Chapter 18 - Exercises and Problems - Page 333: 75

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

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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