College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 15 - Problems - Page 567: 7

Answer

The total work done by the gas during these two processes is $~202~J$

Work Step by Step

The work done by the gas during a constant pressure process is $P\Delta V$. We can find the work done by the gas from A to B: $W = P~\Delta V$ $W = (2~atm)(2~L-1~L)$ $W = (2~atm)(1~L)$ $W = (2~atm)(1.01\times 10^5~Pa/atm)(1~L)(10^{-3}~m^3/L)$ $W = 202~J$ When $\Delta V = 0$, the work done is zero. Therefore, the work done from B to C is zero. The total work done by the gas during these two processes is $~202~J$
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