Vaporization of Mixture at constant temperature

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SUMMARY

The discussion centers on the vaporization of a mixture at constant temperature, specifically addressing the challenges of completely vaporizing a mixture with components that have significantly different boiling points. It is established that while a mixture can reach its boiling point, the component with the higher boiling point may remain in the liquid state, necessitating an increase in temperature to fully vaporize the mixture. The conversation highlights the relationship between boiling point, vapor-liquid equilibrium, and the enthalpy of vaporization (ΔH(vap)), emphasizing that ΔH(vap) values are typically defined at specific temperatures and pressures, particularly at boiling points.

PREREQUISITES
  • Understanding of vapor-liquid equilibrium concepts
  • Familiarity with enthalpy of vaporization (ΔH(vap))
  • Knowledge of boiling point determination in mixtures
  • Experience with heat exchanger processes
NEXT STEPS
  • Research the principles of vapor-liquid equilibrium curves
  • Study the calculation and implications of ΔH(vap) for mixtures
  • Explore the design and operation of heat exchangers in phase change processes
  • Investigate methods for determining boiling points in multi-component mixtures
USEFUL FOR

Chemical engineers, process engineers, and anyone involved in thermodynamic analysis of mixtures and phase transitions will benefit from this discussion.

johnny_b_good
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How is it possible to completely vaporize a mixture at constant temperature? I have seen listed values for ΔH(Vaporization); however, this doesn't quite make sense to me. If a mixture (with two largely diffent boiling points) is at its boiling point in the liquid state and I want to completely vaporize the mixture, won't the final vapor mixture be at a higher temperature than the mixture boiling point?

This is my rationale: boiling is obtained when the combined vapor pressures reach atmospheric pressure. However, following a vapor-liquid equalibrium curve, it becomes apparent that the compenent with the higher boiling point will be left in the liquid state (regardless of how little is left). Therefore, to completely evaporate the mixture, doesn't the temperature need to be increased to evaporate the last bit of high boiling component??

Thanks for the help
 
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You are right, boiling temperature is a function of composition. But I don't see how it is related to ΔHVaporization in the context of your question, please elaborate.
 
For a mixture to vaporize (say at atmospheric pressure) it must be at its boiling point. ΔH(vap) is specified for a specific temp and pressure of a mixture, typically at its boiling point. However, what I am wondering is how it is possible to completely vaporize a mixture at a constant temperature (as is implied by the ΔH(vap) values)? Or is there just an assumed small error in calculations involving ΔH(vap) values?

I'll give an example: I have a process that states that a mixture enters a heat exchanger at its boiling point, while in the liquid state. The liquid exchanges heat with steam and exits the heat exchanger at its boiling point, while in the gaseous state. How is it possible for this process to occur at one temperature??
 
I am not sure I understand what the problem is, but here is my take. As long as the composition is constant, boiling point is constant as well. Does all mixture vaporize, or is it just losing part of the most volatile components? If so, temperature difference can be relatively small, just several degrees C, and the "boiling point" can mean "something close enough to the boiling point".
 

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