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Work by an ideal gas, thermodynamics

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data
    abcd.jpg
    At point D: P=2atm, T=360K, n=2mol
    At point B: V=3VD, P=2Pc
    Paths AB and CD represent isothermal processes
    The gas is carried through a complete cycle along the path DABCD. Determine the total amount of work done by the gas.

    2. Relevant equations
    PV=nRT
    W=nRT*ln(V2/V1) (by integration)

    3. The attempt at a solution
    I only need the work for the two curved sections (because the work by the vertical ones is zero). Using a system of equations, I determined the following (in the form PV=nRT):
    A: Pa*29.6=2*R*721
    B: (4/3)*88.8=2*R*721
    C: (2/3)*88.8=2*R*360
    D: 2*29.6=2*R*360

    Using the work equation, I get Wab=-130J and Wcd=64.9J

    But the solutions in my book say I should have -13.2kJ and 6.58kJ, respctively


    I am pretty sure I solved for the variables correctly; am I not using the correct formula to find work? Or is there something I've missed....
     
  2. jcsd
  3. Nov 18, 2009 #2

    tiny-tim

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    Hi v0id19! :smile:
    Perhaps I'm misunderstanding the problem, but isn't work done = ∫PdV, the area inside the graph? :confused:
     
  4. Nov 18, 2009 #3
    yes, but I don't have any of the equations for the lines on the graph.
    using the energy equation (E=Qin+W). for an isothermal process E=0, so Qin=-W, and W=PdV=∫(nRT)/V*dV=nRT*ln(V2/V1)
     
  5. Nov 18, 2009 #4
    actually--i just redid the problem, it turns out i was using the wrong gas constant R...
     
  6. Nov 19, 2009 #5

    tiny-tim

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    Hi v0id19! :smile:

    (just got up :zzz: …)
    But you are given the width and height of the region as proportions of the (x,y) coordinates of D. :wink:

    (and, assuming AB and DC are meant to be "parallel", the area is width times height)
     
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