Raoult's law is used to calculate the partial pressures of the vapour phases of volatile components of a solution

Given that there are two components, A and B, in the liquid-liquid solution, there are three types of intermolecular forces of attraction, A-A, A-B and B-B

A non-ideal solution disobeys Raoult's law

If A-B is lesser than A-A and B-B, the molecules of the components find it easier to escape the liquid phase. Thus, there is higher partial pressure than that predicted by Raoult's law. The solution is said to exhibit positive deviation

If A-B is higher than A-A and B-B, the solution exhibits negative deviation

My question:

Are there solutions in which A-B is greater than A-A but less than B-B? Would component A exhibit negative deviation, and component B exhibit positive deviation, in the same solution?

Can the magnitudes of these deviations cancel out each other such that the total partial pressure of the solution is still linear (with respect to change in more fractions of the components)? Can the overall solution obey Raoult's law even when the components don't?



  • 1
    $\begingroup$ It's hard to see how you might make this work. For instance, perhaps a complex AB would observe Raoult's law while the dissociated components A and B don't. But if you stray from an associated or stoichiometric balance then you would have A or B in solution, and Raoult's law would not be obeyed. $\endgroup$
    – Buck Thorn
    Jul 26, 2021 at 11:57
  • 2
    $\begingroup$ IMHO, there is no need for A-B to be either higher either lower than both A-A and B-B. Important is relating of A-B to the average of A-A and B-B. $\endgroup$
    – Poutnik
    Jul 26, 2021 at 13:17
  • $\begingroup$ When considering Gibbs free energy comparisons across a (binary) phase diagram, you get to pick the reference states at the extremes. In other words, one can set A-A and B-B to be zero, and A-B is then calculated relative to those reference states. Or you could set B-B to a billion J/mol and ... nothing changes in the analysis at all. $\endgroup$
    – Jon Custer
    Jul 26, 2021 at 17:04
  • $\begingroup$ @JonCuster My textbook has shown the cases where A-B is less than both A-A and B-B or greater than both A-A and B-B. My doubt is whether A-B can be in between them instead $\endgroup$
    – Balu
    Jul 27, 2021 at 3:47


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.