Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Partial derivative in thermodynamics

  1. May 21, 2012 #1
    So I have a proof and I can't follow the process, I think its because I haven't learned how to do partial derivatives or I've forgotten, anyways can someone tell me if this is a rule in calculus

    (∂Cv/∂V)T=0

    I've gotten to
    [(∂/∂V)(∂U/∂T)V]T

    and the proof I have goes to
    [(∂/∂T)(∂U/∂V)T]V
    -is this a rule? they were able to switch the denominators of the derivatives which is can see but then they switched the constants aswell If so what is the name so I can read up on it, I can't seem to find it with google
     
  2. jcsd
  3. May 21, 2012 #2
    That's allowed because U is a function that is in the conditions of Schwarz's theorem (which can be derived from stokes' theorem)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Partial derivative in thermodynamics
  1. Partial Pressure (Replies: 3)

  2. Partial pressure (Replies: 4)

  3. Partial Pressures? (Replies: 5)

  4. Laws of Thermodynamics (Replies: 2)

Loading...