Will evaporation from two immiscible liquids blow up a balloon?

Consider the following setup:


[![enter image description here](https://i.stack.imgur.com/OcOic.jpg)](https://i.stack.imgur.com/OcOic.jpg)


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?](https://chemistry.stackexchange.com/q/112769) 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.


