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Thermodynamics piston problem

  1. May 1, 2015 #1
    1. The problem statement, all variables and given/known data
    An adiabatic piston of mass m equally divides an insulated container of volume V0 and length l filled with Helium.The initial pressures on both sides of the piston is P0 and the piston is connected to a spring of constant k. The container starts moving with acceleration a. Find the stretch in spring when acceleration of piston becomes a. Assume displacement of piston << l.
    Untitled.png
    2. Relevant equations
    For adiabatic process, ##PV^{\gamma}##=constant

    3. The attempt at a solution
    using above equation, if piston is displaced by x towards left, ##P_1=P\frac{V^{\gamma}}{V^{\gamma}-2Ax}##
    similarly, for right portion, ##P_2=P\frac{V^{\gamma}}{V^{\gamma}+2Ax}##
    Now A=V/l
    substituting, ##P_1=P(1-2x/l)^{\gamma}=P(1-\frac{2x\gamma}{l})##
    similarly, ##P_2=P(1+2x/l)^{\gamma}=P(1+\frac{2x\gamma}{l})##
    Now ##\Delta P=4\gamma P/l##
    and from newton's law, ##\Delta PA+kx=ma##
    so ##4x\gamma \frac{P}{l}\frac{V}{l}+kx=ma##
    so $$x=\frac{ma}{K+\frac{4P_0V_0\gamma}{l^2}}$$
    But answer given is: $$x=\frac{ma}{K+\frac{8P_0V_0\gamma}{l^2}}$$
     
  2. jcsd
  3. May 1, 2015 #2

    Delta²

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    There seems to be some mistakes or typos in the algebra involved. The denominator in the first 2 equations should be ##({V+-2Ax})^\gamma## . I seem to get ##P_1=P(1-2x/l)^{-\gamma}## and similar for P2.
     
    Last edited: May 1, 2015
  4. May 1, 2015 #3

    haruspex

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    Do you mean ##P(1-2x/l)^{-\gamma}##?
     
  5. May 1, 2015 #4
    sorry... ##P_1=(1-2x/l)^{-\gamma}=(1+2x\gamma/l)##
    Answer is still same.
     
  6. May 1, 2015 #5

    Delta²

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    You use the approximation ##(1-2x/l)^{-\gamma}\approx(1+2x\gamma/l)## which i think its valid only if the exponent is positive and 2x<<l.
     
  7. May 1, 2015 #6
    yes 2x<<l.
    also, ##(1-x)^{n}=1-\frac{n}{1!}x+\frac{n(n+1)}{2!}x^2-##.This expansion is true for both positive and negative rational numbers and for negetive integers.
    so, if x is small, then you can neglect all terms starting from 3rd term which gives you ##1-nx##
     
  8. May 1, 2015 #7

    haruspex

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    I'm inclined to agree with your answer, it's 4, not 8.
     
  9. May 1, 2015 #8

    Delta²

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    Well the only "easy" thing i see and could fix the result is that the V0 might refer to the volume of the half container, so that it is ##A=2V_0/l##. Maybe check the excersice description again?
     
  10. May 1, 2015 #9
    The question is correct and i checked it again. The answer given is wrong.
     
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