In open air, when vapour pressure reaches 1 atm, boiling takes place.
I read that if we add two immiscible liquids together, the total vapour pressure of the 'mixture' is close to $p = p^*_A + p^*_b$, means that the vapour pressure of the 'mixture' is higher than the vapour pressure of both the constituent liquids, subsequently means that the 'mixture' boils at lower temperature than both of the constituent liquids.
If we have many liquids which are not immiscible with each other, and we add them together to form a 'mixture' such that $p = p^*_A + p^*_B + ... > \pu{1atm}$ , will boiling results at room temperature?
I suspect we have many liquids which are not miscible with each other, since we classify solvents as organic solvent and inorganic solvent only, but let's say we indeed have that (or we have two immiscible liquids volatile enough to make $p>\pu{1atm}$), what will be observed? Boiling at room temperature? In that case, we can use the vapour to do work (e.g. rotate turbine), but when we mix immiscible liquid together there should be no energy transaction right? The particles do not react with each other. If all these are true, we are actually doing work without any input which violates the first law, so there must be something wrong with my reasoning above.
No comments:
Post a Comment