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: 56


a) $211\ J$ b) $12.94 \ L $ c) The answer is below.

Work Step by Step

a) First, we need to find its volume after the adiabatic process: $P_BV_B^{\gamma}=P_AV_A^{\gamma}$ $(150)V_B^{1.67}=(50)(25)^{1.67}\\ V_B=12.94 \ L$ Now, we can use the equation for the work done over an adiabatic process: $W=\frac{(150)(12.94)-(50)(25)}{1.67-1}=1033 \ J$ The next part of the process involves no work being done, for there is no volume change. Thus, we find the work involved in the final isothermal process. Substituting the ideal gas law into the equation for work done during an isothermal process, we find: $W=-P_fV_fln(\frac{V_f}{V_0})$ $W=-(50)(25)ln(\frac{25}{12.94})=-822\ J$ Thus, the net work done is: $211\ J$ b) At the end of all of the processes except the first one, the volume is 25 liters. Thus, we see that the minimum volume is: $12.94 \ L$. c) Start with point A of the right side of the PV diagram. Going from A to B involves a pressure increase and volume decrease, so go up and to the left, making the line between the two points look like an exponential decay function. Next, since the volume remains constant during the next process, draw a line that is straight down to point C. Finally, connect point C and point A.
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