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On phase diagram of water, solid to state border is a straight vertical line. Someone can argue it is only an approximation and nature cannot have vertical lines. For me however, it is clearly vertical. There exist no mention of this limit and it is not given a name. What is the name of this point, if such name exists, and what is the pressure value of it? If it is not measured, why? It seems like a very important point. Is there a measured, exact value of that point, not calculated or approximated?

https://upload.wikimedia.org/wikipedia/commons/thumb/0/08/Phase_diagram_of_water.svg/1214px-Phase_diagram_of_water.svg.png

In response to comments below (website does not reacts to my attempts to send a comment): I wrote 0.0 °C which is one-decimal digit. Anyone can see the line is absolutely straight in range that stretches at least from 10 mbar to approximately 10 bar which is a very big range. The upper limit of this range is not defined or named which is what my question is about.

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    $\begingroup$ It is not clear to me what you are asking here. You have a link to the phase diagram of water, which clearly shows various key triple points. BTW, the ice Ih (ordinary ice) / liquid water fusion curve is not quite vertical. It slopes slightly to the left with increasing pressure. This sloping to the left with increasing pressure is highly unusual but is not unique. The fusion curve for most substances slopes to the right with increasing pressure. This sloping to the left is a consequence of the fact that ice floats. $\endgroup$
    – David Hammen
    Jan 10 at 11:37
  • $\begingroup$ "[...] and nature cannot have vertical lines." Where does that come from? $\endgroup$
    – noah
    Jan 10 at 20:00

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It sounds like you're describing the solid–liquid coexistence line for water. This line isn't vertical but appears that way because its slope is -13.5 MPa/°C and because it's plotted on a logarithmic scale going down to ~600 Pa. What looks like a change in slope to a nonvertical line around 1 MPa is an optical illusion caused by the logarithmic scale:

That slope can be calculated from the Clausius–Clapeyron equation:

$$\frac{\Delta P}{\Delta T}=\frac{L}{T_\mathrm{m}\Delta v},$$

where $\Delta P/\Delta T$ is the change in pressure per unit temperature, $L=334\,\mathrm{J}\,\mathrm{kg}^{-1}\,^\circ\mathrm{C}^{-1}$ is water's latent heat of fusion, $T_\mathrm{m}=273\,\mathrm{K}$ is its melting temperature, and $\Delta v=9.05\times10^{-5}\,\mathrm{m}^3\,\mathrm{kg}^{-1}$ is its change in specific volume upon melting. The volume change $\Delta v$ is small, so a great deal of pressure is required to shift the melting temperature.

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  • $\begingroup$ How can I get my desired upper pressure limit from this graph? What is drawn on your graph? And graph that I attached is not logarithmic, but linear on the X axis on which vertical line of 0.0 continuous temperature of water phase border is visible. On Y axis, range from 1 to about 10 bar do not seems like a "logarithmic optical illusion", it is a range of values from 1 atmosphere to 10 atmospheres which is a lot. $\endgroup$
    – user268587
    Jan 11 at 21:23
  • $\begingroup$ Please annotate your image with a circle showing the region you’re talking about. I don’t know what you mean by “upper pressure limit.” On a linear–linear plot, you would see a nearly straight sloped line from 273 K, 612 Pa to 251 K, 210 MPa. On a log–linear plot, you would see what I plotted and what appears in the real phase diagram. There is no “continuous temperature” or vertical line. Try plotting your own log–linear chart of a slope of -13.5 MPa/°C; I think you’re unfamiliar with how such charts distort linear relationships. $\endgroup$
    – Chemomechanics
    Jan 11 at 22:55

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