Vapor-Liquid Equilibrium Calculations

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Raoult's law and partial pressures are discussed in the context of a problem involving a mixture of methanol and water. The system initially contains 3 moles of methanol and 7 moles of water at 25 degrees Celsius. When the temperature is raised to 88 degrees Celsius, the gas phase reaches 40 mol% methanol, while the liquid phase contains 10% methanol. A mass balance approach is used to determine the remaining methanol in the liquid phase, resulting in approximately 0.33 moles of methanol left. This illustrates the application of Raoult's law in calculating phase compositions.
runningman19
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Hi Everyone,
I am having trouble grasping the concept of Raoult's law, partial pressures etc. One problem in specific is: A system contains 3 moles of methanol and 7 moles of water at 25 degrees celsius. The temperature is raised until there is 40 mol% methanol in the gas phase. How much is left in the liquid phase?
The Txy diagram is attached.
Thanks so much guys,
Nick
 

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This is strictly a mass balance problem. Let L represent the total number of moles in the liquid and W represent the number of moles in the vapor. So, $$L+V = 10$$

From the diagram, there is 40 mol% methanol in the gas phase ( y = 0.4) when the temperature is 88 C. And the mole percent in the liquid at this state is 10 % (x = 0.1). So, the mass balance on the methanol is $$L(0.1)+V(0.4)=3$$So, solving these equations simultaneously for L and V, we obtain:$$L=\frac{10}{3}$$and $$V=\frac{20}{3}$$So the amount of methanol remaining in the liquid is ##\frac{10}{3}(0.1)=\frac{1}{3}\ mole##.
 
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