How do you calculate the vapor pressure of ethanol?

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    You use the Clausius-Clapeyron equation.
    Experiments show that the vapour pressure ##P## of vaporization ##H_vap## and temperature ##T## are related by the equation
    ##lnP = constant (H_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)) = (H_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 = 351.15 K##; ##P_2 = 760 Torr##
    ##ln(P_2/P_1) = (H_vap)/R (1/T_1 1/T_2)##
    ##ln((760 Torr)/P_1) = ((38 560 color(red)(cancel(color(black)(Jmol^(-1)))))/(8.314 color(red)(cancel(color(black)(JK^(-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##

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