# Question 5b292

Sep 16, 2016

The water will boil at 119 °C.

#### Explanation:

Chemists often use the Clausius-Clapeyron equation to estimate the vapour pressures and boiling points 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 = enthalpy of vaporization of the liquid

$R$ = Universal Gas Constant

${P}_{2} = \text{1.90 atm}$; ${P}_{1} = \text{1 atm}$

Δ_"vap"H = "40.66 kJ/mol" at 100 °C

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

${T}_{2} = \text{?}$; ${T}_{1} = \text{100.00 °C" = "373.15 K}$

ln((1.90 color(red)(cancel(color(black)("atm"))))/(1 color(red)(cancel(color(black)("atm"))))) = (40.66 color(red)(cancel(color(black)("kJ·mol"^"-1"))))/("0.008 314" color(red)(cancel(color(black)("kJ")))·"K"^"-1"color(red)(cancel(color(black)("mol"^"-1"))))(1/("373.15 K") - 1/T_2)#

$0.642 = 13.11 - \frac{4891 \text{K}}{T} _ 2$

$\frac{\text{-12.46" = "-4891 K}}{T} _ 2$

${T}_{2} = \text{4891 K"/12.46 = "392 K" = "119 °C}$