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Confused - chemical potential of an ideal classical gas

  1. Jan 17, 2010 #1
    Can anyone help me. I am very confused about the chemical potential.

    In the following equation

    dU = TdS - pdV + u dN, where u is the chemical potential

    it seems to me that if you add particles to a system you are increasing the energy of that system, i.e. the chemical potential is positive.

    Why is it then that the chemical potential is negative above a certain temperature (as can be seen by taken the derivative of the canonical partition function with respect to the number of particles (times -1/kT))?

    This would seem to suggest that as you add particles to a system, the energy of the system decreases!!!!!!
     
  2. jcsd
  3. Jan 18, 2010 #2

    vanesch

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    Why would you say that ?

    There's nothing wrong with that.

    You can put your "zero" for energy wherever you want, what counts are *differences* in energy (at least in the classical setting where we are in this case).

    Consider mechanical potential energy, for instance. Suppose that you put the "0" at floor level. That means that below floor level, in the basement, potential energy is negative. If you add more blocks of lead in your basement, the total potential energy decreases. Nothing wrong with it. With chemical potential, it is the same...
     
  4. Jan 18, 2010 #3

    Mapes

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    Hi Mireno, welcome to PF. Watch out, you're ignoring the other variables in that equation (most significantly, entropy). And entropy has an increasing influence on system behavior with increasing temperature.

    Consider the Gibbs free energy, criterion of spontaneous processes at constant temperature and pressure, defined as [itex]G=U-TS+PV[/itex]. If I add a protruding atom to a formerly flat plane of atoms, for example, the increase in energy could be small compared to the increase in entropy (or rather, [itex]TS[/itex]) because of the many possible equivalent positions of the new atom. The chemical potential [itex]\mu=(\partial G/\partial N)_\mathrm{T,P}[/itex] would therefore be negative for this system. Does this make sense?
     
  5. Jan 18, 2010 #4
    What if you use dU = TdS - pdV + udN.

    Then u = the partial derivative of U w.r.t. N (WITH S FIXED).

    Mireno (aka "Morrie")
     
  6. Jan 18, 2010 #5

    Mapes

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    It's the same chemical potential:

    [tex]\mu=\left(\frac{\partial U}{\partial N}\right)_\mathrm{S,V}=\left(\frac{\partial G}{\partial N}\right)_\mathrm{T,P}[/tex]

    In this case, to maintain a constant entropy, you're going to need to cool the system to counteract the entropy of adding the particle. Depending on the magnitude of this entropy, the energy removed via cooling may be larger than the energy added by the particle, which will make the chemical potential negative again.
     
  7. Jan 18, 2010 #6
    Mapes (is that your name or a pseudonym?)

    Thanks for your help. By way of background, I am a 48 year old revising my university (college) physics - currently 1/2 way through 3rd year. I kept meticulous notes in my university years. This is one problem that has caused me trouble and that I have not been able to nut out using Reif (my text) or the Internet.

    I take it that the chemical potential is also used in circumstances where there are different chemical species or where electrical (potential) or gravitational (potential) effects, etc, need to be taken into account.

    Again

    Many thanks for your help.
     
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