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

Thermodynamics: Reversible adiabatic work and internal energy?

  1. Dec 18, 2011 #1
    When dealing with a process undergoing reversible compression from state 1 (P1,V1,T1) to γstate 2 (P2,V2,T2), the internal energy dU and the work, dw, are the same:

    considering dq= 0. Since dU = CvdT = dw. ΔU = w. ΔU = Cv(T2 - T1)

    w = [(-P1 (V1)^γ/ (-γ+1)][V2^(-γ+1) - V1^(-γ+1)] for reversible adiabatic processes.

    So my question is, w = [(-P1 (V1)^γ/ (-γ+1)][V2^(-γ+1) - V1^(-γ+1)] = ΔU = Cv(T2 - T1) ?
    Are they equal?

    If the state 1 and state 2 are only described with two variables, say (P,V) for both states respectively, that still holds true right?
  2. jcsd
  3. Dec 18, 2011 #2
    Nevermind. I figured it out. It does not equal because. Please delete this post or close it.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook