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

Nov 30, 2016

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

${P}_{1} = \text{100 mmHg}$; ${T}_{1} = \text{34.9 °C" = "308.05 K}$

${P}_{2} = \text{?}$; $\textcolor{w h i t e}{m m m m m} {T}_{2} = \text{55.5 °C" = "328.65 K}$

Δ_"vap"H = "42.3 kJ/mol"

$R = \text{0.008 314 kJ·K"^"-1""mol"^"-1}$

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

${P}_{2} / \left(\text{100 mmHg}\right) = {e}^{1.035} = 2.816$

${P}_{2} = \text{2.816 × 100 mmHg = 282 mmHg}$

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