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Phase Transitions and Chemical Potential

  1. Feb 10, 2010 #1
    Hello,
    right now, I am learning thermodynamics with Reichl: "A modern Course in Statistical Physics"

    In chapter 3.C page 100 "classification of phase transitions", the text says:

    "As we change the independent intensive variables (p, T, x_1,... ,x_l) of a system,
    we reach values of the variables for which a phase change can occur. At such
    points the chemical potentials (which are functions only of intensive variables)
    of the phases must be equal and the phases can coexist."

    I wonder, if there are systems with no phase transition at all. Or is there always a phase transition, if I change for example pressure, or temperature, or the mole fractions of particles of type i ?

    What does the chemical potential have to do with the phase transition? I know, that it is equal in both phases at the transition point, but how does the phase transition occur? What forces the system to change it's phase? What does the chemical potential of the system do before the transition?

    I hope you can help me.

    Regards,
    Mr.Fogg
     
  2. jcsd
  3. Feb 10, 2010 #2

    DrDu

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    When the chemical potential is equal in both phases, then it costs no (free) energy (or enthalpy) to move a particle from one phase to the other. When they are not equal, then you can gain energy by moving particles from the phase with lower potential to the one with higher chemical potential (don't try to nail me down on the sign here) until you eventually reach equilibrium.
    When you plot the chemical potential of your phases as a function of e.g. temperature at p fixed, the curves for the different phases cross at the transition temperature. In principle, nothing spectacular happens, often you can continue on a single curve beyond the crossing point, e.g. you can supercool liquid water below 0 degrees centigrade.
    The classical example to study a phase transition is the van der Waals gas. You can find for the van der Waals equation a phase transition analytically from the condition that p1=p2 and mu1=mu2.
     
  4. Feb 10, 2010 #3
    Thanks DrDu,

    Since in equilibrium the Gibbs' Free Energy is a minimum, the system should prefer the lower chemical potential. So this should be correct, to gain energy from lower to higher potential.

    Mr.Fogg
     
  5. Feb 10, 2010 #4
    Can anybody answer my other questions?
     
  6. Feb 11, 2010 #5

    DrDu

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    Basically all systems become solid or (super-)liquid near absolute zero of temperature and become gasses at sufficiently high temperatures. So there is allways some phase transition.
     
  7. Feb 11, 2010 #6
    Thanks,
    I understand now, one could explain it by taking a look at the p-T diagram.

    [tex] \lim_{T \to 0} \left( \frac{\partial p}{\partial T} \right)_V = 0 [/tex]

    The coexistence curve starts at zero p and T and the fusion curve goes to infinity. So there must be a temperature in between, where a phase transition occurs. At least solid <-> liquid and solid <-> gas should change the phase. The vaporization curve terminates at the critical point.

    But what forces the system to change it's phase? For example, when I heat it up? Or if I increase the pressure?

    Regard,
    Mr.Fogg
     
  8. Feb 11, 2010 #7

    Mapes

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    Systems tend to proceed toward states with the lowest energy; it's a consequence of the Second Law. But we need to consider the correct energy potential; for systems at constant temperature and pressure, it's the Gibbs free energy that's minimized.

    Or are you asking what happens atomistically, e.g., why more atoms would evaporate from a liquid's surface rather than condense there?
     
  9. Feb 11, 2010 #8
    Okay, I see. So from the Second Law, we get for the Gibbs Free Energy [tex] d(U+pV -TS) < 0 \Rightarrow dU +pdV - TdS < 0 [/tex].
    And to deal with the change in entropy, we take the Clausius (In)equation [tex] dS = \frac{\delta Q}{T} [/tex]

    The system prefers the state with lower Gibbs Free Energy. So I have to calculate the Gibbs Free Energy for each phase with G = U + pV - TS and then I know, where the system will transfer to? Or is there another possibility to predict that? How do I handle the entropy in this case? I think, one only can measure changes in entropy, not an absolute entropy?

    Could You explain that? Maybe it helps me to understand.

    Regards,
    Mr.Fogg
     
  10. Feb 11, 2010 #9

    Mapes

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    Right, that's the criterion for phase equilibrium at constant temperature and pressure. The atomistic view is that (between a liquid and gas, for example), atoms are always evaporating and condensing, and the rates are equal when the chemical potentials (i.e., the partial molar Gibbs free energies) of each phase are equal.
     
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