Suppose a long rigid pipe (e.g., 30m) open at both ends is completely submerged into some deep body of water. The pipe gets completely filled with water. Then, one end gets closed and the pipe is slowly vertically lifted out of the water, with the closed end on top.

At some point, the open end stays submerged below but the pipe will start poking out beyond 10m in height above the water, which is the theoretical maximum height of such a water column at atmospheric pressure.

So, what happens as the pipe keeps getting lifted out? Does water start seeping out of the bottom of the tube, and thereby forming and extending a perfect vacuum at the closed end? There doesn't seem to be anything at the top to be taking the place of the water column escaping the pipe, so would actually happen?

Note: I'm ignoring vaporization and material constraints for the purposes of this question ; )


You will "pull a vacuum" which will fill with water vapor (at whatever the saturated vapor pressure is at the temperature of the water).

If you chose to ignore evaporation, you would consider the void above the water to be a perfect vacuum. But you would be wrong to make that assumption. Even at 0 C, liquid water has a saturated vapor pressure of 611 Pa. The column of water that can be supported will be a little bit less than "10 m" (if that is your column corresponding to the prevailing atmospheric pressure, local gravitational acceleration, and the density of the water). In fact 611 Pa will cause the column to be about 6 mm shorter than you expect. At 20 C the error would be about 23 mm.

Note - if you pull the pipe out suddenly, it may take a short while for the vapor to reach equilibrium. The column will "sag" over time.

Because the vapor pressure of mercury is lower, and the density higher, it makes a much better material for a barometer.

  • $\begingroup$ So, ignoring vaporization and material constraints, I'll essentially be expanding a "vacuum" at the top of the pipe? $\endgroup$ – ManRow Sep 10 '17 at 2:54
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    $\begingroup$ I think ignoring vaporization becomes inappropriate when you put a liquid in contact with a vacuum. $\endgroup$ – EL_DON Sep 10 '17 at 3:52
  • $\begingroup$ The water will be vigorously boiling and cooling if you pull it out quickly, it might even form some ice. $\endgroup$ – Rick Aug 30 '19 at 13:34
  • $\begingroup$ @Rick it will boil - whether it does so “vigorously” depends on the surface texture of the straw and how clean the water is - you need nucleation sites. Because of this I doubt you would lose heat quickly enough to freeze. Latent heat of fusion is approx 80 cal/g and latent heat of vaporization is about 500 cal/g. You would have to evaporate approx 1/5 th of the water at 20 C to freeze the rest. That implies a LOT of space above the straw... $\endgroup$ – Floris Aug 30 '19 at 13:40
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    $\begingroup$ Good point, the vapor would fill even 20m of extra straw pretty quickly. I was just thinking of demo I saw at a science museum that pulled a vacuum on some water, and partially froze it. The demo continuously evacuated the vapor though. In this case that 20m of extra tube could only produce 5 microns of ice if the cooling was somehow constrained to the surface... so no ice :p $\endgroup$ – Rick Aug 30 '19 at 14:29

You have described a barometer with water instead of mercury as the working fluid.

Amongst the reasons that mercury is used rather than water is that it is denser and so the barometer tube is much shorter (less than one metre as compared to 10 metres) and the saturated vapour pressure of mercury (0.002 mmHg) at room temperature is much less than that of water (24 mmHg).

So the space above the water in your tube will be occupied by saturated vapour and as you lift the tube up water liquid will change to water vapour to maintain the vapour pressure at a constant value.

So you cannot ignore the vaporisation of water in the context of your question.

  • $\begingroup$ So no matter what kind of "fluid" or liquid is used, vaporization will always occur in this case? $\endgroup$ – ManRow Sep 10 '17 at 5:26
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    $\begingroup$ @ManRow Yes because the liquid and vault must reach a state of dynamic equilibrium.. The space above a mercury column is sometimes called a Torricelli vacuum. en.m.wikipedia.org/wiki/Torricelli%27s_experiment $\endgroup$ – Farcher Sep 10 '17 at 5:32

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