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