Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 5 - Sections 5.1-5.10 - Exercises - Problems by Topic - Page 239: 40

Answer

$P=1.69 atm$

Work Step by Step

We know that $PV=nRT$...........eq(1) But we also know that $n=\frac{m}{M}$ Where $m$ and $M$ are given mass and molar mass respectively Thus eq(1) becomes $PV=\frac{mRT}{M}$ This can be rearranged as $P=\frac{mRT}{VM}$ We plug in the known values to obtain: $P=\frac{32.7\times 0.08206\times 302}{15.0\times 31.9988}=1.69 atm$
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