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If you evacuate water with an oil pump is the heavy bubbling of the water due to the water boiling at room temperature or is the water just degasing ?
Question Date: 2001-05-07
Answer 1:

Boiling occurs (rapid bubbling) when the vapor pressure of a liquid (here it is water) is equal to the total pressure of the surroundings.
There is a simple relationship between the temperature of the liquid and its vapor pressure - simply put, the vapor pressure steadily increases with temperature. The vapor pressure of water is equal to the normal atmospheric pressure (1 atm, 760 torr) at 100 C. When the pressure of the surroundings is lowered, either by going up a mountain, or by putting the liquid in a bell jar and evacuating, the temperature at which the vapor pressure will equal the surroundings pressure lowers - if the surroundings pressure is lowered sufficiently, water will bubble at room temperature and it is truly boiling. On the other hand, if the total pressure is increased, the boiling temperature increases - hence the reasoning behind the pressure cooker.

The reason the boiling stops is because changing the water from a
liquid to a vapor requires energy - the latent heat of vaporization is the technical term, and the magnitude of the latent heat is about the same regardless if the water boils at room temperature or at 100 C. As the water boils in the bell jar, it needs to supply this latent heat of vaporization. One place the energy comes from is lowering the temperature of the remaining liquid water. So the boiling water in the bell jar actually is getting cooler (this is also why evaporating sweat cools you off). The liquid remaining will continue to get cooler and cooler as more liquid is boiled away - as there really isn't a good external source of heat. In principle, you could get ice forming in the water if your bell jar could pull enough vacuum. As the liquid gets colder, its vapor pressure also starts to decrease. At some point, the water has gotten so cold that the vapor pressure is just a little bit less than the surroundings pressure, and the rapid boiling stops. An equilbrium of sorts is established where the rate of water evaporation causes just enough heat loss to be compensated by whatever heat is being conducted into the liquid from its surroundings. Eventually all the liquid will be evaporated.

The second part of the experiment - the fact that the bubbling isn't as vigorous the after the air is entered into the bell jar and then evacuation repeated, depends on letting the water warm back up to room temperature in between experiments. If the water stays cold, then it won't take as much evaporation to lower the temperature back down to that equilibrium state I mentioned earlier.

You can find the vapor pressure of water in a number of handbooks on chemical engineering - the books are known as steam tables.

good luck with the experiments - check the temperature of the water after the boiling. I'll bet its pretty cold.

Answer 2:

According to my basic physics book, the vapor pressure of water at room temperature (20 C) is 2.3 X 10^3 Pa = 0.023 atm = 0.33psi = 17.2 torr (choose your favorite pressure unit).When you start pumping you could easily be first getting dissolved gases out. This is much like the small bubbles you see when heating up water. As you continue pumping down, you would of course expect this activity to slow down as the gases are released and pumped away.
If you continue to pump down, eventually you should get the pressure to less than 17 torr where the water would start to boil.
So here are a couple things to think about:

Did you wait long enough for the pressure in the bell jar to get to
lower than 17 torr? This could take a while depending on stuff like the specs. of the pump, how thin and long the tubing is connecting the pump to the bell jar, how leaky the bell jar is, how big the bell jar is, and whether the pump is functioning as it should.

Answer 3:

I am not completely sure of this explanation that I am giving you, but here's my opinion:
When the water is at very low pressure (when you're pumping), the water boils at room temperature, as you had hoped. However, as the water boils, it disperses vapor (evaporates) and thus forms an atmosphere inside the bell jar. This increases the pressure inside the bell jar and the water no longer boils at room temperature. If you were to completely remove all vapor in the bell jar (including condensation), it would boil just as vigorously as before.

Answer 4:

You probably were just removing dissolved air from the water. If the water had been boiling, it would not have stopped. Do you have a pressure guage reading the pressure? (Is it near the bell jar, if you have a long, narrow tube connecting the jar and the pump, which would slow down the pumping speed and give you a much higher pressure in the jar than near the pump input?) I have seen water boil in a bell jar when the pressure was a few milliTorr. It might boil at a slightly higher pressure, but I suspect your bell jar was not at a low enough pressure.
If so, try cleaning the rim of the bell jar and whatever it is sealing to.
Check for small cracks, and replace any gaskets with small cracks. Also, try to have as short a tube with as large an inner diameter as possible to reduce the load for the pump.

Answer 5:

It's hard to be certain, but everything you wrote seems consistent with the interpretation that you have "out gassed" the water. Assuming that you cannot get your hands on a better vacuum pump or seal your chamber better, you could do the experiment with alcohol instead. Alcohol has a high vapor pressure, and should boil well in your chamber. If you don't have any pure alcohol, you could use something like bourbon or even beer. Beer would be particularly interesting as I believe you would pull all the carbonation out as well (just< dissolved carbon dioxide). You could finish the experiment by letting your kids try some flat alcohol-free beer. Suitable for minors, but probably not very good.

Answer 6:

I always use a small beaker of water. The water bubbles vigorously, and the temperature goes down, showing that boiling is really a vapor pressure phenomenon. Use more water if you want to avoid the problems of removing dissolved gases.

Answer 7:

One thing to try is to first boil some water at 1 atm and this will remove the dissolved gases. Then try your experiment again. If you dont see the "boiling" then I think your friend is correct. You also can look up the Henry's law const for O2 in the CRC handbook and determine the amount of oxygen that dissolves in tap water... also there will be some N2 in there as well. And knowing the dissolved amount and using the ideal gas law you can calculate the VOLUME of oxygen bubbles that are created if you can degas the volume of water you started with. Then you can estimate the size and duration of bubbling to see if the volumes are aprroximately correct.

Answer 8:

The initial bubbling is dissolved oxygen and nitrogen in the water. Water will boil in a vacuum but in doing so it looses heat very rapidly -- limiting the boiling rate. Usually you get a limited fast sublimation. Much more satisfying is ethanol or isopropal alcohol which have a higher vapor pressure and lower melting points. (I have seen benzene freeze-- but it is now classed as a carcinogen so isn't a good choice.)
The boiling of water is more pronounced if the water is held in a large hgeat sink -- i.e. something large enough to prevent the rapid cooling. On the other hand, placing a sealed thermometer in the dish makes the demo more impressive-- but be sure it is sealed! An electronic thermistor based one will work well -- but I have seen inexpensive LCD's damaged by vacuum despite their being sealed...
This is a case where the old red alcohol type works quite well.

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