- #1
wg1337
- 11
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Hi! I have this experiment, that keeps getting me frustrated. The experiment is simple:
We have hot water and a long plastic tube with one end closed. We simply measure how much water is in the tube and from that we get the water volume (water is in cylinder shape).
But there is something I don't understand: when water is hotter then there is less water in tube (about 1cm long cylinder). When the water is less hot, then thee is more water (about 3cm long cylinder).
I can figure that the air which is at the end of tube is giving too much pressure so the volume doesn't add up.
For example, if this is really a isobaric process, then pV = RTm/M (p = RTm/(VM) should be the same all the time, but the numbers aren't the same.
If we want to threat water as an ideal gas, then the mass shouldn't change (could simply say that m/M is 1 mol), but the mass changes, because if the water expands because of heat, then there should be more water volume, not less. So the water actually flows away, can't really threat it as an ideal gas.
What I'm missing here?
It is possible that experiment is supposed to fail, but then I need a good reason. I don't think the mass change reason is a good enough reason.
For example:
length of cylinder - 0.008m
Diameter of tube - 0.0037m
Temperature - 337 K
Volume - 0.86 * 10^-7 m^3
Mass - 1000*0.86 * 10^-7 = 0.86*10^-4 kg
Molmass - 1 * 10^-3 kg/mol
Pressure - p = R*T*m/(V*M) = 8.31*337*0.86*10^-4/(0.86*10^-7*10^-3) = 2.8 * 10^9 Pa
So basically I get that there is 10 000 atm pressure... very possible...
We have hot water and a long plastic tube with one end closed. We simply measure how much water is in the tube and from that we get the water volume (water is in cylinder shape).
But there is something I don't understand: when water is hotter then there is less water in tube (about 1cm long cylinder). When the water is less hot, then thee is more water (about 3cm long cylinder).
I can figure that the air which is at the end of tube is giving too much pressure so the volume doesn't add up.
For example, if this is really a isobaric process, then pV = RTm/M (p = RTm/(VM) should be the same all the time, but the numbers aren't the same.
If we want to threat water as an ideal gas, then the mass shouldn't change (could simply say that m/M is 1 mol), but the mass changes, because if the water expands because of heat, then there should be more water volume, not less. So the water actually flows away, can't really threat it as an ideal gas.
What I'm missing here?
It is possible that experiment is supposed to fail, but then I need a good reason. I don't think the mass change reason is a good enough reason.
For example:
length of cylinder - 0.008m
Diameter of tube - 0.0037m
Temperature - 337 K
Volume - 0.86 * 10^-7 m^3
Mass - 1000*0.86 * 10^-7 = 0.86*10^-4 kg
Molmass - 1 * 10^-3 kg/mol
Pressure - p = R*T*m/(V*M) = 8.31*337*0.86*10^-4/(0.86*10^-7*10^-3) = 2.8 * 10^9 Pa
So basically I get that there is 10 000 atm pressure... very possible...
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