17 Dec 07
1 edit
Originally posted by Great Big SteesAssuming:
I have a pipe that is 0.09m in diameter and the pipe is 0.06km long. Into this pipe I pour 19Kg of chlorine. How many mg/L of the chlorine is in that pipe?
1) The diameter given is the inner diameter of the pipe.
2) The chlorine is pure chlorine gas that dissolves completely in the water.
3) The resulting mixture has the same properties as pure water.
4) The resulting mixture takes up the entire volume of the pipe.
5) Standard temperature and pressure.
The volume of the pipe is:
pi * ((0.09/2)^2) * (0.06 * 1000) = 0.381704 = 0.4 m3 = 400 L
To find the concentration, just divide the mass of chlorine by the volume of water:
(19 * 1,000,000) / (0.381704 * 1000) = 49776 = 50,000 mg/L
However, after browsing Wikipedia, only about 2.4 L of chlorine dissolves in 1 L of water at 20 C (interpolated value). This translates into about 1.4 kg in 400 L of water, much less than the 19 kg given in the question. The pressure would have to be pumped way up and/or the temperature lowered to just above freezing to dissolve all 19 kg.
19 Dec 07
Originally posted by PBE6My assumptions:
Assuming:
1) The diameter given is the inner diameter of the pipe.
2) The chlorine is pure chlorine gas that dissolves completely in the water.
3) The resulting mixture has the same properties as pure water.
4) The resulting mixture takes up the entire volume of the pipe.
5) Standard temperature and pressure.
The volume of the pipe is:
pi * ((0.09/ ...[text shortened]... to be pumped way up and/or the temperature lowered to just above freezing to dissolve all 19 kg.
The density of the chlorine gas is looked for.
Since we have 382 L and 19 kg of chlorine: 49,8 g/L or 49800 mg/l
at Standard consitions we'll have
p*V=n*R*T
p=(n*R*T)/V
3570 Pa or 35 mbar
19 Dec 07
Originally posted by PonderableGood point, I assumed this was a drinking water pipe (sprang to mind because of the chlorine), but of course the question doesn't mention water anywhere.
My assumptions:
The density of the chlorine gas is looked for.
Since we have 382 L and 19 kg of chlorine: 49,8 g/L or 49800 mg/l
at Standard consitions we'll have
p*V=n*R*T
p=(n*R*T)/V
3570 Pa or 35 mbar