Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 18 - Exercises and Problems - Page 342: 43

Answer

a) 300 kPa b) 240 J

Work Step by Step

a) We know that pressure times volume is constant in this case. Thus, it follows: $(60)(5) = P_f (1) \\ P_f=300 \ kPa$ b) We know that the work done from A to B is the same as the sum of the works to go from A to C and from C to B. Going from C to B involves no volume change, so no work is done. Thus, we find the work involved in going from A to C to obtain: $W_{AC}=(60)(5-1)=\fbox{240 J}$ *Note, we have to go against the arrow in the diagram, for the work when going from C to A is the negative work when going from A to B.
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