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Clausius Clapeyron

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devon10
#1
Dec13-09, 12:23 AM
P: 2
It has been years scene I have taken a math class and I need help to see if I am setting the equation up correctly and if I have solved the problem correctly.

Diethyl ether has ∆Hvap of 29.1kJ/mol and a vapor pressure of 0.703 atm at 25.0 C. What is its vapor pressure at 95.0 C?

I converted the temp to K and kJ to J

In P2/P1=delta H vaporization/R (1/T2- 1/T1)

rearranged to

P2/P1= antilog (delta H vap/R (1/T2- 1/T1))

Then muliply by P1
P2= P1 antilog (delta H vap/R (1/T2- 1/T1))

antilog is also exp so...

P2= P1 exp(delta H vap/R (1/T2- 1/T1))

P2= .703 atm exp (29100/8.3145 (1/368-1/298)

P2= 2.19 atm
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Borek
#2
Dec13-09, 04:30 AM
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P: 23,592
Looks OK. I have not checked the final result.

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