Yahoo Answers: Answers and Comments for What is the normal boiling point of the substance? [Chemistry]
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From Alexis
enUS
Tue, 19 Nov 2019 14:33:35 +0000
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Yahoo Answers: Answers and Comments for What is the normal boiling point of the substance? [Chemistry]
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https://answers.yahoo.com/question/index?qid=20191119143335AAJCblY
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From Dr W: so you have two data points with T and vapor p...
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https://answers.yahoo.com/question/index?qid=20191119143335AAJCblY
Tue, 19 Nov 2019 16:40:14 +0000
so you have two data points with T and vapor pressure info. and you want T for a 3rd point. That should immediately lead you to the clausius clapeyron equation.
let's start there
.. ln(P1/P2) = dHvap/R) * (1/T2  1/T1)
rearranging
.. (dHvap/R) = ln(P1/P2) / (1/T2  1/T1)
now you want a third data point.. .so let's use point 1 and set up another CC
type equation to solve for T3
.. ln(P1/P3) = (dHvap/R) * (1/T3  1/T1)
rearranging
.. 1/T3 = ln(P1/P3) / (dHvap/R) + 1/T1
subbing in that previous equation for dHvap/R
.. 1/T3 = ln(P1/P3) / (ln(P1/P2) / (1/T21/T1)) + 1/T1
or better yet..
.. 1/T3 = ln(P1/P3) * (1/T2  1/T1) / ln(P1/P2) + 1/T1
now.. the normal mp gives use the P3 = 1.00atm
(that was the point of throwing in the normal melting point fyi)
and we can sub in that and the rest to solve this
.. .. . .. . .ln(0.107atm / 1.00atm) * (1/(273+548) + 1/(273+114)).. . ... .1
.. 1/T3 =               +  
.. .. .. . .. .. .. ... .. .. .. . . . .ln(0.107atm / 117.9atm) .. ... ... ... .. .. . (273+114)
.. 1/T3 = 0.002148 / K
.. T3 = 465K = 192°C

From hcbiochem: Use the ClausiusClapeyron equation and the t...
https://answers.yahoo.com/question/index?qid=20191119143335AAJCblY
https://answers.yahoo.com/question/index?qid=20191119143335AAJCblY
Tue, 19 Nov 2019 15:48:46 +0000
Use the ClausiusClapeyron equation and the temperatures and pressures that you have for the triple and critical points to calculate Delta Hvap. Then, using that value and one of the points, calculate the Temperature at 1 atm pressure:
ln (P2/P1) = (Delta Hvap / R) (1/T1 1/T2)
ln (117.9 / 0.107) = Delta Hvap / 8.314 J/molK (1/387 K  1/823 K)
7.0048= Delta Hvap (1.6565X10^4)
Delta Hvap = 4.23X10^4 J/mol
Then,
ln (117.9 / 1 ) = (4.23X10^4 / 8.314) (1/T1  1/823)
T1 = 464 K = 191 C