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Consider the following setup:

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Two immiscible liquids A and B are placed inside a balloon at a temperature where neither one of them boils. The inside of the balloon contains some argon, and the outside is air at atmospheric pressure. The balloon is made of mylar or some other similar material (flexible but not elastic).

I expect that the volume of the balloon would increase a bit due to evaporation of A and B into the argon. If we start without any gas in the balloon, is there a scenario (choice of liquids, choice of temperature) where the combined evaporation of A and B will be sufficient to fill up the balloon?

This question is related to Is it possible to boil a liquid by just mixing many immiscible liquids together? It addresses the question how to get both liquids into contact with the gas phase without worrying that the more dense liquid would be covered by the less dense one. It specifically does not ask about boiling, though, but rather about combined vapor pressure higher than ambient pressure, i.e. sufficiently high to displace gas at atmospheric pressure.

If this setup works, I wonder if it is possible to build a balloon pump where the balloon blows up "automatically", i.e. by heat transfer from the environment to drive evaporation of the two liquids.

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    $\begingroup$ The 'system' is possible and true according to what my (physical chemistry) teacher has taught us regarding Raoult's Law and corresponding Vapour pressure for a two-immiscible-liquids system. Is it correct that I understand 'automatically' does not refer to 'perpetual'? As liquids would be consumed. $\endgroup$
    – glucose
    Apr 16, 2019 at 12:40
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    $\begingroup$ It is surely possible to blow up a balloon by evaporation of one liquid (provided it boils low enough; liquid nitrogen will do). Two liquids hardly provide any new insight. $\endgroup$
    – Ivan Neretin
    Apr 16, 2019 at 12:46
  • $\begingroup$ @Ivan_Neretin Well, the liquid nitrogen would boil. I am asking about a temperature below the boiling point where the liquid would evaporate. Evaporation of a single liquid would not blow up the balloon. $\endgroup$
    – Karsten ♦
    Apr 16, 2019 at 13:16
  • $\begingroup$ @glucose No, not 'perpetual', just until the balloon is full. $\endgroup$
    – Karsten ♦
    Apr 16, 2019 at 13:17
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    $\begingroup$ If each liquid had a vapor pressure of 0.5 atmosphere at some temperature T, then combined they would blow up the balloon, and neither would be boiling. Water for instance has a vapor pressure of 0.5 atm at about 82 C. $\endgroup$
    – MaxW
    Apr 16, 2019 at 16:29

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