Vapor-Liquid Equilibrium Calculations

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SUMMARY

The discussion focuses on vapor-liquid equilibrium calculations, specifically applying Raoult's law to a system containing 3 moles of methanol and 7 moles of water at 25 degrees Celsius. The problem involves determining the amount of methanol remaining in the liquid phase when the gas phase reaches 40 mol% methanol at 88 degrees Celsius. Through mass balance equations, it is established that 1/3 mole of methanol remains in the liquid phase after the temperature increase.

PREREQUISITES
  • Understanding of Raoult's law and its application in vapor-liquid equilibrium.
  • Familiarity with mass balance equations in chemical systems.
  • Knowledge of Txy diagrams for phase equilibrium analysis.
  • Basic thermodynamics, particularly relating to phase changes and temperature effects.
NEXT STEPS
  • Study Raoult's law and its implications in vapor-liquid equilibrium calculations.
  • Learn to interpret Txy diagrams for different binary mixtures.
  • Explore advanced mass balance techniques in chemical engineering.
  • Investigate the effects of temperature on phase compositions in vapor-liquid systems.
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Chemical engineers, students studying thermodynamics, and professionals involved in process design and optimization will benefit from this discussion on vapor-liquid equilibrium calculations.

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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