phase - What happens if you cool water in a container too small for it to freeze?

Freezing a full bottle of water tends to shatter the glass bottle. What if you used something tougher than glass, like diamond? What would happen if you kept dropping the temperature, but restrained the liquid volume so it couldn't freeze and what sort of force would the liquid exert on the container's walls?

Answer

Good question. Let's assume the container is infinitely strong, non-deformable, and constant in volume. Let's also assume that cooling the water is an equilibrium process -- that way, we won't have any supercooling.

At equilibrium, the first tiny bit of ice that freezes will take up more volume than the water it froze from. This will raise the pressure on the rest of the water. Eventually the pressure may get so high that additional freezing of more water is not thermodynamically favored.

Of course, as the pressure is raised, even the solid ice compresses a bit, freeing up a bit more volume for the liquid water. According to this paper from 2004, ice is less compressible than water, so as a starting assumption, it may be approximately true to neglect the ice compression effect.

Figure 4 from that same paper gives the freezing point depression of water as a function of pressure:

To fully answer your question, in addition to that data, an equation that gives pressure as a function of ice volume would also be needed. If we make the assumption I was talking about above -- i.e. that ice is incompressible, then from the data point that water has a constant compressibility of 46.4 ppm per atm we can come up with a very simple version of that equation.

$\frac{\Delta V_{water}}{V_{water}}=46.4 \times 10^{-6} \times P$, where P is the pressure in atmospheres.

Before freezing of a fraction $X$ of the water:

$$V_{ice} = X V_{tot}$$ $$V_{water} = (1-X) V_{tot}$$

After freezing:

$$ V_{ice} = X V_{tot} 1.11 $$ $$ V_{water} = (1-X) V_{tot} - \Delta V_{ice} $$

Combining those equations, you can get

$$0.11 \frac{X}{k(1-X)}= P$$, where $k$ is the compressibility of water. If even 1% of the water in the container freezes (and all our assumptions are true), then the pressure will be 24 atmospheres! Freezing 10% of the water would mean a pressure of 260 atmospheres. Looking at the chart above, reaching this point would require a temperature of only 271 or 272 K, i.e. only -1 °C or -2°C. Freezing 45% of the water would reach a pressure of 2000 atm, already off the chart above -- but the temperature required to reach that point would only be 253K or -20 °C, the setting of the average home residential freezer! ((Of course, at these extreme pressures, (i) ice is actually compressible, and (ii) the compressibility of liquid water is not constant but also a function of pressure, so the calculations would get quite a bit more complicated.))

The lesson is that for even moderate degrees of cooling, you'd need a very, very strong container.