41

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'll soon see, even though diamond is slightly ...


23

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.


18

The two units torr, and mm of Hg were the same until they were redefined. The torr was named after the Italian Evangelista Torricelli. 1 atmosphere is $101325\ \mathrm{Pa}$. The torr is defined as $1/760$ of an atmosphere. This is equal to $133.322\overline{368421052631578947}~\mathrm{Pa}$, which periodically infinitely repeats. The mm of Hg was defined as ...


16

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

The change in internal energy $U$ is $$\Delta U=Q+W$$ where $Q$ is amount of heat transferred to the system and $W$ is work done on the system. Since the process is adiabatic, no heat is transferred into or out of the system, i.e. $Q=0$ and thus $$\Delta U=W$$ The reversible expansion is performed continuously at equilibrium by means of infinitesimal ...


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

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 ...


7

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.


7

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

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 ...


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 ...


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

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 ...


6

Since the drink is not carbonated, no liquid has been escaping the bottle, and the components don't seem to be able to participate in any kind of chemical reactions involving gaseous products, I assume the gas that's been leaked is air. I also cannot imagine any kind of chemical reaction involving consumption of this amount of air leading to such a severe ...


6

As temperature increases, it activates the rotational and vibrational degrees of freedom, so $\ce{Cv}$ increases with increase in temperature. The graph $\ce{CO2}$ would be:


5

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 ! ;)


5

In general no, the pressure will not be the same, although there will be intersection points where (by coincidence) they will be the same, and there will be regions of the phase diagram for some equations of state (EOS) and for some substances that might be close enough that it doesn't matter. For example, the van der Waals EOS gives pressure values at high ...


5

This is a bad question. But the answer is probably supposed to be (d) because: Water is more dense than ice. Thus when ice melts, the total volume of "substance" goes down. This means that atmospheric pressure did PV work on the melting ice. Of course, heat transfer would also be required to melt the ice but the question isn't asking about that.


5

If exceeding the "liquid–liquid critical point" for a solution suffices, then perhaps you could survive contact with methane hydrate for a short while. Though He is supercritical at ~two atmospheres, running you hand through it would be a chilly experience at about 5 K.


5

Quick googling got me to $\ce{CF3Cl}$, that has cr.p. at $\pu{28.8^\circ C}$ / $\pu{38.6 bar}$. $\ce{C1}$-$\ce{C2}$ freons look to have critical point in $\pu{30 .. 60 bar}$ / $\pu{0 .. 100^\circ C}$ range, so some of them probably are survivable. However, some compounds of this class are used for general anesthesia, so $\ce{He/O2}$ mixture respirator would ...


5

ChemGuide has a good introductory article here. The effects of increasing pressure and temperature are, to an extent, equivalent. Increased pressure leads to increased collisions and increased collision strength between molecules, allowing the (usually high) activation energy barrier to be overcome at a noticeable rate; at standard temperatures and ...


5

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, ...


5

Another interesting paper (arXiv:1310.4718 [cond-mat.mtrl-sci]) on this topic. I think you should be able to access this one, but I found three passages worth noting in the article. Peculiarities in structural behaviour of K under pressure are based on specific features of its electron configuration: K opens in the Periodic table the first long period ...


4

Yes. Pressure, as a concept, is simply the force exerted by random molecular motion per unit area, and since all matter is composed of molecules, and all matter above absolute zero has some random motion, everything has molecules exerting some force over some area. (This pressure is the vapor pressure of the solid, which is typically extremely low. If the ...


4

When we say $$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$ we are referring to $\Delta G^\circ$, $\Delta H^\circ$ and $\Delta S^\circ$ of the system. As such, the variable $T$ in the equation refers to the temperature of the system. The condition for constant pressure $p$ is implicit; the equation above does not directly show why $p$ has to be ...


4

You need to consider the $b$ term, as well as $a$. Rearranging the van der Waals equation to solve for $p$ gives $$p = \frac{RT}{V_\mathrm{m} - b} - \frac{a}{V_\mathrm{m}^2}$$ You are right that a positive $a$ will always decrease the pressure, compared to the ideal gas result. But a positive $b$ will act to increase the pressure compared to the ideal ...


4

$\ce{NC-CN <=> 2CN}$ Which side is favored at low pressure?


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