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 344: 75

Answer

$W = \frac{4P_1V_1}{3}$

Work Step by Step

$ W = \int p dV$ Thus, the work is given by the area under the given graph. Unlike in exercise 20, we are given an expression for the pressure this time. Thus, we find: $W=\int_{V_1}^{2V_1}P_1[1+(\frac{(V-V_1)^2}{V_1^2})]dV$ This gives: $W = \frac{4P_1V_1}{3}$
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