= Solution
For liquid-vapor equilibrium, use molar quantities. At low vapor <pressure> the gas is dilute, approximately ideal, and has a much larger <molar volume> than the liquid. Hence
$$
\Delta v_m=v_{m,g}-v_{m,l}\simeq v_{m,g}\simeq\frac{RT}{p}.
$$
Substitution into the <Clapeyron equation> gives
$$
\frac{dp}{dT}\simeq\frac{Lp}{RT^2},\qquad
\frac{d\ln p}{dT}\simeq\frac{L}{RT^2}.
$$
If the molar <latent heat> varies sufficiently slowly to be treated as constant over the <temperature> interval, integration gives
$$
\ln p=\text{constant}-\frac{L}{RT},\qquad
\boxed{p=p_0\exp\left(-\frac{L}{RT}\right).}
$$
The positive constant $p_0$ has <pressure> units and is an integration constant. It is not necessarily a real infinite-temperature <pressure>, since the approximations fail before that limit.
For <mercury (element)> between $300$ and $500\,\mathrm K$, the original plot places this range above the <triple point> and below the normal boiling and critical <temperatures>. Its vapor <pressure> is low relative to the critical <pressure>, making a dilute-gas description reasonable; the vapor's <molar volume> greatly exceeds that of the dense liquid. The interval is also far from the critical region where $L$ tends to zero, so a roughly constant <latent heat> is a useful first approximation. The approximations are not exact: <Kirchhoff's law of thermochemistry> gives, at fixed <pressure>, the <temperature> <derivative> of a transformation enthalpy as the difference between the two molar <heat capacities>, so a precise fit over a wide interval would allow $L$ to vary.
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