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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.)

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  • $\begingroup$ 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 May 23 at 8:26
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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.

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