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Find equilibrium position of piston

  1. May 1, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    A portion of helium gas in a vertical cylindrical container is in thermodynamic equilibrium with the surroundings. The gas is confined by a movable heavy piston. The piston is slowly elevated a distance H from its equilibrium position and then kept in the elevated position long enough for the thermodynamic equilibrium to be reestablished. After that, the container is insulated and then the piston is released. After the piston comes to rest, what is the new equilibrium position of the piston?


    3. The attempt at a solution

    Let the piston initially be situated at a distance l from bottom of cylinder and let P1 be the pressure of the gas inside. Let M be the mass of piston. Doing force balance on piston gives

    Mg+P0A = P1A

    where A =area of cross-section and P0 = atmospheric pressure

    When the piston is taken to height H let pressure inside be P2.

    [itex]\dfrac{P_1 l}{T_0} = \dfrac{P_2 (l+H)}{T_2} [/itex]

    After cylinder is insulated and piston is released, the process will be adiabatic
    Let final height of piston from bottom of cylinder be l'.

    [itex]P_2 (l+H)^{\gamma} = P_1 l'^{\gamma} [/itex]

    But how do I eliminate T2 and l?
     
  2. jcsd
  3. May 1, 2014 #2
    Errm, i am myself a student, but...

    Why do you take temperature into account. Isothermal process means ΔT=0, but since they first stated that process is infinitesimally slow, it is an approximate isothermal process. So you apply combination of Boyle's law and Net force equilibrium.

    In second set up you use adiabatic process formula.
     
  4. May 1, 2014 #3
    Sankalpmittal is right about the first part of the problem. The required relationship should be
    [tex]P_1l=P_2(l+H)[/tex]

    The second part of the problem is more complicated. Let P(t) be the pressure exerted by the gas on the bottom face of the piston at time t during the deformation and let m be the mass of the piston. Apply Newton's 2nd law to determine the equation of motion of the piston during this deformation (in terms of P(t), m, P0, g, A, and x (the distance that the piston moves downward from position l + H. What do you get? We are next going to use this equation to determine the cumulative amount of work done by the piston on the gas when the system reaches final equilibrium.

    Chet
     
  5. May 1, 2014 #4
    I am new here, so I do not know how things work, so forgive my stupidity.

    I am getting 0.6H
     
  6. May 1, 2014 #5

    utkarshakash

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    It's the correct answer.
     
  7. May 1, 2014 #6

    utkarshakash

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    Can't we directly use the adiabatic formula?
     
  8. May 1, 2014 #7
    No. This is an irreversible compression. The "adiabatic formula" applies only to reversible expansions and compressions.

    Chet
     
  9. May 1, 2014 #8

    utkarshakash

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    Thanks for clarifying this. OK, here's what I'm getting

    [itex]mg+(P_0-P(t))A = mv \frac{dv}{dx} [/itex]
     
  10. May 1, 2014 #9
    Nice. Now, if we rearrange this equation, we get:

    [itex]P(t)A=mg+P_0A - mv \frac{dv}{dx} [/itex]

    If we integrate this equation with respect to x, we get:
    [tex]-W(t)=\int{PAdx}=(mg + P_0A)x(t) + \frac{mv^2(t)}{2}[/tex]
    where W(t) is the cumulative amount of work that the gas does on the piston between time t = 0 and time t=t, x(t) is the downward displacement of the piston at time t, and v(t) is the downward velocity of the piston at time t.

    After a very long period of time, the piston will stop oscillating and will come to rest. This will tell us the cumulative amount of work that has been done by the gas on the piston at final steady state (infinite time).
    [tex]W(∞)=-(mg + P_0A)x(∞)[/tex]
    The reason that the piston comes to rest (i.e., loses all its kinetic energy) is that viscous dissipation has been occurring within the cylinder.

    Since this compression takes place adiabatically, Q = 0. Therefore,
    [tex]ΔU = MC_v(T(∞)-T_0)=(mg + P_0A)x(∞)[/tex]
    where M is the number of moles of gas in the cylinder. From the ideal gas law, in terms of P1, T0, l, and A, what is M?
    Combine this with the previous equation and with the equation for P1 in terms of mg, A, and P0 to solve for T as a function of T0 and x(∞)/l.
     
  11. May 1, 2014 #10
    Hi Yellowflash. Welcome to Physics Forums!!!

    The way things work here is that, rather than giving the answer (or every detail of how to solve the problem) we try to help the original poster (OP) work through the problem on his own, by keeping him pointed in the right direction. (I personally have not worked through the problem to completion, but, my gut feeling is that yours is the correct answer: x = H/γ)

    Chet
     
  12. May 1, 2014 #11

    utkarshakash

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    [itex]T_{\infty} = T_0 \left\{ (\gamma -1) \lambda +1 \right\} [/itex]
    where λ=x(∞)/l

    If this is correct, what should be my next step?
     
  13. May 1, 2014 #12
    Nice work.

    The hard part is over. You know the final temperature in terms of x, you can express the final volume in terms of A, l, H, and x, and, you know that the final pressure has to be the same as the original pressure P1 (since the same equilibrium force balance applies again). You also know that the number of moles haven't changed. This is enough information to calculate x from the ideal gas law.

    Chet
     
  14. May 2, 2014 #13
    Hello Chet!

    I am at a loss in concluding the above from the problem statement. Can you please explain how it is irreversible? Thanks!
     
  15. May 2, 2014 #14

    haruspex

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    It's because heat was allowed to flow in after the piston moved up. That increases entropy.
     
  16. May 2, 2014 #15

    utkarshakash

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    Thanks a lot! :smile: I get x=H/γ ≈ 0.6H which is correct.
     
  17. May 2, 2014 #16
    Because the problem statement says "and then the (heavy) piston is released." Before the piston was released, you were supporting it (and it was in force equilibrium). Once you let go, it is no longer in force equilibrium, and it rapidly drops. This results in an irreversible adiabatic compression of the gas. In order for the compression to be reversible, you would have to lower the piston gradually (i.e., quasistatically).

    Chet
     
  18. May 2, 2014 #17

    utkarshakash

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    @Chestermiller: What do you have to say about this?
     
  19. May 2, 2014 #18
    I would say that the process of raising the piston was done close to reversibly, and the process of dropping the piston took place irreversibly. Of course, it doesn't matter how raising the piston took place, because you are told the initial and final equilibrium conditions of the system for this case. However, to determine the final equilibrium conditions for dropping the piston, you need to take into account that the piston was released spontaneously, and the compression took place irreversibly.

    Chet
     
  20. May 2, 2014 #19

    haruspex

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    Sorry, Chet, but I believe you have it backwards. Adiabatic processes, i.e. where heat does not flow down a temperature gradient, are reversible. The entropy does not change (much). The slow rise of the piston and the subsequent pause allowing temperatures to equalise are the irreversible steps. Entropy has increased and cannot be made to go back down again.
     
  21. May 2, 2014 #20
    Hi haruspex. I guess we have a disagreement between experts here. My understanding is that adiabatic processes are reversible only if they are carried out reversibly. Spontaneously allowing the piston to drop in the compression step is not a reversible process.

    The wording for the expansion step (raising the piston) is not too clear. Whether it was carried out reversibly or irreversibly is not an issue, since, in either case the entropy of the system increased between the initial and final states. However, just because the entropy of the system increased does not mean that the process was irreversible; the entropy of a system increasing or decreasing is not a criterion for reversibility or irreversibility of a process. Irrespective of whether raising the piston was reversible or irreversible, the change in entropy of the gas was MR ln[(H+l)/l], where M is the number of moles of gas. The process was reversible if the sum of the entropy changes for the system plus surroundings was zero.

    The problem statement is ambiguous regarding the expansion step: "The piston is slowly elevated a distance H from its equilibrium position and then kept in the elevated position long enough for the thermodynamic equilibrium to be reestablished." Raising the piston slowly has a suggestion that the step was close to reversible, but saying that it was kept in the elevated position long enough for the thermodynamic equilibirum to be reestablished suggests a certain degree of irreversibility. That's why, in my earlier post, I said "the process of raising the piston was done close to reversibly.

    In any event, this has nothing to do with the compression step which is clearly irreversible. The final answer to the problem is consistent with this.

    Chet
     
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