= Ideal-vapor Clausius-Clapeyron approximation
{c}
{title2=$d\ln p/dT\simeq L_m/(RT^2)$}
For <liquid>–<gas> <phase coexistence>, make two distinct approximations: the <gas> obeys the <ideal gas law>, $v_{m,g}=RT/p$, and its <molar volume> greatly exceeds the <liquid> <molar volume>, $v_{m,g}\gg v_{m,l}$. Here $R$ is the <molar gas constant> and $p$ is the <saturation pressure> of the pure substance. Substitution into the <Clapeyron equation> gives
$$
\frac{dp}{dT}\simeq\frac{L_m(T)p}{RT^2},\qquad \frac{d\ln(p/p_*)}{dT}\simeq\frac{L_m(T)}{RT^2},
$$
where $p_*>0$ is any fixed reference <pressure>, making the <logarithm> dimensionless. Constant <latent heat> is a further approximation, needed for the elementary <Integrated Clausius-Clapeyron relation> below.<Statistical Physics by David Tong>[§5.1.2, “An Approximate Solution to the Clausius-Clapeyron Equation”: the three assumptions and the exponential solution]{ref}
The neglected <liquid> <molar volume> has a transparent error estimate. If the <gas> is ideal but $\epsilon=v_{m,l}/v_{m,g}$ is retained, then
$$
\left(\frac{dp}{dT}\right)_{\rm full}=\frac{L_m p}{RT^2(1-\epsilon)},\qquad \frac{(dp/dT)_{\rm approx}}{(dp/dT)_{\rm full}}=1-\epsilon.
$$
Thus ignoring the <liquid> <molar volume> underestimates this slope by a relative amount $\epsilon$. This estimate does not bound errors from treating a <real gas> as ideal or from freezing $L_m(T)$. Near a <thermodynamic critical point>, the <gas> and <liquid> <molar volumes> approach each other and these dilute-vapor assumptions fail.
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