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Homework Help: Work Done By Gas

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data
    An ideal gas is taken clockwise around the circular path shown in the figure.

    http://img529.imageshack.us/img529/5513/rw1864bp8.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

    How much work does the gas do?
    If there are 1.3 mol of gas, what is the maximum temperature reached?

    2. Relevant equations

    Not sure.

    3. The attempt at a solution

    The problem does not specify what type of process this is, so I don't know which equations to apply and how to handle this problem at all. Please help.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Nov 8, 2008 #2


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    It's a way of plotting a continual change in pressure (or volume if you prefer)
    Imagine the gas is in a piston and the plunger is pushed in then out so that plotted against time it forms a sine wave - this is just that plotted in circular form.

    The important point is that you return to exactly the same pressure/volume as at the start!
  4. Nov 8, 2008 #3
    I see that as pressure increases, volume also increases. So temperature, too, must increase. So there process is neither isothermal, nor isometric nor isobaric. Therefore, it has to be adiabatic. But the adiabatic formulas require a y value: (p_1 V_1 - p_2 V_2) / (y - 1). The value is not provided. I cannot calculate it from the graph because p V ^ y = const. does not appear to be true. What do I do?
  5. Nov 9, 2008 #4
    How do you calculate the work done by an ideal gas?

    [tex] W=\int P dV [/tex]

    For a complete cycle, the integral is a closed path, so

    [tex] W=\oint P dV [/tex]

    Now, what does the line integral over a closed path give you? You could break it up into an upper hemisphere and lower hemisphere, then find the area below each. No calculus needed, just geometry. . .
  6. Nov 10, 2008 #5
    I don't know how to calculate work from that equation, please explain that to me further. Thanks.
  7. Nov 19, 2008 #6
    This is just like your other question with the pressure and volume.

    Once again, this requires NO calculus at all (except maybe some theory). Normally you could divide the circle into an upper and lower hemisphere. Intergrate to find the area under the 2 hemispheres and take the difference to get the area of the circle.

    Essentially, work equals the area of the circle. I'm sure you know how to find the area of a circle. (Hint: You may think that there are 2 different units so its impossible to find area since you don't know what radius to use, but if you just work out all the math with the units then you'll figure it out)
  8. Nov 19, 2008 #7
    I have managed to solve this problem already, just haven't updated. Thank you.
  9. Nov 19, 2008 #8
    So the work is the area under the curve, so isn't that 4pi?
    As far the temp change...
    pv = nRT
    we know n, R, pinitial, vinitial, so we can figure out Tinitial:
    Ti = PiVi/nR = 93.2 Kelvin. How would I know if it were in Celsius otherwise?

    We know...
    PiVi/Ti = PmaxVmax/Tmax

    Pmax = 550 kPa, so Vmax = 7 L,
    Tmax = PmaxVmax(Ti)/PiVi = 341.73 Kelvin.

    Did I do this right?
  10. Nov 19, 2008 #9
    Actually, I'm not sure how to calculate the Pmax and Vmax...
  11. Apr 27, 2009 #10
    Think of it as an ellipse. The hints given above are pretty vague considering this is a really easy problem.

    The PV in this case ( kilopascals * liters) yields an answer in Joules, but the problem wants the answer in Kilojoules.

    What's the area of an ellipse?

    Pi * A * B


    A = the radius for the kilopascal side

    B = radius of the Volume side

    pi * (200) (4) = 800 pi

    divide that by 1000 and you get 2.51 kJ
  12. Apr 27, 2009 #11


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    Welcome to Physics Forums :smile:

    Please note, giving hints is how we give homework help here. We don't give out the solution, because we believe that does not help the student as much as getting them to think and understand the problem.

    Since the question was asked 5 months ago, no harm done.

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