= Solution
The <phase diagram> must contain three coexistence curves meeting at the <triple point>: a solid-gas sublimation curve, a solid-liquid melting curve and a liquid-gas vaporization curve. The vaporization curve ends at a <liquid-gas critical point>; beyond it liquid and gas are continuously connected as a <supercritical fluid>.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2021/paper-2-c8-mercury-phases.png]
{title=Schematic pressure-temperature phase diagram of mercury with positive melting-line slope}
The imposed contraction on freezing means $v_{m,s}<v_{m,l}$. Melting therefore has $\Delta v_m=v_{m,l}-v_{m,s}>0$ and $L_{\rm fusion}>0$. The <Clapeyron equation> then requires
$$
\boxed{\left(\frac{dp}{dT}\right)_{\rm melting}=\frac{L_{\rm fusion}}{T(v_{m,l}-v_{m,s})}>0.}
$$
Thus raising the <pressure> raises the melting <temperature>. The liquid-gas and solid-gas slopes are also positive because the transformations into gas increase both <entropy> and <volume>. The liquid region lies between the positively sloping melting line and the vaporization curve, while gas occupies the low-pressure side. The figure is a qualitative solution sketch, not a reproduction or fit of the paper's vapor-pressure measurements; its axes are schematic.
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