If liquid water is constrained in vessel then the temperature is dropped below freezing what pressure would be measured assuming the vessel is unbreakable? Or put another way, will ice form if the water is not allowed to expand?
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1$\begingroup$ You can checkout phase diagram on en.wikipedia.org/wiki/Ice which is very complex $\endgroup$– jw_Commented Mar 27, 2020 at 2:27
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4 Answers
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Yes, the water will freeze, but not into the normal hexagonal crystalline ice form that we are familiar with. Assuming that the container is capable of withstanding an internal pressure of at least 300 Mega Pascals (about 43,500 pounds of force per square inch), the water will freeze into its Ice II rhombohedral crystalline form.
This also assumes that the water was introduced into the container at 4 degrees Celsius, at which liquid water is most dense. If it was warmer, if would initially shrink a bit and create some vapor space within which some normal ice would first form.
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Depends on how fast it freezes, if it is very fast it will form a disordered solid which will fill the vessel without excess pressure.
If you freeze it slowly ice crystals with lower density will form and expand. Eventually ice would stop forming and there would be water below its freezing point because of the pressure increase.
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water confined in an enclosed space like the plumbing in a house or the engine block in an automobile and then forced to freeze will expand, and in so doing will generate sufficient pressure to burst those pipes or crack that engine block.
The pressure actually required to inhibit water from freezing far exceeds that which can be developed in everyday circumstances, and can be determined by reading a phase diagram for the ice/water/water vapor system.
answered Mar 27, 2020 at 3:45
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Freezing point is dependent on external pressure acting on a liquid (see this link How does the freezing temperature of water vary with respect to pressure?)
So the pressure from vessel will cause a depression in freezing point (assuming vessel to have 0 coefficient of volumetric expansion($\alpha_v$) so it does not change its volume and does not break under high pressure). As you continue to drop below 4°C water will expand and causing pressure on container which in turn applies pressure to water causing more and more depression in freezing point.
So if you keep volume unchanged and reduce temperature you will reach unreasonably high pressures.
So in theoretical ideality
No water will never freeze
But any practical is bound to have limitation and the cointaner will burst eventually.
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$\begingroup$ A non zero $\alpha_v$ will cause cointainer to have a tendency reduce its volume which will lead to thermal stress on cointainer complicating the problem. So I have neglected it. $\endgroup$ Commented Mar 27, 2020 at 6:48
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$\begingroup$ As I was thinking about freezing water and how it works I thought the answer to my question would be relatively simple. I find out that is not so. Water is complex stuff. I did not realize there are a number forms of solid water, each with its own characteristics. I need to a bit more research. I do thank you for your response. $\endgroup$ Commented Mar 27, 2020 at 15:32
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$\begingroup$ Yes, water is indeed very complex mainly due to its exceptional property to form large number hydrogen bonds (especially in ice where you have cage like tetrahedral units). I happen to leave out this rather chemical phenomenon in my answer, but it is $hidden$ in the fact that freezing point of water changes with pressure. $\endgroup$ Commented Mar 27, 2020 at 15:38