I'm reading section Experiment 13 Vapor Pressure of Pure Liquid in Experiments in Physical Chemistry [1, p. 200]:

For the case of vapor–liquid equilibria in the range of vapor pressures less than $\pu{1 atm},$ one may assume that the molar volume of the liquid $\widetilde{V}_\mathrm{l}$ is negligible in comparison with that of the gas $\widetilde{V}_\mathrm{g}$, so that $\Delta\widetilde{V} = \widetilde{V}_\mathrm{g}.$ This assumption is very good in the low-pressure region, since $\widetilde{V}_\mathrm{l}$ is usually only a few tenths of a percent of $\widetilde{V}_\mathrm{g}$. Thus we obtain

$$\frac{\mathrm dp}{\mathrm dT} = \frac{\Delta_\mathrm{v}\widetilde{H}}{T\widetilde{V}_\mathrm{g}}\label{eqn:4}\tag{4}$$

Since $\mathrm d\ln p = \mathrm dp/p$ and $\mathrm d(1/T) = -\mathrm dT/T^2,$ we can rewrite Eq. \eqref{eqn:4} in the form

$$\frac{\mathrm d\ln p}{\mathrm d(1/T)} = -\frac{\Delta_\mathrm{v}\widetilde{H}}{R}\frac{RT}{p\widetilde{V}_\mathrm{g}} = -\frac{\Delta_\mathrm{v}\widetilde{H}}{RZ}\label{eqn:5}\tag{5}$$

where we have introduced the compressibility factor $Z$ for the vapor:

$$Z = \frac{p\widetilde{V}_\mathrm{g}}{RT}\label{eqn:6}\tag{6}$$

Equation \eqref{eqn:5} is a convenient form of the Clapeyron equation. We can see that, if the vapor were a perfect gas $(Z \equiv 1)$ and $\Delta_\mathrm{v}\widetilde{H}$ were independent of temperature, then a plot of $\ln p$ versus $1/T$ would be a straight line, the slope of which would determine $\Delta_\mathrm{v}\widetilde{H}.$ Indeed, for many liquids, $\ln p$ is almost a linear function of $1/T,$ which implies at least that $\Delta_\mathrm{v}\widetilde{H}/Z$ is almost constant.

How is it possible to rewrite equation \eqref{eqn:4} into equation \eqref{eqn:5} in order to introduce the compressibility factor?


  1. Garland, C. W.; Nibler, J. W.; Shoemaker, D. P. Experiments in Physical Chemistry, 8th ed.; McGraw-Hill Higher Education: Boston, 2009. ISBN 978-0-07-282842-9.
  • 1
    $\begingroup$ @Edgardo In the future, please cite your sources, avoid posting screenshots of text, and please try to formulate the question as clearly as possible. $\endgroup$
    – andselisk
    Commented Jan 27, 2022 at 0:49

1 Answer 1


The quoted transformation is an application of the chain rule:

$$\begin{align} \frac{\mathrm{d} (\ln p)}{\mathrm{d} (1/T)} &= \frac{\mathrm{d} (\ln p)}{\mathrm{d}p} \cdot \frac{\mathrm{d}T}{\mathrm{d} (1/T)} \cdot \frac{\mathrm{d}p}{\mathrm{d}T} \\[5pt] &= \frac{\mathrm{d} (\ln p)}{\mathrm{d}p} \cdot \left(\frac{\mathrm{d}(1/T)}{\mathrm{d}T}\right)^{-1} \cdot \frac{\mathrm{d}p}{\mathrm{d}T} \\[5pt] &= \frac{1}{p} \cdot (-T^2) \cdot \frac{\mathrm{d}p}{\mathrm{d}T} \end{align}$$

Substituting in the expression for $\mathrm{d}p/\mathrm{d}T$ will yield the desired equation.

If you're not comfortable with the chain rule (i.e. $\mathrm{d}z/\mathrm{d}y = (\mathrm{d}z/\mathrm{d}x)(\mathrm{d}x/\mathrm{d}y)$), or with the fact that $(\mathrm{d}x/\mathrm{d}y)^{-1} = \mathrm{d}y/\mathrm{d}x$, I'd suggest picking up a book on calculus, which will explain these things.

Note that, although this looks as though we are treating derivatives as fractions and 'cancelling out top and bottom', this is not the correct approach and will fail in other cases. Again, I refer you to a calculus book for a discussion of this.


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