It would be useful to be able to predict relative evaporation rates of liquids. For small reduced temperatures the heat of vaporization is relatively constant, but it is not one of the more common properties cited for substances.

Is there any sort of functional relationship between enthalpy of vaporization and common properties like viscosity, melting point, and boiling point?

  • $\begingroup$ I don't believe there necessarily is one, and I considered making a plot to confirm that suspicion, but finding the data is going to be a pain since both of those properties vary with temperature. I think you would get substantially different answers with viscosity at the melting point of individual compounds vs. at a fixed temperature across all compounds. I'd be interested if someone did have the data to plot such. $\endgroup$ – William Kappler Aug 14 '15 at 8:31
  • $\begingroup$ @WilliamKappler: Good point. Yes, we probably have to specify it at some standardized temperature. Melting point is a good one. Or perhaps the "liquid-midpoint": i.e., the temperature halfway between boiling and freezing. $\endgroup$ – feetwet Aug 14 '15 at 19:14

Yes. The phase envelope (or equilibrium curve) of most liquids have a similar shape and can be approximated, to a reasonable degree, by using the Clausius-Clapeyron equation:

$$ \Delta h_v=T(v^V-v^L)\frac{dP^s}{dT} $$

which requires finding ${dP^s}/{dT}$ from e.g. a vapor-pressure-temperature correlation like the Antoine equation, and also a separate estimate for $\Delta v$ before $\Delta h_v$ can be obtained. This is an exact thermodynamic relationship, but is often simplified by using the ideal gas law for $v^V$ and neglecting the liquid volume $v^L$ i.e. $v^V \gg v^L$, giving

$$ \Delta h_v=\frac{RT^2}{P^s}\frac{dP^s}{dT} $$

However, it is very important to note that the term $(v^V-v^L)$ becomes increasingly decisive as you go above 1 bar, where the assumption $v^V \gg v^L$ is no longer valid. In such a case, separate correlations for $v^L$ and $v^V$ are required before the correct curvature is obtained. $v^L$, for instance, can be calculated from e.g. a density correlation like the Rackett equation, and $v^v$ from an equation of state like Peng-Robinson.

  • $\begingroup$ Wow, so this looks like relatively cutting-edge thermodynamics: Do the Rackett and Peng-Robinson equations really date only from the 1970s? $\endgroup$ – feetwet Apr 19 '16 at 21:23
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    $\begingroup$ $\Delta h_v$ is the specific enthalpy of vaporization [J/mol], $v$ the pure component molar volume for the vapor ($V$) and liquid ($L$) phases, while $P^s$ is the saturated vapor pressure ($T$ is in Kelvin). $\endgroup$ – ChE Junkie Apr 19 '16 at 21:30
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    $\begingroup$ Cutting-edge? Google the names Rudolf Clausius and Clapeyron... Not so much as cutting-edge as it is still very relevant. New correlations for $P^s$, $v^V$, $v^L$ or any other thermodynamic property are constantly being developed. Most of the time, however, you are only interested in answers that are sufficiently precise and that follow known trends i.e. 5 decimal place accuracy is typically not required (which is why you find many older correlations still in active use). $\endgroup$ – ChE Junkie Apr 19 '16 at 21:41
  • $\begingroup$ I'm still trying to digest this. The differential equation form is needed to find an equilibrium, but before solving for equilibrium I like to be able to find instantaneous values. So given a T and P what determines a chemical's instantaneous evaporation rate? Or, put another way, holding the conditions constant and changing the chemical, what do these equations tell us? For "everyday" T & P it looks like it reduces to saying, "Evaporation rate is proportional to saturated vapor pressure," so Clausius-Clapeyron doesn't actually relate it to other characteristics? $\endgroup$ – feetwet Apr 19 '16 at 21:55
  • $\begingroup$ Thermodynamic properties are most definitely influenced by molecular characteristics, which is why different molecules have different saturated vapor pressure curves and normal boiling points (just to name a few). Therefore it is not necessarily correct to say that the Clausius-Clapeyron equation does not relate to other molecular characteristics. $\endgroup$ – ChE Junkie Apr 19 '16 at 22:59

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