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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: Calculate the pressure exerted by 1.00 mol of Ne in a box that is 0.300 L and 298 K. For Ne, a = 0.211 L2 atm/mol2 and b = 0.0171 L/mol.

Problem

Calculate the pressure exerted by 1.00 mol of Ne in a box that is 0.300 L and 298 K. For Ne, a = 0.211 L2 atm/mol2 and b = 0.0171 L/mol.

Solution

We’re being asked to calculate the pressure exerted by a Ne in a box using 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


Let’s first isolate the pressure in the Van der Waals Equation:

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