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PV diagram net work and heat

  1. Mar 9, 2010 #1
    1. The problem statement, all variables and given/known data

    A sample of an ideal gas is taken through the thermodynamic cycle
    shown on the P-V diagram at attached. Two of the “legs” of the cycle are
    isobaric and the other two are isothermal.


    a) Find the net work done by the gas along the high pressure isobaric
    leg in terms of p0 and Vo.


    b) Find the net work done by the gas in one complete cycle in terms of
    p0 and Vo.

    c) Find the net heat supplied to the gas in one cycle in terms of p0 and
    Vo.



    2. Relevant equations



    3. The attempt at a solution

    Part A)

    W = integral fr v to 3v of Pdv = P(3v - v) = 2PV

    Part B) Not sure if this is crrect do i just find the work for each leg of the of the cycle and then sum them up

    Ill start with the high pressure isbaric leg and work around the cycle clockwise

    Leg 1 = 2PV (Part a)

    Leg 2 = not sure how to integrate this if i dont knw what the v final is.... i guess im doing this wrong
     

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  3. Mar 9, 2010 #2

    ideasrule

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    Homework Helper

    Kind of, but you have to consider that sometimes the work is positive and sometimes it's negative. Remember that work is equal to the area enclosed by the PV curve and you can adjust your positives and negatives accordingly.

    You do know what the final v is because PV=nRT, and if temperature is constant, then P=C/V.
     
  4. Mar 9, 2010 #3
    so to find the area, i woud add the first two integrals then subracts the second two.

    for the second integral my book says fr a isothermal process W = nRT(ln V_B/V_A), so does that simplify to ln V_B/V_A because it is constant and nRT is essentially 1

    for the third leg, the pressure does nt change but the volue does so it is nt constand, how do i find the v final fr this integral
     
  5. Mar 9, 2010 #4

    ideasrule

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    Homework Helper

    Yes.

    How do you know nRT is 1? nRT is equal to PV, and you know PV.

    You're supposed to read it from the graph. It looks like 2Vo to me.
     
  6. Mar 9, 2010 #5
    of first off V_b is the secnd point in the cycle and V_c is the third point

    W = nRT ln V_c/V_b = PV ln Vc/Vb

    P(3V) = .5P(Vc) therefore Vc = 6V and Vb = 3V

    W = PV ln Vc/Vb = PV ln 6V/3V = PV ln2
     
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