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Thermodynamics thought experiment

  1. Jan 20, 2015 #1
    There is some ideal gas in a container moving with some velocity on a smooth surface and you suddenly stop it( say by using your hands) , will the temperature of the gas increase? It seems to me that since you're suddenly stopping it there is no work done (because of no displacement ) so the kinetic energy must change to internal and hence temperature of the gas will increase. Is this right?

    Now consider the same container but it is now moving on a rough surface . Assume there is a constant frictional force which stops it after travelling a certain distance. Now , will the temperature of the gas increase ? Now here I think since friction does work , the work will be equal to change in kinetic energy so the temperature will not increase . This seems strange. I think I'm not right.

    What is the difference between these two cases ? (Or , is there a difference?) Will the temperature change be same in both these cases ? (If at all it takes place).
     
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  3. Jan 20, 2015 #2

    Quantum Defect

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    Assume a monatomic ideal gas. The total energy of the gas in the container is 3/2 nRT + 1/2 Mnv^2 (M is the molar mass, n = number of moles, R is the gas constant). When you stop the container, the energy of the gas is still conserved, but now the total energy is 3/2 nRT'. Is T' > T?

    If you think about the energy, like above, is anything different here? [You probably need to assume that the container is a perfect insulator -- no heat generated by friction with the surface is transferred into the gas] Remember, Energy is a state function. What does this mean, with respect to the path?

    What do you think?
     
  4. Jan 20, 2015 #3
    Yes.
    Yes. Here's how it plays out:

    The gas has 3 important characteristics:

    1. It has mass.
    2. It has springiness, because it can compress and expand
    3. It has is dissipative (viscous), which has the effect of degrading mechanical energy to thermal energy. It's like a damper.

    When you suddenly stop the cylinder, the gas will initially slosh back and forth within the cylinder as a result of the interaction between the mass and the springiness. When this is occurring, there is an ongoing interchange of kinetic energy with the potential energy of the springy compression and expansion. However, as this is going on, the viscous dissipation will slowly degrade these forms of mechanical energy to thermal energy. In the end, the gas will no longer be sloshing, it will stop moving, and its temperature will be higher.

    No. You are not right. You need to focus on the gas. As the container slows down, its walls are doing some work to also slow down the gas, and the gas is doing the opposite work to try to keep the cylinder moving against the frictional drag. But, if the container has significant mass and associated kinetic energy to begin with, the work that the gas does to keep the container moving will only be a small fraction of the work that friction does to stop the container. And, if the container is insulated, none of the frictional heat will reach the gas. So, as far as the gas is concerned, it will do some small amount of work on the container as the container is slowing down. But, we have no basis for calculating how much this work is (at least not using thermo). The net result will be that the gas heats up (as when the cylinder comes to an instant stop), but it won't heat up quite as much (because the gas has done some work).

    Chet
     
  5. Jan 21, 2015 #4

    Quantum Defect

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    Another way to look at it...

    Think about a monatomic gas as a microscopic bunch of spheres undergoing collsiions with the walls of the cylinder. [Assume the gas is isolated.]

    The total energy of an ensemble of gas molecules at thermal equilibrium in an inertial rest frame is E = 3/2 nRT (n = number of moles, R = gas constant)

    If this inertial frame (the cylinder) is traveling at constant velocity, a very busy person making measurements from within the cylinder would measure all of the kinetic energies of all of the atoms and get the answer above. E_cyl = 3/2 nRT

    Now suppose a person in the laboratory (rest) frame measures the kinetic energies of all of the gas molecules, as the cylinder whizzes by. She would measure a different set of velocities -- if the cylinder is moving in the +x direction with velocity +x, she would obtain measurements of the total velocity that have a +v bias in the x direction. If she added up all of the kinetic energies, she would get: E_LAB = 3/2 nRT + 1/2 Mn v^2 (Her brother inside the cylinder phoned her with the temperature that he measured).

    The cylinder stops moving suddenly. All of the molecules keep bouncing around, their kinetic energy in the laboratory frame is conserved (no PV work is done, no heat is exchanged with the surroundings). At some later time, brother in the cylinder measures a new temperature T'. Brother (at rest in the lab frame) and sister also now agree about what the velocity distribution is: E_LAB = 3/2 nRT' (T' > T) -- essentially the motion that was initially in the +x direction has been redistributed into the other directions.

    The interesting thing here is that gas molecules with more degrees of freedom would see a smaller temperature rise. E.g. an ideal gas of diatomic molecules should have
    E = 5/2 nRT, polyatomic molecules would have even higher numbers in front (the front is C_v == heat capacity at constant volume).

    In theory, this could be turned into a way to measure the heat capacity of the gas in the cylinder -- (except the rise in temperature would be quite small for most situations)

    As Chet notes, you should be able to extract some of the COM energy from the gas (I had this bit wrong) in which case the temperature rise inside the container would be less. e.g. if the cylinder was not there, we could use a windmill to extract useful work from this packet of moving air.
     
    Last edited: Jan 21, 2015
  6. Jan 21, 2015 #5

    anorlunda

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    It seems to me that you should be able to solve this via conservation of energy without reference to thermodynamics.r

    A container has mass, shape, and size. It obeys Newton's laws. It matters not whether the contents of the container are solid, liquid, or gas, or a vacuum. Therefore, I think you can treat the container as a particle for purposes of your thought experiment.

    If the temperature changes if it contains a gas, then it should also change if it contains a solid.
     
  7. Jan 21, 2015 #6
    How would you go about calculating the temperature change of the gas using Newton's laws of mechanics exclusively, without any consideration of thermodynamics?

    Chet
     
  8. Jan 21, 2015 #7

    anorlunda

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    Temperature is merely a way to express energy in a quasi steady state.

    E=kT where k is Boltzman's constant.

    Particles in a gas have KE and PE individually, in the average, and in aggregate. My point (if I'm correct) is that putting them in a container makes it easy to think about their properties in aggregate.

    We are not accustomed to thinking this way because we use Newton's law on a couple of particles at a time, but temperature is defined only for thermal equilibrium. It is thermal equilibrium, not energy bookkeeping that is missing from Newton's laws.

    If a moving solid block hits a barrier and stops, its KE must be converted to thermal energy.
     
  9. Jan 21, 2015 #8
    So are you saying that, to quantify what happens the gas (in terms of its actual temperature change), you need to solve Newton's law for each and every particle in the container?
     
  10. Jan 21, 2015 #9

    anorlunda

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    I'm saying that you could tread the container and its contents as a particle, or as a solid, when viewed from outside the container.

    Think of it like Einstein's elevator. What experiment could you do to a sealed container to tell if the interior of the container was solid or gaseous?

    I neglected to mention, that if the container was partially filled with liquid, I could feel the liquid sloshing. It is harder to do that with a gaseous interior.
     
    Last edited: Jan 21, 2015
  11. Jan 21, 2015 #10
    We're obviously on different wavelengths, and I'm withdrawing from this discussion. I have no idea what you are talking about. But, whatever it is, at least in my judgement, it is not relevant to the question being posed by the OP.

    Chet
     
  12. Jan 22, 2015 #11

    anorlunda

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    \

    Uh Oh. When a PF staff mentor says that, I fear that I may have gone off the deep end. But reviewing my point, I think I'm right. Please let me know. Sould I hang my head in shame?

    The OP asked about a container that was accelerated and declerated, and the question had to do with changes in temperature. Not magnitude of the temperature, but the sign of change plus, minus, or zero.

    First, I reasoned that we have a sealed container. Fixed mass fixed volume. The container and contents have mass, and a capacity to store termal energy. At the outset I presumed that we don't know about the contents of the container; solid or gas.

    Second, the sign of change in temperature is identical to the sign of change in internal energy.

    Third, we can manipulate the container. It can be a closed system or an open system exchanging energy with its surroundings. In any case, using Newtons laws of motion, and foremost simple conservation of energy, we should be able to do the energy bookkeeping and calculate whether the internal thermal energy of the container and contents increases or stays the same. For this step, it is irrelevant if the contents of the container are gas or solid.

    Fourth, someone comes along and reveals that the contents are an ideal gas. Therefore, what you did in step three is invalid. You must use a Boltzman distribution and the ideal gas law to answer the question. I say "not so."
     
  13. Jan 22, 2015 #12
    No, he was being much more quantitative than this. He was asking whether the change in internal energy was equal to the change in kinetic energy. (I have had other interactions with this OP in other threads, so I know that he is knowledgeable in thermo and was asking how to determine the temperature increase).
    The question stated that it was an ideal gas, and, the OP's implicit assumption of no heat transfer from the gas was equivalent to saying that the container was adiabatically isolated from the gas.

    The definition of a closed system is one that exchanges no mass with its surroundings. The definition of an open system is one that exchanges mass with its surroundings. I think you are referring here to the possibility of the system exchanging heat with its surroundings. The clear implication of the problem statement is that the gas does not exchange heat with either the container or the region outside the container.
    If, as you are suggesting, the gas is able to exchange heat with the container and then with the surroundings, then after everything equilibrates, the gas will be at the same temperature it started with. So, no temperature change will have occurred.

    That wasn't I. I never mentioned the Boltzmann distribution or the ideal gas law.

    Chet
     
  14. Jan 22, 2015 #13

    anorlunda

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    In that case, I apologize. I was indeed out in left field. It was Quantum Defect, not you, who started talking about the gas law, and my entire point in post #5 was to contest the need for that.



    No that's not what I was thinking. I was thinking that the content of the container will experience positive internal energy change, even if the contents were solid instead of gas. Solids too have mass, springiness, and are dissipative, so I think it irrelevant to OP whether the contents are gas or solid.


    If the nature of the OP was quantitative, not qualitative, then I did misread it and I apologize.

    Thank you for your devotion to PF Chet, aiding us laymen to improve our understanding.
     
  15. Jan 22, 2015 #14

    Quantum Defect

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    Look at the original post: The OP says: "There is some ideal gas in a container moving with some velocity on a smooth surface and you suddenly stop it (say by using your hands) , will the temperature of the gas increase?"

    Neither Chet nor I introduced an ideal gas where there was none. The ideal gas (with all of its assumptions, ramifications, etc. ) was there at the birth. I missed at first seeing how a gas in a constant volume container could do work, but after seeing Chet's explanation, I can.

    Long story short: In both cases, the T will increase. It will increase most in the first case. To calculate the actual rise in temperature in the first case, you will need to know the velocity of the cylinder and Cv (J/K) of the ideal gas in the cylinder.
     
  16. Jan 22, 2015 #15

    No hard feelings. Thanks for your kind words of appreciation.

    Chet
     
  17. Jan 22, 2015 #16
    Hi guys. It looks like we're all in agreement now. I think this is an appropriate time to close this thread.

    Chet
     
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