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Thermodynamics: T-S slope at constant volume

  1. Feb 22, 2010 #1
    1. Prove that the slope of an isochoric process in a T-S diagram is T/Cv where T is the temperature and Cv is the heat capacity at constant volume



    2. dQ = T * dS (I understand this)
    dQ = Cv * dT for isochoric processes (I don't understand this)




    3. Since the two equations share dQ I set them equal to each other:
    Cv * dT = T * dS
    Then I just rewrote the equation as:
    T/Cv = dT/dS
    And dT/dS describes the slope of the line which is T/Cv
    My main problem is that I don't understand why dQ = Cv*dT. The book just gave it to us, but I'm thinking that I need to be able to derive it or the problem is just too easy.

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 23, 2010 #2

    Mapes

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    For isochoric (constant-volume, [itex]dV=0[/itex]) processes, no work is done (work being [itex]P\,dV[/itex]), so any energy change [itex]dU[/itex] in the system must be in the form of heat transfer [itex]q[/itex]. Additionally, we can use the differential energy equation

    [tex]dU(=q)=T\,dS-P\,dV=T\,dS=T\left(\frac{\partial S}{\partial T}\right)_V\,dT=C_V\,dT[/tex]

    which uses the definition of [itex]C_V[/itex]. Does this make sense?
     
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