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Moving cylinder with gas and piston

  1. Dec 12, 2014 #1
    1. The problem statement, all variables and given/known data
    A thin-walled cylinder of mass m is filled with monatomic ideal gas and is lying on a horizontal frictionless surface. The initial internal energy of gas is U = cT. There is a thin insulating piston in the middle of the cylinder of mass M. The piston and the cylinder are heat-insulating and there is no friction between them. By a short impulse the cylinder is given a velocity v along its axis. Find the temperatures of the gases on different sides of piston when the piston stops oscillating.

    2. Relevant equations
    No idea.

    3. The attempt at a solution
    I am seriously lost. I was thinking about energy conservation, but since we have some external impulse acting on the cylinder and there is no way of measuring that we can't really write the conservation down I suppose. Also I don't see a thermodynamic approach to the gas processes? I do feel intuitively that the piston will not be in the middle of the cylinder when moving, but I can't really explain it, although possibly that might lead to the solution. Any hints appreciated!
     
  2. jcsd
  3. Dec 12, 2014 #2

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    That's a good start.
    "Insulated." Cannot exchange heat with surroundings. What can the two IG volumes do?
     
  4. Dec 12, 2014 #3
    I'm not sure what is IG volume - I'm guessing, insulated gas volume? Adiabatic processes you mean?
     
  5. Dec 12, 2014 #4

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    Pardon me --- IG means "ideal gas."
     
  6. Dec 12, 2014 #5
    :) okay, so if there is no heat exchange, it means that the internal energy change of the gas will be equal to the work done by gas. But it is still quite complicated, since both volume and pressure may change (referring to work done by gas) and we do not know anything about those..
     
  7. Dec 12, 2014 #6

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    Correct with one addition, "minus work done on the gas."
     
  8. Dec 12, 2014 #7
    How much external work do the two gases do in combination, if the wall of the cylinder is rigid?

    Chet
     
  9. Dec 13, 2014 #8
    Here's another hint. This is one of those rare problems where you need to take into account the change in kinetic energy of the system, and include it in your first law of thermo equation.

    Chet
     
  10. Dec 13, 2014 #9

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    This problem statement is "too clever by half."
     
  11. Dec 13, 2014 #10
    Rugile, are you still out there? The thing to do on this problem is to start out with an overall momentum balance. After the piston stops oscillating, is it traveling (a) faster than the cylinder, (b) the same speed as the cylinder, or (c) slower than the cylinder? Will the cylinder be traveling at the same speed at the end as when it started? Neglecting the momentum of the gases, what is the initial momentum of the overall system? Based on your answers to the previous questions, what is the final momentum of the overall system (algebraically)?

    Chet
     
  12. Dec 14, 2014 #11
    Sorry for my late reply, yes, I'm still here :)
    And thank you for all your replies.

    Minus work done on the gas is the same as plus work done by gas, right?

    Zero? Gas in one side expands as much as the other one contracts (don't know the right word for opposite of expand ?

    So $$ \Delta U = Q + W + \Delta E_k = W + \frac{(M+m)v^2}{2} = \frac{(M+m)v^2}{2}$$? But then I'm not sure why does the internal energy of gas depend on kinetic energy of the system?

    As in, the oscillations won't really cease without friction? Didn't even notice that before you pointed it out! :) I suppose they really mean that there is some friction, but we don't have to take it into account.

    I think it will be traveling at the same speed as the cylinder, because otherwise, it would be compressing gas more, against what the gas resists?
    Initial momentum is 0, and the final momentum is (M+m)v ?
     
  13. Dec 14, 2014 #12

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    Correct. There is no mechanism for dissipation. It is possible the author intended you to determine two temperatures at the extremes of oscillation when the piston was not moving relative to the cylinder.
     
  14. Dec 14, 2014 #13
    Dissipation is definitely a key part of this problem. What do the words, "when the piston stops oscillating" mean to you.

    The dissipation is caused by viscous damping within the deforming gas. This is how the kinetic energy of the piston and cylinder get converted into internal energy.

    Here's what happens. Just before the problem begins, an impulsive force is applied to the cylinder. This involves an insignificant displacement of the cylinder over a very short distance, and over a very short time. This impulsive work imparts kinetic energy to the cylinder. Our problem starts at time t = 0 when the cylinder already has velocity v and kinetic energy mv2/2; neither the piston nor the cylinder has moved significantly yet, and the gas has not yet deformed.

    During the process that ensues, both the cylinder and the piston oscillate back and forth (in addition to translating). Initially, the oscillations are large, but, as time progresses, as a result of viscous dissipation, the oscillations decay, and eventually vanish. At the end, the piston and cylinder are moving together at the same velocity V.

    Since the cylinder is rigid, no work is done on the combined system of piston and cylinder (W = 0) and, since the cylinder is adiabatic, no heat enters of leaves the system (Q = 0). So ΔU + ΔKE = 0.

    The first step in solving this problem is to apply an overall momentum balance to the cylinder/piston system, to determine the final velocity V. Once this is done, one can calculate the change in kinetic energy of the system. Once this is done, we know the change in internal energy that results from viscous dissipation of the portion of the initial kinetic energy that was lost.

    Chet
     
  15. Dec 16, 2014 #14
    But how can we apply the balance if we know nothing about the impulse? Initial momentum will be (m+M)v I suppose, so is the final momentum?
     
  16. Dec 16, 2014 #15
    You're applying the balance after the impulse has taken place. The initial velocity of the cylinder is v immediately after the impulse, and the initial velocity of the piston is zero.
    No. The initial momentum will be mv. This is also the final momentum. The question is, if the piston and cylinder are finally moving at the same velocity (and momentum is conserved), what is the final velocity of the piston and cylinder (after the piston has stopped oscillating)?

    Chet
     
    Last edited: Dec 16, 2014
  17. Dec 16, 2014 #16
    Let me clarify further. What we are dealing with here are two time intervals.

    Time Interval 1: This time interval is between t = 0- to t = 0+

    Time Interval 2: This time interval is between t = 0+ to t = ∞

    Interval 1: This is the time interval over which the large impulsive force is applied. At the beginning of this time interval, nothing is moving. At the end of this time interval, the cylinder has a velocity v. Nothing else is moving (yet). The impulse of the force is given by I = mv, where m is the mass of the cylinder and v is its velocity at the end of the time interval.

    Interval 2: This is the time interval of interest in our problem. At the beginning of this time interval, only the cylinder is moving. Its velocity is v. At the end of this time interval, the cylinder and the piston are both moving at the same velocity. The change in total momentum of the cylinder and piston combination is zero during this time interval (since they are not experiencing any external force, and the mass of the gas is negligible).

    Chet
     
  18. Dec 19, 2014 #17
    Okay, so here is the final solution:

    the change in momentum (in the interval 0+ to ∞) will be from mv to (m+M)V, thus $$V = \frac{mv}{m+M}$$.
    Change in kinetic energy thus will be $$\Delta E = \frac{(m+M)V^2}{2} - \frac{mv^2}{2} = \frac{m^2 v^2}{2(M+m)}- \frac{mv^2}{2} = \frac{-Mmv^2}{2}$$.
    Since $$\Delta U + \Delta E = 0$$, we get that $$\frac{Mmv^2}{2} = c \Delta T$$ and ΔT is what we are looking for!

    Thank you all for the help once again!
     
  19. Dec 19, 2014 #18
    Very excellent job. Just one minor algebraic correction:
    [tex]\Delta E = -\frac{Mmv^2}{2(M+m)}[/tex]
    And this should be applied to the ΔT equation also.

    Chet
     
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