= Solution
Three <phase coexistence curves> meet at the <triple point>, where <ice>, <liquid> <water> and <water> vapour have equal <chemical potential>. In the ordinary-pressure region, the low-temperature phase is <ice>, the high-pressure moderate-temperature phase is <liquid>, and the low-pressure high-temperature phase is vapour. The <triple point> is approximately
$$
\boxed{T_t=273.16\,\mathrm K,\qquad p_t=611.7\,\mathrm{Pa}.}
$$
Below $p_t$, warming stable ordinary <ice> crosses the <sublimation> boundary directly into vapour; there is no intervening stable <liquid>. Above $p_t$, <melting> and then boiling can occur. The <liquid>–vapour curve terminates at the <liquid-gas critical point>, approximately $647.1\,\mathrm K$ and $22.06\,\mathrm{MPa}$, beyond which there is no <liquid>–vapour phase boundary.
The distinctive feature is the negative slope of the ordinary <ice>–<water> <melting> line. The <Clapeyron equation> gives
$$
\frac{dp}{dT}=\frac{\Delta s}{\Delta v}.
$$
<Melting> increases <entropy> but reduces <volume (thermodynamics)> because ordinary <ice> is less dense than <liquid> <water>, so $\Delta s>0$ and $\Delta v<0$. Increasing <pressure> consequently lowers the <melting> <temperature> locally. <Sublimation> and <vaporization> both increase <volume (thermodynamics)> and <entropy>, giving positive slopes. The following original schematic shows these signs; it is not a quantitative fit and omits the additional high-pressure <ice> polymorphs.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2019/paper-2-a2-water-phases.png]
{title=Schematic water pressure–temperature phase diagram with ordinary ice, liquid, vapour, triple point and critical point}
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