Chemistry: The Molecular Nature of Matter and Change 7th Edition

Published by McGraw-Hill Education
ISBN 10: 007351117X
ISBN 13: 978-0-07351-117-7

Chapter 5 - Problems - Page 247: 5.113


See explanation below.

Work Step by Step

First, let's convert all the pressures to the same units: (0.456 atm) (760 torr/atm) = 346.6 torr (25 kPa) (7.5 torr/kPa) = 187.5 torr To solve the problem, we use both Boyle's Law and Dalton's Law. When the stopcocks are opened, the volume of the gases are increased, and their pressures are decreased. (Boyle's Law). However, these individual partial pressures sum to the total pressure in the system (Dalton's Law). It does not matter if the gases are the same or different, we can still handle them as partial pressures. The total volume of the system after the stopcocks are opened is: 1.0 L + 1.5 L + 2.0 L = 4.5 L After the stopcocks are opened, the partial pressure of the gas from the first container is found by Boyle's law: P1 V1 = P2 V2; (185 torr) (1.0 L) = P (4.5 L); P = 41.1 torr Similarly, for the gas in the second container: (346.6 torr) (1.5 L) = P (4.5 L); P = 115.5 torr For the gas in the third container: (187.5 torr) (2.0 L) = P (4.5 L); P = 83.4 torr So the final total pressure is (by Dalton's Law): 41.1 torr + 115.5 torr + 83.4 torr = 240.0 torr
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