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Blowing up a balloon (thermodynamics and ideal gases)

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data
    A balloon behaves such that the pressure is P = CV3
    where C
    = 100 kPa/m3
    . The balloon is blown up with air from a starting volume of 1 m3
    to a volume
    of 3 m3
    . Find the work done by the air.


    2. Relevant equations
    W=PdV



    3. The attempt at a solution

    What I tried is simply trying the above solution by multiplying the P which is given by P=CV^3 by the change in volume

    so i got

    W=(CV^3)(Vf-Vi)

    I then realized I had no idea what would the V in this equation be, can someone tell me if I am even doing this question right and if I am how I can determine the V?

    Thanks
     
  2. jcsd
  3. Oct 8, 2012 #2
    Hint: This will be an integration problem because P is a function of V.
     
  4. Oct 8, 2012 #3
    Do I use the formula W=∫VfV1PdV?

    That's what I have been trying to do but i'm not sure how to do this with two different P because you can usually move the P out of the integral

    edit: nvm i think i figured it out, should it looook something like this

    PiVi = PfVf

    so

    Vf/Vi = Pi/Pf

    and then replace Vf/Vi with Pi/Pf to get

    W = nR*ln(Pi/Pf)

    ?
     
  5. Oct 8, 2012 #4
    The pressure is a constant times volume. So the integral would be

    W = integral(C * V * dV) from 1 m^3 to 3 m^3.

    Check the units and you'll see that the units are work.
     
  6. Oct 8, 2012 #5
    could I instead just find the initial and final pressures and then directly plug them into the modified version of the ideal gas formula like in my edit above?

    Also i tried to do the integral you told me to do above, should I end up with 31| C*3V2?
     
    Last edited: Oct 8, 2012
  7. Oct 8, 2012 #6
    How do you determine n without knowing the temperature?
     
  8. Oct 8, 2012 #7
    yeah I realizd that when i started doing the calculations

    i tried to do the integral you told me to do above, should I end up with 31| C*3V2?

    however when I plug everything in I end up with units in kPa instead of joules... unless the integral of V would change it's units?
     
  9. Oct 8, 2012 #8
    also, my answer of 7800J seems unrealistically high so i think i might be doing something wrong.
     
  10. Oct 8, 2012 #9
    You have

    W = integral ( (kPa/Volume) * Volume * dVolume) = kPa * Volume after integration

    kPa is Newtons/area

    So you get Newtons-meters
     
  11. Oct 8, 2012 #10
    Hope this has helped. I have to leave my computer now.
     
  12. Oct 8, 2012 #11
    just one last question, can you give me an approximate value of Joules that would be needed in this question? I just need a reference to determine whether or not my answer is plausible and i have no idea how much 1 joule is.

    Thanks.
     
  13. Oct 8, 2012 #12
    ok nvm then, thanks alot for the help!
     
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