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Also in the article On Scale and Conformal Invariance in Four Dimensions there's appendix about such phenomena.

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You can go from a gas to a liquid by raising maneuvering at high pressure and temperature to go around the critical point. That is what the phase diagram (whose qualitative form can be derived, by applying the Maxwell construction, from the van der Waals equation of state) indicates, since there are paths from the gas phase to the liquid phase that do not ...

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You are correct. There is instead a substance-specific "enthalpy of fusion" to account for the heat gained/lost during the constant-temperature phase change.

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Let $\phi(\beta)$ denote the free energy density at inverse temperature $\beta$ (and with no magnetic field). Then, the argument you sketch (a rigorous version of which can be found in Section 3.10.1 of this book) implies that $$\phi(\beta) = \phi(\beta^*) - \log \sinh(2\beta^*), \tag{\star}$$ where $$\beta^*=\mathrm{arctanh}\,(e^{-2\beta}).$$ This ...

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First part is answered by @CGS. Answering second part in a strictly ideal sense, with following assumptions: no heat sources around (ice doesn't melt on its own, except due to pressure). metal is at same temperature as ice. We can find the volume of ice molten after the metal cube crosses a distance $x$. With this, we can find the extra pressure ...

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The answer to your first question is contained in the phase diagram for $H_2O$ shown here. Note how the line separating ice from water, line A to D, is negatively sloped. This means that if you start in the solid phase and apply enough pressure (move upward on the Y axis), you will incur a phase change to liquid. So to answer you question, one just needs ...

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For "persistent structure" read "bound structure by quantum mechanical potential sollutions". A the atomic and molecular level structures arise because the number of atoms can settle at a lower energy level, than when free. This means there is a binding energy that has to be payed for the atoms to be freed from the structure. the atoms ...

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First order phase transitions are characterized by discontinuous first derivatives of the free energy. Specifically, the derivative of Gibbs free energy with respect to pressure determines the density of the gas ($\Delta N/\Delta V$) and this may be discontinuous. The Early Universe was nearly but not quite homogeneous, and near the phase transition ...

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First, the expression given in the OP is not the expression for the actual free energy, only what comes out of the naive heuristic energy/entropy argument. In reality, renormalization group computations lead to the following predictions: first, the correlation length should blow up at the transition as $$\xi \simeq A\exp\bigl( B/\sqrt{t} \bigr)$$ for $t>... 0 It is determined entirely by the rate of heat loss of the flask, which is determined by the difference in temperature of the Dewar flask with its surroundings. So long as there is a mix of water and ice, the temperature is the same, but if you put in ice much colder than 0deg, it will be able to warm until the ice starts to melt. At that point, the situation ... 1 If the Dewar flask is an ideal insulator then in Scenario 1, no ice will melt at all. But in Scenario 2, adding some water to the ice will make some of it melt (the more water added, the more ice will melt) So for a perfect insulating Dewar flask, Scenario 1 will ALWAYS leave the most ice. It's trivial, really... But if the Dewar isn't perfect, the ... 0 first case: the dewar is ideal, so almost no ice will melt, only the air in the dewar will go du 32°F second case, if the water ist not to much, it will cool down to 32°F, how much ice is remaining depends only on the amount of water compared to ice. 1 The typical heuristic argument here is to look at the case for an infinite system, i.e. at the limit$R\rightarrow \infty$. For$T<\pi J/(2 k_{\mathrm{B}})$, the first term ($E$) dominates and the free energy will diverge$F\rightarrow \color{red}{+} \infty$. It can only lower$F$by having the lowest$E$and hence no vortices. For$T>\pi J/(2 k_{\...

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