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Forum » SpaceEngine » Science and Astronomy Discussions » Science and Astronomy Questions
Science and Astronomy Questions
SalvoDate: Monday, 23.11.2015, 19:59 | Message # 361
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Quote NikolaAnicic007 ()
I am aware that these three values are specific for each molecule

Very interesting question.

Actually I know that the melting temperature of "something" not only depends on the bond length but also on the molecular geometry, but I'm not sure, because what I find on internet is pretty confusing.
Then another very important thing is pressure, but in that case you can take whatever value you want.

The state of each molecule depends on the "stability" of the chemical bonds, if the atoms are in a very stable geometry then the whole thing is at solid state, if their bonds can be broken and they can move on each other then it's liquid, at least if there is not a real stable structure it's gaseous.

I'm afraid that there is not a simple formula that can give you the state at a certain temperature/pressure.
What I know is that there are some graphs like this, that are called Phase Diagrams:






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(still don't know why everyone is doing this...)
 
NikolaAnicic007Date: Monday, 23.11.2015, 23:07 | Message # 362
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I need numbers, mate. I do have the geometry, but not the bond strength. I have the total energy, doe.

Do you posses a formula I'd need to fulfill to obtain the numbers I need?
 
WatsisnameDate: Tuesday, 24.11.2015, 05:16 | Message # 363
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Quote NikolaAnicic007 ()
How does one calculate the Latent heat of Fusion, the Latent heat of Vaporization and the Latent heat of Sublimation for a molecule?


Generally, you don't. These are almost always determined experimentally.

The reason is that deriving them is hard. It requires extremely involved quantum mechanical calculations, and there is no simple analytic formula that you can apply to any arbitrary molecule. To get an idea of what is involved, try doing a google scholar search for "ab initio calculations of enthalpy" and read some abstracts that turn up, e.g. this one on vaporization enthalpy of silicon.

"Enthalpy" vs. "latent heat" may be another reason why you have difficulty finding information on this online. They mean essentially the same thing, but enthalpy is the more commonly used term in the jargon of materials science and thermodynamics.





 
NikolaAnicic007Date: Tuesday, 24.11.2015, 12:43 | Message # 364
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Quote Watsisname ()
Generally, you don't. These are almost always determined experimentally.

The reason is that deriving them is hard. It requires extremely involved quantum mechanical calculations, and there is no simple analytic formula that you can apply to any arbitrary molecule. To get an idea of what is involved, try doing a google scholar search for "ab initio calculations of enthalpy" and read some abstracts that turn up, e.g. this one on vaporization enthalpy of silicon.

"Enthalpy" vs. "latent heat" may be another reason why you have difficulty finding information on this online. They mean essentially the same thing, but enthalpy is the more commonly used term in the jargon of materials science and thermodynamics.


Thank you for taking the time to properly answer my question with quoted sources that I can further explore, and additional information you've supplied me with in your post. I was always aware that quantum mechanical effects would have an influence on the calculation and that they would be involved, but I guessed it was worth a shot. The example you supplied me with has showed me that the calculations exist and are possible, and can thus be learned.

Now, will it be hard? Probably. But a truly determined explorer never backs down from a challenge!

Pertaining to the subject at hand, the usage of Enthalpy instead of Latent Heat completely slipped my mind, which sounds in character for me since I tend to overlook small details.

I have another question.

Earlier, you mentioned that SE calculates the speed of sound in a given atmosphere based on the square root of temperature over molar mass. That makes no sense, at least if you are treating temperature as temperature and not heat. The way you explained it, it seemed to me that c = sqrt(T/M), which doesn't add up since temperature and molar mass don't exactly produce any velocity.

Using the following formula:

c = PM/(RT)

c = 101,325 * 28.97 / (8.314459848 * 288.45) = 1,223.9411149 km/h = 339.984 m/s

I received high-accuracy speed for air at 15.3 Degrees Celsius, where my calculated value differed from the known value (343 m/s) by around 3 or 4 m/s, which can be accounted to small changes such as a higher accuracy determination of air temperature on average for Planet Earth.

Your usage of the formula indicated that:

c = sqrt(T/M) = sqrt(288.45 / 28.97)

Which leaves me thinking I misunderstood your calculation, since you seem to have a strong foothold in these matters. Your usage of the word "Temperature" leaves me thinking you might have wanted to use the term "Heat" which would then change the calculation to a one that requires specific heat, which is a constant gained from scientific experimentation, as is phase transition energy, which, again, seems unlikely.

Could you further clarify your statement and explore the formula I supplied to you for errors? It would mean a lot.

Thank you for your time.

Edit: Oh God I've been using the density formula. Does the density formula get close to the real value because it approximates the compressibility of the atmosphere if it were an ideal gas?

Edit: Tried Mars, the numbers don't add up. Abandoning aforementioned formula.


Edited by NikolaAnicic007 - Tuesday, 24.11.2015, 16:04
 
SalvoDate: Tuesday, 24.11.2015, 18:33 | Message # 365
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Quote NikolaAnicic007 ()
Abandoning aforementioned formula.

Unfortunately it's very very complex to do these calculus. But if you want to get somewhere I suggest you to use Microsoft Mathematics (one of the few Microsoft programs that are really well made) that is also able to plot 2D and 3D functions.

Anyway, my Math knowledge is horrible so I can't really help you, but I can tell you (looking at Phase Diagrams) that what you're looking for are not Formulae but probably conditioned Systems of Inequations. So... good luck.

If you just want to have the answers you're looking for or to play around a little, you can use TermoCalc, but unfortunately it's a 30 days demo. If you search on the internet I think you can find freeware alternatives.





The universe is not required to be in perfect harmony with human ambition.

CPU: Intel Core i7 4770 GPU: ASUS Radeon R9 270 RAM: 8 GBs

(still don't know why everyone is doing this...)
 
WatsisnameDate: Wednesday, 25.11.2015, 03:06 | Message # 366
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Quote NikolaAnicic007 ()
Thank you for taking the time to properly answer my question with quoted sources that I can further explore, and additional information you've supplied me with in your post.


Hey, my pleasure. smile I'm excited that you're so interested in pursuing the deeper maths and physics of this problem, and I'm sorry I can't help you further with it. The hard problems of computational thermochemistry go far beyond my own knowledge and solving ability.

Let me clear up that confusion with the sound speed problem:

Quote NikolaAnicic007 ()
Your usage of the formula indicated that:

c = sqrt(T/M) = sqrt(288.45 / 28.97)


In the chatbox I was describing the proportionality of sound speed to temperature and molar mass (this is an interesting and counter-intuitive fact: that it can be derived in terms that are not explicitly dependent of the gas pressure or density). But that is not what the full formula actually is. As you saw its dimensions don't even equate to velocity.

The actual formula is this:



where gamma is the adiabatic index (ratio of the specific heats Cp/Cv of the gas), and is thus a dimensionless quantity.

R is the gas constant (with dimensions J/mol/K) (and recall joules are kg*m^2/s^2)
T is the absolute temperature (Kelvin)
and M is molar mass of the gas (kg/mol)

This all comes out as m/s = sqrt(m^2/s^2). So it works. That was the confusion, I think. The sound speed is proportional to the square root of temperature over molar mass, but it is not equal to it. There are some constants involved, one of which (the gas constant) has dimensions which make it come out to velocity.

Let's also apply dimensional analysis to the PM/RT formula you were using:

Quote
c = 101,325 * 28.97 / (8.314459848 * 288.45) = 1,223.9411149 km/h = 339.984 m/s


This has dimensions of density, with SI units of kg/m^3. Be very careful here. You do not get km/hr out of this (hours are not even SI!). And the number (~1224) you are computing is actually in units of grams per cubic meter, since you used M in terms of grams/mole.

That 1224g/m^3 "converts" to ~340m/s is therefore pure coincidence. It has no basis in physics or even in units.

Sorry for all the confusion; I hope that helped clarify it for you! Dimensional analysis is a very powerful tool in your physics arsenal. Use it well. smile





 
NikolaAnicic007Date: Wednesday, 25.11.2015, 12:18 | Message # 367
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Quote Salvo ()
Unfortunately it's very very complex to do these calculus. But if you want to get somewhere I suggest you to use Microsoft Mathematics (one of the few Microsoft programs that are really well made) that is also able to plot 2D and 3D functions.

Anyway, my Math knowledge is horrible so I can't really help you, but I can tell you (looking at Phase Diagrams) that what you're looking for are not Formulae but probably conditioned Systems of Inequations. So... good luck.

If you just want to have the answers you're looking for or to play around a little, you can use TermoCalc, but unfortunately it's a 30 days demo. If you search on the internet I think you can find freeware alternatives.


I will check it out, thank you!

:B

Quote Watsisname ()
Hey, my pleasure. I'm excited that you're so interested in pursuing the deeper maths and physics of this problem, and I'm sorry I can't help you further with it. The hard problems of computational thermochemistry go far beyond my own knowledge and solving ability.

Let me clear up that confusion with the sound speed problem:


Fire away.

Quote Watsisname ()
This has dimensions of density, with SI units of kg/m^3. Be very careful here. You do not get km/hr out of this (hours are not even SI!). And the number (~1224) you are computing is actually in units of grams per cubic meter, since you used M in terms of grams/mole.

That 1224g/m^3 "converts" to ~340m/s is therefore pure coincidence. It has no basis in physics or even in units.

Sorry for all the confusion; I hope that helped clarify it for you! Dimensional analysis is a very powerful tool in your physics arsenal. Use it well.


I...may or may not have ignored this fact, while seeing it anyways... ._.'

I was like: "The numbers work so why not?", but, alas, I was wrong. They only work for Earth (sorta) which is a little peculiar. That's a fun fact I guess, that Earth's air density is approximately it's speed of sound in kilometers per hour. :B

Now...

Quote Watsisname ()

The actual formula is this:

where gamma is the adiabatic index (ratio of the specific heats Cp/Cv of the gas), and is thus a dimensionless quantity.

R is the gas constant (with dimensions J/mol/K) (and recall joules are kg*m^2/s^2)
T is the absolute temperature (Kelvin)
and M is molar mass of the gas (kg/mol)

This all comes out as m/s = sqrt(m^2/s^2). So it works. That was the confusion, I think. The sound speed is proportional to the square root of temperature over molar mass, but it is not equal to it. There are some constants involved, one of which (the gas constant) has dimensions which make it come out to velocity.


Aha, so yeah, my doubts were just confirmed. I found this formula on wikipedia earlier, but I saw that adiabatic pressure would ALSO need some calculations about the gas, which brings up some more issues. I thought that SE used a system where only the temperature, density and molar mass were needed (I know the dimensions don't add up, but hey, shortcuts are a thing :P). The only issue I have with the formula is the adiabatic pressure.

I guess my question in this case is: How do I calculate adiabatic pressure for a randomly selected molecule? Is it possible or is it, like phase transition, something that's hard to determine using calculus?


Edited by NikolaAnicic007 - Wednesday, 25.11.2015, 12:20
 
Tac1017Date: Wednesday, 25.11.2015, 12:45 | Message # 368
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Quote Salvo ()
I'm afraid that there is not a simple formula that can give you the state at a certain temperature/pressure.What I know is that there are some graphs like this, that are called Phase Diagrams:


So... there are temperatures when liquids or gasses can't exist?





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NikolaAnicic007Date: Wednesday, 25.11.2015, 13:06 | Message # 369
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Quote Tac1017 ()

So... there are temperatures when liquids or gasses can't exist?



No, however, the highest known temperature that I know of is that capable of transforming matter into a quark-gluon plasma.

That and the theoretical Big Bang.

I also found this:

The most straightforward candidate for an upper limit is the Planck Temperature, or 142 nonillion (1.42 x 10^32) Kelvin (K)—the highest temperature allowable under the Standard Model of particle physics.

Source: http://www.popsci.com/article....ossible


Edited by NikolaAnicic007 - Wednesday, 25.11.2015, 13:39
 
WatsisnameDate: Wednesday, 25.11.2015, 21:13 | Message # 370
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Quote Tac1017 ()
So... there are temperatures when liquids or gasses can't exist?


A phase diagram shows what phase a particular substance will be in at a given pressure and temperature. As you can see from such a diagram (say, for water), the liquid phase ceases to exist below a certain threshold of temperature or pressure. Under those conditions, water will skip the liquid phase entirely and transition directly between solid and gas.

You can see this process with dry ice (frozen CO2). CO2 cannot be liquid at standard atmospheric pressure, and instead sublimates directly into a gas. But if you increase the pressure, then the liquid phase for CO2 appears. (Do not try that at home, the pressure inside the bottle can increase to pretty dangerous levels.) There are even conditions where the liquid and gas phases merge together and you get what's called a "supercritical fluid".

Different substances have different phase diagrams. For example, helium can be liquid at far colder temperatures than water can. But are there temperatures where no substance can exist as a liquid? Certainly, like Nikola says, if you go to sufficiently high temperatures such as at the center of a star or soon after the Big Bang, then the usual three matter phases can't exist because all atoms are completely ionized into plasma.

Quote NikolaAnicic007 ()
I was like: "The numbers work so why not?", but, alas, I was wrong. They only work for Earth (sorta) which is a little peculiar.


Always check your units. smile If you are doing a calculation where the dimensions are in MKS system (meters, kilograms, seconds), then you cannot say the answer comes out as kilometers per hour, or pounds per square inch, or whatever seems convenient. Your answer must be in the exact same units as what went into the calculation itself. An equation with mismatched dimensions, or a calculation with mismatched units, cannot be correct.

It is kind of neat though that the density of air at Earth's surface (in grams/m^3) is approximately equal to the sound speed in meters per second. When I first read the result of your calculation, I was like "wait, what?" before I recognized what you were doing. :p

Quote NikolaAnicic007 ()
Aha, so yeah, my doubts were just confirmed. I found this formula on wikipedia earlier, but I saw that adiabatic pressure would ALSO need some calculations about the gas, which brings up some more issues. I thought that SE used a system where only the temperature, density and molar mass were needed (I know the dimensions don't add up, but hey, shortcuts are a thing :P). The only issue I have with the formula is the adiabatic pressure.


Yeah, there's no formulation for sound speed which does not invoke the adiabatic index in some way or another. But it is just a number, and you can find it for most common atmospheric gases. For monatomic gases you can treat it as 5/3. For diatomic, 7/5. Of course, if your atmosphere of interest has a significant concentration of more complex molecules, then you have to do more work to figure it out.

Then the next question to examine is whether your calculation is more convenient to do in terms of pressure and density, or as temperature and molar mass? SE gives you all this information, so it is mainly a matter of preference.





 
NikolaAnicic007Date: Thursday, 26.11.2015, 12:17 | Message # 371
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Quote Watsisname ()

Different substances have different phase diagrams. For example, helium can be liquid at far colder temperatures than water can. But are there temperatures where no substance can exist as a liquid? Certainly, like Nikola says, if you go to sufficiently high temperatures such as at the center of a star or soon after the Big Bang, then the usual three matter phases can't exist because all atoms are completely ionized into plasma.


Not to be a bitch but plasma is an ordinary phase transition. Degenerate matter (such as matter found on neutron stars) is an extreme example, as is quark gluon plasma (I think it's degenerate, but really, degenerate phase states are only extreme variations of the basic ones.)

Quote Watsisname ()

Always check your units. If you are doing a calculation where the dimensions are in MKS system (meters, kilograms, seconds), then you cannot say the answer comes out as kilometers per hour, or pounds per square inch, or whatever seems convenient. Your answer must be in the exact same units as what went into the calculation itself. An equation with mismatched dimensions, or a calculation with mismatched units, cannot be correct.

It is kind of neat though that the density of air at Earth's surface (in grams/m^3) is approximately equal to the sound speed in meters per second. When I first read the result of your calculation, I was like "wait, what?" before I recognized what you were doing. :p


Of course, from now on I'll be more strict about the matter.

Nice to see I'm not the only one who got carried away there. :B
Quote Watsisname ()
Yeah, there's no formulation for sound speed which does not invoke the adiabatic index in some way or another. But it is just a number, and you can find it for most common atmospheric gases. For monatomic gases you can treat it as 5/3. For diatomic, 7/5. Of course, if your atmosphere of interest has a significant concentration of more complex molecules, then you have to do more work to figure it out.


How would I "figure it out"?

I could google it but I trust people directly more than a wide technical analysis.

Quote Watsisname ()
Then the next question to examine is whether your calculation is more convenient to do in terms of pressure and density, or as temperature and molar mass? SE gives you all this information, so it is mainly a matter of preference.


Because why not? I want to be able to simulate conditions that I can't in-game, for what is my brain if not a programmable simulator? :P

It's also a fun thing to test on random planets and can be useful in some of my wilder mind games. :B


Edited by NikolaAnicic007 - Thursday, 26.11.2015, 12:19
 
WatsisnameDate: Thursday, 26.11.2015, 20:36 | Message # 372
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Of course plasma is just another phase of matter. So are Bose-Einstein Condensates. I distinguish it from "the usual three" because Tac was asking of the conditions for liquid and gas phases, not plasmas. smile Plasmas aren't common at STP and usually are not shown on phase diagrams.




 
AlekDate: Friday, 27.11.2015, 01:09 | Message # 373
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Quote Watsisname ()
usually are not shown on phase diagrams.


Why not?





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WatsisnameDate: Friday, 27.11.2015, 04:06 | Message # 374
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Phase diagrams generally show the solid-liquid-gas phase transitions since they are what we deal with most often. There's not much use in showing where various substances become plasmas unless you're in the field of working with them. (When was the last time you needed to know the ionization conditions for H2O, or CO2, for example?)

Also, plasma is not a precise phase transition like the others, since it involves successive levels of ionization.





 
WatsisnameDate: Friday, 27.11.2015, 06:45 | Message # 375
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Quote NikolaAnicic007 ()
How would I "figure it out" [the adiabatic index]?


Personally, I would look it up from a table. tongue

Calculating it is only convenient for simple ideal gases. Otherwise and in general, it's pretty complicated, like the latent heat problem. In intro thermodynamics they teach you about equipartition of energy and the degrees of freedom of the molecule (which is why we get those nice rational numbers for monatomic and diatomic ideal gases). But reality is more complex than that. These are not classical systems, so we have to apply quantum mechanics. Equipartition of energy isn't always valid. You can get significant error if you assume it, even for something as "simple" as water vapor. The specific heats are temperature dependent, too.

So, yeah. Unless you're really motivated to do those calculations yourself (in which case the best I can do is help you choose textbooks or something like that), then just use data from experiments. Other people already did the work to figure out the thermodynamic properties of materials. Why not use it? smile





 
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