If we assume H vap is independent of temperaturethe following equation can be de
ID: 686916 • Letter: I
Question
If we assume Hvap is independent of temperaturethe following equation can be derived from the Clausius-ClapeyronEquation:ln(p2/p1) = -(Hvap/R) [(1/T2)- (1/T1) ]
where p1 is the vapor pressure of the liquidT1 and p2 is the vapor pressure of the liquidat T2.
a) Using the above equation and the data given below to estimatethe value for the vapor pressure of water (H2O) at T =50.0oC. Give your final answer in the units of torr.
Hovap(H2O) = 40656.J/mol
To(H2O) = 100.00oC
Cp,m(H2O) = -41.71 J/mol x K
Explanation / Answer
ln(p2/p1) =-(Hvap/R) [ (1/T2)- (1/T1)]At boiling temperature of water, T2 = 373.2K , vapor pressureis (P2) 1atm = 760torr At 500C = (50+273.2)K = 323.2K , vapor pressure, P1= ? Given vapH = 40565J/mol ln(760torr/p1) = -(40565Jmol^-1/8.314Jmol^-1K^-1) [ (1/ 373.2K) - (1/ 323.2K) ]
ln(760torr/p1) = -(40565Jmol^-1/8.314Jmol^-1K^-1) [-4.15x10^-4 K^-1] = 2.02 760torr/ P1 = e^2.02 = 7.54 760torr/ 7.54 = P1 P1 = 100.80torr Hence, at vapor pressure of water (H2O) at T =50.0oC is 100.80torr