Low-pressure Water

Fact: If you have a 30x30x30 foot sealed container half full of liquid water / H2O, which exists in normal gravity, the liquid will rest at the bottom of the container. If all the air within the container, is sucked out, creating a near 100% vacuum environment within the container, will the A) Environment within the container become a weightless environment. If no, disregard C, & D below. B) When the air within the container is removed, approaching the 100% vacuum environment, will the position / level of the bulk liquid be affected by the ‘airless’ environment and rise in bulk form toward the top of the container. If so, once the container is at 100% vacuum, will the bulk liquid rise to an equidistant level from the bottom as from the top of the container, or will, C) The liquid become segmented, and disperse itself into separate individual molecules of H2O, with even droplet distribution throughout the entire container, or will, D) The liquid disperse into separate Hydrogen and Oxygen atoms and disappear during the process of sucking the air out of the container, when creating the near 100% vacuum. Thank you for your help and response to this/these answers. Pending your response, I may have a sequential follow-up question. Regards, Randall
- Randall (age 55)
Irving, TX USA

Say you suck almost all the air out, and inevitably in the process suck a bit of water out too. You'll still have almost enough water to half-fill the container with liquid. Water molecules will fly off into the open space until the pressure there equals the equilibrium vapor pressure of water at the temperature of the liquid. That will use up only a small fraction of the liquid, if it's near normal room temperature.

There won't be any significant fraction of the material in the form of little droplets. It will essentially all be in the main liquid pool or in the isolated molecules. The reason is that if a few molecules join up they lose almost as much entropy as if they joined the pool, but don't lose nearly as much energy.

There will also be almost no isolated H, H₂, O or O₂ atoms or molecules. They have much higher energy than the H₂O form. Only under really extreme low-pressure vacuum conditions would you start to get an appreciable fraction of isolated atoms. The vapor pressure of liquid water is far higher than that.

Mike W.

(published on 12/28/2011)

Relative to my prior question and your response: A) At room temperature, how close could I get to an airless envirnment? B) Would an airless environment intrepret to less pressure, which means I would have to manipulate the temperature to enable the liquid solution to remain within its current liquid state of existance rather than turn solid or to vapor. Regards, Randall
- Randall (age 50)
Irving, TX. USA

A) Near room temperature the vapor pressure of water is around 1/40^th of atmospheric pressure. It's strongly temperature dependent, as you can easily find on tables on the Web. Unless you keep pumping, the pressure will stay at that equilibrium value.

B) I'm not quite sure what you mean by "an airless environment". If you just mean no air, but still water vapor, then the vapor pressure is as described in (A). The presence of air has very little effect on the water equilibrium vapor pressure. If you mean no gas of any sort, then you'd have to keep pumping to get close to that, and gradually you'd pump all the water away.

Mike W.

(published on 01/01/2012)

Mike W. In response to your answer, my question may need to be further clarified. Let’s presume the sealed tank is 65% full of liquid. For this continued example we are using water. By airless environment, I want the remaining 35% of ‘open’ space within the container to be void of air. I’ve been using the words ‘airless’, and within a ‘vacuum’ to previously imply this ‘empty’ space. My concern is – as I remove the air, via use of a suction pump, to create this near vacuum / airless environment -will the water remain in bulk ‘liquid’ form, or will it turn to water vapor. I do not want to lose the 75% bulk water liquid that exists within the sealed tank – nor do I want the 25% ‘open’ space within the container to sequentially fill with water vapor. Would there be an additive or additional element which I could add to the water that would alter the molecular structure, weight, effect of gravity, etc, to prevent water vapor from occurring – or at best alter the ‘point of change? Is there a formula I can use to determine how much ‘air’ can be removed and/or ATM pressure and temperature required, to ensure the bulk water will remain in liquid form and not transition to water vapor – while maintaining the 35% near ‘airless’ environment. This is truly a brain twister for me. I have a specific application where I need to adhere to details described within the prior paragraph. Regards, Randall
- Randall (age 50)
Irving, TX. USA

Your question has now come into sharper focus for me. There is going to be an unavoidable tradeoff between keeping the pressure low and keeping the liquid around for a long time.  Molecules will always leave the liquid for any evacuated space above. Depending on your constraints, there are things you can do to reduce that effect.

Let's say that you would like to keep the pressure low but really need to avoid gradually pumping the water out. Then the best you can do is to lower the equilibrium vapor pressure of water as much as possible. Here are the obvious steps to take, although I don't know which ones are practical in your application:

1. Lower the temperature of the liquid as much as possible, say to right near the freezing point.

2. Add non-volatile solutes (salt, sugar, antifreeze, ...) to the liquid. That both lowers the vapor pressure directly and allows you to lower the temperature further without freezing. That comes closest to the sort of step you were considering, but it only lowers the vapor pressure, not reducing it to zero.

Now if you need lower pressure than you can reach with these methods, you'll have to leave the pump on and live with the gradual water loss. In that case you might want to take some step not only to adjust the equilibrium vapor pressure but also to reduce the rate at which water leaves and enters the liquid. A layer of non-volatile oil on the surface would help to do that.  There will be a basic limit making it hard to reduce the pressure much below the equilibrium vapor pressure, however, since at lower pressures the liquid will boil, releasing vapor regardless of what's on the surface.

Mike W.

(published on 01/02/2012)

Mike W. I believe you have pretty much answered my questions, yet with your answers you have triggered a number of additional questions – if I may ask. A) In your # 1 paragraph answer you suggest to “1. Lower the temperature of the liquid as much as possible, say to right near the freezing point.� – yet thereafter you suggest in the paragraph sequential to your # 2 paragraph, “however, since at lower pressures the liquid will boil,�. I am wondering, if during the process of lowering the temperature – how do I keep the water at this lower ‘near freezing’ temperature – if the water will boil at lower pressure - which is what will happen as I reduce the amount of air within the tank. I feel I am not clearly thinking this properly. Can you clarify for me? I will refrain from sending sequential quesitons B & C unitl I can absorb your answer.
- Randall (age 50)
Irving, TX. USA

Very nice question.

Water can boil at low temperature if the pressure is low enough. The boiling process actually soaks up heat and cools the water further. You can freeze water just by pumping on it with a vacuum pump. So if you pump a little too strenuously and start to boil the water, it won't heat things up and lead to rapid boil-off. Instead it will start to freeze the water and thus, if you're lucky, be self-limiting.

Mike W.

(published on 01/02/2012)

Thanks Mike W. While communicating our many questions and answers, I've taken liberty to try to extend my education and thus have come across a number of “Triple Point� and “Critical Point� ‘charts'. If I am understanding correctly, assimilating your information to these charts, it seems your solution can be best accomplished by negotiating ATM pressure somewhere above the .00603 ATM pressure point and by adjusting temperature to a point above .0098 Celsius to manipulate the 'system' to a ratio allowing the liquid to remain in a liquid state above the critical point, preventing a transition from Liquid to Vapor or to a Solid. (If you would like me to e-mail you the charts, with respective explanations, I will have to send them to you via another adjunct e-mail address of yours. Your website does not allow me to copy and paste graphs, charts, etc. to your site) B)Moreover is there a formula that will allow me to – 1) enter a specific ATM pressure ‘number’ and show the resulting temperature of the water and the resulting amount of water vapor potential --- and vice versa, is there a formula that will allow me to 2) insert the temperature of the water I desire - to arrive at the required pressure and which will also show the resulting amount of water vapor potential. Example: I want temperature to be 5 degrees Celsius, thus the resulting pressure will need to be at xxx, which will give me a % water vapor prox. potential of …. Additionally if I 3) enter the ultimate desired water vapor potential of say .00005, is there a formula to tell me what I should attempt to set the pressure and temperature at – so I might be able to determine the required T & P to attain a .00001 potential water vapor level. MY ULTIMATE GOAL IS TO CREATE AN ENVIORMMENT WITHTHE LEAST AMOUNT OF WATER VAPOR WITH THE LEAST AMOUNT OF AIR REMAINING WITHIN THE ENCLOSED CONTAINER, WITHOUT THE BULK WATER TURNING TO WATER VAPOR OR TO A SOLID. If so, can you provide these. I know this is a mouthful – and beyond. C)Moreover, to extend my scenarios …what can you tell me about a further reduction in the combination process of reducing ATM pressure below .00603 ATM point and as well reducing water temperature below .0098 Celsius to create supercooled water. Presuming this would also allow for a near airless environment - what would be the advantages or disadvantages of creating or using supercooled water. Mike, please stay with me. I am soon to reach the end of my questions.
- Randall (age 50)
Irving, TX. USA

Randall- You're definitely on track.

A) Yes, if you're going to use pure water the thing to do is to pump for a while until essentially all the air is gone, then leave the temperature just barely above the triple point. You'll have liquid in equilibrium with a minimal amount of vapor.

B) For calculating the vapor pressure-temperature relation, the Clausius-Clapeyron equation is what you're looking for. (See for an account, with some slight corrections.)

C) Especially if you use well-filtered water, you can supercool a few degrees for a pretty long time, reducing the vapor pressure further. I think you'll just have to try experimentally to see how far that can work.

Again, I'm not sure what your application is, but if you can use any sort of non-volatile solute in the water, that directly reduces the vapor pressure and lets you cool further without freezing, further lowering the vapor pressure.

Mike W.

(published on 01/03/2012)

If you were to take let's say a very solid steel container one mile down in the ocean using the oceans pressure ! Then managed to fill and seal the container with heavy water. Then bring it back to the surface at high speed , would the outcome be that saying the container could take the pressure would the water permanently boil inside or turn to gas. Or explode.
- Simon (age 52)
Uk

The sealed container might explode if it wasn't strong enough to withstand the pressure difference between the high-pressure inside water and the low-pressure outside water. So long as the container didn't explode, the water inside would be held too compressed to boil. It would stay liquid.

Mike W.

(published on 10/03/2015)

Ok so I'll just outright ask.... If you are trying to move water continuously through a closed system and you want to maintain as much of the original mass as possible (throughout each "cycle") is increasing or decreasing pressure the only way? To clarify, the only reason for the closed system in my illustration is to reduce degradation of mass. Is a closed system the only way..?
- Forrest (age 37)
Georgia, USA

You can pump water through a closed cycle with virtually no leaks. A sealed liquid pump can be used at one or more points of the cycle. That's a different problem from Randall's, because he wanted to keep the vapor pressure low right above the water.

Mike W.

(published on 10/31/2015)