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Thermodynamics adiabate

  1. Mar 20, 2008 #1
    Thermodynamics adiabat - help

    1. The problem statement, all variables and given/known data
    Vertical cylindrical bottle is closed by a fixed piston (mass m). Under the piston in volume [itex]V_1[/itex] there is 1 mole of ideal 1-atomic gas with pressure [itex]P_1[/itex]. Over the piston there is vacuum, cross section area is equal to S. The piston becomes free and begins moving. There goes an adiabatic process. Friction between piston and walls of bottle is very small. Calculate gas temperature in the moment of first stop of the piston.


    2. Relevant equations
    We have a system of three equations with three variables.
    [tex] p(V_1+Sh)=RT [/tex]
    [tex] mgh=c_v(T-T_1)[/tex]
    [tex] p_1V_1^\frac {5} {3} = pV^ \frac {5} {3}[/tex]
    First is equation of state, second is energy conservation, third - adiabatic process. h - difference in height of piston before moving and in the moment of first stop. [itex]T_1[/itex] is gas temperature in the beginning. [itex]c_v[/itex] - heat capacity in constant volume process (it's equal in this case to [itex]\frac {3} {2}R[/itex]).


    3. The attempt at a solution
    After some algebra I got a single equation determining T:
    [tex]T^3(aT+b)^2=c[/tex]
    where a, b, c - constants equal to:
    [tex]a=3RS[/tex]
    [tex]b=(2mg-3p_1S)V_1[/tex]
    [tex]c=T_1^3V_1^2m^2g^2[/tex]
    It seems to me that this equation has one real solution but I can't find it.

    Please help me.

    Thank you.
     
    Last edited: Mar 20, 2008
  2. jcsd
  3. Mar 20, 2008 #2

    Mapes

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    You can write another equation addressing mechanical equilibrium.

    EDIT: Also, I would write [tex] mgh=-c_v(T-T_1)[/tex] instead.
     
    Last edited: Mar 20, 2008
  4. Mar 21, 2008 #3
    What mechanical equilibrium do you refer to? If you suggest something like
    [itex]p_1S=mg[/itex], I don't think it's right because in the moment of first stop we don't have mechanical equilibrium. If it was the case then the first stop would be the final stop because nothing would force the piston to move again. My previous problem was just the same but I was obliged to find gas temperature after the final stop - it's quite simple and actually there is mechanical equilibrium.
     
  5. Mar 21, 2008 #4

    Mapes

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    Got it. Are you envisioning the piston gaining speed, passing the [itex]
    p_1S=mg
    [/itex] point, and decelerating due to gravity?
     
  6. Mar 21, 2008 #5
    Yes, I think the piston will gain speed, then pass the equilibrium point but because it has a non-zero kinetic energy, it will continue to move after that point - so in fact it will be a nonlinear oscillation. The moment of first stop will be the moment when the piston turns around to move back to an equilibrium point after passing that point in the first time.
     
  7. Mar 21, 2008 #6

    Mapes

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    Wouldn't c be four times the value you have, from when you take care of [itex]\left(\frac{3}{2}R\right)^2[/itex]? Also, what do you think about the earlier comment about switching the sign? I can have Mathematica pick out the real solution.
     
  8. Mar 24, 2008 #7
    Maybe, I've got wrong idea but I think I should reconsider energy conservation equation.
    It seems to me that in the moment of first stop of the piston during its oscillations we should consider not only its potential energy but also its kinetic energy which has maximum at this point. Am I right? If yes, then it requires more change than just switching the sign (I agree with it).
     
  9. Mar 24, 2008 #8

    Mapes

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    Wouldn't the kinetic energy be zero when the piston comes to a stop? I think you were on the right track originally.

    Incidentally, Mathematica can solve your equations but the real solution ends up being quite complex and many lines long.
     
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