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Q & A: Baloney with solutions

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Most recent answer: 03/19/2011
Q:
http://answers.yahoo.com/question/index;_ylt=ArdCP.z_O3u.s2GU.ZQ5iqXIDH1G;_ylv=3?qid=20101227084750AAe9Owt
Yahoo's physics Q&A site is quite good for a very open site, but as might be expected it's not as reliable as one would like. Here's an example of an error, although not an extreme error. Question: If you have a solution of 0.52A+0.48B (by mole fraction), and the vapor pressures of A and B are 114.9mmHg & 238.3 mmHg respectively, what's the total vapor pressure?
- Mike W (age 61)
A:

The answer given was "0,52*114,9 + 0,48*238,3 = 174,1 mmHg" (the commas were Euro-style). This assume a completely ideal solution of the components A and B. Solutions are rarely ideal to nearly the precision given in the answer. The reason is, to put it in crude terms, that A and B don't stick to each other exactly as well as each sticks to itself. As a result, the vapor pressure of the solution is almost always noticeably higher or lower than the simple weighted average.

There's a nice discussion of the rule that they used and the deviations from it on http://en.wikipedia.org/wiki/Raoult%27s_law. 

One very familiar example of a solution with strong deviations from the simple rule is the water-ethanol combination, described here: :http://www.chemguide.co.uk/physical/phaseeqia/nonideal.html.

I'm being a little picky here. The reason is that in thermodynamics there are some amazing exact relations, so it's important not to get various rules-of-thumb mixed up with the exact laws. One of the surprising exact laws in this specific area (vapor pressure of solutions) is discussed on another of our answers.http://van.physics.illinois.edu/qa/listing.php?id=1470

Mike W.


(published on 01/19/2011)

Follow-Up #1: comma vs.period

Q:
"The answer given was "0,52*114,9 + 0,48*238,3 = 174,1 mmHg" (the commas were British-style)." The British don't use a comma. http://en.wikipedia.org/wiki/Decimal_mark
- Dan
UK
A:
fixed, thanks

(published on 03/19/2011)

Follow-up on this answer.