Two beakers with volatile solvents in closed system

AI Thread Summary
In a closed system with two beakers, one containing a volatile solvent and the other a 10% solution of a non-volatile solute, equilibrium results in all solvent moving to the beaker with the solution. This occurs because the vapor pressure of the solution in beaker B is lower than that of the pure solvent in beaker A, leading to a greater tendency for the solvent to evaporate from beaker A. The discussion addresses concerns about thermodynamics and entropy, clarifying that the system does not violate these principles. Instead, the movement of solvent aligns with the concept of vapor pressure lowering due to the presence of the solute. Understanding these principles helps clarify the behavior of volatile solvents in solutions.
black_squirrel
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Homework Statement


Beaker A contains a volatile solvent and beaker B contains 10% solution of a non-volatile solute in the same solvent. Both beakers are placed in a closed environment as shown. The question asks what will happen when the system has reached equilibrium. The answer is given that all the solvent will be in beaker B after equilibrium because beaker B will have lower vapor pressure than beaker A. I don't understand why this is true. Isnt this against thermodynamics? isn't entropy decreasing here if that was the case?

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black_squirrel said:

Homework Statement


Beaker A contains a volatile solvent and beaker B contains 10% solution of a non-volatile solute in the same solvent. Both beakers are placed in a closed environment as shown. The question asks what will happen when the system has reached equilibrium. The answer is given that all the solvent will be in beaker B after equilibrium because beaker B will have lower vapor pressure than beaker A. I don't understand why this is true. Isnt this against thermodynamics? isn't entropy decreasing here if that was the case?

http://pics-hosting.com/files/w4zv6fuoasz6vxth3y7.jpg
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Perhaps this link can help you understand the vapor pressure lowering aspect of the solute on the solvent?
http://en.wikipedia.org/wiki/Raoult's_law

Given that the vapor pressure would be lowered with the solute, are you on board with understanding that the liquid on the left would have a greater tendency toward evaporation than the one on the right?

In which case what do you think happens when the pressure of a gas at the surface of a liquid is greater than the vapor pressure of the liquid itself at that temperature and pressure?

For further reading about the colligative properties of solutions like boiling points, and vapor pressure and osmosis see also:
http://en.wikipedia.org/wiki/Colligative_properties
 
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