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Thermodynamics- please help

  1. May 18, 2009 #1

    C.E

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    A thermally insulating horizontal tube of cross-sectional area A is filled with a
    volume 2V of gas at a pressure p. At the centre of the tube is a freely sliding piston
    of mass M which seals the left and right hand sides of the tube from each other (i.e.
    at equilibrium there is a volume V either side of the piston).

    The piston is given a small displacement x to the left and then released. Assuming
    that friction has a negligible effect over many cycles:

    (i) Show that the force F on the piston is given by:

    F=[tex]\frac{-2xpA^2}{\gamma V}[/tex]

    [tex]\gamma[/tex]= C_p/ C_v

    ii) and comment on the nature of the motion. {1}

    (iii) Derive the expression for [tex]\omega[/tex] the frequency of oscillation.

    3. I don't know how to start this, as the gas is not Ideal I have no idea what formulas to apply to it. I know that Q=0 and think the work done =0. Is this right? Can somebody please help me get started?
     
    Last edited: May 18, 2009
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  3. May 18, 2009 #2

    Mapes

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    If the piston is moved through a displacement against a resisting pressure, I wouldn't say that the work done is zero. But I agree that Q = 0 is a reasonable assumption. What would we call such a process, and what are the applicable equations?
     
  4. May 18, 2009 #3

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    Is the process is adiabatic? I have only been taught equations which apply to ideal gasses in adiabatic processes, are there others? Oh, I have learn't that change in internal energy= - work done though, I don't see how that is much help. Do I need to assume the gas is ideal?
     
  5. May 18, 2009 #4

    Mapes

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    Yes, assume an adiabatic process on an ideal gas.
     
  6. May 18, 2009 #5

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    I have been shown the following.

    [tex]pv^{\gamma}=constant[/tex] - 1

    [tex]pT^{\gamma -1} =constant[/tex] - 2

    Work=n[tex]C_v(T_1-T_2)[/tex] -3

    and

    Work= [tex]\frac{C_v}{R}(P_1V_1-P_2V_2)[/tex] - 4

    I don't think 2, 3, 4 are relevant as we do not know anything about the temperature/ the temperature change, is this right? I have tried playing around with 1 substituting it into the following. Force = A(P1-p2) (where P1 is the pressure in the left side of the tube and P2 is the pressure in the right hand side) but I did not get anywhere, what do you suggest I try now?
     
  7. May 18, 2009 #6

    Mapes

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    How would you approximate

    [tex](V-xA)^\gamma[/tex]

    when [itex]x[/itex] is small?
     
  8. May 18, 2009 #7

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    I would do the following, is this what you are thinking of?

    [tex](v-Ax)^{\gamma} = v^{\gamma}(1-\frac{Ax}{v})^{\gamma}[/tex]

    Therefore [tex](v-Ax)^{\gamma}[/tex] can be approximated by the following:

    [tex]v^{\gamma}(1 + \gamma \times A \times x/v)[/tex]

    =[tex]v^{\gamma}[/tex] + gamma [tex]v^{\gamma}[/tex]Ax/v
     
  9. May 18, 2009 #8

    Mapes

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    You need to get the [itex]\gamma[/itex] as a multiplicative factor rather than an exponent. Are you familiar with expanding a function as a Taylor series?

    (Hint: A Taylor series would give [itex](v-Ax)^{\gamma}\approx v^\gamma-Ax\gamma[/itex] above.)
     
  10. May 18, 2009 #9

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    I have seen Taylor series but not in much detail. Which point are you expanding the Taylor series about?
     
  11. May 18, 2009 #10

    Mapes

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    V ("small displacement").
     
  12. May 18, 2009 #11

    C.E

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    Is this the way to progress with the question?

    [tex]\ (v-Ax)^{\gamma}\approx v^\gamma-Ax\gamma[/tex]

    and [tex]\ (v+Ax)^{\gamma}\approx v^\gamma+Ax\gamma[/tex]

    Therefore v2-v1 [tex]\approx[/tex] -2Ax[tex]\gamma[/tex]

    Is there a way to directly relate p1 to v1 and p2 to v2 given that I don't know the temperature or the number of moles of gas?
     
    Last edited: May 18, 2009
  13. May 18, 2009 #12

    Mapes

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    Yes, your equation (1) above.
     
  14. May 18, 2009 #13

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    Using (1) I get that:

    p1v1^(gamma)=p2v2^(gamma)

    How can I use this?
     
  15. May 18, 2009 #14

    Mapes

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    Plug in the volume after a small displacement. Use Taylor series approximation. What is the new pressure? How is this related to the force?
     
  16. May 18, 2009 #15

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    Am I plugging in the volume of the left or right hand side of the cylinder after the displacement?
    force= pressure x area in this case A x pressure.
     
  17. May 18, 2009 #16

    Mapes

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    Generally you would calculate the force related to the left side, calculate the force related to the right side, and add them together. However, you may notice that the system is symmetric for small displacements (i.e., the forces are equal from both sides and additive), which simplifies things.
     
  18. May 18, 2009 #17

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    for the left hand side I get:

    p1= [tex]\ v^{\gamma}[/tex]p/[tex]\ ( v^{\gamma}-Ax\gamma[/tex])


    for the right hand side I get:

    p2= [tex]\ v^{\gamma}[/tex]p/[tex]\ (v^{\gamma}+Ax\gamma[/tex])


    Is this right? Do I just add these and multiply by A?
     
  19. May 18, 2009 #18

    Mapes

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    It would be a little easier if the equations were arranged so that the two-term parts are in the numerator rather than the denominator. Adding them as they are doesn't really get you anywhere, I don't think.
     
  20. May 18, 2009 #19

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    Adding the two equations I get:
    p1([tex]\ v^{\gamma}-Ax \gamma[/tex]) + p2([tex]\ v^{\gamma}+Ax \gamma[/tex])=2p[tex]v^{\gamma}[/tex]
    Is this right? How do I show this is equivalent to the expression given in the question?
     
  21. May 18, 2009 #20

    Mapes

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    I recommend solving for p1 and p2 in a way that has the two-term parts in the numerators, rather than the denominators. This will make it easier to find their difference.
     
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