46
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 ...
22
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
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 ...
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
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
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
Well, it seems that diamond is not forever because upon exposure to sunlight it loses atoms.
7
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 ! ;)
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
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
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:
6
In ideal gases no intermolecular forces, therefore no potential energy. Thus, internal energy is equal to total kinetic energy (KE) of the system. Consider $N$ monoatomic particles in a cubical box of side $\ell$ (assumption: ideal gasses consist of monoatomic point particles). The amount of ideal gas in the box is $\frac{N}{N_A} = n \ \pu{mol}$ where $N_A$ ...
5
At equilibrium (that is, if volume/temperature/mass is stable and not changing over time), the external pressure applied to the gas will equal the internal pressure of the gas.
This is (more or less) a consequence of Newton's third law. The external environment pushes on the gas with a certain force, and the gas pushes back with an equal and opposite force. ...
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
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
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 ...
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 ...
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