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how long would it take for this super-material to convert to the stuff
I scribble with?
No, despite the fact that James Bond said "Diamonds are Forever", that is not exactly the case. Although Bond's statement is a fair approximation of reality it is not a scientifically accurate description of reality.
As we will soon see, even though diamond is ...
26
$\ce{H2}$ 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.
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The critical temperature of Hydrogen 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 ...
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Interestingly, nobody addressed the reason why diamonds are hard in the first place. The pressure (and temperature) are not the reason why they're hard, only the reason why they are formed. The diamonds are hard because the carbon atoms are bonded together by sigma ($sp^3$) bonds, which are the strongest chemical bonds. Other materials exhibiting the same ...
21
Here are some compounds that have other structures, followed by their hardest structure (based on Moh's Scale).
Titanium dioxide: Rutile structure or Cotunnite structure
Aluminum oxide: Corundum
Silicon Oxide: Stishovite
Boron Nitride: Wurtzite Boron Nitride
There are many, many more.
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I didn't know that balloons expanded during the fly because of thermodynamics, and I didn't know how high they can fly, but a rapid search tells that a partially unfilled regular balloon can fly until an altitude of around $\pu{25 km}$.
Now, $\pu{25 km}$ means that it reaches the first part of the stratosphere, with temperatures of $\pu{-60 ^\circ C}$, that ...
11
External and internal pressure
To study the effect of pressure on properties of a solid, is equivalently to learn how changes in volume transform physical parameters. For external pressure at constant temperature, this relationship manifests through compressibility $\kappa$.
$$\kappa =-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$$
An ...
10
Here's your confusion:
You need to consider two different things:
The momentum transfer per particle per collision. There, since we assume an instantaneous collision, it doesn't make sense to try to figure out force from acceleration. [I suppose you could do this using limits, and maybe there are applications in which that does make sense, but adding that ...
9
The study Explosion Characteristics of Hydrogen-Air and Hydrogen-Oxygen Mixtures at Elevated Pressures includes data for pressures up to 200 bar.
Data were collected for Hydrogen-Oxygen mixtures at both 20 and 80 degrees C at pressures ranging from 1 to 200 bar. A high voltage spark was still needed to cause explosion.
Storing a compressed mixture of ...
9
If the piston is frictonless and massless, then, if you do a force balance on the piston, you must have that the force per unit area that the gas exerts on the inside face of the piston will always be equal to the external force per unit area that one imposes on the outside face of the piston. The sudden drop in pressure on the outside face of the piston ...
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{...
answered Jun 28 '20 at 18:00
Mathew Mahindaratne
34.1k18 gold badges45 silver badges95 bronze badges
9
There is one importing principle to consider.
Air does not store vapor. Space does. Less space means less vapor capacity, no matter how much of air is there. (*)
If air is being compressed, the partial vapor pressure increases. It it reaches the saturated vapor pressure at given temperature, water vapor starts to condensate. When air is then expanded, its ...
8
As you can see on this data page on Wikipedia helium-4 has critical pressure of 2.24 atm, but helium-3 has 1.13 atm, only slightly above atmospheric pressure. Hyperbaric chambers work with 1,3-1,5 atm, so you could try there.
There is, unfortunately, question if it has sense, as you wouldn't notice any serious difference in properties between subcritical ...
8
You are exactly correct that it is a matter of atmospheric pressure decreasing at a rate great enough to overcome the contraction due to decrease in temperature. On a nice, clear, dry 25°C day at sea level, atmospheric pressure decreases by about 12% per km, where the air temperature decreases by about 3% per km.
This is very similar to the process that ...
8
I have calculated pressure in container one and that is p1 = 342817.92 Pa, and in container two p2 = 326979.2275 Pa. Then I have substracted this two values which gives me 15838.6925 Pa. After that I have divided this value with two, to figure out how much do I have to decrease pressure in container 1, and that gives me 7919.34625 Pa.
You cannot generally ...
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The term “negative pressure” is often used in engineering to refer to a situation in which an enclosed volume has lower pressure than its surroundings. If the a region is surrounded by more pressurized area around it would cause substances to flow inwards and thus the term negative pressure.
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Well, it seems that diamond is not forever because upon exposure to sunlight it loses atoms.
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I add an other answer : it depends what kind of people you are, if you are as crazy as them, of course not !
2012 Christmas Lectures - Burning a Diamond
But ron's answer is better sure ! ;)
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First, a point that is more like a comment: I'm not sure that just because electrons get put in to d-orbitals (and thinking of crystal band structures in terms of atomic orbitals is of questionable value anyway) that it would not be a metal. Many elements (one could even say most elements) with d-electrons are quite happy as metals.
You might want to start ...
7
The answer by camd92 is correct, but there is no reason to calculate amounts. According to the chemical equation, one third of the particles in the gas phase will be dinitrogen oxide, and two thirds will be water. So you know the mole fractions, and you can directly calculate the partial pressures. (And yes, camd92 is also correct in pointing out the ...
7
Of course, we're starting by acknowledging a perpetual motion machine is impossible. The question, then, is how do we understand, through chemical thermodynamics, why your specific set of steps can't constitute a perpetual motion machine.
The answer is that what you're not accounting for is that the electrical energy required for your step 2 is greater ...
7
According to German Wikipedia water potential is defined as
$$\psi := \frac{\mu - \mu_0}{\bar{V}},\tag{1}$$
where $\mu_0$ is the standard chemical potential (usually pure water at atmospheric pressure at a specified reference height), and $\bar{V} \approx \pu{18 cm3 mol-1}$ is the molar volume of pure liquid water.
This means that water potential $\psi$ is ...
7
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 ...
6
A number of long carbon chain molecules with large numbers of fluoride atom, have $P_c < \pu{10 atm}$ and $T_c < \pu{1000 K}$. That is within the reach of a good bicycle pump and a burner. In particular $\ce{C12F26}$, $\ce{C15H4F28O}$ and $\ce{C20F42}$:
\begin{array}{lll}
&\ce{C12F26} &P_c = \pu{912 kPa} &T_c = \pu{417 K} \\
&\ce{...
6
Let me begin my discussion by first introducing the notion, using some intuitive model building, and will try to derive, or perhaps guess the van der Waals equation of state.
The perfect gas equation, $ PV = nRT $ doesn’t allow for any interactions between molecules. The van der Waals equation is a refinement of this model, in the sense that it introduces ...
6
If we maintain nitrogen at 220 bar and 400 °C, what will be the state of $\ce{N2}$? Will it be still supercritical?
Yes, it will still be supercritical. The only reasonable definition of "supercritical" is something along the lines of "a fluid phase without surface tension wherein the fluid pressure and temperature exceed the critical point".
Thus, ...
6
It is not true that heat exchanged at constant pressure is always reversible.
But, if you want to determine the change in entropy from thermodynamic equilibrium state 1 at $(T_1,P)$ to state 2 at $(T_2,P)$, you need to forget entirely about the actual irreversible process path that took you from state 1 to state 2. It is of no further use. You instead need ...
6
First off, "chemical pressure" (CP) is often portrayed as an empirical concept or effect for linking chemical composition and physical properties in the first place, and not a physical quantity -- even though it is one. That might be a reason for certain confusion.
Doing some research I encountered "CP" term being used in other fields with substantially ...
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