# How compressibility factor is introduced in Clapeyron equation?

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?

### Reference

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.
• @Edgardo In the future, please cite your sources, avoid posting screenshots of text, and please try to formulate the question as clearly as possible. Commented Jan 27, 2022 at 0:49

\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.