The enthalpy of vaporization for methanol is 35.2 kJ/mol. Methanol has a vapor pressure of 1 atm at 64.7 oC. Using the Clausius-Clapeyron equation, what is the vapor pressure for methanol at 34.1 oC? Give your answer in atmospheres, to the third decimal point.

Respuesta :

The vapor pressure is obtained as 3.7 atm.

What is the  Clausius-Clapeyron equation?

The Clausius-Clapeyron equation is the equation that we could use to obtain the vapor pressure of a solution at any temperature. Of course we would always have two different temperatures to work with.

Hence;

ln(P1/P2) = ΔHvap/R (1/T2 - 1/T1)

P1 = initial pressure

P2 = final pressure

ΔHvap =  enthalpy of vaporization

T2 = final temperature

T1 = initial temperature

ln(1/P2) = 35.2 * 10^3/8.314 (1/338 - 1/307)

ln(1/P2) = 4234(0.00295 - 0.00326)

ln(1/P2) = -1.31

1/P2 = e^-1.31

P2 = 1/e^-1.31

P2 = 3.7 atm

Learn more about Clausius-Clapeyron equation:https://brainly.com/question/13162576

#SPJ1