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: A certain substance has a heat of vaporization of 47.70 kJ/mol. At what kelvin temperature will the vapor pressure be 3.50 times higher than it was at 293K?

Solution: A certain substance has a heat of vaporization of 47.70 kJ/mol. At what kelvin temperature will the vapor pressure be 3.50 times higher than it was at 293K?

Problem

A certain substance has a heat of vaporization of 47.70 kJ/mol. At what kelvin temperature will the vapor pressure be 3.50 times higher than it was at 293K?

Solution

We’re being asked to determine the temperature at which the vapor pressure is 3.50 times higher than it was at 293 K, given the heat of vaporization


For this problem, we can use the Clausius-Clapeyron Equation:



where: 

P1 = vapor pressure at T1

P2 = vapor pressure at T2

ΔHvap = heat of vaporization (in J/mol)

R = gas constant (8.314 J/mol•K)

T1 and T2 = temperature (in K).


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