What is Raoult's Law of vapor-pressure? Could someone please explain with diagrams?

1 Answer
Truong-Son N.
Jul 30, 2018

Raoult's law just says that the vapor pressure #P_A^"*"# above a pure liquid will decrease to #P_A < P_A^"*"# when solvent is added into it.

For ideal mixtures (no change in intermolecular forces after mixing), it is based on the mole fraction #chi_(A(l))# of solvent in the solution phase:

#P_A = chi_(A(l))P_A^"*"#

where #A# is the solvent.

Since #0 < chi_(A(l)) < 1#, it follows that the vapor pressure of the solvent must decrease. It starts out as #P_A = P_A^"*"#, and then as #chi_(A(l))# decreases, #P_A# decreases.

https://chemistryonline.guru/

The solute blocks the solvent from vaporizing, so it is harder to boil, and thus the vapor pressure is lower than desired; it is harder to reach the atmospheric pressure, so the boiling point is also higher.

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