Question

The vapor pressure of a liquid doubles when the temperature is raised from 78C to 86C. At what temperature will the vapor pressure be four times the value at 78C?

Answer 1

via the clausius clapeyron equation
ln(P1/P2) = (dHvap/R) x (1/T2 - 1/T1)

where..
P1 = P1
P2 = 2 x P1
dHvap = heavt of vaporization
R = gas constant = 8.314 J/molK
T2 = 86+273.15 = 359.15K
T1 = 78+273.15 = 351.15K

rearrange the equation for dHvap
dHvap = R x ln(P1/P2) / (1/T2 - 1/T1)

plug and chug.. .
btw.. notice ln(P1/P2) = ln( P1/(2P1)) = ln(1/2) = ln(1) - ln(2) = 0 - ln(2) = -ln(2)

dHvap= R *(-ln2) /(1/359.15 -1/351.15)

SO,

now

using the same equation again

ln(P1/P2) = (dHvap/R) x (1/T2 - 1/T1)

ln(P1/4*P1) =   (-ln2) /(1/359.15 -1/351.15) x (1/T2 - 1/(273.15+78))

-2ln2=   (-ln2) *(1/359.15 -1/351.15) x (1/T2 - 1/(273.15+78))

2= 1/(1/359.15 -1/351.15) x (1/T2 - 1/(351.15))

-16/351.15*359.15 = (351.15 -T2 )/ 351.15*T2

-16/359.15= (351.15 -T2 )/ *T2

-16T2 = 359.15*351.15 - 359.15T2

T2 = 367.523

so,final temperature = 94.37C