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I Kinetic Gas Theory Puzzle

  1. Jun 20, 2017 #1
    If I place an insulated container of gas on the edge of a turntable and then rotate
    the container as a constant angular velocity, what happens to the gas?

    Does the temperature of the gas increase? If so, how was work done on the gas?
     
  2. jcsd
  3. Jun 20, 2017 #2
    No. Ordered motion of the entire sample of gas as a whole does not contribute to the temperature. Temperature is related to random disordered energy of the molecules.
     
  4. Jun 20, 2017 #3
    The total energy of the molecule wouldn't change. But because the pression on the border would be quickly higher than the pression in the centre, can we assume that, in a first moment, the temperature at the border increases and in the centre decreases? And then return to a state of equilibrium?
    Am I completely wrong?
     
  5. Jun 20, 2017 #4
    Well, then how does the ordred motion imparted by a piston doing work on the gas lead to an increase in temperature?

    Pressure-volume work, though a piston, imparts an odered (linear) motion to the molecules. This ordered motion is then redistributed with a resulting increase in temperature. The opposite happens when the gas expands against a piston.

    In the case of the revolving container of gas, if we look at things from the reference frame of the gas, the only change is the presence of a force field (due to the turntable rotation). This force field is constant (after a transient build up to constant angular velocity) and should cause a density gradient to occur in the container, with gas as the far end being more dense. This density gradient will persist but I don't know if it will cause an increase in temperature. For some reason I cannot adequately conceptualize this problem in terms either of kinetic theory or thermodynamics and that's why I present it here.
     
  6. Jun 20, 2017 #5

    russ_watters

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    Since you answered your own question, I guess it was rhetorical?
    A slight increase on one side and decrease on the other, but it will equalize after some time.
     
  7. Jun 20, 2017 #6
    I didn't answer my own question. I was responding to the post by Chandra.

    But does the force field cause an increase in temperature?

    Why should the density gradient equalize? The force field will remain as long as the turntable keeps spinning.

    The situation is essentially a kind of artificial gravity. Does the earth's atmosphere, which also has a pressure gradient due to the force field of gravity, equalize in time? No, it doesn't. So why should the revolving container of gas equalize?

    Maybe this question should have been posted to a more advanced level?
     
  8. Jun 20, 2017 #7

    russ_watters

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    Chandra didn't ask a question. You asked: "Well, then how does the ordred motion imparted by a piston doing work on the gas lead to an increase in temperature?"

    I was checking to make sure you were comfortable with your own answer.
    Force field? The compression of the gas on one side causes an increase in temperature on that side and the expansion on the other causes a decrease on that side.
    No, the temperature equalizes.
    Earth's atmosphere is an open thermodynamic cycle. We're assuming your container of gas is closed.
    No, it's fine here. It's just a matter of peeling back the potential layers of the question and scenario you are presenting. It takes a few posts to get it clear.
     
  9. Jun 20, 2017 #8
    In the case of a piston pushing on a gas, we normally assume that the piston is moving so slowly that at every instant, the gas is in equilibrium (quasistatic process). It is only in such a case that we can define a temperature and pressure of the gas. In this case, we are giving enough time for the gas to convert each infinitesimal amount of "organized" work done into disordered energy of the molecules. Any sudden motion of the piston will cause the gas to go into a non-equilibrium state, and we then have to wait for the gas to equilibrate. Until it does equilibrate, the gas does not have a defined temperature or pressure.
    If on the other hand, you take a closed sample of gas, and the entire container is moving with constant velocity, the kinetic energy of that motion is superposed on the random motion of the molecules. The random motion of the molecules is still described by a Maxwell-Boltzmann distribution, which determines the temperature.
     
  10. Jun 20, 2017 #9
    Questions to be asked are:
    1, Why should that be so?
    2. Can a hydrostatic equation be set up to confirm?
    3. What does a profile of graph of a density vs height look like? Pressure vs height? temperature vs height?
    4 Will assumptions need to be made about the fluid? If so, which assumptions?
    5. Will changing assumptions lead to a different profile?
    6. How are other variables in the state of the fluid affected by the assumptions?

    Of course that is pretty much what you are asking in the first place, so really nothing new.
    But, rather than trying to explain it to yourself, how would you explain in to another person, perhaps younger and learning, keen in mathematics and science, Sometimes easier to see the problem by stepping back a little.

    By the way, for an incompressible fluid, such as water, we assume that density does not vary with pressure. The graph of density vs distance under the surface of a lake is linear, a vertical straight line, and that of P vs h is also linear, with P increasing with depth.
     
  11. Jun 21, 2017 #10

    vanhees71

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    Setting a container with a fluid into rotation, will increase the temperature since the entire fluid is first at rest and the container walls already rotate. This leads to friction between the fluid and the container walls and thus dissipation, entropy and temperature increase. The friction finally leads to a new equilibrium state with the fluid rotating with the container. In this state you can describe the fluid motion with ideal hydrodynamics leading to paraboloidal surface.
     
  12. Jun 21, 2017 #11

    A.T.

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    Yes, but the OP seems more about effects of introducing a g-field ("edge of a turntable"), rather then the rotation. Maybe a linearly accelerating rocket would be a better example.
     
  13. Jun 21, 2017 #12

    Nidum

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    If so then we just have a centrifuge . Radial pressure gradient . Possible friction heating during start up . No further heating at steady state .
     
  14. Jun 21, 2017 #13
    First, I want to sincerely thank all those who responed.

    I suppose that this answer is the one I was seeking. The system _is_ like a centrifuge. But my goal was to know the concepts that lead to the conclusion that temperature does not increase.

    Looking at the turntable system, energy is imparted to this system to make it spin. Once it reaches the final angular velocity no more energy is required and it will spin forever (assuming a vacuum environment and frictionlass bearings).

    I suppose that I was wondering if any of this energy input went to increasing the internal energy, and hence the temperature, of the gas. I suppose that the answer is that it does not.

    I was perhaps unsure of the concepts, thermodynamic or otherwise, that would cause me to automatically conclude that there is no temperature increase.

    Thanks again.
     
  15. Jun 22, 2017 #14

    vanhees71

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    Well, as I said, despite the little temperature increase from friction with the wall in the transient state, the temperature won't change much. In the final static state the energy put into the fluid is in its potential energy due to the potential of the centrifugal force (in the local restframe of the rotating fluid elements).
     
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