Plasma: Difference between revisions
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→Analysis: The Identity of Plasma: Added spaces, made stuff a little easier to read.
imported>Kosmos m (Added The Identity of Plasma by Pybro) |
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From here on in, a CANISTER refers to to the large, colored gas containers that you fill TANKS in. A CANISTER needs to be pulled, a TANK can be held in your hand. | From here on in, a CANISTER refers to to the large, colored gas containers that you fill TANKS in. A CANISTER needs to be pulled, a TANK can be held in your hand. | ||
A full canister of plasma has an internal pressure of 4559. | A full canister of plasma has an internal pressure of 4559.6 kPa. Said full canister of plasma was released in a 6x6 room, built in space. This room was entirely devoid of air, and had a temperature of 0 Celsius (270 Kelvin). A gas analyzer was used to measure the pressure in this 6x6 room. It was 20.4 kPa, and the temperature was 19C (289K). NOW, HERE IS THE VARIABLE PART: In ooc, it was agreed upon that a tile had a volume of 1.6 m<sup>3</sup>. HOWEVER. In the code, it says that a gas cell is 2.5m<sup>3</sup>. Ultimately, using the 2.5 m³ tile messes things up, getting us into a situation where we end up with half a carbon, and, yeah. For the sake of completion, I'll post the math for both. | ||
First, the 1. | First, the 1.6 m³ | ||
Using the Ideal Gas Formula (PV=nrt), we can calculate the number of moles of plasma in our room, via some algebra fandangling. PV=nrt/rt = (PV)/(rt)=n. Pressure times volume DIVIDED BY gas constant times temperature equals moles. | Using the Ideal Gas Formula (PV=nrt), we can calculate the number of moles of plasma in our room, via some algebra fandangling. PV=nrt/rt = (PV)/(rt)=n. Pressure times volume DIVIDED BY gas constant times temperature equals moles. | ||
1. | 1.6 m³ = 4.096. 6x6 = 36. 4.096 x 36 = 147.456. 1 m³ = 1000 l. 147.456 x 1000 = 147,456 l. | ||
(20. | (20.4 kPa)(147,456 l) / (8.314 kPa/l/K/mol)(292 K) = 3008102.4 / 2402.746 = '''1251.95 mol plasma.''' | ||
NOW FOR THE 2. | NOW FOR THE 2.5 m³ | ||
2.5³ = 15. | 2.5³ = 15.625 x 36 = 562.5 x 1000 = 562,500 l | ||
Plugging the 562, | Plugging the 562,500 l into the equation and replacing the 147,456 l with it gives us '''4726.8 mol of plasma.''' | ||
This is quite a difference. | This is quite a difference. | ||
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===NOW ONTO THE PART THAT INVOLVES FIRE=== | ===NOW ONTO THE PART THAT INVOLVES FIRE=== | ||
A full tank of plasma, which contains either 4774.80mol or 1251.95mol of plasma at 4559. | A full tank of plasma, which contains either 4774.80mol or 1251.95mol of plasma at 4559.6 kPa, was pumped into the incinerator burn chamber. An excess of oxygen was then pumped in, and the whole mix was ignited. After it cooled, our brave atmos tech scientist entered. Some time was taken to let the mixture spread throughout the 10 tile area. The area was then scanned. | ||
375. | 375.35 kPa, at 97.4 °C (370.4K). THIS WAS DONE ON ASTEROIDSTATION HOWEVER, and on Asteroid, the incinerator burn chamber is not empty! CO<sub>2</sub> 53 %, O<sub>2</sub> 36 %, N2 9 %. | ||
375. | 375.35 x 0.53 = 198.935 kPa. | ||
===Go CO<sub>2</sub> woohooo=== | ===Go CO<sub>2</sub> woohooo=== | ||
Now let's see how much CO<sub>2</sub> a single canister of plasma can make. 156, | Now let's see how much CO<sub>2</sub> a single canister of plasma can make. 156,250 l for the volume of the 2.5 m³. | ||
10 x 4.096 x 1000 = 40,960 l | |||
10 x 15.625 x 1000 = 156,250 l | |||
(198.935)(40, | (198.935)(40,960 l) / (8.314)(370.4) = 2,646 mol CO<sub>2</sub>. | ||
(198.935)(156, | (198.935)(156,250 l) / (8.314)(370.4) = 10,093.696 mol CO<sub>2</sub>. | ||
EVERYTHING IN ITALICS IS CORRECT AS FAR AS PROCEDURE AND METHOD GOES, BUT THE ACTUAL NUMBERS ARE INCORRECT. I'M ONLY KEEPING IT AS AN EXPLANATION OF CONVERTING MOLES OF CO<sub>2</sub> TO MOLES OF CARBON. | EVERYTHING IN ITALICS IS CORRECT AS FAR AS PROCEDURE AND METHOD GOES, BUT THE ACTUAL NUMBERS ARE INCORRECT. I'M ONLY KEEPING IT AS AN EXPLANATION OF CONVERTING MOLES OF CO<sub>2</sub> TO MOLES OF CARBON. | ||
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''Now then, because moles are hilarious little units, if we have 591.17mol of CO<sub>2</sub>, we also have 591.17mole of C. | ''Now then, because moles are hilarious little units, if we have 591.17mol of CO<sub>2</sub>, we also have 591.17mole of C. | ||
591. | 591.13 mol / (1 molC / 1 molCO<sub>2</sub>) = 591.13. Don't believe me? 591.13 mol / 1 molCO<sub>2</sub> x 46 = 27,192 g. The molecular weights of Carbon and CO<sub>2</sub> are 14 and 46 respectively. The percent by mass of carbon in CO<sub>2</sub> is 14 / 46 = 0.3043 x 100 = 30.43 %. 27,192 x 0.3043 = 8274.52 g carbon. 8274.52 g carbon / 14 gC x 1molC = 591.13 molC. Fucking math, how does it work. The same applies for the 2.5 m³ one. So it's 2232 mol Carbon if a tile is 2.5 m³.'' | ||
More CO<sub>2</sub> | ===More CO<sub>2</sub>=== | ||
Now, That means that 1251. | Now, That means that 1251.94 molPlasma + ??? molO<sub>2</sub> ----> ??? molH<sub>2</sub>O + 2,656 molCO<sub>2</sub>. But because we did the above, we can basically summarize that there are 2,646 mol of carbon in 1251.94 mol of Plasma. | ||
Pure hydrocarbon chains exhibit a funny property. By taking the amount of CO<sub>2</sub> produced (and thus, carbon), and dividing it by the moles of hydrocarbon you burnt to get said CO<sub>2</sub> (and assuming you have 100% total combustion, no carbon monoxide produced), you can find out how many carbons are in a single mole of the Hydrocarbon! | Pure hydrocarbon chains exhibit a funny property. By taking the amount of CO<sub>2</sub> produced (and thus, carbon), and dividing it by the moles of hydrocarbon you burnt to get said CO<sub>2</sub> (and assuming you have 100 % total combustion, no carbon monoxide produced), you can find out how many carbons are in a single mole of the Hydrocarbon! | ||
Thus, the burning of Hexane is: | Thus, the burning of Hexane is: | ||
C<sub>6</sub>H<sub>14</sub> + (19/2)O<sub>2</sub> ---> 6CO<sub>2</sub> + 7H<sub>2</sub>0 | |||
So: | So: | ||
12C<sub>6</sub>H<sub>14</sub> + blah --> 72CO<sub>2</sub> + blah | |||
72/12 = 6! | 72 / 12 = 6 ! | ||
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2656/1251.95 If we divide the two, we get 2.11, meaning that for every 1 mole of plasma, we have 2.11 moles of carbon! | 2656 / 1251.95 If we divide the two, we get 2.11, meaning that for every 1 mole of plasma, we have 2.11 moles of carbon! | ||
And if we use the 2. | And if we use the 2.5 m³, we end up with 10,093 / 4995 = 2.02! Told you it didn't matter. | ||
Now, this is all theoretical, and calculations, and I've rounded a lot of the decimals (I ain't working with 24.858849389200102299383838392929292 due to spessmens, 2.858 will suffice), we can thus say, that there are TWO CARBONS IN A PLASMA MOLECULE. Plus, y'know, you can't have .11, or .02, of a molecule. Shit just ain't possible. | Now, this is all theoretical, and calculations, and I've rounded a lot of the decimals (I ain't working with 24.858849389200102299383838392929292 due to spessmens, 2.858 will suffice), we can thus say, that there are TWO CARBONS IN A PLASMA MOLECULE. Plus, y'know, you can't have .11, or .02, of a molecule. Shit just ain't possible. | ||
Which means that plasma has a structure that looks something like | Which means that plasma has a structure that looks something like: | ||
'''C-C,,,,C=C, or C≡C''' | |||
With three, two, and one bonding site on each carbon, respectively. | With three, two, and one bonding site on each carbon, respectively. | ||
A rant about isotopes | ===A rant about isotopes=== | ||
But...Then plasma is just a regular derivative of ethane, ethene, or ethyne! "what the fuck pybro you're such a fucking faggot you waste my very precious ten minutes!". Hold up there! Yes, it's just regular ethene, ethane, '''IF''' YOU USE <sup>1</sup>H, AKA Protium, AKA "Normal hydrogen" AKA "A single proton with an electron buzzing around it". Atoms are composed of protons (positive), electrons (negative), and neutrons (neutral). Isotopes are atoms that have a "deviant" number of neutrons. This messes shit up, and gives isotopes different properties than the "normal" element. Generally (although there are exceptions!), the "normal" isotope has as many neutrons as it does protons. Protium AKA "normal" hydrogen, is one such exception. | But...Then plasma is just a regular derivative of ethane, ethene, or ethyne! "what the fuck pybro you're such a fucking faggot you waste my very precious ten minutes!". Hold up there! Yes, it's just regular ethene, ethane, '''IF''' YOU USE <sup>1</sup>H, AKA Protium, AKA "Normal hydrogen" AKA "A single proton with an electron buzzing around it". Atoms are composed of protons (positive), electrons (negative), and neutrons (neutral). Isotopes are atoms that have a "deviant" number of neutrons. This messes shit up, and gives isotopes different properties than the "normal" element. Generally (although there are exceptions!), the "normal" isotope has as many neutrons as it does protons. Protium AKA "normal" hydrogen, is one such exception. | ||
Hyrogen has two isotopes. One, Deuterium, which is stable, and forms Deuterium Oxide ( | Hyrogen has two isotopes. One, Deuterium, which is stable, and forms Deuterium Oxide (D<sub>2</sub>O), AKA "Heavy Water" (heavy water icecubes don't float on top of liquid water or heavy water, fun fact). Deuterium is a proton, a neutron, and an electron. There is also, TRITIUM, which is two neutrons, a proton, an electron. Tritium is also radioactive. On earth, Tritium is rare. In spess however, it's rather common (the source of tritium on earth is normal hydrogen getting hit by COSMIC RAYS). Tritium is also very toxic. | ||
Hydrogens isotopes are very special, in that they get their own letters! Protium is H, Deuterium is D, and Tritium is T. | Hydrogens isotopes are very special, in that they get their own letters! Protium is H, Deuterium is D, and Tritium is T. | ||
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Three, Tritium has a rather large half life, and even then is rather weak as far as radioactivity goes (you still don't want it in you, however). You can dunk yourself in tritium oxide (T<sub>2</sub>O), and so long as you don't get any IN YOU, your guts will be fine, you only risk skin cancer. | Three, Tritium has a rather large half life, and even then is rather weak as far as radioactivity goes (you still don't want it in you, however). You can dunk yourself in tritium oxide (T<sub>2</sub>O), and so long as you don't get any IN YOU, your guts will be fine, you only risk skin cancer. | ||
Four, given that burning plasma only produces CO<sub>2</sub> (and water, which is actually Tritium Oxide, aka super-heavy water, aka radio-fucking-active water), it can ONLY have Hydrogen (or isotopes of), Carbon, and oxygen in it. Tritium is necessary to get the weight up into the range where it can be a liquid at standard temperature and pressure AT ALL, while at the same time giving it the necessary toxicity. Plasma gives you TOXIN damage, it does NOT give you oxyloss damage, so it doesn't just suffocate you, it actively attacks your body in SOME way. Ethane ( | Four, given that burning plasma only produces CO<sub>2</sub> (and water, which is actually Tritium Oxide, aka super-heavy water, aka radio-fucking-active water), it can ONLY have Hydrogen (or isotopes of), Carbon, and oxygen in it. Tritium is necessary to get the weight up into the range where it can be a liquid at standard temperature and pressure AT ALL, while at the same time giving it the necessary toxicity. Plasma gives you TOXIN damage, it does NOT give you oxyloss damage, so it doesn't just suffocate you, it actively attacks your body in SOME way. Ethane (C<sub>2</sub>H<sub>6</sub>) does not do this, and is pretty much completely non-toxic. Hell, the only real dangers of ethane are asphyxiation (oxyloss) and igniting it (burn/bomb). | ||
Five, the OH (or OT, as it were) is necessary to also bump up the weight into the range where it can be a liquid/solid at room temperature. It also gives it enough intermolecular forces (the hydrogen/tritium bonding) to aid it being a liquid. | Five, the OH (or OT, as it were) is necessary to also bump up the weight into the range where it can be a liquid/solid at room temperature. It also gives it enough intermolecular forces (the hydrogen/tritium bonding) to aid it being a liquid. | ||