# Tag Info

14

I will try to answer these questions from different views. Macroscopic view The "quantitative" rather than qualitative difference in a liquid-gas phase transition is due to the fact that the molecules arrangement does not change so much (there is no qualitative difference) but the value of the compressibility changes a lot (quantitative difference). This ...

9

It's certainly possible for ice to sink in water under the right conditions. The diagram this section of Wikipedia's ice page will show you the conditions under which the various types of ice can form. Most of the "exotic" ones such as XII will form only at pressures greater than around 200MPa. These high-pressure forms are all denser than water, so they ...

8

For a pure substance that can exist in the solid, liquid, and vapor states (i.e., wood is not in this category), let's assume that a closed container is half full of liquid and half full of vapor. As the temperature rises, the liquid expands and the liquid density falls. Also, as the temperature rises, the pressure in the container rises due to the vapor ...

7

Consider that the liquid was saturated, for a density of ~808 kg/m^3. Conservation of mass requires the density of the vapor to be the same. Using NIST's property tables you can find the pressure to be about 43,000 psi or 2,900 atm. Never do this... Both 700 atm and 82 atm are very low... NIST Data Table: http://1.usa.gov/12fCuhv

6

I am curious to know under what conditions of the air pressure(atm), temperature, solute density in the water would cause the Niagara fall frozen? In general, the answer is "a bit lower than 32 Fahrenheit". Here's two things which one might think would come into play, but actually do not to an appreciable extend. Solute concentation The major solutes ...

6

The situation is well represented in the following very pictorial picture but this is a very active field of study. It is interesting to note that a real proof of existence for the critical endpoint (CEP, indicated as a critical point in the figure), both from a theoretical and numerical point of view, does not exist yet. The reason, at least for the ...

5

From the densities of liquid nitrogen and nitrogen gas at standard pressure the volume ratio is about 1:700. For an ideal gas in a closed 1L container this would result in a pressure of 700 atm according to $$P V = n R T$$ From the phase diagram nitrogen is a gas at standard pressure and becomes supercritical at approximately 100 atm. The ideal gas law can ...

5

You are right, these terms are related. Metastability usually comes about in systems which are described by a Landau free energy which contains a cubic or power 6 term on top of the usual $\phi^4$-theory. E.g. the Landau free energy with a cubic term is shown below. There are three special temperatures: $T^{**}$ at which an additional local minimum forms ...

5

(1) Classifying "Phase Structure of (Quantum) Gauge Theory" (with a gap) is roughly the same as classifying phase structure of topologically ordered states. Some topologically ordered states are described by a group and can be related to a gauge theory. Some other topologically ordered states are not related to gauge theory. (2) One way to classify "Phase ...

3

No doubt that a must read on this topic is the classic work by Fradkin and Shenker: http://journals.aps.org/prd/abstract/10.1103/PhysRevD.19.3682 In particular, it was pointed out that for $Z_2$ gauge theories (and I believe for all $Z_n$) the confined phase and the Higgs phase are in fact smoothly connected. There is no sharp phase boundary between the ...

3

We normally consider the various states of matter to be separated by a phase transition, and generally this is a first order phase transition (an exception is the second order glass-liqid transition). So for example the solid to liquid transition is (usually) a first order phase transition, and likewise the liquid to gas transition. However if we move from ...

3

consider the conditions of nearly completely frozen falls: which means there are water branches from the very top of the falls are frozen connected to the bottom of the falls. The picture you shown suggests the stuffs connecting from top of waterfall to the bottom is icicle, which can only be formed with not so fast flowing of water, otherwise high speed ...

3

I'll make an attempt of partial answer here, and perhaps extend the question a bit : I think liquid water-gas water are already phases that spontaneously break symmetry of say your "water Hamiltonian". Since you can go continuously (without any phase transition) from one phase to the other phase (by going around the critical point at high temperature and ...

3

yes, hydrogen (and anything) evaporates/sublimes in vacuum

2

If the ice floats (and not falls to the bottom), then it is completely irrelevant what density it has. It displaces exactly the volume of the water that corresponds to its mass, and since ice melts into the water it will not change the level. Let me just supplement you that the reported density of ice XII 1.29 g/cm$^3$ is for the temperature of 127 K, I ...

2

No. As @Jim said, the heat would weaken the rock, which would cause a tunnel collapse before any sublimation could occur. Also, remember that the air in the tunnel would generally be at the same temperature as the rock (unless a large cooling system was put in), so thermal equilibrium would be maintained without any sublimation.

2

The way to distinguish intermetallics (or any other phases) is to used the commonly accepted phase description, such as Mg$_{2}$Pb as you did above. This phase descriptor than points people to information on the crystal structure, thermodynamics of the phase, etc. The point is that these are thermodynamically distinct phases - a first order phase transition ...

2

Estimation: I want the two densities of vater and vapour to become approximately equal. the density of water is nearly constant the vapour pressure (you can derive this from the above mentioned Clausius-Clapeiron-equation) is approximately exponential in $1/T$. This means, that if you increase pressure by a factor, the inverse of the evaporation ...

2

Consider a variant of the FitzHugh–Nagumo system: \begin{align}\dot{x} &= x (a - x) (x - 1) - y \\ \dot{y} &= bx - cy \end{align} with parameters like $a=-0.02; b=0.01; c=0.02$. The eigenvalues of the Jacobian at $(0,0)$ (which is a fix point) are $±\frac{\sqrt{6}}{25}i$. The attractor for this system looks like this: Obviously, this is ...

2

As you can see from the phase diagram plot in the first link you provided, the only other ice phase which is stable at atmospheric pressure is ice XI, and its density is about the same as that of the most familiar ice phase (ice Ih). The other denser ice phases that you see on the phase diagram are only stable at pressures significantly above 1 atmosphere. ...

2

Good question. I don't have my Widom around, but I'll try to answer from memory. I think the consensus is to say a substance is at its gas state if it could be a liquid at the same temperature. This, as opposed to same pressure, same volume, etc. If the temperature is supercritical, there is no transition between liquid and gas, and the generic term "fluid"...

2

Note that the sum of two phasors is another phasor: $Ae^{j(wt+\phi_1)} + Be^{j(wt+\phi_2)} = Ce^{j(wt+\phi_3)}$ Where A,B,C are real. The only way for the real part of the right side to be equal to zero at all times is if $C=0$. In which case the whole thing is $0$, both the real and complex parts are 0. So a sum of phasors (of the full sinusoidal form ...

2

Attempting to answer the "why" question intuitively: In a liquid, the molecules experience significant intermolecular force - so much so, that the average energy of the molecules is insufficient to escape the attractive force of the surrounding materials. The result is that it energetically favorable for them to remain close together, even if that means ...

2

I have no idea what you are asking in (1). (2) During adiabatic compression, the temperature of the system does change. Starting from the first law of thermodynamics: $$\mathrm{d}U = \mathrm{d}Q + \mathrm{d}W$$ For an adiabatic process $\mathrm{d}Q=0$. If we assume an ideal gas, then the internal energy is $\mathrm{d}U = C_V\, \mathrm{d}T$ and the work ...

1

From the diagrams on the webpages you linked, it appears that other ice phases begin to form at around 200 MPa of pressure, and about $-20\text{C}^{\circ}$. Keep in mind normal water freezes at $0\text{C}^{\circ}$, and air pressure at sea level is around 0.101MPa. That means an ice cube made of one of these phases would sublimate or explode very quickly. If ...

1

@FasEtNefas presents us with this quote: Now suppose a uniform binary mixture [e.g. a CuCr metal alloy] at a high temperature and concentration u∗ is suddenly quenced to a given lower temperature. A commonly occuring situation is that there is a pair of values of u [which is the local concentration of one of the components], say u1 and u2, and a ...

1

Have a look at the following idealised binary phase diagram: The vertical axis is temperature, the horizontal is composition (in mole fraction but weight % would work too here). Say we started from point 1. where the alloy is fully liquid with a well defined composition, say $u$. Now we cool down, following the red line. At some point we hit the black ...

1

The melting point of nitrogen is greater than that of oxygen. Is gaseous oxygen really liquid oxygen dissolved in gaseous nitrogen? No. It is gaseous oxygen. "Water vapour in the air" (your phrase) is just that - vapour. Not liquid droplets "dissolved" in air. You may be confused by the fact that wet steam does contain water droplets, and the steam from a ...

1

I haven't done a calculation yet, but I would use an extrapolation based on the Clausius-Clapeyron formula: $$\frac{dP}{dT} = \frac{L}{T\Delta V}$$ You then take any two known thermodynamic quantities of water and water vapor and linearly extrapolate to that point where the difference is zero. A good choice could be the entropy, the entropy of water vapor ...

1

Please note the number $2$ in the phase rule indicates freedom in $p$ and $T$, but for the critical point, those parameters are fixed.

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