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

AI Thread Summary
The discussion centers on the latent heat of fusion at the triple point for ammonia, with a focus on the distinctions between triple points and critical points in phase diagrams. It is clarified that while the triple point allows for phase transitions without external heat transfer, it does not imply that latent heat is zero; rather, latent heat exists at the triple point but is not zero. In contrast, at the critical point, phase changes occur without latent heat, marking a transition from first-order to second-order phase changes. The divergence of heat capacity at first-order transitions leads to latent heat, while second-order transitions, such as those at critical points, do not exhibit this characteristic. Overall, the conversation emphasizes the nuanced understanding of phase transitions and their thermodynamic implications.
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
Wiki The Great said:
At that point, it is possible to change all of the substance to ice, water, or vapor by making arbitrarily small changes in pressure and temperature.
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?
 
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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
At that point, it is possible to change all of the substance to ice, [or] water,[STRIKE] or vapor[/STRIKE] by making arbitrarily small changes in pressure and temperature.
? But does that mean there is no latent heat?
 
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Jolb said:
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.
I see, that's basically the same graph I had in mind with the same definitions.
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?
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?
 
fluidistic said:
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;
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.

unlike what happens at the triple point apparently.
Are you sure?
 
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Jolb said:
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?

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?
 
fluidistic said:
But at a critical point it is worth 0J, right?
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)

Phase transition
Transition between two equilibrium phases of matter whose signature is a singularity or discontinuity in some observable quantity. First-order transitions are characterized by a discontinuity in a first derivative of a thermodynamic potential. In particular, entropy S, which is the temperature derivative of a free energy (S = -∂F/∂T ) has a discontinuity ΔS leading to a latent heat L = TΔS. For second-order transitions, first derivatives of thermodynamic potentials are continuous.

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.
 
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I see, thank you very much!
 
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