Question

How do you calculate the vapor pressure of ethanol?

Answer 1

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

`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"`