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Is this a simple energy conservation problem?

  1. Apr 17, 2016 #1
    1. The problem statement, all variables and given/known data

    ?temp_hash=fd06c3d9e4379f4be4f51e657f230a59.png
    2. Relevant equations


    3. The attempt at a solution

    The answer to this problem can be obtained by equating ##\frac{1}{2}mv^2 = ΔU## .

    But I am not sure why this is to be done . In fact I think this is not quite right .

    1) The Kinetic energy ##\frac{1}{2}mv^2## is the energy due to bulk motion whereas ΔU relates the microscopic motion .Can we really equate the two ?

    2) Is this an application of 1st law of thermodynamics . I do not think so .

    3) Is this a simple energy conservation problem . Again I doubt that is the case .

    Any help is appreciated .

    Thanks
     

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  2. jcsd
  3. Apr 17, 2016 #2

    Titan97

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    Why isn't it a simple energy conservation?

    Initial energy is ##U_1+\frac{1}{2}mv^2##
    Final energy is ##U_2##
     
  4. Apr 17, 2016 #3
    Hi ,

    What is U1 and U2 ?
     
  5. Apr 18, 2016 #4

    Titan97

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    Initial and final internal energy.
     
  6. Apr 18, 2016 #5
    Ok .

    And what is the mechanism by which this energy is transferred from macroscopic kinetic energy to microscopic internal energy , since work done on the gas is zero ?

    Do you agree work done on the gas is zero ?
     
  7. Apr 18, 2016 #6

    ehild

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    The gas does not gain energy. Initially, all molecules had a velocity component equal to the velocity of the vessel and a random component corresponding to the velocity distribution of the initial thermodynamic state of the gas. When the vessel stops, that regular component is randomized through collisions between the molecules. The internal energy is the energy of random motions. It is increased when the macroscopic KE, the KE of the regular motion, transforms to the KE of the random motions of the molecules.
     
  8. Apr 18, 2016 #7
    Hello ,

    An increase in internal energy means an increase in temperature of the gas . But I have studied that the bulk motion of the vessel does not affect the random motion of the gas . Irrespective of whether the vessel accelerates /decelerates/moves with uniform speed , the temperature does not change . But this is not the case here .An increase in internal energy means an increase in the temperature .

    I am seeing a contradiction in the concepts .
     
  9. Apr 18, 2016 #8

    ehild

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    Initially, the average velocity of the molecules was equal to Vm, the velocity of the vessel, as the random component of the velocities (Vr) summed up to zero. When colliding with the wall, only the random component of the molecule changed. Assume one dimensional motion: after the vessel stopped, a molecule colliding with a wall with velocity Vm+Vr will rebounce with velocity -(vm+vr). At the end there will be equal amount of molecules moving with -(vm+vr) as those moving with vm+vr so the average velocity becomes zero. And the velocities randomize further by the collisions between molecules, and even other degrees of freedom (mainly rotational) are excited during the collisions. At the end all initial macroscopic energy transforms into the energy of random motion.
     
  10. Apr 18, 2016 #9
    Ok .

    So you are suggesting that accelerating/decelerating a gas container increases the temperature of gas . Right ??
     
  11. Apr 18, 2016 #10

    ehild

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    Not quite. You can speak of temperature of the gas when it is in steady state. Till the molecules interact with an accelerating wall, the gas is not in a state which is called a "state" by Thermodynamics.
     
  12. Apr 18, 2016 #11
    But after the gas container stops accelerating/decelerating temperature of gas does increase . Right ??

    Isn't this exactly what is happening in the original problem (the temperature increases by ΔT when the vessel stops) ??
     
  13. Apr 18, 2016 #12
    For this closed system, the 1st law of thermodynamics tells us that:
    $$\Delta U+\Delta(KE)+\Delta(PE)=Q-W$$
    For this particular problem, the equation reduces to:$$\Delta U+\Delta(KE)=0$$
    But, mechanistically, how can the kinetic energy of the gas get converted to internal energy. The gas within the container has inertia, and when the container is brought to a stop, a flow and compression/expansion action takes place within the gas. This dynamic movement and deformation of the gas is damped out by viscous stresses. In the end, all the non-random kinetic energy KE is converted to random kinetic energy U (i.e., internal energy) by viscous dissipation.
     
  14. Apr 18, 2016 #13
    Would you agree that irrespective of whether the vessel(thermally insulated) is stopped suddenly or gradually ( decelerated ) , the temperature of gas would increase ??
     
  15. Apr 18, 2016 #14
    No.

    So what are your thoughts on why I answer this question with a "no?"
     
  16. Apr 18, 2016 #15
    I think if gas container is stopped gradually ,the container would be doing work on the external agent ( whatever it may be ) . So by doing work , the non random translational kinetic energy of the gas transfers to the external agent , leaving the random kinetic energy unchanged . Hence temperature of gas remains same as before .

    Makes sense ???
     
  17. Apr 18, 2016 #16
    Extremely well-analyzed!!!!

    So, in this case, even though the volume of the gas does not change, work is still done on the gas by its surroundings (the container) by unbalanced forces applied to the gas through a displacement. This is combined with the fact that there are negligible viscous stresses and viscous dissipation, since these depend on the rate at which the gas is deforming. In this case, the 1st law energy balance approaches:
    $$\Delta (KE)=-W$$
    which is basically the work energy theorem. The gas in this case behaves analogous to a rigid body.
     
  18. Apr 18, 2016 #17
    I am having a little trouble understanding the work term"W" in the 1st law of thermodynamics as applied in setup where we are gradually stopping the vessel .

    Is "pdV work " (due to volume change) and "bulk work" ( work done by the force applied by the container ) two different types of work we need to sum in the "W" term while applying the 1st law of thermodynamics ?
     
    Last edited: Apr 18, 2016
  19. Apr 18, 2016 #18
    Basically, yes. But recognize that both of these involve a force at the boundary of the gas integrated over a displacement.
     
  20. Apr 18, 2016 #19
    Ok .

    Suppose the vessel is initially at rest . Now it is accelerated till it reaches a speed 'v' .

    Would you agree that same reasoning applies here as well i.e when the vessel is suddenly moved ,the temperature of gas rises whereas if it is accelerated gradually the temperature remains unchanged ??
     
  21. Apr 18, 2016 #20
    Yes.
     
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