|Mar20-13, 05:42 PM||#1|
The behaviour of a vacuum without the influence of air pressure?
I am curious to know if anyone knows if the following experiment has ever been conducted? If not, can anyone theorise what the result may be? 2 flasks A,B are of equal size. Flask A contains a vacuum. Flask B is entirely full of mercury, with no air present. To ensure that the effect of external air pressure has no influence on the experiment,both flasks A and B are contained within a third larger flask C, which itself contains a vacuum.(Apart from the presence of the flasks within obviously). A valve is opened between flasks A and B. Does the mercury in B flow into A? If so, it can be assumed that the consequential vacuum(ous) space so created in flask B, will necessarily always be 'harder' than that contained in flask A, therefore preventing the possibility that the mercury levels in both flasks will equalise. Alternatively, does the mercury simply remain in flask B? If so, this contradicts the statement that 'nature abhors a vacuum, does it not?
|Mar21-13, 06:15 AM||#2|
Welcome to PF!
I think you can understand the situation better if you do not think of the vaccum as a thing. When flask A is open, mercury vapours will start floating around and some will end up in flask B. Equilibrium should be reached when the vapour pressure of mercury is the same in both flasks. But I don't see why there would be any flow of liquid mercury.
|Mar21-13, 06:50 AM||#3|
Assuming that the flasks are level with each other then there is a tiny pressure differential. The pressure of the mercury vapor at the bottom of the empty flask is slightly higher than it is at the top of the full flask. This is due to gravity.
If the vapor at the top of the full flask is in equilibrium with the liquid at the top of the full flask and if the vapor at the bottom of the empty flask is at a higher pressure than that then one would expect condensation at the bottom of the empty flask.
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