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Thermodynamics: Work Done by heating a cylinder and Heat transferred in an isobaric

  1. Oct 7, 2012 #1
    1. The problem statement, all variables and given/known data
    A cylinder with a frictionless piston contains 0.05 m3 of gas at 60 kPa. The spring holding the piston is now in tension. The cylinder is heated until the volume rises to 0.2 m3 and the pressure rises to 180 kPa.

    Assuming no losses in the system, and the force on the spring ot vary linearly with length, compute:

    a) The amount of work done by the gas
    b) The amount of work done if the system consists of the gas, the piston, and the spring


    2. Relevant equations
    W= [itex]\int pdv[/itex]
    dw = du + dq


    3. The attempt at a solution
    I used dw= du + dq, because I had trouble figuring out which pressure to use when integrating, because I got different answers depending on whether I calculated work with Po or Pf.

    dw = [itex](3/2)[/itex][itex]RdT[/itex] (from [itex]dU/dT = Cv = 3/2R[/itex]) + dT
    I then integrated this, from To to Tf. Using the ideal gas law, I solved for T in terms of the given P and V values, and assumed n=1. I got an answer of 45524 J, but this seems rather odd to be.

    b) I am confused where to start here, because you're putting energy into the system by adding heat, but then you're releasing it because the system is doing work on the surroundings...So would this just be 0, because of no losses?

    1. The problem statement, all variables and given/known data
    Calculate the heat energy transferred for an isobaric expansion.


    2. Relevant equations
    [itex] Q = CvΔT + nRΔT [/itex] (substituting in for ΔU and W)


    3. The attempt at a solution
    = [itex] ΔT(Cv+nR)[/itex]
    [itex] = ΔT[(3/2)(Rn)+(Rn)][/itex]
    [itex] = ΔT(5/2)(Rn)[/itex]
    [itex] = (5/2)(Rn) (Tf-To)[/itex]
    [itex] = (5/2)(Rn) (zTo-T0)[/itex]
    [itex] = (5/2)(Rn)(To)(z-1) [/itex]

    However, it says the answer is [itex] (3/2)(Rn)(To)(z-1) [/itex]

    Where am I messing up at?
     
  2. jcsd
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