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Internal Energy as a function of U(S,V,A,Ni)

  1. Jan 15, 2013 #1
    A general thermo question...
    for the function describing internal energy U(S,V,A,N)


    please explain how the total differential is

    dU=TdS-PdV+γdA+μdN (for a one component system)

    Basically why is dT=dP=dγ=dμ=0? Is it because they are intensive or potentials?

    Thank you,


    Attached Files:

  2. jcsd
  3. Jan 15, 2013 #2
    Does this help?

    U = U(S,V,.N,A.......) \\
    dU = {\left( {\frac{{\partial U}}{{\partial S}}} \right)_{V,N,A...}}dS + {\left( {\frac{{\partial U}}{{\partial V}}} \right)_{S,N,A}}dV......... \\
    T = {\left( {\frac{{\partial U}}{{\partial S}}} \right)_{V,N,A...}} \\
    P = {\left( {\frac{{\partial U}}{{\partial V}}} \right)_{S,N,A}} \\
    {\rm{etc}} \\

    I will leave you to fill in the bits for moles and area or other quantities.
  4. Jan 15, 2013 #3
    Thank you for your reply...what is this technique called? I am working with Legendre transforms and this is similar...

    Another thing...how can you derive T=partialU/partialS or others without knowledge of the internal energy equation?
  5. Jan 15, 2013 #4
    This is the Gibbs Formulation.

    Carington : Basic Thermodynamics P 187ff : Oxford University Press

    Also in many Physical Chemistry texts.
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