Re: question on solution thermodynamics
- From: "A.H." <steamdoc@xxxxxxx>
- Date: 18 Sep 2006 09:21:15 -0700
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?
To a first approximation, you would look at this question by examining
the vapor-pressure curves of the 2 components. The partial pressure of
each component will go down with temperature approximately in
proportion to its pure-component vapor pressure.
So, if you go down to a certain T and the vapor pressure of each drops
by the same proportion, then the equilibrium vapor composition will
stay the same (ignoring the temperature dependence of nonideality in
the solution, which might not be negligible for this system). But for
example if a drop in T reduced water's vapor pressure by a factor of 4
and ethanol's by a factor of only 2, then the mole fraction of water in
the equilbirium vapor would be reduced.
.
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