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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 23 Chapter 8

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 23

23E:

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CH 111 Notes: Week 9 Chapter 6: Properties of Gas 1 atm: (1 atmosphere) the pressure of air at sea level on a “normal” day o 1 atm = 760 mmHg (millimeters mercury) o Table 6.2 in textbook has list of the different units and conversions for pressure Ideal Gas Law: PV = nRT *memorize* o P = pressure (atm or sometimes bar) o V = volume (L) o n = moles of gas (mol) o R = gas constant (usually atm*L/mol*K)  R = 0.08206 atm*L/mol*K  (R constant will be given to you, no need to memorize) o T = temperature (K)  Conditions for an Ideal Gas o The volume of gas particles is minimal relative to the volume of the gas sample o Gas particles have no attraction to each other o *We will always assume gases are ideal in this unit*  Laws Derived from Ideal Gas Law (don’t need to memorize) o Boyle’s Law: relationship between volume and pressure  Assumes n and T are constant  P is inversely proportional to V  P1V1= P 2 2 (units don’t matter as long as they are the same on both sides)  There is a gas sample with a volume of 2.0 L and a pressure of 4.0 atm, what will the new volume be if the pressure is increased to 8.0 atm (2.0 L)(4.0 atm) = V2(8.0 atm) V2= 1.0 L o Charles’s Law: relationship between volume

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Chapter 8, Problem 23 is Solved
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Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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