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Finding the work

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  1. Sep 5, 2015 #1
    1. Find the work done by a gas if both pressure and volume increase at the same time, where the pressure rises as the square root of specific volume rises.


    I know that work done is the integral of pdV, but i'm not sure how to set up this same integral when there is a specific relationship where both pressure and volume increase at the same time.

    can someone help me out?

    the answer will be in terms of p and v, there are no numbers in the problem.
     
  2. jcsd
  3. Sep 5, 2015 #2

    haruspex

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    If the pressure starts at P0 and the volume starts at V0, what, according to the given information, is the pressure when the volume is V?
     
  4. Sep 5, 2015 #3
    The numbers are arbitrary, the volume and pressure are both expressed as P2 and V2 when it reaches it's final state
     
  5. Sep 5, 2015 #4

    haruspex

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    I don't see how that stops you from answering my question.
     
  6. Sep 5, 2015 #5
    The pressure would be P^2 when the volume is V
     
  7. Sep 5, 2015 #6

    haruspex

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    No, the pressure is P.
    If the volume quadruples, say, what happens to the pressure?
     
  8. Sep 6, 2015 #7
    Would the pressure be 2P?
     
  9. Sep 6, 2015 #8

    haruspex

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    The pressure would double, yes, but it is confusing to write "the pressure is 2P". We need P to be a variable, standing for different pressures at different volumes. If the initial pressure and volume are P0 and V0 then when V=4V0 we will have P = 2P0.

    So, in general for this question, if the initial pressure and volume are P0 and V0, what will P be when the volume is V?
     
  10. Sep 6, 2015 #9
    P will be equal to the square root of V?
     
  11. Sep 6, 2015 #10
    The term "pressure rises as the square root of specific volume rises" means that the pressure is proportional to the square root of V (not equal to it). If, for an arbitrary state of the system, the pressure is P and the volume is V, what is the relationship between P and V (given P is proportional to the square root of V; you can use a proportionality constant k in your equation)?

    Chet
     
  12. Sep 6, 2015 #11
    Well, the proportionality constant would equal the square root of k
     
  13. Sep 6, 2015 #12
    Well, the proportionality constant would equal the square root of k
     
  14. Sep 6, 2015 #13

    haruspex

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    No, k is the proportionality constant. It cannot be equal to the square root of itself (unless it is 1).
    You wrote that the pressure would be equal to the square root of V. That cannot be right because V is a volume and P is a pressure, and the square root of a volume is not a pressure. The proportionality constant fixes that up. So instead of ##P=\sqrt V##, what would you write? You just need a factor k.
     
  15. Sep 6, 2015 #14
    Would it be P= (P0/√V0)*V?
     
  16. Sep 6, 2015 #15

    haruspex

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    Not quite. If you were to quadruple V in that equation, would P double?
     
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