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Work done by gas

  1. Sep 3, 2007 #1
    1. The problem statement, all variables and given/known data

    There is some gas in a baloon. The volume of the baloon is V and the gas is under a pressure of 2P0 ---where P0 is atmospheric pressure. The gas is allowed to come very slowly out of the ballon and eventually the gas pressure decrease to P0. How much work is done?
    Where does the energy required to do work come from?
    Thx


    2. Relevant equations

    PV=nRT

    3. The attempt at a solution

    My solution: gas comes out slowly so we can assume the process to be isothermal. So boyles law it applicable. So when gas pressure is halved the volume is doubled (2V).
    so work= integration of PdV
    = integration of K/V*dV (k=2PoV=boyle's constant)
    = K*In2 =2PoV*In2

    But the answer is PoV. Can you explain why?
     
  2. jcsd
  3. Sep 3, 2007 #2
    Don't forget the balloon.
     
  4. Sep 3, 2007 #3
    I can't get it. Someone please help!
     
  5. Sep 3, 2007 #4
    You are correct that at the end, the volume of the balloon is half of what it used to be. However, the source of the work isn't the gas inside the balloon! The balloon is under tension when it's filled, and it is this tension that does work. Draw a diagram, write some words...
     
  6. Sep 3, 2007 #5
    The tension of the ballon decreases as you let some gas out--- I understand that. But how that makes you get the work PoV (V2-V1=V)?
    It seems that the gas has expanded under a constant pressure of Po. W=Po*(del V) when net pressure is constant and the process is isothermal. Can you show that these are the case here?
     
    Last edited: Sep 3, 2007
  7. Sep 3, 2007 #6
    Remember that there is already a gas outside of the balloon, and that it's at P0.
     
  8. Oct 4, 2007 #7
    Is it the correct soulution to the problem:-
    W=delta(PV)
    =V delta P (since the volume remains unchanged)
    =V (2Po-Po)
    =PoV
     
  9. Oct 4, 2007 #8

    learningphysics

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    No, that doesn't look right to me. work is integral of PdV

    For work done on a system... as integral of PdV, P is the external pressure. hence you need to use 2Po as the P here.
     
    Last edited: Oct 4, 2007
  10. Oct 7, 2007 #9
    Any of you please show me the right way to solve it :-(.
     
  11. Oct 8, 2007 #10

    learningphysics

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    :( looks like I misunderstood the question... seems the volume stays the same while the gas leaks...

    Since work is -Pexternal*change in volume, and since change in volume is 0... wouldn't work be 0?

    I don't know how to get PoV.
     
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