# Do liquids typically solidify under enough pressure?

Looking at the phase diagram of water,

it looks like increasing pressure of a liquid at any temperature eventually produces a solid. Liquid water becomes Ice VII at some pressure over 1 GPa and 350-600 K, or Ice VI at 280-350 K.

Is this true for most liquids? Thinking mainly of simple organics like methanol or benzene, do they have a solid phase that forms at increased pressures in the GPa range? Is it a general rule for all liquids, and is there derivation of this from first principles?

It seems the liquid-Ice Ih transition upon cooling happens at roughly the same temperature over a very wide pressure range 1 kPa to 0.2 GPa. So the "left" (low temperature) boundary of the liquid region is largely pressure-independent, while the "upper" (high pressure) boundary is pressure dependent and also ends up in a different solid phase. Why is that?

The example you have chosen to start your question (water) is an atypical case. The phase diagram has a logarithmic scale for the pressure, somewhat hiding the negative slope of the transition line between the liquid and solid phases. That negative slope says that by increasing pressure, a solid in a state close to the melting curve will melt, and a liquid also close to the phase transition line will remain liquid.

However, by further increasing pressure, liquid water eventually solidifies. The reason is that the negative slope of the melting curve is connected to the lower density of normal ice Ih relative to the liquid. This is possible if the crystalline structure is open with low coordination. At high pressure, more dense solid phases with higher coordination appear, thus making a negative slope of the melting curve over the whole phase diagram unlikely.

The relation between the slope of the melting curve in the $$P-T$$ plane and differences in entropy $$(\Delta s$$ and molar volume $$(\Delta v$$ at the transition is ruled by Clapeyron's equation : $$\frac{{\mathrm d}P}{{\mathrm d}T} = \frac{s_{liq}-s_{sol}}{v_{liq}-v_{sol}}.$$ The numerator on the right-hand side can be rewritten as $$\frac{{\cal l}}{T}$$, where $${\cal l}$$ is the latent heat of the liquid-solid transition (positive). Therefore, the sign of $$v_{liq}-v_{sol}$$ controls the sign of the melting curve.

As far as I know, by increasing the pressure at a constant temperature, eventually, all materials have a fluid-crystal transition. This is consistent with the idea that by increasing pressure, the solid becomes more and more compact, making it unlikely that at very high pressure $$v_{liq}-v_{sol}<0$$. A final word of caution is about the possibility that, in some cases, kinetic effects may influence the practical observability of the transition to a crystalline solid.

• Thanks! If I understand the explanation, when the slope of the melting curve is positive is because the solid is more dense than the liquid? So at some pressure, the pressure x change in volume becomes favorable enough to overcome entropy? In other words, in dG = dU + p dV - T dS, the high pressure solidification is driven by dV? Nov 21 at 17:44
• @AlexI, The connection between the slope of the melting curve and changes in density (and entropy) is much more direct. I'll add a new paragraph to clarify such a point. Nov 21 at 22:58

No, there are some substances which never become solids. One example: He. If we have 3He, it will form a Bose-Einstein condensate if I remember correctly.

• Thanks! I was kinda expecting helium to be anomalous because it has superfluidity, but it seems it actually does solidify at not-crazy pressures hal.science/jpa-00231630/file/ajp-jphyslet_1979_40_13_307_0.pdf That's just 4-He though Nov 20 at 21:51
• Now that you have bring it up , any superfluid never becomes a solid. Nov 20 at 21:52
• Helium does not freeze into a solid at normal pressure. Under pressure, both $He^4$ and $He^3$ undergo a liquid-solid transition. Nov 21 at 0:33