
#1
Dec1309, 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 (1/3681/298) P2= 2.19 atm 


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