I am wondering what would happen in a situation where there is some liquid in a closed container
A is vacuum and B is the liquid.
I am also interested if there are any differences in behaviour when the ratio A/B changes.
$\begingroup$
$\endgroup$
9
-
$\begingroup$ in vacuum the water will come into the shape of a spherical ball due to surface tension $\endgroup$– AnniCommented Jan 18, 2016 at 9:57
-
$\begingroup$ A lot depends on what the liquid is, what the initial pressure is, what the temperature is and whether you are maintaining the vacuum by some mechanism (otherwise it might quickly become non-vacuum). $\endgroup$– RedGrittyBrickCommented Jan 18, 2016 at 10:22
-
$\begingroup$ @RedGrittyBrick, I really wouldn't want this to be very specific, but the liquid could be juice (water seems a bit too common), the initial pressure could be 1 atm, room temperature. And regarding the vacuum, I don't want to maintain it. $\endgroup$– Raul MCommented Jan 18, 2016 at 10:32
-
1$\begingroup$ This is a fairly broad, because there is a lot that could happen and depends on the initial conditions previously mentioned. Is there anything specific you are looking for? $\endgroup$– Kyle KanosCommented Jan 18, 2016 at 11:52
-
2$\begingroup$ @Anni: Physics is not my expertise, but aren't you ignoring gravity? $\endgroup$– phresnelCommented Jan 18, 2016 at 15:08
|
Show 4 more comments
2 Answers
$\begingroup$
$\endgroup$
3
A liquid-vapor interface is not a static interface, there is a so-called liquid-vapor equilibrium where molecules in the liquid phase are continuously escaping from the liquid into the vapor phase and vice versa vapor molecules are continuously captured by the liquid. In equilibrium, the number of molecules leaving the liquid into the vapor and leaving the vapor into the liquid are equal.
The liquid-vapor equilibrium is a function of pressure and temperature. When you heat a liquid more molecules will leave the liquid into the vapor (due to higher kinetic energies) than molecules from the vapor phase into the liquid and the interface is no longer in equilibrium; the liquid phase will slowly vaporize until it has disappeared. When you increase the pressure, the molecules require more kinetic energy to leave the liquid phase because the liquid molecules are closer together and their interaction energies are higher. This effectively increases the boiling point of the liquid and it takes a higher temperature to evaporate all the liquid (e.g. in pressure cookers).
The opposite effect occurs when you decrease the pressure, i.e. you lower the boiling point of the liquid. It requires less energy (lower temperatures) to vaporize all the liquid. So by using a vaccuum pump you will first suck out some of the vapor reducing the pressure, causing the liquid to start vaporizing, creating vapor which is again sucked out by the pump. Only after all the liquid has vaporized, will the production of vapor stop and will you be able to reach a vaccuum.
So to answer your questions; If the situation can be created, some of the liquid will vaporize to fill section $A$ and the liquid will reduce in mass. The ratio $A/B$ dictates if only some of the liquid is vaporized or all of it.
-
$\begingroup$ What will happen to the body of liquid? Will it "disperse"(I don't think it's the appropriate word) in the whole space of A+B? $\endgroup$– Raul MCommented Jan 18, 2016 at 18:38
-
$\begingroup$ @RaulM: If there was to exist a sudden vaccuum in $A$ then there would be a higher pressure in $B$ driving all the molecules toward $A$. These free molecules is what is known as the vapor. But this is a highly theoretical situation with little practical use because it is not realizable. $\endgroup$– nluigiCommented Jan 18, 2016 at 21:04
-
$\begingroup$ Thank you for taking time to look at my question and answer it! $\endgroup$– Raul MCommented Jan 19, 2016 at 9:54
$\begingroup$
$\endgroup$
5
The liquid will vaporize. You can just keep sucking. Will not be able to create a vacuum until all the liquid is gone.
If you had a barrier between the two and created a vacuum and then removed the barrier you would have vapor of the vapor pressure of the liquid at that temperature.
-
$\begingroup$ As I said in a comment. I don't intend in „keep sucking”. $\endgroup$– Raul MCommented Jan 18, 2016 at 18:29
-
1$\begingroup$ As I said in my answer then you are not going to achieve a vacuum. $\endgroup$ Commented Jan 18, 2016 at 18:30
-
$\begingroup$ My question was more like the second paragraph of your answer $\endgroup$– Raul MCommented Jan 18, 2016 at 18:39
-
$\begingroup$ The question lacks that detail and inconsistent with the comment initial pressure could be 1 atm. $\endgroup$ Commented Jan 18, 2016 at 18:41
-
1$\begingroup$ Thank you for taking time to look at my question and answer it! $\endgroup$– Raul MCommented Jan 19, 2016 at 9:54