= Stipulated vapor-pressure calculation
Consider an unnamed model substance with molar <latent heat> $L_m=40.0\,\mathrm{kJ\,mol^{-1}}$ and <saturation pressure> $p_1=10.0\,\mathrm{kPa}$ at $T_1=300\,\mathrm K$. Stipulate the <ideal-vapor Clausius-Clapeyron approximation> throughout $280\leq T\leq340\,\mathrm K$, constant $L_m$, and $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$. These parameters define an illustrative model, not a measured dataset or a fit to <water>. At $T_2=330\,\mathrm K$,
$$
\ln\frac{p_2}{p_1}=\frac{40000}{8.314}\left(\frac{1}{300}-\frac{1}{330}\right)\simeq1.458,\qquad p_2\simeq42.97\,\mathrm{kPa}.
$$
At the reference state the <gas> <molar volume> is $RT_1/p_1=0.24942\,\mathrm{m^3\,mol^{-1}}$, and the local slope is $dp/dT\simeq0.5346\,\mathrm{kPa\,K^{-1}}$. The tangent estimate $p_2\approx10.0+30(0.5346)=26.04\,\mathrm{kPa}$ is much lower than the integrated result: a local <derivative> should not be extrapolated over this whole interval. By contrast, over a small <temperature> increment, the fractional sensitivity is $d\ln(p/p_1)/dT=L_m/(RT_1^2)\simeq0.05346\,\mathrm{K^{-1}}$.
\Image[/ideal-vapor-pressure-and-inverse-temperature.png]
{title=Stipulated <saturation pressure> versus <temperature> and its linear <logarithm> versus <inverse temperature>}
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