A small amount of liquid ethanol (.26g, C2H5OH) is placed in a 4.7-L bottle and
ID: 783331 • Letter: A
Question
A small amount of liquid ethanol (.26g, C2H5OH) is placed in a 4.7-L bottle and tightly sealed. The bottle is placed in a refrigerator/freezer where the liquid reaches equilibrum, with its vapor at -11 degrees C.
A) What mass of ethanol is present in the vapor? (the vapor pressure of ethanol is 10.0 torr at -2.3 degrees C and 40.0 torr at 19 degrees C.)
B)When the container is removed and warmed to room temperature, 20 degrees C will all teh ethanol vaporze?
a)No, all the ethanol will not vaporize
b) Yes, all the ethanol will vaporize
Explanation / Answer
first. if we assume ethanol is ideal...
PV = nRT
and since n = mass / mw
PV = (mass / mw) RT
mass = mw x PV/(RT)
mw ethanol (C2H5OH) = 2x12.01 + 6x1.008 + 1x16.00 = 46.07 g/mole
P = vapor pressure of ethanol at -11 C
V = 4.7L - volume of liquid = 4.7L - less than 0.29mL = 4.7 L - 0.00029L = 4.7L... ie we assume volume of liquid is negligible
R = 0.08206 Latm/moleK
T = -11C = 262.15K
all we need is vapor pressure of ethanol at -11C
we know these two points
(P1,T1) = (10 torr, 270.85K)
(P2, T2) = (40 torr, 292.15K)
and we're looking for this one
(P3, T3) = (? torr, 262.15K)
if we use the clausius clapeyron equation
ln(P1/P2) = (dHvap/R) x (1/T2 - 1/T1)
(dHvap/R) = ln(P1/P2) / (1/T2 - 1/T1)
and if use it again to find (P3, T3).. and pick either point.. say (P1,T1)
ln(P3/P1) = (dHvap/R) x (1/T1 - 1/T3)
substituting in that (dHvap/R) from above...
ln(P3/P1) = [ln(P1/P2) / (1/T2 - 1/T1)] x (1/T1 - 1/T3)
rearranging..
ln(P3/P1) = ln(P1/P2) x [(1/T1 - 1/T3) / (1/T2 - 1/T1)]
taking exp of both sides... and noting exp(AxB) = exp(a) + exp(b).. and exp(ln(A)) = A
P3 / P1 = (P1/P2) + exp[(1/T1 - 1/T3) / (1/T2 - 1/T1)]
one last rearrangement
P3 = P1 x (P1/P2) + exp[(1/T1 - 1/T3) / (1/T2 - 1/T1)]
solving..
P3 = 10torr x (10torr / 40torr) + exp[ (1/270.85K - 1/262.15K) / (1/292.15K - 1/270.85K)]
P3 = 4.08 torr
and back to the ideal gas law to wrap this up
mass vapor = mw x PV/(RT)
mass vapor = (46.07 g/mole) x (4.08 torr x 1atm/760torr) x (4.7L) / ((0.08206 Latm/moleK) x (262.15K))
mass vapor = 0.054g
and if you're interested..
mass remaining in liquid phase = .26g - 0.054g = 0.206g