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Combined gas law

  1. Dec 18, 2007 #1

    OmCheeto

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    I'm a bit confused about the cgl.

    If I pump 3 liters of air into a 3 liter cylinder already filled with 3 liters of air, what will be the temperature rise? Assuming an initial temperature of 300'K. It seems like from practice, the temperature goes up. Is the temperature rise due strictly to the friction in the pump?

    Should I break this down into two separate problems? the physics of the pump and that of the tank?
     
    Last edited: Dec 19, 2007
  2. jcsd
  3. Dec 19, 2007 #2
    You'll sorta have to describe your pumping process a little bit more.
     
  4. Dec 19, 2007 #3

    OmCheeto

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    hmmm... those german rotary compressors seem a bit spendy, so I was thinking in the line of the piston pumpers you can buy for $14.99 at shucks.

    I looked in my physics book and they described the adiabatic process where there is no temperature rise, but that seemed a bit too perfect.

    I guess I'm looking for a real world temperature vs pressure curve.
     
  5. Dec 19, 2007 #4

    stewartcs

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    What are you confused about exactly? The CGL is pretty straight forward.

    Yes the temperature goes up.

    No, the temperature increase is due to the addition of energy as a result of the compression of the gas. Since the gas is being added by a pump, you can think of it as a piston compressing a given quantity of gas. So as the piston compresses the gas in the reservoir (tank), negative work is done and the internal energy of the gas is increased.

    Be aware that there are various ways to model the change from the intial state to final state of the gas depending on the assumptions made.

    Not really, it's probably not necessary. I don't think the seals of the piston would add that much heat from friction.
     
  6. Dec 19, 2007 #5

    stewartcs

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    In an adiabatic process, there can be a temperature rise internally. An adiabatic process is one in which there is no transfer of energy as heat between the system and the enviornment (i.e. one in which the expansion or compression is very rapid, or the system is well insulated).
     
  7. Dec 19, 2007 #6

    OmCheeto

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    I guess that's why I asked for a real world answer, and why TVP45 asked for the pumping process. I will go back to my text and try and figure out where a temperature rise can occur in the perfect system.
     
  8. Dec 20, 2007 #7

    stewartcs

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    I just explained where the temperature rise comes from in post #4. Was that not clear enough? If not we can try again.

    Your physics book should have some good examples also. If not, a google search should pull up some more info.
     
  9. Dec 20, 2007 #8

    OmCheeto

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    I'm starting to feel like Will Smiths character in I Robot.
    I seem to not know what the correct question is.

    I re-read the section on adiabatic processes and see that you are correct.

    I think my confusion comes from reading my book. They speak either of constant temperature or of constant pressure, or in the real world scenario they give initial and final states of the system where both change, but not why they changed as they did.

    Googling actually brought me to this forums archives where Pengwuino was asking basically the same thing: https://www.physicsforums.com/archive/index.php/t-81104.html
    (though I'm not interested in creating liquid nitrogen)
    There I found the requirement of a second equation: PV^(gamma)=k
    where gamma = 7/5 in my case
    They reference: http://en.wikipedia.org/wiki/Adiabatic_process
    and http://en.wikipedia.org/wiki/Ideal_gas_law
    where we get pV=nRT

    The combined gas law doesn't really tell me much as indicated by this calculator: http://www.1728.com/combined.htm
    It just tells you that you have to know the value of 5 variables to get the 6th.
    Which is totally useless. I can even use it to prove you wrong: Cut the volume in half and double the pressure and there is no temperature change.

    Sooner or later I will determine what my question should be.
    Anyways, I'm late for work. I'll work on the solution later.
     
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