1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question on adiabatic expansion

  1. Dec 8, 2003 #1
    I've been struggling with this problem for an hour or so now, and can't seem to find the right answer. Perhaps someone here can help? I would be very grateful. ;)

    Two moles of an ideal gas ( = 1.40) expands slowly and adiabatically from a pressure of 5.09 atm and a volume of 12.7 L to a final volume of 30.0 L.

    I had to find this stuff first, and I know I have all of it right.

    final pressure: 1.528
    initial temp: 393.5K
    final temp: 279

    Find Q, W, dEint.

    I know Q is 0, and I know that Work and change in internal Energy are the opposite of eachother, but I can't seem to find the right value for them. I thought work = nRdT, which in this case is:


    However, that answer is wrong according to webassign. There is also a problem in my text book that is the same, except with different numbers, and i tried that one and got it wrong too. Anyone know what I'm doing wrong? Your help is greatly appreciated. :)
  2. jcsd
  3. Dec 8, 2003 #2
    Formula for work done in adiabatic process is wrong
  4. Dec 8, 2003 #3
    so what is the proper formula? The textbook goes through a proof of why PV^y is constanst, but doesn't really go any further from that... I'm guessing that's somehow incorperated into the formula...
  5. Dec 8, 2003 #4
    Write the differential equation for for work done for gas

    Note [tex]PV^\gamma= K[/tex]

    substitute P from above in workdone equation and calculate the work done by taking limits from V1 to v2
  6. Dec 8, 2003 #5
    eh... P from above what? P is not constant.
  7. Dec 8, 2003 #6
    Refining it
    [tex] dW=kV^-^\gamma dV [/tex]

    I hope u will got it now
    Last edited: Dec 8, 2003
  8. Dec 8, 2003 #7
    Okay, I thought maybe that's what you meant, but I'm sorry I don't see what good that does if I don't know what k is. Maybe I'm just slow tonight as I only got about 4 hours of sleep last night and am rather tired... Sorry. [zz)]

    Ah well, it was due at midnight. So I'll just ask my teacher how to do it tomorrow...
    Last edited: Dec 8, 2003
  9. Dec 8, 2003 #8
    k is a constant which is given by

    [tex]PV^\gamma= k[/tex]

    [tex] P_iV_i^\gamma = P_fV_f^\gamma= k[/tex]
    Last edited: Dec 8, 2003
  10. Dec 10, 2003 #9
    Oh duh. Sorry, I guess I was just really tired last night. ;) Anyway thanks for the help.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Question on adiabatic expansion
  1. Adiabatic Expansion (Replies: 3)

  2. Adiabatic expansion (Replies: 9)

  3. Adiabatic expansion (Replies: 1)