physical chemistry - Vapor Pressure Paradox

This is quite an interesting situation I had thought of. Consider two beakers, each containing $\pu{180 g}$ of water. Also, suppose one of the two has $\pu{1.8 g}$ of glucose dissolved in it. Let's call the one without glucose as A, and the one with glucose as B. I keep a small bell-jar that is large enough to just cover both of the two vessels. Now I will try to measure the vapor pressure inside the jar. Here's a little diagram to show what I mean:

I do not have access to bell-jars or manometers so I'll try to use Raoult's law to find the vapor pressure inside the vessel. The law in a simplified way would be:

$$P_{\text{solution}}=P_{\text{solvent}}x_{\text{solvent}}$$

Where $x$ denotes mole fraction, and solute is considered non-volatile. Keeping vessel A in mind, I get the vapor pressure as 23.8 mmHg, while applying Raoult's law for vessel B, I get the vapor pressure as:

$$P=23.8\left(\frac{10}{10.01}\right)=23.7 \text{mmHg}$$

Now, both of the values can't be true at the same time, and that's what I believe is the paradox.

How can this paradox be resolved?

My thoughts are that perhaps water from vessel A condenses back onto B, thus slowly increasing water content in B, increasing the vapor pressure. However, since theres solute already in B, the vapor pressure can never reach the value provided by the pure water.

What are your thoughts? Does the paradox ever resolve or does it remain in an unbalanced situation?

Answer

Although "paradox" is not quite the right term, what you have discussed is actually a simple, yet interesting and important phenomenon.

Given the ideal situation as you have presented, your thoughts on what would happen are correct. If the system were to achieve $\pu{100\%}$ humidity with respect to the pure water, that would always be slightly over $\pu{100\%}$ humidity with respect to solution B, so water would condense onto B. This of course would cause the humidity with respect to pure water to drop, so more would evaporate. All things being ideal, eventually all of the water in A would transfer to the solution in B (plus the gas phase). The rate that would happen would depend strongly on the concentration of the solute in B.

An example of this situation playing out in the environment can be seen in the growth of water and ice particles in clouds. The vapor pressure of one particle can differ from that of another due to solute concentration, particle size (small droplets have a greater vapor pressure than larger droplets) and phase (supercooled water droplets have a greater vapor pressure than ice particles at the same temperature). Regardless of the cause of the vapor pressure difference, whether it's solute concentration or one of the other factors, the situation is the same as depicted in your question. And the result is the same; the particles having greater vapor pressure will end up evaporating and then condensing onto the particles having lower vapor pressure (of course we're talking about vapor pressure of water/ice only, not any solute). This is generally referred to as the Wegener–Bergeron–Findeisen process, although this technically only refers to water-to-ice vapor transfer.