Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 10 - Section 10.3 - The Ideal Gas Equation - Checkpoint - Page 441: 10.3.1

Answer

a) 196 L

Work Step by Step

According to ideal gas law, pV = nRT Then, V = $\frac{nRT}{p} = \frac{(8.75 mol)(0.08206\frac{L.atm}{K.mol})(273.15K)}{1 atm}$ = 196 L
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