Two beakers with volatile solvents in closed system

[black_squirrel]

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

 

[black_squirrel]

anyone?

 

[LowlyPion]

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