How do you calculate the vapor pressure of ethanol?

Jul 1, 2014

You use the Clausius-Clapeyron equation.

Explanation:

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

$\ln P = \text{constant" – (ΔH_"vap")/"RT}$

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

${T}_{1} = \text{(50.0+ 273.15) K = 323.15 K}$; ${P}_{1} = \text{?}$
${T}_{2} = \text{(78.4 + 273.15) K = 351.55 K}$; ${P}_{2} = \text{760 Torr}$

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

$\frac{\text{760 Torr}}{P} _ 1 = {e}^{1.159} = 3.188$

${P}_{1}$ = ("760 Torr")/3.188 = "238.3 Torr"