Vaporization of Mixture at constant temperature

[johnny_b_good]

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

 

[Borek]

You are right, boiling temperature is a function of composition. But I don't see how it is related to ΔH_Vaporization in the context of your question, please elaborate.

 

[johnny_b_good]

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??

 

[Borek]

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".

 

[LudusRex]

I can provide some insight into the process of vaporization of a mixture at constant temperature. First, let's define vaporization as the process of converting a liquid into a gas or vapor. This process typically occurs when the temperature of a liquid reaches its boiling point, at which point the vapor pressure of the liquid is equal to the atmospheric pressure.

In the case of a mixture with two largely different boiling points, the process of vaporization can be a bit more complex. This is because the components of the mixture will have different vapor pressures at the same temperature. The component with the lower boiling point will have a higher vapor pressure and will therefore vaporize more readily, while the component with the higher boiling point will have a lower vapor pressure and may remain in the liquid state.

However, at constant temperature, the total vapor pressure of the mixture will still be equal to the atmospheric pressure. This means that even if one component is vaporizing more readily, the total amount of vapor produced will remain constant. As a result, the final vapor mixture will still be at the same temperature as the boiling point of the mixture, even if one component remains in the liquid state.

The value of ΔH(Vaporization) represents the energy required to completely vaporize a certain amount of a substance at its boiling point. This does not necessarily mean that all components of a mixture will vaporize at the same rate, but rather that the total amount of vapor produced will be equivalent to the energy required for complete vaporization.

In summary, it is possible to completely vaporize a mixture at constant temperature by maintaining a balance between the vapor pressures of the different components. The final vapor mixture will still be at the same temperature as the boiling point of the mixture, even if one component remains in the liquid state. I hope this helps clarify any confusion.