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: 21

Answer

Please see the work below.

Work Step by Step

(a) We know that $PV=nRT$ This can be rearranged as: $V=\frac{nRT}{P}$ We plug in the known values to obtain: $V=\frac{(2.0)(8.314)(250)}{(1.5)(1.01\times 10^5)}$ $V=0.027m^3$ (b) As $PV=nRT$ This can be rearranged as: $T=\frac{PV}{nR}$ We plug in the known values to obtain: $T=\frac{(4.0)(1.01\times 10^5)(\frac{0.027}{2})}{(2.0)(8.314)}$ $T=330K$
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