jdurg Posted September 24, 2004 Posted September 24, 2004 Hey guys. I'm hoping someone here can help me out. I've been looking at my halogens and have been paying close attention to my bromine ampoule. I did some calculations and figured out that if all of the bromine vaporized, the pressure in the tube would be 99 atmospheres. heh. It would go boom pretty violently. However, what I can't seem to figure out is how to calculate the pressure present inside a tube when there is still liquid present. I.E. at about 105 degrees Fahrenheit, what would the pressure inside the bromine ampoule be? (That would be about 313.7 K). The cylinder has a total volume of about 24.333 mL, and there is about 4.5 mL of liquid bromine inside the cylinder. The one thing I don't know is the total mass of the bromine that was added, so I really don't know what the pressure is inside the tube right now. Is there any way that a rough estimate could be made. Arrggh. It's late and I'm having a brain fart. lol
jdurg Posted September 24, 2004 Author Posted September 24, 2004 Okay. I think I figured it out. I just made an assumption that all the space in the tube not occupied by the liquid bromine was occupied by bromine vapor. I then was able to use the density of bromine vapor to calculate how much bromine was in the gaseous state. With that information I calculated the pressure at room temperature to be about 1.06 atmospheres. Not too bad. I then did the calculation at a temperature of 313.7 Kelvin and figured out that the pressure would rise to ~1.22 atmospheres. Again, that's not terribly high but it didn't take into acount that at the elevated temperature more bromine would be vaporized. However, I don't believe that the amount of extra bromine in a gaseous state would be all that much in terms of the overall pressure. I also remembered that as the pressure inside the tube increases, the amount of bromine that would be in the vapor state would decrease as well. The boiling point of the bromine is also elevated at a higher pressure since it's harder for the bromine molecules to separate from each other. So really, I have nothing to worry about unless I throw this ampoule in a fire. Especially since the ampoule was formed at an elevated temperature. (We had to melt the borosilicate glass in order to seal it, so the bromine was at a pretty high temperature while the glass melting was occurring.)
r1dermon Posted September 24, 2004 Posted September 24, 2004 was the tube a vaccum when you put the bromine in it?
jdurg Posted September 24, 2004 Author Posted September 24, 2004 No. The tube was in the open atmosphere, but it was heated to a fairly high temperature in order for the ampoule to get sealed. That is why I'm making the assumption that all the gas inside the tube is bromine vapor.
Professio Posted September 26, 2004 Posted September 26, 2004 That makes sense. How long was it heated?
jdurg Posted September 26, 2004 Author Posted September 26, 2004 The tube was inside the fire until the borosilicate glass melted evenly all around, then the tube was "pulled" until it sealed off in a perfect cone shape at the top. The tube itself was VERY hot to the touch, and the inside of the glass was filled with a dark bromine vapor. So I'd say that while the liquid bromine itself wasn't at its boiling point, the "empty space" inside the tube was pretty warm. With all the calculations I've done, having the tube at about 105 degrees should be okay as long as more of the bromine inside doesn't vaporize. Again, I don't think an appreciable amount will vaporize as the gas pressure inside will reduce it's ability to evaporate, and the equillibrium with the vapor already in there will prevent more from vaporizing. This is a case where really the only problem is the increase in temperature which will increase the pressure of the gas inside there. I just wish I knew how many atmospheres the vials can withstand.
Professio Posted September 26, 2004 Posted September 26, 2004 Well do the expreriment agian, but this time keep a "log", and calculate how many atmospheres you get in the vials. I will try to find some information on this for you, then I will post it.
jdurg Posted September 27, 2004 Author Posted September 27, 2004 I would if I could. heh. When we did this, it was simply taking some liquid bromine and putting it in a test tube. We then heated the test tube to warm the bromine and make a dense vapor, then we put an intense trio of propane flames to the top part of the tube until it melted shut, thus making an ampoule. The tube was then slowly cooled down. So the pressure inside the tube is probably close to one atmosphere. After reading through my chemistry textbooks, I also realized that the bromine cannot boil in the sealed tube. Boiling cannot occur in a closed container, because as the vapor pressure increases, less vapor will be able to form. I did a rough calculation, and the difference in vapor pressure at room temperature and at about 105 degrees Fahrenheit would result in perhaps 0.005 mL of bromine vaporizing inside the tube. So this thing is pretty safe.
Woxor Posted September 27, 2004 Posted September 27, 2004 I don't know if it still matters, but I have been trying to find Antoine coefficients for bromine, but I can't seem to do so. Perhaps it's no good describing bromine with that equation, or something. I have no idea whether I'm talking down to you by explaining, but to be safe rather than sorry: the Antoine equation gives the log of the vapor pressure (i.e. the pressure when liquid and gas are present) of a given substance as a function of the temperature. Specifically, log(P*) = A - B/(T+C), where A, B, and C are constants for the substance. I can't find A, B, and C for bromine anywhere on the web or in my textbooks, but if you can find them, they might help in theory. I also don't know how much this equation will help, given that it only applies when both phases are present, i.e. it doesn't predict the point at which vaporization occurs, since, as you say, the event depends on the pressure as well. In sum, I may be entirely useless. But good luck.
jdurg Posted September 27, 2004 Author Posted September 27, 2004 Actually, I have heard of that equation too. In one of my chemistry Textbooks, they showed how if you know the Heat of Vaporization for a compound and it's vapor pressure at one temperature, you can figure out its vapor pressure at any other temperature without having to use any of those constants. So that's what I did and got a result I was happy with.
Woxor Posted September 28, 2004 Posted September 28, 2004 I should probably know how to do that, but I seem to have no idea. My thermo test next week seems more intimidating now.
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