Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 17 - Exercises and Problems - Page 313: 19

Answer

Please see the work below.

Work Step by Step

We know that $PV=nRT$ This can be rearranged as: $P=\frac{nRT}{V}$ We plug in the known values to obtain: $P=\frac{(3.5)(8.314)(273.15-150)}{0.002}$ $P=1.8\times 10^6Pa$
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