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Work done on a gas

  1. Apr 29, 2007 #1
    1. The problem statement, all variables and given/known data
    A cylinder with initial volume V contains a sample of gas at pressure p. The gas is heated in such a way that its pressure is directly proportional to its volume. After the gas reaches the volume 3V and pressure 3p, it is cooled isobarically to its original volume V. The gas is then cooled isochorically until it returns to the original volume and pressure.
    Find the work W done on the gas during the entire process.

    2. Relevant equations
    W = 0 (for the isochoric cooling)
    W = -p(V_f - V_i) for the isobaric cooling
    W = -nRT * ln (V_f/V_i) <---- this is what I'm unsure about.

    3. The attempt at a solution

    Total Work is going to equal the sum of the 3 work on the 3 processes:
    the expansion
    the cooling
    the further cooling

    Expansion process (isothermal?)
    W_1 = -nRT * ln (V_f/V_i)
    pV = nRT
    W_1 = -pV * ln (V_f/V_i)
    This is wrong but I don't know why. I don't know for sure that this an isothermal process. But it's not isochoric (because V changes from V to 3V). It's not isobaric because pressure increasing from p to 3p. It's not adiabatic because Q > 0, I think; it's being heated after all. So isothermal seems to be the right answer.

    For the second process
    W_2 = -p(V_f - V_i) = -3p(V - 3V)
    That I'm sure is right

    For the third process
    It's isochoric, so 0 work is being done.
  2. jcsd
  3. Apr 29, 2007 #2

    Andrew Mason

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    This is the problem, as you suspected. It would help to draw a PV diagram.

    dW = PdV where dW is the incremental work done BY the gas.

    Since PV=nRT, and since V/P = k (constant), what does the P-V graph look like? How do you calculate the area under that part of the graph? What does the area represent?


  4. Apr 30, 2007 #3
    The problem gave me that equation.
    But when you integrate it, it comes out to be:
    W = -p_i*V_i * ln(V_f/V_i) = -p_f*V_f * ln(V_f/V_i)
    I even have a page in my physics book that derives it for me. But according to masteringphysics, that's wrong.

    But if I draw a P-V graph, it doesn't look like that at all, assuming I did it right. V/P is constant, so it has a constant slope, so it'll just be a triangle. Hence I think the equation is...
    -(1/2)(pf - pi)(Vf-Vi) = -2pV
    Is that right?
  5. Apr 30, 2007 #4

    Andrew Mason

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    Not quite. If V/P = k then PdV = VdV/k . If you integrate that, you get:

    [tex]W = \frac{1}{2k}(V_f^2 - V_0^2) [/tex]

    Since [itex]V = kP[/itex] this works out to:

    [tex]W = \frac{1}{2}(P_fV_f - P_0V_0) [/tex]

    The work done by the gas is all of the area under the graph for each path (when it compresses, the work done is negative (ie. done on the gas) so you have to subtract that area, leaving the area in between the paths as the net work in the cycle.

    You are quite right that you can see this from the straight line graph. That is why you should always do a PV graph!

    Last edited: Apr 30, 2007
  6. Apr 30, 2007 #5
    Thanks for the solution. But I don't understand why your integral is so much different from my textbooks. I'll have to bring it up in discussion. Thank you though.
  7. Apr 30, 2007 #6

    Andrew Mason

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    You have to determine W2 and W3 and add them to W1 (applying the signs correctly - you actually subtract the area under the second path). This gives you the net area - the area between the path lines. It is the area between the paths that gives you the work done on/by the gas.

    If you draw the path on a PV diagram, you get a triangle between the paths. The area of that triangle is negative (since the area under the compression phase, which is negative work by the gas, is greater the area under the expansion phase, which is positive).

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