Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 18 - Exercises and Problems - Page 332: 54


$3.4KJ$ and the work is done by gas.

Work Step by Step

We know that $W_{AB}=-nRT\ln\frac{V_B}{V_A}=-P_aV_a\ln\frac{V_B}{V_A}$ $W_{AB}=-50Kpa\times 25L\times \ln(\frac{25}{\frac{25}{3}})=-1373J$ We also know that $P_CV_C^{\gamma}=P_AV_A^{\gamma}$ This can be rearranged as: $P_C=P_A(\frac{V_A}{V_C})^{\gamma}$ $P_C=50KPa(\frac{25}{\frac{25}{3}})^{1.67}=313KPa$ $W_{CA}=\frac{50KPa\times 25L-313KPa\times \frac{25L}{3}}{1.67-1}$ $W_{cycle}=W_{AB}+W_{CA}$ We plug in the known values to obtian: $W_{cycle}=-1373+(-2027)=-3.4KJ$ The negative sign shows that work is done by gas.
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