Question

The vapor pressure of ethanol is 100 mmHg at 34.9°C. What is its vapor pressure at 55.5°C?

Answer 1

The vapour pressure at 55.5 °C is 282 mmHg.

Explanation:

Chemists often use the Clausius-Clapeyron equation to estimate the vapour pressures of pure liquids:

`color(blue)(bar(ul(|color(white)(a/a) ln(P_2/P_1) = (Δ_"vap"H)/R(1/T_1- 1/T_2)color(white)(a/a)|)))" "`

where

`P_1` and `P_2` are the vapour pressures at temperatures `T_1` and `T_2`

`Δ_"vap"H` = the enthalpy of vaporization of the liquid

`R` = the Universal Gas Constant

In your problem,

`P_1 = "100 mmHg"`; `T_1 = "34.9 °C" = "308.05 K"`

`P_2 = "?"`; `color(white)(mmmmm)T_2 = "55.5 °C" = "328.65 K"`

`Δ_"vap"H = "42.3 kJ/mol"`

`R = "0.008 314 kJ·K"^"-1""mol"^"-1"`

`ln(P_2/("100 mmHg")) = (42.3 color(red)(cancel(color(black)("kJ·mol"^"-1"))))/("0.008 314" color(red)(cancel(color(black)("kJ·K"^"-1""mol"^"-1"))))(1/(308.05 color(red)(cancel(color(black)("K")))) - 1/(328.65 color(red)(cancel(color(black)("K")))))`

`ln(P_2/("100 mmHg")) = 5088 × 2.035 × 10^"-4" = 1.035`

`P_2/("100 mmHg") = e^1.035 = 2.816`

`P_2 = "2.816 × 100 mmHg = 282 mmHg"`

The on-line calculator here predicts a value of 286 mmHg. Close enough!