# How much pressure does Gallium exert when it freezes?

Assuming the liquid Gallium is in a sealed container, how much pressure will it exert on its container when it freezes?

• I suggest to reopen this question. Its intention is quite clear, and it can be answered meaningfully. – Ivan Neretin May 29 '16 at 12:14
• – Mithoron May 29 '16 at 23:15
• What's the volume of the container? We'll need to know that, as pressure is force per unit area. The coefficient of thermal expansion for gallium can be used here, but only if we're assuming the container doesn't yield to the expansion forces - is that part of your query, or do you have a material in mind for the container? – Todd Minehardt May 30 '16 at 0:52

## 2 Answers

Enough to break a glass container.

Gallium is one of the few substances that expands on freezing (like water). The solid is about 3% less dense than the liquid. This is enough to shatter a glass bottle full of gallium on freezing (see this video).

Solids that expand on freezing can exert a lot of pressure (clearly more than enough to break glass) but the exact amount of pressure probably depends on the vessel and the circumstances of freezing.

Suppose we have a substance capable of a phase transition in which it expands. Suppose we put it in a sealed container and change the temperature past the transition point. What would happen? Apparently, some portion of the substance would undergo the transition, which would result in pressure increase, which in turn (according to the Le Chatelier's principle) would attempt to drive the transition back, and so the new equilibrium will be reached.

As to the extent of the effect, let us consider $\Delta G$ of the said transition. When the system is at equilibrium, $\Delta G=0$. If we change our pressure and temperature (not by much, so the linear approximation still applies), $\Delta G=\left({\partial(\Delta G)\over\partial T}\right)_PdT+\left({\partial(\Delta G)\over\partial P}\right)_TdP=\Delta V dP - \Delta S dT$. Now, this new state is also an equilibrium, so this new $\Delta G$ is also $0$. This gives us certain relation between $dT$ and $dP$, namely: $dP = {\Delta S\over\Delta V}dT={\Delta H\over T\Delta V}dT$. By looking up the molar volumes of solid and liquid gallium, and also its enthalpy of melting, you may find out exactly how many hundreds of atmospheres of pressure would we get for each degree of temperature.

Note that this description applies to any phase transition that goes with expansion, be it freezing of gallium (or water, or one of the few more exceptions), or melting of nearly any other substance. The signs of $dT$ and $\Delta H$ in case of melting are opposite to those in case of freezing, but the overall expression is still the same.

Of course, at some point all gallium would freeze, so further cooling will not increase pressure any more. To calculate this point, we need the compressibility figures which are not that easy to come by. Also, at pressures that great we can hardly ignore that the container walls must give in to some extent, no matter what they are made of.