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: 53

Answer

$930J$

Work Step by Step

We know that $P_C=P_B(\frac{V_B}{V_C})^{\gamma}$ $P_C=200(\frac{4}{1})^{1.4}=1393KPa$ Now $W_{BC}=\frac{P_CV_C-P_BV_B}{\gamma-1}$ We plug in the known values to obtain: $W_{BC}=\frac{1393\times 1-200\times 4}{1.4-1}=1.5KJ=1483J$ Now $W_{DA}=-nRTln(\frac{V_A}{V_D})$ We plug in the known values to obtain: $W_{DA}=-100\times 4\times \frac{4}{1}=-555J$ Thus, $W_{cycle}=W_{BC}+W_{DA}=1483-555=930J$
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