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Question including ideal gas law, change in energy/work.

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data

    This question would be impossible to solve without the diagram, which I can't post so instead i'll just tell you what I have so far.

    gas springs are commonly used on trunk lids of cars. They are a closed system consisting of a pressure tube and piston rod with piston, as well as nitrogen gas as an energy source.
    if the nitrogen gas is compressed from initial volume Vi and initial pressure Pi to final volume Vf, as the trunk lid is slowly closed find the formula for how much energy is stored in the gas spring?

    2. Relevant equations

    pV=nRT
    W=pV


    3. The attempt at a solution

    what I have so far is

    dE(energy) = dW = -p(dV) Why is this -pdV?
    pV=nRT

    T= constant ---> pV = constant = PiVi Not sure what this means either

    E=-∫(Vf-Vi)pdV = -PiVi∫(Vf-Vi)(dV/V) = -PiVi[ln(Vf)-ln(Vi)]

    again i know this is to integrate dE into just E, but I don't really understand this process, especially E=-∫(Vf-Vi)pdV = -PiVi∫(Vf-Vi)(dV/V)

    therefore E=-PiVi*ln(Vf/Vi)

    So yes i'm not looking for the answer since i have it im just confused about the work to get to the answer

    PS ∫(Vf-Vi) just means ∫(Vf over Vi)
     
  2. jcsd
  3. Oct 6, 2012 #2

    rude man

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    You need to inform us if the piston was thermally insulated, whether the piston was maintained at a constant temperature, or whatever.
     
  4. Oct 6, 2012 #3
    There is no mention of that in the question, though it does say the gas spring is a closed system, wouldn't that mean constant temperature of the piston once equilibrium is reached?

    And my main problem is just what exactly the formulas im using mean, not so much how to solve it
     
  5. Oct 7, 2012 #4

    rude man

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    because pdV is the work done BY the system (the gas). So that work must result in a REDUCTION in energy of the remaining gas.

    And btw that equation tells me this is an adiabatic process, i.e. a thermally insulated one. Why? Look at the first law with Q = 0. Q=0 defines an adiabatic (also called isentropic) process.

    That being the case, what formula involving p and V did you learn that applies to such a process? Hint: T is NOT constant.
    It would mean the product pV idoes not change as pressure and volume change. Like, if the volume is squished to half the original volume, the pressure doubles. Etc. But that applies to an isothermal (constant temperature process. That is not your case here.

    E=-∫(Vf-Vi)pdV = -PiVi∫(Vf-Vi)(dV/V) = -PiVi[ln(Vf)-ln(Vi)]
    [/quote]
    That formula makes no sense. Where did you dig that up?
     
  6. Oct 7, 2012 #5
    That formula makes no sense. Where did you dig that up?[/QUOTE]

    I'm very confused because this is what was the work that was provided to solve the equation and I copied it down hoping to understand it better but it isn't really helping to just read ovver it.
     
  7. Oct 7, 2012 #6

    rude man

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    I'm very confused because this is what was the work that was provided to solve the equation and I copied it down hoping to understand it better but it isn't really helping to just read ovver it.[/QUOTE]

    You wrote:
    E=-∫(Vf-Vi)pdV = -PiVi∫(Vf-Vi)(dV/V) = -PiVi[ln(Vf)-ln(Vi)]

    OK I see the problem. You meant for Vf and Vi to be the limits of integration. I would suggest using "go advanced" and write ∫ViVfpdV instead. (The smart ones on this forum use itex or whatever but personally I'm too lazy to pick it up.)

    Unfortunately, this is an adiabatic process (dQ = 0), so you can't write pV = constant, which you did in going from your first to your second integral above.

    I ask again: what relation between p and V might you have learned exists for an adiabatic process?
     
  8. Oct 7, 2012 #7
    You wrote:
    E=-∫(Vf-Vi)pdV = -PiVi∫(Vf-Vi)(dV/V) = -PiVi[ln(Vf)-ln(Vi)]

    OK I see the problem. You meant for Vf and Vi to be the limits of integration. I would suggest using "go advanced" and write ∫ViVfpdV instead. (The smart ones on this forum use itex or whatever but personally I'm too lazy to pick it up.)

    Unfortunately, this is an adiabatic process (dQ = 0), so you can't write pV = constant, which you did in going from your first to your second integral above.

    I ask again: what relation between p and V might you have learned exists for an adiabatic process?[/QUOTE]

    I know of the equation

    PiVi^γ=PfVf^y

    However I do not think there is anything wrong with the solution because this is the solution given, im just confused as to how the solution was obtained.

    In addition, this question was before we learned adiabatic processes so im a bit skeptical about needing to account for it.
     
  9. Oct 7, 2012 #8

    rude man

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    Then you must be right, and the gas is kept at a constant temperature in going from Vi to Vf. But I noticed you gave dE = -dW at the beginning, and that equation holds only for an adiabatic process if E is the internal energy U of the system. Possibly your prof/text used E = U + Q where U is internal energy channge and Q is heat added, so that would explain things.

    The thing is, you never defined if the gas/spring system was thermally insulated or whether it was kept at a constant temperature in going from Vi to Vf. That has nothing to do with "settling" of the system's temperature. It has to do whether heat Q is added/subtracted, or not.

    So it looks to me like you're in good shape - anything still puzzling you?
     
  10. Oct 7, 2012 #9
    Yeah I guess this question is nearly impossible to figure out for you without the diagram unfortunately. I guess I will just go find a TA to figure it out. Thanks alot for the help.
     
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