## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 15 - Problems - Page 567: 8

#### Answer

The total work done by the gas during these two processes is $~101~J$

#### Work Step by Step

When $\Delta V = 0$, the work done is zero. Therefore, the work done from A to B is zero. 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 B to C: $W = P~\Delta V$ $W = (1~atm)(2~L-1~L)$ $W = (1~atm)(1~L)$ $W = (1~atm)(1.01\times 10^5~Pa/atm)(1~L)(10^{-3}~m^3/L)$ $W = 101~J$ The total work done by the gas during these two processes is $~101~J$

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