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I Is the definition of a phase arbitrary?

  1. Nov 1, 2016 #1
    Gibbs phase rule says f = r-M+2

    with f: thermodynamic degrees of freedom; r: number of components; M: number of phases

    I wonder whether the defintion of "phase" is restricted or almost arbitrary. For example, consider a system of H2O, O2 and H2 in a closed vessel. Let there be the contstraint that there is a gaseous state which contains all three components (H2O, O2 and H2) and a liquid state which contains only H2O and O2. From my understanding of what a phase is, one would define two phases, a gas g and a liquid l, as follows:

    phase 1: H2O(g), O2(g), H2(g)
    phase 2: H2O(l), O2(l)

    Gibbs rule would give f = 3-2+2=3 degrees of freedom.


    Couldn't we have also defined the phases as follows:

    phase 1: H2O(g), H2(g)
    phase 2: O2(g)
    phase 3: H2O(l)

    Gibbs rule would give f = 3-3+2=2 degrees of freedom.

    My motivation for this definition is: O2 might consists of macroscopically tiny gas regions, perhaps invisible in the liquid, but ultimately they are thermodynamic subsystems and a constant pressure and temperature is applied to these regions, no matter if they are in the "gas" or in the "liquid" or if they are tiny (invisible) or large.

    There might be other defintions of phases, for example:

    phase 1: H2O(g)
    phase 2: O2(g)
    phase 2: H2(g)
    phase 3: H2O(l)

    Gibbs rule would give f = 3-4+2=1 degree of freedom.

    Or how about this:

    phase 1: H2O(g)
    phase 2: O2(g)
    phase 3: O2(l)
    phase 4: H2(g)
    phase 5: H2O(l)

    Gibbs rule would give f = 3-5+2=0 degrees of freedom.

    Are all of these definitions of phases valid and can we almost arbitrary assign the term "phase" to a homogeneous subregion of a bigger system, or am I missing a critical point here?
     
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  3. Nov 1, 2016 #2

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    Yes.
    Where's the H2?
     
  4. Nov 1, 2016 #3
    Oh, you mean each phase must include all components? Because of the derivation of Gibb's rule, which includes chemical potential functions µ(T,p,x1,...,xr-1) (with xi: mole fraction of component i) for each phase?
    If this example is not possible, how about a simplified version with H2O as the only component?

    phase 1: H2O(g)
    phase 2: H2O(l)
    Gibbs rule would give f = 1-2+2 = 1 degree of freedom. -> Coexistence curve of H2O(g) and H2O(l) with p = p(T).

    But, if I was not interested in water being solid or liquid, it's just water, could I define ONE water phase?
    phase 1: H2O
    Gibbs rule would give f = 1-1+2 = 2 degrees of freedom. -> All p-T-combinations would give water.

    So, what makes a phase a phase? Is it about convenience?

    And: What can be said about the initial example? Can this be analyzed by thermodynamics at all?
     
    Last edited: Nov 1, 2016
  5. Nov 1, 2016 #4
    Phase boundaries are characterized by a sudden change of physical properties. Thus you just need to check if there are such boundaries (in space or in a phase diagramm).
     
  6. Nov 1, 2016 #5

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    Yes.
    Yes.
    Not in the sense of being a dependent variable. This is pretty much the way it's been handled for you to this point.
     
  7. Nov 1, 2016 #6
    Do I need to check for all conceivable changes? Or is the selection of properties application dependend? In the example above: If I'm not interested in water being either gaseous or liquid, I would say I have only one "water-phase".


    I think Gibbs phase rule is applicable here, because µ(T,p) ist equal everywhere?
    When f = 2 in Gibbs phase rule I would say it tells me that whatever the values of p and T are, I get water, no matter which properties it has.
     
  8. Nov 1, 2016 #7
    Yes.

    Gases and liquids usually have very different properties (e.g. density or viscosity) - no matter if you are interested in them or not.
    But the scale may depend on the application: liquid and gas can be two different phases on small scales but a single pase (foam or fog) on larg scales.
     
  9. Nov 1, 2016 #8
    Thanks so far.
    For the simple example it's now becoming clearer.

    But to be sure to really understand it: For the system with H2O(l) in which O2(g)-bubbles are dissolved in contact with H2O(g) where also O2(g) exists:
    Are the phases
    phase 1: H2O(g)
    phase 2: O2(g)
    phase 3: H2O(l)

    or

    phase 1: H2O(g) + O2(g)
    phase 2: H2O(l)+ O2(g)
    ?
     
  10. Nov 1, 2016 #9
    Do you think that there is a phase boundary between H2O(g) and O2(g)?

    This makes sense on sufficiently large scales, where the dispersion of O2-bubbles in water can be assumed to be homogeneous. Otherwhise the liquid phase would be just water (with dissolved oxygen).
     
  11. Nov 1, 2016 #10
    Okay, I see my error in reasoning here, thanks.

    So, on small scales one has
    phase 1: H2O(g) + O2(g)
    phase 2: O2(g)
    phase 3: H2O(l)
    ?
    And this situation is not describable by Gibbs rule, because not each component is contained in each phase (no H2O in phase 2 and no O2 in phase 3)?
     
  12. Nov 1, 2016 #11
    Gibbs rule applies to an equilibrium only and in equilibrium there would be no pure water or oxygen phases.
     
  13. Nov 1, 2016 #12
    The water-example might be nonsense. But there are equilibriums with completely different chemical compositions of the phases. For example, a platinum bar immersed in a copper solution. Does Gibbs phase rule (or classical thermodynamics at all) makes a statement about such situations?
     
  14. Nov 1, 2016 #13
    I don't see why Gibbs rule shouldn't hold for this cases.
     
  15. Nov 2, 2016 #14
    .. because in the derivation of Gibbs rule I know the chemical potential of each component is set equal. µPt(in Pt bar) = µPt(in copper solution) (and the same of the copper solution). But there is not Pt in the copper solution, so this kind of derivation does not make sense, I guess.
     
  16. Nov 2, 2016 #15
    That's just an artefact resulting from your assumption that the phases are pure in equilibrium. To solve this problem you just need to change the concentrations of Pt in Cu and vice versa from zero to almost zero (infinite dilution).
     
  17. Nov 2, 2016 #16
    Ah, okay: Regarding a Pt-Cu-system: With specifying the concentrations in both phases as beeing pure phases I have already "used up" both of two degrees of freedom (f = 2-2+2=2 here) and T and p would need to take on very specific values to make the system consisting of two pure phases (if such a state exists at all).
    And bei choosing T and p, e.g., I would generally get two nonpure phases. If this is the case, I've finally understood it. Thanks a lot!
     
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