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23

$\ce{H2}$ cannot be liquified at room temperature, whatever the pressure. Generally speaking, all gases can only be liquified when the temperature is under its critical value.


22

Hydrogen critical temperature is $\pu{32.938 K, resp. -240.21 ^{\circ}C}$. Above this temperature, it cannot be liquified. So to answer your question, you can get as high pressure as you can produce and the container can withstand, as there is no condensation reducing the pressure. WARNING: An accidental explosive container rupture can easily cause severe ...


13

Yes. Any fluid with a temperature is above critical temperature and the pressure above the critical pressure is by defintion a supercritical fluid. Don't be mislead by all the claims that supercritical fluids are special and wonky with all sorts of amazing, bizarre properties. This is true of some supercritical fluids near the critical point, but the ...


10

There are two factors to consider. Certainly, the dominant factor is that the degrees of freedom available for the molecule to disperse energy into increases as the carbon chain extends. Thinking along these lines, one would expect a monotonic increase of entropy with carbon chain length. However, we must also consider phase. As the carbon chain increases in ...


9

Question: Can $\ce{SiO2}$ melt at $\pu{20 ^\circ C}$? According to experimental and calculated data values, my answer is no. See the phase diagram of pure silica based on the experimental and calculated data given in Ref.1: Reference 1 states that: An internally consistent data set on the thermodynamic properties of the silica polymorphs stable up to $\pu{...


8

It is a problem of phase. The standard entropy of butane is $\pu{310 J mol-1 K-1}$ in the gaseous phase. The standard entropy of hexane is $\pu{296 J mol-1 K-1}$ in the liquid phase. But it is $\pu{389 J mol-1 K-1}$ in the gaseous phase. So the entropy increases from butane to hexane in the gaseous phase, as expected.


7

This is the phase diagram from a better source: The pressure axis is the vertical one at the left. One bar (which equals 0.986923 atmospheres) is near the bottom, as shown on the diagram. So carbon dioxide is a gas at all temperatures above −78.5 °C when the pressure is one bar. Solid carbon dioxide (dry ice) sublimes at one bar (about one atmosphere), i.e.,...


7

Ionic liquids are not magical liquids. They are just typical organic compounds with certain properties. These compounds generated a storm in the tea cup in the late 1990s and 2000s, but they are as toxic as other solvents. Certainly they have niche applications especially as a gas chromatography stationary phase, where they can be used at high temperature (...


6

Ionic liquids have very special properties, the most important of those of interest is that they are liquids at low temperature with extremely low vapour pressure. This is in fact the main reason they have caused excitement among chemists. Quoting (1) from a report comparing the volatility of various ILs: An extremely low vapor pressure (e.g., ca. 100 pPa ...


6

I think what you are asking is this: Equilibria for chemical reactions typically* (see note at end) require specific ratios of products to reactants (as expressed by the equilibrium constant). By contrast, equilibria for phase transitions don't require specific ratios of products to reactants. [For instance, at the phase transition between ice and water, ...


6

As others have said, hydrogen cannot be liquified above its critical temperature, which my source (Wolfram Alpha chemical database*) says is $\pu{32.97 K} = \pu{-240.18 ^\circ C}$ However, with sufficient pressure, the molecules can be squeezed together until they have a liquid-like density**, and are thus no longer considered a gas, but rather a ...


5

Calderon & Mohazzabi give an excellent summation of the various theories proposed through the years to explain why ice is so slippery in their 2018 paper "Premelting, Pressure Melting, and Regelation of ice revisited" in the Journal of Applied Mathematics and Physics. They offer both theoretical and experimental evidence that neither pressure ...


5

The phase equilibrium diagram is for a system consisting of just water. The diagram pressure is the total system pressure. It may be equal to the saturated vapor pressure, if no external pressure is imposed on the system. Or, it can be ( much ) higher, if some external pressure higher than the saturated vapour pressure exists. In such a case all vapor is ...


5

The other answer already shows that this is a solid fact. Carbon dioxide is not (actually cannot stay) liquid at 100 kPa, period. Why that is the case is much less obvious. I cannot give you a clear, rigorous, quantitative explanation. But the general reasoning about this and similar observations is that once the crystal structure has broken up under ...


4

Imagine filling the e.g. water into a cylinder with a piston on top. No air inside. Now the experiment is to control the temperature inside the cylinder, and move the piston to get the wanted pressure inside. If the pressure is above the vapour pressure, your cylinder is completely filled with liquid water, if it is below, all the water will be evaporated. ...


4

You can liquify ethane at standard pressure (1 bar) simply by lowering the temperature (cooling), as indicated by its phase diagram (see for instance here). On the other hand, if you look at the phase diagram of $\ce{CO2}$ you find that starting at standard pressure (1 bar) and room temperature, it cannot be cooled into the liquid state. Rather it will ...


3

Melting is a process. A process is not a state. Since only states can be "at equilibrium", it doesn't make sense to speak of a real melting substance as being in an equilibrium state. However, melting is deemed reversible if performed infinitely slowly, and during such a transition the system passes through a series of equilibrium (or "quasi-...


3

There's a very nice—and fairly accessible—discussion of your third question in Nature Magazine's News and Views Section[1]. In this article, Bonn discusses the results of a fairly recent work that appeared in Physical Review X[2]. Bonn summarizes the key points as follows: The idea that a thin film of meltwater wets the surface of ice has been accepted ...


3

This is a summary of the equations to use to calculate phase transitions. The Clapeyron equation $\displaystyle p_2-p_1=\frac{\Delta H}{\Delta V}\ln\left( \frac{T_2}{T_1} \right)$ is used for a solid-liquid transition. The changes in enthalpy and volume relate therefore to changes occurring in fusion. The Clausius-Clapeyron equation describes solid-vapour ...


3

[OP] Why are melting and boiling considered equilibrium processes [...] They should not be considered equilibrium processes. If melting is defined as the process where there is a net change from solid to liquid phase, this is not an equilibrium. If boiling is defined as the process where liquid turns into vapor (rolling boil with bubbles forming below the ...


3

Two different phases of a substance in contact with each other in a closed system at some uniform temperature and pressure (thermal and mechanical equilibrium) will be in equilibrium if the chemical potential of the substance is the same in both phases. It turns out that at its boiling point, a liquid has the same chemical potential as its vapor at that ...


3

The way to understand the effect of the pressure of an inert gas (oxygen here assumed to be inert) on the vapor pressure of a liquid is to consider the effect of the extra pressure on the chemical potential of the liquid: $$\left(\frac{\partial \mu}{\partial P} \right)_T=V_m$$ The additional pressure compresses the liquid and thereby increases its chemical ...


3

There is really not much to it. Basically you have a metal and an oxide phase, and never the twain shall mix significantly even after melting the oxide above 2000°C [1]. The suboxide that is sometimes mentioned, $\ce{Al2O}$, is not indicated here. From Ref. 1 Reference 1. S. Das, "The Al-O-Ti System", J. Phase Equilibria 23(6) (2002), 525-536.


3

Latent heat of vaporization/evaporation: It's the amount of heat required for liquid ---> gas phase change of a substance at it's Boiling point. Not exactly. It's the amount of heat required for liquid ---> gas phase change of a substance at the particular temperature, there is usually and implicitly the substance boiling point. But generally, it is ...


2

Linde Gas contains a vapor pressure curve. Note that this does not extend to Venusian surface temperatures which are above the critical point. Hydrogen fluoride may not lift as well as you think. Its gas phase may associate through hydrogen bonding so molecules could be heavier than just HF. See the note near the end of the safety information here and ...


2

Air, fresh water, sea water, whiskey are fluid phases, aside of solid phases. A phase is a space region of homogenous properties of matter, possibly with gradient of theses properties (like sea water profile, or solution of statically dissolved solid), with strong discontinuity of these properties at the region boundary. The gradient can be at special ...


2

The evaporation of water does not depend on the depth of the lake. It depends on the temperature and of the wind. Here are some values, valid for a lake of infinitive surface, with an "usual humidity" At $10°$C, the level of the water decreases by $0.9$ mm/day if the wind is zero, by $1.2$ mm/day with a $2$ m/s wind, and by $1.7$ mm/day with a $9$ ...


2

I got that you are trying to visualize what is actually happening. I have not done the compression myself, but this is what I think will be observed. It will be good if someone with real experience can confirm the answer. At the beginning you are in the region marked 'vapour'. When you are approaching the liquid-vapour boundary from below, before you touch ...


2

The question: $\pu{1kg}$ of a solution of cane sugar is cooled and maintained at $\pu{-4 ^\circ C}$. How much ice will separate out if molality of solution is $0.75$? Given: $K_\mathrm{f} \text{ (Water)}= \pu{1.86 K kg/mol}$ The solution is given assuming you understand the concept. It seems you did not completely understand the colligative properties of ...


2

Can we justify that "For sublimation of a solid at 1 atm $\Delta U>0$ at low temperature and $\Delta U<0$ at high temperature?" No. $\Delta U>0$, always, for sublimation, because of the energy needed to separate the atoms or molecules in changing from the solid to the gas phase. As for the enthalpy, $$H = U +PV \Rightarrow \Delta H = \...


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