Answer
$2.27 \space kJ$
Work Step by Step
The net work done of the rectangular area is
$ W = (V-V_o) (P-P_o)$
Where $V = 2V_o$
and $P = 2P_o$
$ W = (2V_o-V_o) ( 2P_o-P_o)$
$ W = (V_o) ( P_o)$
Where $V_o = 0.0225 m^3$
and $P_o = 1.01 \times 10^5 Pa$
Substitute all values into equation
$ W = ( 0.0225 m^3) (1.01 \times 10^5 Pa)$
$ W = 2272.5 J$ or $2.27 kJ$
