# Water phase diagram

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I notice that water phase diagrams provided online always seem to show a rather linear behaviour for the solid-liquid boundary (and an extremely steep slope).

How is this modelled mathematically? Say we use the Clapeyron equation with ΔH and ΔV_m being constant, as online example problems (meant for students) do. Integration with this yields a ln(T2/T1) for example--not the equation of a straight line. So where does the almost straight line, which suggests that P=kT for some very large negative k, come from? And how valid is it really to model ΔH and ΔV_m as constant, if this doesn't produce a line that looks much like the diagrams show?

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mfb
Mentor
It is not linear. Here is a phase diagram: It's complicated.
Integration with this yields a ln(T2/T1) for example
For what, and what are T1 and T2?

The line is so steep because both a few hundred kPa is a low pressure in the context of water and ice. You see larger deviations at tens of MPa.

Many other diagrams show a line, e.g. https://uh.edu/~jbutler/physical/chapter6notes.html, https://scholar.harvard.edu/files/schwartz/files/9-phases.pdf. I saw the diagram you mentioned too.

I would like to know the mathematical model that leads to either of these. It doesn't appear to be constant ΔH and ΔV_m since neither the shape in the diagram you linked, nor a straight line, corresponds well to

$$P = P_0 + \frac{\Delta H}{\Delta V_m} \rm{ln} \frac{T}{T_0}$$

which is the result from integrating Clapeyron equation. (T0,P0) can be any known point, e.g. (273.15 K, 1 atm) for water solidus

Yet all the problems/examples one finds online seems to treat ΔH and ΔV_m as constant. If it is valid to do so I would like to know how the boundary can have the shape we see on the diagrams.

Locally (if T/T0 doesn't deviate too much from 1) a logarithm looks like a straight line.
Good point! That may explain the linear shape.

What about the non-linear curved bits at high pressure (that you can see on the diagram you linked originally, or the one here https://scholar.harvard.edu/files/schwartz/files/9-phases.pdf on p4)? They don't necessarily appear to follow the equation I gave...

Edit: actually it looks like there might be a cusp and another independent solidus?

Chestermiller
Mentor
The ##\Delta V## is tiny, and that accounts for the huge slope.