How do you calculate the vapor pressure of ethanol?

Experiments show that the vapour pressure #P#, enthalpy of vaporization, #ΔH_"vap"#, and temperature #T# are related by the equation

#lnP = "constant" – (ΔH_"vap")/"RT"#

where #R# is the ideal gas constant. This equation is the Clausius- Clapeyron equation.

If #P_1# and #P_2# are the vapour pressures at two temperatures #T_1# and #T_2#, the equation takes the form:

#ln(P_2/(P_1)) = (ΔH_"vap")/R(1/T_1 – 1/T_2)#

The Clausius-Clapeyron equation allows us to estimate the vapour pressure at another temperature, if we know the enthalpy of vaporization and the vapor pressure at some temperature.

Example

Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 50.0 °C?

Solution

#T_1 = "(50.0+ 273.15) K = 323.15 K"#; #P_1 = "?"#
#T_2 = "(78.4 + 273.15) K = 351.55 K"#; #P_2 = "760 Torr"#

#ln(P_2/P_1) = (ΔH_"vap")/R (1/T_1 – 1/T_2)#

#ln(("760 Torr")/P_1) = ((38 560 color(red)(cancel(color(black)("J·mol"^(-1)))))/(8.314 color(red)(cancel(color(black)("J·K"^(-1)"mol"^-1))))) (1/(323.15color(red)(cancel(color(black)("K")))) – 1/(351.55 color(red)(cancel(color(black)("K")))))#

#ln(("760 Torr")/P_1) = 4638 × 2.500 × 10^(-4) = 1.159#

#("760 Torr")/P_1 = e^1.159 = 3.188#

#P_1# = #("760 Torr")/3.188 = "238.3 Torr"#