamic process during which the volume of the closed system undergoing such a process remains constant.

    The isochoric process does not include the non-equilibrium process? Why?
    The isochoric process must be a quasi-static process? I don't think so.
    I think that the non-equilibrium process on floor #1 and #3 belongs to the isochoric process too.
    Last edited: Aug 10, 2014
  13. Simon Bridge

    Simon Bridge 15,471
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    Wikipedia is not a good place to learn thermodynamics.
    The system has to be describable by a state equation. This means it has to have the same temperature and pressure throughout the volume at any particular time. This means that it has to be in thermal equilibrium with itself.

    You will notice that the wikipedia article you cite describe the "isochoric process" in terms of a P and T which does not vary across the volume.

    It is not surprising that a gas in a fixed enclosed volume may do work - we see it everyday around us: it's called "weather". Nobody wonders at the wind moving stuff about, thermals lifting things, etc. Therefore it makes no sense to talk about every possible fixed volume doing no work does it?

    It is always possible to put a fixed volume around any thermodynamic process - that does not magically make everything an isochoric process, even if you do drill a small hole in the container. If I put a car in a sealed shipping container and turn on the engine, are the engine gasses undergoing constant-volume processes?

    [/quote]Then your thinking is mistaken ;)

    I'm not saying you are wrong to say that a gas enclosed in a fixed volume container may do work. What I am saying is that you have misunderstood the definition of an isochoric process. To be fair - most books are quite relaxed about desribing it.
  14. To Simon Bridge:
    We both know no work can be done in a quasistatic isochoric process. However, a few (perhaps a lot of) persons generalize the result to all fixed volume processes including non-equilibrium processes.
  15. Simon Bridge

    Simon Bridge 15,471
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    Well quite a lot of people make mistakes.
    Can you provide a reference to a reputable source that makes that generalization?
    Then we can see better what you are trying to talk about.
  16. Hi Simon,

    I think you need to clarify what you're saying a little.
    I may be interpreting you incorrectly, but I think you are talking about fixed mass here, not fixed volume. To me, the term "fixed volume" means isochoric. If you take the entire atmosphere as your system, then this system does no work on its surroundings (outer space).
    The parcels of gas inside the shipping container are not undergoing constant-volume processes, but, if your system is the entire contents of the shipping container, and the container is air tight and insulated, then the overall combined contents undergo a constant volume process, there is no work done by the system on the surroundings, and the change in internal energy from the initial equilibrium state (car not yet started, and gasoline and air unreacted) to the final equilibrium state (gasoline or oxygen fully consumed and contents equilbrated) is zero. Is this your understanding?

  17. Simon Bridge

    Simon Bridge 15,471
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    For this discussion I am making a distinction between things that occur within a fixed geometric volume and an isochoric process.

    Some additional care being made to be sure that the volume of gas inside this fixed geometric volume is also fixed - although the density of the gas through the volume need not be. The geometric volume can also contain anything else. My example was a sailcar.

    That sounds about right. This is probably a better way of putting it too.

    I think there is a fundamental and common misunderstanding here and I'm trying to express it in terms the OP already is familiar with.

    The concept of mechanical work in thermodynamics is something a lot of students wrestle with.
    I suspect OP has become confused by sources saying that the work "done by the gas" (implicitly: "on anything") is zero.

    A classical closed system cannot do work on anything external to the system: it's closed! Thermodynamic "systems" tend not to be like that. We may be able to add heat to them and extract mechanical work, for instance. In order to extract mechanical work, the volume of the system must change.

    In OPs examples, one part of the system loses energy and another part gains it - since the part that loses energy is the gas (goes to KE in the mass), the assertion is that the gas is doing work. But, then, the gas is not undergoing an isochoric process. (The entire system may be though.)

    Is that better?

    Executive version - it is not surprising that OP finds the ability to get a mass in a fixed enclosed volume to move a mass about inside the same volume.
  18. OK. Good. So it looks like we are dealing with terminology here. When you refer to isochoric, you are referring to no volume change for a gas. When I think isochoric, I am thinking of a fixed geometric volume.

    Also, when you refer to a closed system, you are referring to what I call an isolated system. When I refer to a closed system, I am referring to a system for which no mass enters or leaves, but one which can do work and exchange heat with the surroundings.

    Engineers and Physicists on the Tower of Babel.

  19. Say you have a closed system, and you have an constant-volume heat addition. The increase of the internal energy (delta u) = qin. But enthalpy (a thermodynamic property) increases by an amount that is greater than the internal energy. delta (h) > delta (u).

    In a control volume, Pv represents flow energy (or flow work); that is, it represents the energy required to push fluid into the control volume. But in a closed system, Pv is not doing any kind of work. Can we call it a flow potential energy (not to be confused with gravitational potential energy or other kinds of potential energy)? Clearly it is some kind of energy, but where did the increase of Pv come from? All we added was qin to the system, and all of it went into u. u2 - u1 = qin. h2 - h1 > qin. P2v2 - P1v1 > 0. What does this mean?
  20. You are talking about the difference between the closed system version of the first law of thermodynamics and the open system version of the first law. In the development of the open system version of the first law (see Smith and van Ness, Introduction to Chemical Engineering Thermodynamics), the work is divided into two parts: work to push material into and out of the system (control volume) and all other work. The all other work is called the "shaft work." The work to push material into and out of the system is lumped together with the internal energy of the material entering and leaving to give the enthalpy of the material entering and leaving. You need to study some additional descriptions of the development of the open system version until you get comfortable with it and can reconcile it with the closed system version.

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