Answer
$3.4KJ$ and the work is done by gas.
Work Step by Step
We know that
$W_{AB}=-nRT\ln\frac{V_B}{V_A}=-P_aV_a\ln\frac{V_B}{V_A}$
$W_{AB}=-50Kpa\times 25L\times \ln(\frac{25}{\frac{25}{3}})=-1373J$
We also know that
$P_CV_C^{\gamma}=P_AV_A^{\gamma}$
This can be rearranged as:
$P_C=P_A(\frac{V_A}{V_C})^{\gamma}$
$P_C=50KPa(\frac{25}{\frac{25}{3}})^{1.67}=313KPa$
$W_{CA}=\frac{50KPa\times 25L-313KPa\times \frac{25L}{3}}{1.67-1}$
$W_{cycle}=W_{AB}+W_{CA}$
We plug in the known values to obtian:
$W_{cycle}=-1373+(-2027)=-3.4KJ$
The negative sign shows that work is done by gas.
