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A zeroth-order phase transition defined as a finite jump of some free energy or any other fundamental equation is incompatible with the requirement of convexity (concavity) for such state functions. It is a convex analysis theorem that every convex function must be continuous. Moreover, a convex function must have left and right derivatives almost everywhere....


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If the change of phase is happening slowly, with the whole system in equilibrium at each stage, then the temperature is uniform throughout the system. As the heat is supplied, liquid changes to vapour. If heat is removed, vapour changes to liquid. The system is moving along an isotherm; its temperature is both uniform throughout liquid and vapour, and also ...


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A piece of wet cloth from which water evaporates will have a lower temperature than the surrounding air. As long as the relative humidity is below 100 %, there is no equilibrium. Evaporation of sweat is how humans can maintain a physiological body temperature even when the air temperature is higher.


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Indeed, binary phase diagrams (not to mention ternary) are a bit difficult to 'get' when you first encounter them. While I dearly love Porter and Easterling's book, this figure is not the best presentation. First, the final diagram, (f), should really be considered as a map, not curves. If you look at the map of a country, the cities and roads are not ...


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Ah I see! Your plot is the saturation curve not an isotherm! My mistake... So the answer is "Yes" as Gert says. Indeed you have pointed out why it is tricky to demostrate critical opalescence. If you seal so liquid and vapour in a tube and heat towrds the crical point, the mesicus will go up if you have too much liquid , and down if you have too ...


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There is no violation of the second law here -- you're neglecting that a glass of water is an open system. In other words, it interacts with the environment not only via "heat transfer" but also via mass transfer as well. So, while heat is transferred from a "colder/liquid" phase to a "hotter/vapor" phase, the entropy lost by ...


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But now, if only the most energetic molecules were able to escape the inter-molecular forces and end up as water vapour, does this not mean that the water vapour above the surface now has a higher temperature than that of the water beneath it since the vapour is composed only of the most energetic molecules and the remaining liquid contains only the least ...


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I'll address your questions individually. Why is metallic hydrogen a big deal? Because it demonstrates very exotic properties compared to any other material we know of. What is the point if we can't make a lot of it? The same could be said about high purity silicon, until we figured out how to make a lot of it in the 50s. Now you find it in every digital ...


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Absolutely. On a clear, windless, dry night, get one of those IR thermometers and point it up at the sky. Temperatures below -40C can be seen in lots of places if it's dry enough. The less water vapor in the air, the cooler it can appear. The thermal environment for the water is: conduction with the tray (chosen for low conductivity) conduction with the ...


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Possible but requires very specific circumstances. Suppose we put the same bowl of water on the outside of a space satellite orbiting in the thermosphere and position huge mirrors in nearby orbit to block the radiation of the sun. The near-vacuum atmosphere there is at temperatures above 2000 K (1). The water will never equilibrate to 2000K. The thermosphere ...


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The molecular mechanics of particles are happening all the time as you go from one phase to another. Nothing about the processes changes as you lower the temperature, just the rate they occur. Assuming you have some appropriate nucleation sites, then even at temperatures above freezing, some of the water molecules will bond to each other and join up. There'...


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Nothing would happen. You could even cool to $-1 ^\circ$C without a phase change. Then, if nucleated maybe by a bump, a small fraction of the water would crystallise to a solid.


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short answer: yes. Entropic forces are "effective" forces that spawn out of entropy maximization: if your system is a in a low entropy state and can evolve into a higher-entropy one, there will be a push ("force") towards a higher entropy state. Interestingly, this can arise even in the absence of long-range interaction between particles. ...


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When you refer to SPTO transitions as being 'nonlocal symmetry-breaking transitions', I guess you are referring to the idea of 'hidden symmetry-breaking'? This indeed the old way of looking at SPTOs (i.e., 80s and 90s), before the more modern understanding in terms of topological invariants was developed. For clarity, let me briefly review this with an ...


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