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PV diagram and circuit process!

  1. Mar 17, 2009 #1
    A circuit process?

    A gas consisting of two atoms have gamma = 7/5. It is used as "worksubstance" for the following steps:
    I: Isobar expansion from V1 = 1.00m3 to V2 = 2.00m3 with P1 = 3.00*105Pa
    II: adiabatic expansion to (P3, V3), and
    III: compression back to (P1,V1), as a parabola with bottompoint where (Pmin = 1.00*105 Pa, and V= V2 = 2.00m3)

    First of all: draw the "circuit" in a PV-diagram. (I use circuit in lack of a better word, suggestions?) :) This goes okey, and then: Find an equation for the "adiabatic" in step II, and for the parabola in step III. Then find the volume V3 and the pressure P3.

    I cant get past this point.. Excuse my poor english, as this assignment is given in a different language and I therefore had to translate it :) Shout if there are things you dont understand.

    2. Relevant equations:
    For isobar process: W = p(V2 - V1) and for abdiatic processes: P1*V1gamma = P2*V2gamma, is this all I really need?

    3. The attempt at a solution:

    My attempt is really poor really. Ive painted a quick scetch of the diagram, and it is attached to the thread. My guesses is that the work W < 0 in II and III, and W > 0 in I?

    I really dont know how to find the equations for the abdiatic, or the parabola. My guesses is that I should end up with two equations with V3 and P3 as unknowns, and solve.

    It would be really helpful if you could at least point me in the right direction! :)
     

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    Last edited: Mar 17, 2009
  2. jcsd
  3. Mar 18, 2009 #2

    Redbelly98

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    Welcome to PF :smile:

    In English, "cycle" is the term usually used.

    Almost.

    The adiabatic path does not actually contain the point (V1,P1) as defined in the problem statement, so that equation needs to be modified. But you do have the right idea.

    Also, you'll need the equation for the parabola. They tell you where it's minimum value is, and we also know it contains the point (V1,P1). You can use that information to find it's equation.
     
  4. Mar 18, 2009 #3
    Thank you sir! :blushing: And how could I forget the word cycle, as it "fits" way better. :smile:

    At the moment I'm focusing on the first problem (that really should be an easy nut to crack, but it got me stuck).

    Let me rephrase the statemen then: I know that the abdiatic path gives pVgamma = constant, but how does this help me? I'm really intrigued that this gives me such a hard time..:grumpy:
     
  5. Mar 18, 2009 #4

    Redbelly98

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    You're welcome.
    Since we know P=P1 and V=V2 at the beginning of the adiabatic path, we can calculate what the constant is in the equation

    pVgamma = constant
     
  6. Mar 18, 2009 #5
    So the constant equals 6.0*105, but how does this help me further? Should I set P3*V3gamma = "the constant"? But then I have an equation with two unknowns.

    I really appreciate your help RedBelly :smile:
     
  7. Mar 18, 2009 #6

    Redbelly98

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    At this point, we're working on the 2nd question, "Find an equation for the adiabatic in step II". So don't worry about P3 and V3 yet, it's just P and V at this point.
     
  8. Mar 19, 2009 #7
    Okey, with that in mind I'll try to solve problem two.

    Equation for parabola: y = a(x - h)2 + k.
    h and k are respectively the coordinates for the vertex, that gives:

    y = a(x - Pmin)2 + V2.

    Since the P1 and V1 is a point on the graph of the line it gives that:
    V1 = a(P1 - Pmin)2 + V2.
    This gives:

    a = (V1 - V2)/(P1 - Pmin)2

    Now that I have a, I can set up a similar equation for P3, V3, since it is a point on the line of the graph? Sorry for my messy maths, as I was on my way out the door :shy:

    V3 = a(P3 - Pmin)2 + V2.
    Edit: no, this can't be correct? After som trying and failing, when I substitute either V3, or P3 with p3*V3gamma = constant, I get a really really messy equation =/
     
    Last edited: Mar 19, 2009
  9. Mar 19, 2009 #8

    Redbelly98

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    We should be plotting P along the vertical y-axis, and V along the horizontal x-axis. You have P and V reversed.

    Except for that, it looks like you have the right idea.
     
  10. Apr 21, 2009 #9
    Well, I'm sorry that I'm reply'ing this late, but I just wanted to give this thread some closure. I managed to do the assignment (with your held Redbelly, and some other students help). After finding the equation for the parabola, we had to solve the equation with either Newtons Method or graphically. This assignment was pretty much over our head really :) But thanks for helping out!
     
  11. Apr 21, 2009 #10

    Redbelly98

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    Glad it worked out :smile:
     
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