[CHM 112] Question

[Andrew J. Pounds]

On 02/02/14 08:46,  Towe wrote:
>
> Hello Dr. Pounds,
>
> I had two questions. 1. Could you explain how you got your answer on 
> number 1 part B on quiz 3?
>

Avogadro's law (p 426 of your text) states that, for gases, volumes are 
proportional to the moles of gas.  In the problem on the quiz the 
percent by volume was give for each gas.  That then, because of 
Avogadro's law, translates to a mole percentage.  On the quiz I 
converted the percentage to a fraction and multipled by the total 
pressure to come up with the partial pressure of oxygen.


> 2. Could you explain how I enter in the log problem on number 14 B in 
> the additional problems for chapter 11?
>
> Thanks,
>

$\log P = 7.3605 - \frac{1617.9}{t+240.17}$

Let's rearrange for t

$\log P - 7.3605 = - \frac{1617.9}{t+240.17}$

$\frac{log P - 7.3605}{-1617.9} = \frac{1}{t+240.17}$

$\frac{-1617.9}{log P - 7.3605} = t+240.17$

$\frac{-1617.9}{log P - 7.3605} - 240.17 = t$

Since it is the normal boiling point, the pressure is 1 atm -- but this 
problem wants the pressure expressed in
mmHg -- so it is 760 mmHg.

$\frac{-1617.9}{\log(760) - 7.3605} - 240.17 = t = 121$

So, since 121 is the temperature at which the vapor pressure is 760 
mmHg, it is the boiling point.


-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

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