Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Suppose you dissolve an amount of salt in water. Because of the attraction between the ions and the water molecules, you'd expect lower vapor pressure and higher enthalpie, both depending on the concentration of the salt and material properties of the salt..

I have found no way to calculate this however?

share|improve this question
    
To calculate the change in enthalpy, you'll want to look into enthalpy of solution (the link has the value for sodium chloride). For the change in vapour pressure I'm less sure. If I think of a way to calculate it I'll post it as an answer. –  Nathaniel Jan 31 '13 at 11:00
    
I remember the vapor pressure beeing lower proportional to the mole fraction - but that can't be right because it would treat different enthalpic salts the same –  mart Jan 31 '13 at 11:02
    
I've a feeling the change in vapour pressure will be closely related to the osmotic pressure. –  Nathaniel Jan 31 '13 at 11:03
add comment

1 Answer 1

For ideal solutes the vapour pressure is given by Raoult's Law. There's an excellent introduction to it here. The law states that:

$$P = P_{solvent} \times X_{solvent} $$

where $P_{solvent}$ is the vapour pressure of the solvent and $X_{solvent}$ is the mole fraction of the solvent. Note that for solutions like sodium chloride in water the solute dissociates so take this into account when calculating the mole fraction of the water.

The ideal behaviour is restricted to very dilute solutions. To calculate the vapour pressure of concentrated solutions you'd usually use an empirical equation like the ones described in this paper (NB behind a paywall).

share|improve this answer
    
Thanks so far, I'll need to find a non-paywalled source for more concentrated solutions since I'm interested in the behaviour near the saturation point. –  mart Jan 31 '13 at 12:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.