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Is the triple point a kind of critical point? (thermodynamics)

  1. Jun 20, 2013 #1

    fluidistic

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    In a problem in Callen's book I was asked to say what was the latent heat of fusion at the triple point for ammonia. I answered "without performing any algebra, 0J". Because I remember a video I saw on youtube about the triple point () and now I read on wikipedia
    which seem to confirm that indeed, the system can undergo a phase transition without any external heat transfer.
    So would the triple point be equivalent to a critical point in some way? In the sense that if I understood well, in a critical point the system can change its phase without any external heat, just like in the triple point?
     
    Last edited by a moderator: Sep 25, 2014
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  3. Jun 20, 2013 #2
    I guess the answer depends on your definitions. But typically the triple point and the critical point mark two specific (and distinct) points on a P-T phase diagram. Take a look at this phase diagram for water: =Phase_Diagram_H2O.jpg

    According to the definitions I use (Chaikin and Lubensky, Principles of Condensed Matter Physics), the triple point would be point D while the critical point would be point E. Notice that the liquid-gas phase border stops at the point E--by circling around point E, one could go from liquid to vapor without any discontinuous changes or latent heat, so this is a second-order phase change. If you go through the gas-vapor phase boundary below E, there is latent heat, so this is a first-order phase change. First order phase changes are the typical ones. The critical point marks the distinction between first- and second-order phase changes.

    Anyway, just because the phases coexist at some point (P,T) doesn't mean there is no latent heat. Take for example water and ice at 0°C and 1atm (or whatever freezing point you'd prefer). It can be in either liquid or solid form. Isn't it true that
    ? But does that mean there is no latent heat?
     
    Last edited: Jun 20, 2013
  4. Jun 20, 2013 #3

    fluidistic

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    I see, that's basically the same graph I had in mind with the same definitions.
    If I have water at 0°C and 1 atm and I add an arbitrary small heat transfer, the whole system won't convert into all solid or liquid phase; unlike what happens at the triple point apparently. So in the 1atm/0°C state, of course there's a latent heat different from 0J. But I am not sure for the triple point of water. What do you think?
     
  5. Jun 20, 2013 #4
    Correct. But remember what happens at a first-order phase change and why it gives rise to latent heat: C=dQ/dT diverges.

    C=dQ/dT→∞

    This means that, if you're at the phase boundary and you try crossing it, a small transfer of heat energy doesn't change the temperature/pressure at all. All it does is go into "assembling"/"disassembling" the crystal. So the entire sample doesn't change phase with an infinitesimal heat transfer, only a small portion does, but the thermodynamic variables T and P stay constant during this "latent heat" transfer. On the other hand, the equation also seems to say that an infinitesimal change in T would require dQ to be infinite--so it's plausible that any tiny change in T or P would cause the whole sample to change phase since there's a big change in Q.

    Are you sure?
     
    Last edited: Jun 20, 2013
  6. Jun 20, 2013 #5

    fluidistic

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    Nevermind I see my misunderstanding.
    A small change in temperature or pressure at the triple point would require more than an arbitrary small dQ.
    So ok, the latent heat of fusion is not 0J at the triple point.
    But at a critical point it is worth 0J, right?
     
  7. Jun 20, 2013 #6
    Yes indeed. The critical point marks the change from a first-order (latent heat) transition to a second-order (no latent heat) transition. The definition of these transitions is: (From Chaikin and Lubensky)

    From this definition, it follows that C diverging only happens for first-order transitions, so C is finite for second order transitions (including the one that goes through the critical point) and thus there is no latent heat. That is why the phase boundary line stops at the critical point: really what the phase boundary is showing you is where there are discontinuities/divergences which correspond to latent heat. If you don't cross a phase boundary, then no latent heat, i.e. 0J.

    Edit: Gave the Chaikin and Lubensky definition.
     
    Last edited: Jun 20, 2013
  8. Jun 20, 2013 #7

    fluidistic

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    I see, thank you very much!
     
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