vapour pressure above 2 separate beakers containing liquids which form ideal solution

Suppose we have A moles of liquid 1 in one beaker and B moles of liquid 2 in the other. Beakers are placed together in a biger closed container. Both have different vapour pressure when isloated. Now both components can't have their individual vapour pressures as it would disrupt the equilibrium of the other. It turns out the compensated vapour pressure would be the same as both liquids paced in one beaker. Why so? Is liquid from one beaker moving to the other and making the concentration equal in both?

• Well the title leads to a wrong impression for the mixture in one beaker. Ideal liquids in this case doesn't just mean that the individual liquids that are ideal, but rather that their mixture that ideal also.
– MaxW
Sep 6 '16 at 21:02
• Liquid might or might not be moving, but not quite for that reason. Say, you have a beaker of water and a beaker of mercury. Inside your container there would be some amount of water vapor and some amount of mercury vapor (and probably some air too, unless you've bothered to vacuumize the container beforehand, which is not important anyway). Just why and how would they disrupt each other? Sep 6 '16 at 21:02
• I just know they would disrupt each other, don't know much in detail how they do, although I would love to know that too. But my real question is that why is the vapour pressure same as a mixture of both liquids in the same beaker? Does having them in different beakers change anything? Sep 6 '16 at 21:16
• Define what is an ideal liquid. (This is a much less universal concept than an "ideal gas" or "ideal solution"). Chances are that your question is hinged on a flawed premise. Sep 6 '16 at 21:26
• Then the vapor pressure over two different beakers is not the same as that over a mixture. Sep 6 '16 at 21:35