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About Gibbs energy

  1. May 7, 2009 #1

    KFC

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    In a book about stat. mech., I read the following relaiton for magnetic system

    [tex]A = G + hM[/tex]

    where A is the Helmholtz energy, G is the Gibbs energy and h is the external magnetic field, M is the magetization. I know from thermodynamic, we have

    [tex]A = U - TS[/tex]

    or

    [tex]A = G - PV[/tex]

    so, [tex]hM = -PV[/tex] ?

    I don't understand what is [tex]hM[/tex]. If I know magnetization, external field and Gibbs energy, how to get Helmholtz energy?
     
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  3. May 7, 2009 #2

    Mapes

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    This goes back to that issue I believe you mentioned earlier: magnetization is usually ignored in thermodynamics, so authors can be inconsistent when adding the relevant terms. When we're including magnetization in an open system, energy is

    [tex]U=TS-PV+\mu N+hM[/tex]

    The relevant potential when temperature, pressure, and field are kept constant is acquired by the Legendre transform

    [tex]\Phi=U-TS+PV-hM[/tex]

    Some authors may call [itex]\Phi[/itex] the Gibbs energy, which risks great confusion. (I'm using the symbol [itex]\Phi[/itex] as a dummy variable here.)

    To figure out the potentials, just remember that you need to remove (by Legendre transform) any conjugate pairs associated with constant variables. Don't rely on the consistency of names like Helmholtz, Gibbs, etc. If someone's working with a system at constant temperature and field, for example, you know you need to consider the potential

    [tex]\Lambda=U-TS-mH[/tex]

    whatever it might be called. (Again, [itex]\Lambda[/itex] is just a dummy variable.) I hope it's helpful to see the method to constructing these potentials. Does this make sense?

    (To reiterate, it's definitely not true that [itex]hM=-PV[/itex].)
     
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