#### Answer

$W = 1.5 p_1V_1$

#### Work Step by Step

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:
$W = 1.5 p_1V_1$