Do azeotropes evaporate, without boiling, in proportionate amounts, or does this property describe their behaviour only at boiling point?

Concretely, if 91% isopropyl alcohol/water (by volume, i.e. an azeotropic mixture) is left open at say 25 °C, will it remain in that proportion as it evaporates, or become more dilute (until just water) as the alcohol evaporates faster?

Further, above this point, would the alcohol evaporate faster 'degrading' it until 91% at which point it behaves as the answer to above? (Which would incline me to think the answer is no, if not boiling the water and alcohol would evaporate 'independently' and the alcohol concentration would fall no matter the starting point, it's not magic.)

  • $\begingroup$ Only at the boiling point as only here is the mole fraction of the two species the same. The boiling point has a max or min at constant $p$ and the total vpr. pressure a max or min at constant $T$. $\endgroup$
    – porphyrin
    Commented May 23, 2021 at 8:26

2 Answers 2


Azeotropic composition is a function of the boiling pressure and therefore of the boiling temperature. This can be seen in looking up the composition of the constant boiling $\ce{HCl}$ azeotrope with changing atmospheric pressure. However, information on this is hard to find and the effect seems to be small in most cases. Wikipedia does discuss this a bit. Just thinking, this might be a possible research if one had access to an FTIR with a ATR trough cell so evaporation could be observed in situ.


This is a reasonable question. Basically it asked whether the pre- or below-boiling-point vapor composition can be different from the liquid composition, although they are known to be the same at the boiling point according to the property of azeotropes. We can come up with a right answer by the following careful thinking.

You are looking at a bottle of azeotrope open to air. There is always a vapor phase even below the boiling point due to the existence of a dynamic equilibrium between the vapor phase and the liquid phase. Suppose you look at the $P-x,y$ phase diagram at $T=$ 25 °C temperature, since the system is left open to the environment, the external pressure is the atmospheric pressure $P$ = 1 atm. The reason that the system is not boiling is because the vapor pressure is too low compared to the external pressure. Only if the external pressure is reduced to be equal to the vapor pressure under this temperature, the system can be brought into boiling state (note that here we are only changing pressure $P$ because temperature is kept at $T=$ 25 °C). The key question is how much the vapor pressure is for the system at $T=$ 25 °C. In fact, it can be directly read from the phase diagram at $T=$ 25 °C, which contains the full information for the vapor-liquid equilibrium. This tells us that even if the system is not boiling, the vapor pressure is still the one that can be read from the azeotropic point on the phase diagram. Accordingly, the vapor composition is determined by the azeotropic composition, meaning that it is the same as the liquid composition, regardless of whether the system is boiling or not. Finally, leaving the system open simply means that the vapor can keep evaporating to decrease the amount of the mixture, which however would not change the vapor or liquid composition. This is a fundamental reason for taking advantage of the equal vapor and liquid compositions of azeotropes to make perfume and topical mosquito repellent, both used under their boiling temperatures.

To conclude, azeotropes should evaporate, without boiling, in proportionate amounts. However, there is a caveat about the temperature and pressure, which can fluctuate a little bit from time to time. For instance, a mixture put in a bottle at $T=$ 25 °C may have its temperature elevated to $T=$ 37 °C when rubbed or sprayed on skin, and pressure $P$ of a sunny day can be different from a rainy day. Therefore, one composition that is azeotrope under one set of $T$,$P$ may not be azeotrope under another set of $T$,$P$.


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