General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 5 - The Gaseous State - Questions and Problems - Page 220: 5.57


$8.022 atm$

Work Step by Step

We can calculate the number of moles for $O_2$ as $n=91.3\times \frac{1 mol\space O_2}{32.00g\space O_2}=2.853mol O_2$ Now according to ideal gas law $PV=nRT$ $\implies P=\frac{nRT}{V}$ We plug in the known values to obtain: $P=\frac{2.853\times 0.08206\times 294}{8.58}=8.022atm$
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