Answer
$W = 4.7 \space kJ$
Work Step by Step
To find the net work, we must first find out the $Q_3$ which is an isobaric process. Note that the gas is compressed to its initial state.
$Q_3 = n C_P \Delta T$ Where $C_P = \frac{5}{2}R $
$Q_3 = n \frac{5}{2}R (T - T') $
$Q_3 = (2.0 mol) \frac{5}{2}(8.31 J/mol.K) (300K - 800K ) $
$Q_3 = -20.78 kJ$
So, the work is the total of heat energy in $Q_1, Q_2 \space and \space Q_3$
$W = Q_1 + Q_2 + Q_3 $
$W = 25.5 kJ + (-20.8 kJ)$
$W = 4.7 \space kJ$
