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Helmholtz free energy

  1. Dec 7, 2004 #1
    Let’s say we have a 1 mol system in a state A (Pa,Va,Ta are given) and we take it to a state B(Pb,Vb are given) . We want to know what’s the maximum work we can give to a reversible work source. Considering the process is carried out attached to a 150 K heat source, the only data available is this:

    1. equation of adiabatic reversible curve
    2. Cp (heat capacity at a given constant pressure (Pa )
    3.  (expansion coefficient at a given constant pressure (Pa )

    I’ve translated the problem into this: find A Helmholtz free energy change between points A and B. However I don’t know how to get dF=-SdT-PdV from the experimental coefficients mentioned above.

    I’d appreciate your help very much.
  2. jcsd
  3. Dec 8, 2004 #2
    Just a hint would be enough. I don't really know where to start from.
  4. Dec 8, 2004 #3
    It's been a long time since i studied thermo but here i go. If we have only a heat source and the initial and final states are in equilibrium, they have to be at the same temperature of the source. If the process is reversible every single step is a equilibrium state and so it has to be at the same temperature too. So we can conclude that the process is isotherm and then dT=0.

    If Cp is provided you'll need Cv to compute the adiabatic constant. There's a equation relating Cp-Cv and the diferent coefficients of expansion and such, Generalized Mayer's Rule i think.

    I don't know if the process is adiabatic, if the temperature remains constant we'll need some mechanism to achieve this.

    Hope this helps you a bit. Feel free to correct me if you find any mistake please.
  5. Dec 11, 2004 #4
    First of all, let me stress that I've not been solving this kind of problems for over twenty years, so I'm not very confident I'm right. I'll do my best and try and enjoy it. I hope it helps anyway.

    Now, a couple of remarks:
    1- is it an adiabatic or an isotherm transformation? If it is adiabatic, the reservoir temperature is irrelevant; if it is isotherm, I don't see how the adiabatic curve can help;
    2- the change in Helmholtz free energy equals the work done only in the case of adiabatic quasi-static transformations - it does not help much in the isotherm case and might not be necessary to us in the adiabatic case (see below).

    (1) In case the transformation is adiabatic, having the curve allows us to compute the work as

    [tex] dL=PdV [/tex].

    For example, in the case of a perfect classic gas the adiabatic change follows the path along the equation

    [tex] PV^\gamma=cost=C=P_A V_A^\gamma [/tex]

    that is

    [tex] P=C/V^\gamma [/tex]

    and we can compute the work as

    [tex] dL=(C/V^\gamma)dV [/tex]

    [tex] L=\int(C/V^\gamma)dV [/tex]

    [tex] L=(C/\gamma)(V_B^{1-\gamma}-V_A^{1-\gamma}) [/tex]

    I don't see any use of [tex] c_P [/tex] and [tex] \beta [/tex]. On the other hand, having those two values at hand reminds me of a relationship that is valid for any adiabatic reversible transformation:

    [tex] dT=\frac{TV\beta}{c_P}dP [/tex].

    (2) In case the transformation is isotherm, I can't see how to compute the work without further information.
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