Answer
$930J$
Work Step by Step
We know that
$P_C=P_B(\frac{V_B}{V_C})^{\gamma}$
$P_C=200(\frac{4}{1})^{1.4}=1393KPa$
Now $W_{BC}=\frac{P_CV_C-P_BV_B}{\gamma-1}$
We plug in the known values to obtain:
$W_{BC}=\frac{1393\times 1-200\times 4}{1.4-1}=1.5KJ=1483J$
Now $W_{DA}=-nRTln(\frac{V_A}{V_D})$
We plug in the known values to obtain:
$W_{DA}=-100\times 4\times \frac{4}{1}=-555J$
Thus, $W_{cycle}=W_{BC}+W_{DA}=1483-555=930J$
