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Need help in solving this =c

  1. Dec 19, 2008 #1
    1. Ethane, initially at 35 kPa with a volume of .12 m³, is
    compressed without friction in a cylinder until its volume is halved
    in such a manner that its pressure and volume are linearly related,
    p = a + bV. The final pressure is 80 kPa and the change in internal
    energy of the ethane is 3.22 kJ. Determine the heat transfer.

    2. Ethane in a closed system is compressed without friction from 95
    to 190 kPa in such a manner that pV = constant. Initially, the density
    is 1.11 kg / m³ and the volume is 0.045 m³. During the compression,
    2.96 kJ of heat is removed from the ethane. Determine the change in
    internal energy of the ethane.

    3. A closed vertical cylinder containing 0.3 kg of nitrogen at 90ºC is
    fitted with a weighted, frictionless piston so that a constant pressure
    of 275 kPa is maintained on the gas. The nitrogen is stirred by a
    paddle wheel inserted through the cylinder wall until the absolute
    temperature of the gas is doubled. During the process, 20 kJ of heat
    is transferred from the nitrogen to the surroundings. Determine the
    amount of paddle-wheel work required for this process.
  2. jcsd
  3. Dec 19, 2008 #2
    We do not provide answers at PF, you are required to show some effort in order to receive help.

    For# 1 I would start by writing out the energy balance for a closed system.

    same for #2 and #3. We'll go from there.
  4. Dec 19, 2008 #3
    I'm having trouble solving... i've try alot but still can't get it.
    Could anyone write the solutions and with steps on how they did it so I can possibly learn! Thanks a lot
  5. Dec 19, 2008 #4
    When you created an account I am sure you read this. Post some work and we will post some help.

    I will start you off for #1.

    For a closed system, the energy balance reads: [itex]\Delta U=Q-W[/itex]

    You are given [itex]\Delta U[/itex] and you need to find Q which means that all that is left is to calculate the Work done on the gas (i.e., it is being compressed, so work should be negative).

    You are given initial pressure, final pressure and the fact that they are linearly related so you can find the work.

    For work done in a piston-cylinder assembly, can you at least tell me how Work, pressure and Volume are related to each other in general?
    Last edited: Dec 19, 2008
  6. Dec 19, 2008 #5
    weee, i'm done with #1 but i dont know my answer if it's right. = -0.25 kJ
  7. Dec 19, 2008 #6
    And neither do I since you still have not posted any work.
  8. Dec 19, 2008 #7
    Wnf = pV lnP1/P2
    = (35 N / m²)(0.12 m³)[ln (35 kPa / 80 kPa
    = -3.4692 x 10ˉ³ N-m
    = -3.4692 kJ

    Q = ΔU + W
    = 3.22 kJ + (-3.47 kJ)
    = - 0.25 kJ
  9. Dec 19, 2008 #8
    Why, out of curiosity, do you think that that is the correct formula for the work done on the gas?

    I asked you this relatively simple question in post #4 but, as I assumed it would, it went unanswered:
  10. Dec 19, 2008 #9
    /sad. ok, i'm out. i can't solve it, but thnx for the help. goodluck..
  11. Dec 19, 2008 #10
    So you do not know that [itex]W=\int P\, dV[/itex] ?? You haven't seen that anywhere?
  12. Dec 19, 2008 #11
    hmmm, is that the correct formula?
  13. Dec 19, 2008 #12
    What class is this for? You are trying to tell me that you have ever seen the expression [itex]Work=\int P\, dV[/itex]?
  14. Dec 19, 2008 #13
    thermodynamics sir.
  15. Dec 19, 2008 #14
    I feel like I have to ask you everything twice... Do you want help with this problem or not? Believe it or not I do have other things I could be doing.... like nothing.
  16. Dec 19, 2008 #15
    i know that formula...
  17. Dec 19, 2008 #16
    I give up. Clearly you don't want to make any effort towards solving this problem. Best of luck to you.
  18. Dec 19, 2008 #17
  19. Dec 19, 2008 #18
    and tnx for nothing
  20. Dec 19, 2008 #19
    Your are very welcome :smile:

    Maybe if you took the time to participate instead of waiting for others to do your work for you, you would have enjoyed your time here better.
  21. Dec 20, 2008 #20
    W = ∫ P dV
    = P ∫ dV
    = p(V2 – V1)

    what kind of effort do you want me to do? i'll try to solve it but still i'm wrong..
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