Ch.11 - Liquids, Solids & Intermolecular ForcesWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: The enthalpy of vaporization for acetone is 32.0 kJ/mol. The normal boiling point for acetone is 56.5 ˚C. What is the vapor pressure of acetone at 23.5 ˚C?

Solution: The enthalpy of vaporization for acetone is 32.0 kJ/mol. The normal boiling point for acetone is 56.5 ˚C. What is the vapor pressure of acetone at 23.5 ˚C?

Problem

The enthalpy of vaporization for acetone is 32.0 kJ/mol. The normal boiling point for acetone is 56.5 ˚C. What is the vapor pressure of acetone at 23.5 ˚C?

Solution

We need an equation that relates vapor pressure, temperature and enthalpy of vaporization. We're going to use the Clausius-Clapeyron Equation to solve this problem:

Given: 

Normal boiling point of acetone (T1) = 56.5°C + 273.15 = 329.65 K
Normal pressure (P1) = 1 atm
T2 = 23.5°C + 273.15 = 296.65 K
 vapor pressure (P2) = ?
ΔHvap = 32.0 kJ/mol

Substitute values and calculate for P2:

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