## Essential University Physics: Volume 1 (3rd Edition)

$W = 1.5 p_1V_1$
The work done by the gas is given by: $W = \int p dV$ Thus, the work is given by the area under the given graph. We will break it up into a triangle and a rectangle to find: $W=p_1(2v_1-v_1)+(.5)(2p_1-p_1)(2v_1-v_1)$ This gives: $W = 1.5 p_1V_1$