How do you calculate the vapor pressure of ethanol?

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How do you calculate the vapor pressure of ethanol?

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Answer 1 · 2014-07-01T20:52:01

You use the Clausius-Clapeyron equation.

Explanation:

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

$ln P = constant - \frac{Δ H_{vap}}{RT}$ $lnP = "constant" – (ΔH_"vap")/"RT"$

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

If $P_{1}$ $P_1$ and $P_{2}$ $P_2$ are the vapour pressures at two temperatures $T_{1}$ $T_1$ and $T_{2}$ $T_2$ , the equation takes the form:

$ln (\frac{P_{2}}{P_{1}}) = \frac{Δ H_{vap}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$ $ln(P_2/(P_1)) = (ΔH_"vap")/R(1/T_1 – 1/T_2)$

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

$T_{1} = (50.0+ 273.15) K = 323.15 K$ $T_1 = "(50.0+ 273.15) K = 323.15 K"$ ; $P_{1} = ?$ $P_1 = "?"$
$T_{2} = (78.4 + 273.15) K = 351.55 K$ $T_2 = "(78.4 + 273.15) K = 351.55 K"$ ; $P_{2} = 760 Torr$ $P_2 = "760 Torr"$

$ln (\frac{P_{2}}{P_{1}}) = \frac{Δ H_{vap}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$ $ln(P_2/P_1) = (ΔH_"vap")/R (1/T_1 – 1/T_2)$

$ln (\frac{760 Torr}{P_{1}}) = (\frac{38560 {J\cdotmol}^{- 1}}{8.314 {J\cdotK}^{- 1} {mol}^{- 1}}) (\frac{1}{323.15 K} - \frac{1}{351.55 K})$ $ln(("760 Torr")/P_1) = ((38 560 color(red)(cancel(color(black)("J·mol"^(-1)))))/(8.314 color(red)(cancel(color(black)("J·K"^(-1)"mol"^-1))))) (1/(323.15color(red)(cancel(color(black)("K")))) – 1/(351.55 color(red)(cancel(color(black)("K")))))$

$ln (\frac{760 Torr}{P_{1}}) = 4638 \times 2.500 \times 10^{- 4} = 1.159$ $ln(("760 Torr")/P_1) = 4638 × 2.500 × 10^(-4) = 1.159$

$\frac{760 Torr}{P_{1}} = e^{1.159} = 3.188$ $("760 Torr")/P_1 = e^1.159 = 3.188$

$P_{1}$ $P_1$ = $\frac{760 Torr}{3.188} = 238.3 Torr$ $("760 Torr")/3.188 = "238.3 Torr"$

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How do you calculate the vapor pressure of ethanol?

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