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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: In Sample Exercise 10.16 in the textbook, we found that one mole of Cl2 confined to 22.41 L at 0 oC deviated slightly from ideal behavior. Calculate the pressure exerted by 1.00 mol  Cl2 confined to a

Problem

In Sample Exercise 10.16 in the textbook, we found that one mole of Cl2 confined to 22.41 L at 0 oC deviated slightly from ideal behavior. Calculate the pressure exerted by 1.00 mol  Cl2 confined to a smaller volume, 6.00 L , at 25 oC.

Why is the difference between the result for an ideal gas and that calculated using van der Waals equation greater when the gas is confined to 6.00 L compared to 22.4 L?

Solution

Part A. We’re being asked to calculate the pressure exerted by a chlorine gas that deviates from ideal behavior. For this, we shall use the Van der Waal’s equation


The Van der Waals equation is shown below:

P+an2V2V-nb=nRT

P = pressure, atm
V = volume, L
n = # of moles, mol
R = gas constant = 0.08206 (Latm)/(molK)
T = temperature, K
a = polarity coefficient
= size coefficient

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