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A question about position of piston after achieving thermal equilibrium

  1. Aug 18, 2012 #1
    A vertical cylindrical container contains some helium gas that 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?

    This question is quite complex to me and i am totally at a loss.
    i know the second process is an adiabatic process but not sure about the first,the answer is the piston dropped by 0.6H to 0.4H at the new equilibrium,but who can show me how to arrive at it?
     
  2. jcsd
  3. Aug 18, 2012 #2

    Andrew Mason

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    You are correct that the second part (after insulating the cylinder) is adiabatic.

    Assume an initial temperature T0. So what is the temperature of the gas after the first part? (Hint: easy). How do you determine the state of the gas (P, V, T) after the second part? Hint: this involves the [itex]\gamma[/itex] for Helium.

    AM
     
  4. Aug 18, 2012 #3
    during the first part,is the pressure constant? if it is,T1 would be T0(h+H)/h.
     
  5. Aug 18, 2012 #4
    After the adiabatic process,P1V1γ=P2V2γ,for monoatomic helium,γ is 1.67,but still not sure what process the first is,i think i am stuck at this point
     
  6. Aug 18, 2012 #5
    Thank you for the help,i presume the first part is an isothermal process hence T1=T0 and After the adiabatic process,P1V1γ=P2V2γ,for monoatomic helium,γ is 1.67,but still not sure how to calculate the volume change
     
  7. Aug 19, 2012 #6

    Andrew Mason

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    What is the initial pressure in terms of T0 and H (ie. after the isothermal expansion but before the adiabatic compression)? What is the condition that causes the piston to stop after it is released? (what are the forces on the piston? if the piston is stopped - not accelerating, what does that tell you about the sum of all the forces?) How is that related to pressure? How is volume related to H?

    Once you figure that out, all you have to do is apply [tex]P_1V_1^\gamma = P_2V_2^\gamma[/tex] to determine what the new volume will be in terms of H.

    AM
     
  8. Aug 19, 2012 #7
    assume the initial length of the container to be h,
    by p0v0=p1v1,i can get
    p1=p0h/h+H
    as p0=p2=patm,
    i have p0h/(h+H)*(h+H)1.67=p0*(h+kH)1.67,assuming new equilibrium is kH away from the old equilibrium position
    simplify the above eqn i have h3(h+H)2=(h+kH)5
    i think i would be able to get k=0.4 from this equation,but actually i am not very sure about how to solve it,could you give me hints on this final step? thank you
     
  9. Aug 19, 2012 #8

    Andrew Mason

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    In looking at this problem again, the adiabatic condition does not apply here as this is not a quasi-static adiabatic compression.

    Since the cylinder is insulated, no energy is lost from the system. The work done by the atmosphere and the piston ( by the change in gravitational potential energy of the piston at height H to its final position ) is converted into heat flow into the gas. I'll take a closer look at this and see if this is solvable.

    AM
     
  10. Aug 20, 2012 #9

    ehild

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    The final pressure is equal to the original one: pi=po+mg/A, the atmospheric pressure + the pressure due to the weight of the piston. (A is the cross section area of the piston).

    The force F=Apo+mg does the work when the piston is released from the elevated position. The work done by the external force when the piston moves from height H to height h is W=(Apo+mg)(H-h)=Api(H-h). This work is converted entirely into the internal energy of the helium gas. ΔU=W.

    The internal energy of n mol He is 3/2 nRT at temperature T. The change of internal energy is ΔU=3/2nR(Tf-Ti)

    Initially the gas had volume Vi, temperature Ti and pressure pi: Ti=PiVi/(nR). In the final equilibrium state, the volume is Vf=Vi+Ah, the pressure is equal to pi, so Tf=pi(Vi+Ah)/(nR).

    Write up the equation ΔU=W in terms of H, h, Vi, and pi and solve for h.

    ehild
     
  11. Aug 20, 2012 #10
    Thank you for suggesting a new method,i tried it,at first the equation looked promising but it turned out to be hugely complex and almost unsolvable as there were to many unknowns...
     
  12. Aug 20, 2012 #11

    ehild

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    What have you done? It is very simple. ΔU=3/2nR(Tf-Ti), W=Api(H-h)
    Ti=PiVi/(nR), Tf=pi(Vi+Ah)/(nR).

    Plug in and all unknowns cancel, except H and h.

    ehild
     
  13. Aug 20, 2012 #12
    oh,ok,i always thought i needed to assume the initial length of the cylinder to be ho or something...

    so actually the information about the first isothermal expansion is not useful except telling me pressure is not changed?

    why for the adiabatic process pi=pf,i thought pi should be lower than pf?
     
    Last edited: Aug 20, 2012
  14. Aug 20, 2012 #13

    ehild

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    The pressure changes in the first isothermal process, but it does not matter.
    The initial volume of the cylinder is assumed to be Vi, but it cancels at the end.

    ehild
     
  15. Aug 20, 2012 #14
    sorry one more question,if pressure before the isothermal expansion is the same as pressure after the adiabatic compression,and pressure changes in the first isothermal process,shouldn't pressure before and after the adiabatic process be different? why pi=pf ?
     
  16. Aug 20, 2012 #15

    ehild

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    The pressure is different before and after the final compression step, but the process is not quasi-static. The gas is not in equilibrium when the piston moves downward by its own weight, and as pressure is defined only for equilibrium, you can not speak about pressure of the gas during the process. The work on the gas is done by external forces, those caused by the atmospheric pressure and gravity.
    pi means the pressure before anything happened to the gas. pf is the final pressure after the piston has been released.

    ehild
     
    Last edited: Aug 20, 2012
  17. Aug 20, 2012 #16
    Oh,so your initial state actually refers to the state before the isothermal process,not the state just before the adiabatic process,i misunderstood it. now i understand your method.

    Thank you for answering so many questions, i think i learned a lot.
     
  18. Aug 20, 2012 #17

    ehild

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    You are welcome.

    Remember the change of internal energy is equal to the heat added and the work done on the gas: ΔU=Q+W. The work W is done by the external force, and it is -∫PdV only for quasi-static processes, when the pressure of the gas balances the external force exerted on the piston.

    ehild
     
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