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Thermaldynamics - Adiabatic system W = delta(U)

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data

    After integrating the pressure formula of an adiabatic system, I have to show how this is equal to the change in energy. I know that my integral is correct (it was very straight forward), but I'm having trouble showing that it is equal to [itex]\Delta[/itex]U.

    2. Relevant equations

    [itex]\Delta[/itex]U = W + Q (Q = 0)
    PV = NkT
    [itex]\Delta[/itex]U = 1/2 * N * k * f * [itex]\Delta[/itex]T

    Where:
    k = 1.381 * 10-23
    f = degrees of freedom
    [itex]\Delta[/itex]T = change in temperature

    3. The attempt at a solution

    W = -PiVi(Vf1-[itex]\varphi[/itex] - Vi1 - [itex]\varphi[/itex]) / 1 - [itex]\varphi[/itex]

    Given that, I need to somehow get that to be
    [itex]\Delta[/itex]U = 1/2 * N * k * f * [itex]\Delta[/itex]T

    I managed to reduce W to:

    -NkTi((Vf / Vi)1 - [itex]\varphi[/itex] - 1) / 1 - [itex]\varphi[/itex]

    But I'm stuck from there.
    (Note: [itex]\varphi[/itex] = (f + 2) / f )
    Any hints on what to do next would be very helpful.
    Thanks
     
  2. jcsd
  3. Sep 25, 2011 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Ideal gas:
    pV = NkT
    pV^(γ-1) = constant = C for adiabatic proc.
    where γ = (f+1)/f

    1. write expression for V(T) using both the above eq.
    2. get dV = dV(N,k,γ,dT)
    3. dW = p(V)dV(N,k,γ,dT). Note that c and V do not appear in this.
    4. dU = -dW adiabatic
    5. use γ = (f+1)/f and get your answer.
     
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