Different volatilities in ideal solutions?

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In summary, when comparing two liquids, if the intermolecular forces between molecules of one liquid are greater than the other, then the first liquid is more volatile. When mixing these liquids to create a solution, the solution is considered ideal if the tendency for molecules to leave the solution to the gas phase remains unchanged. This can be achieved if the new interactions between the two types of molecules are equal to their respective individual intermolecular forces. However, this is not the only way for an ideal solution to form, as long as the sum of the intermolecular forces between the two types of molecules is equal to twice the new interaction force. So, the volatilities of the two liquids may not necessarily be the same in this case.
  • #1
Suppose ##A## and ##B## are two liquids, and intermolecular forces between molecules of ##A## are of magnitude ##f_{AA}## and between molecules of ##B##, ##f_{BB}##. If ##f_{AA}>f_{BB}##, then the pure liquid ##A## is volatile than ##BB##, i.e., the tendencies that the molecules have to leave liquid phase to the gas phase depends on the intermolecular forces between them.

Say we mix ##A## and ##B## to make a solution. From my understanding, such a solution is said to be ideal, if the tendencies that the molecules of ##A## and ##B## have to leave the solution to the gas phase remain unchanged upon mixing.

Now certainly, once the liquids are mixed, molecules of ##A## will interact with those of ##B##, and new intermolecular interactions will form between them; let us say of magnitude ##f_{AB}##.

If the tendency of molecules of ##A## to leave the liquid must remain unchanged, the new interactions it forms with molecules of ##B## must be of the same magnitude as ##f_{AA}##, so ##f_{AA}=f_{AB}##, or the tendency should change. Similarly, for the tendency of molecules of ##B## to leave the solution to not change, ##f_{BB}=f_{AB}##.

But then ##f_{AA}=f_{BB}##, so the liquids must have the same volatilities. Is this so? Have I misunderstood anything?
 
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  • #2
SilverSoldier said:
If the tendency of molecules of ##A## to leave the liquid must remain unchanged, the new interactions it forms with molecules of ##B## must be of the same magnitude as ##f_{AA}##, so ##f_{AA}=f_{AB}##, or the tendency should change. Similarly, for the tendency of molecules of ##B## to leave the solution to not change, ##f_{BB}=f_{AB}##.

But then ##f_{AA}=f_{BB}##, so the liquids must have the same volatilities. Is this so? Have I misunderstood anything?
Yes, you misunderstood something
The situation ##f_{AA}=f_{BB}=f_{AB}## is one option to have ideal solution, but not the only one.
Consider four molecules - two A and two B molecules.
When the pairs change, they form two pairs of AB.
Therefore, it is sufficient that
##f_{AA}+f_{BB}=2*f_{AB}##
 

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