[chrism008]

A 4.7 L sealed bottle containing 0.33 g of ethanol is cooled to -11C where it reaches equilibrium with its vapor. Given the vapor pressure of ethanol at -2.3C is 10. torr and at 19C is 40 torr, what is the mass of ethanol vapor?

I began by finding the heat of vaporization of ethanol using the Clapeyron equation and  the two given sets of temperature and pressure which I calculated to be 42.82 KJ/mol. I then rearranged the Clapyeron equation again to calculate the vapor pressure at -11C, which was 5.32 torr. Using the Ideal Gas Law I calculated the number of moles of ethanol in the vapor to be .0015295 mol. Multiplying the number of moles of ethanol by the molar mass of ethanol I got 0.071 g ethanol in the vapor pressure. Unfortunately the answer in the back of the book states that there should have been 0.088 g of ethanol in the vapor. The difference between my answer and the textbook's seemed to be too large to have anything to do with rounding.

[chrism008]

P1 = 10.0 torr                   P2 = 40.0 torr
T1 = -2.3C = 270.85K        T2 = 19.0C = 292.15K

Calculating heat of vaporization from the Clapeyron equation
H is the heat of vaporization and R = 8.314 J/mol K

H = -Rln(P2/P1)(1/T2-1/T1)^-1 = 4.282x10^4 J/mol

Now that I have the heat of vaporization I use the Clapeyron equation again but this time, using P1 and T1 from above and T2 = 262.15.15K to calculate P2

P2 = P1e[(-H/R)(1/T2-1/T1)] = 5.3204 torr

Using the Ideal Gas Law to calculate the number of moles of ethanol

P = 5.3204 = .007 atm R = .08206 (atm L) / (mol K)
V = 4.7 L     T = 262.15 K
n = PV/RT = .0015295 mol ethanol

Calculating the mass of the ethanol
Molar Mass of Ethanol = 46.07 g/mol
mass = (.0015295 mol ethanol)(46.07 g/mol) = .070 g ethanol