Re: question on solution thermodynamics

rekuci@xxxxxxxxx wrote:
mrdarrett@xxxxxxxxx wrote:
If I have a vessel filled with a liquid: 0.019 mole fraction ethanol
and 0.981 mole fraction water (1.9 mole % ethanol, 98.1 mole % water),
thermodynamic tables tell me that at 95.5 degrees C, the vapor will be
17 mole % ethanol, 83 mole % water.

But what about if the temperature is not 95.5 degrees C?

Between room temperature (20 C, for instance) and 95.5 C, will the
vapor in equilibrium with the previously mentioned liquid still have 17
mole % ethanol?

No, it definitely won't be the same. You'd expect that the percentage
of ethanol in the vapor will increase with decreasing temperature, due
to its higher volatility.

If you need quantities, the temperature-composition phase diagram
(called the "T-x" diagram) for ethanol should be readily available in
chemical reference works in your local university library, or here, but
can't guarantee correctness:
http://www.physics.rutgers.edu/ugrad/351/Lecture%2016.ppt#267,16,Water-Ethanol
Mixture


I have to study how those T-x-y diagrams are made... maybe that would
give me better insight. It's been a long time since I've actually had
to use those diagrams...

I'm especially interested in the vapor compositions for temperatures
between room temperature and the boiling point of the ethanol...
between 30C and 78C. It was within this region that I was wondering
what the vapor composition would be, given a liquid with a known
mixture of EtOH and H2O... would it be a simple matter of calculating
the partial pressures of the two components...?

Thanks,

Michael

.

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