Plasma: Difference between revisions
m
→Analysis: The Identity of Plasma: Small tweaks.
imported>Kosmos m (→Analysis: The Identity of Plasma: Added spaces, made stuff a little easier to read.) |
imported>Kosmos m (→Analysis: The Identity of Plasma: Small tweaks.) |
||
| Line 36: | Line 36: | ||
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. | 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 1.6 m³ | '''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. | ||
| Line 44: | Line 44: | ||
(20.4 kPa)(147,456 l) / (8.314 kPa/l/K/mol)(292 K) = 3008102.4 / 2402.746 = '''1251.95 mol plasma.''' | (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 2.5 m³ | '''NOW FOR THE 2.5 m³:''' | ||
2.5³ = 15.625 x 36 = 562.5 x 1000 = 562,500 l | 2.5³ = 15.625 x 36 = 562.5 x 1000 = 562,500 l | ||
| Line 55: | Line 55: | ||
===NOW ONTO THE PART THAT INVOLVES FIRE=== | ===NOW ONTO THE PART THAT INVOLVES FIRE=== | ||
A full tank of plasma, which contains either 4774. | A full tank of plasma, which contains either 4774.80 mol or 1251.95 mol 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. | ||
375.35 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 %. | 375.35 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 %. | ||
| Line 74: | Line 74: | ||
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. | ||
''Now then, because moles are hilarious little units, if we have 591. | ''Now then, because moles are hilarious little units, if we have 591.17 mol of CO<sub>2</sub>, we also have 591.17 mole of C. | ||
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 %. | 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 1 molC = 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 1251.94 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. | 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! | ||