Answer
$q_{in}=1275\text{ kJ/kg}$
$q_{out}=212.5\text{ kJ/kg}$
$w_{net}=1063\text{ kJ/kg}$
Work Step by Step
Applying the ideal gas equation to the isothermal process 3-4 gives $$
P_4=P_3 \frac{\boldsymbol{v}_3}{\boldsymbol{v}_4}=(50 \mathrm{kPa})(12)=600\ \mathrm{kPa}
$$ Since process 4-1 is one of constant volume, $$
T_1=T_4\left(\frac{P_1}{P_4}\right)=(298 \mathrm{~K})\left(\frac{3600 \mathrm{kPa}}{600 \mathrm{kPa}}\right)=1788 \mathrm{~K}
$$ Adapting the first law and work integral to the heat addition process gives $$
q_{\text {in }}=w_{1-2}=R T_1 \ln \frac{\boldsymbol{v}_2}{\boldsymbol{v}_1}=(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(1788 \mathrm{~K}) \ln (12)=\mathbf{1 2 7 5}\ \mathbf{k J} / \mathbf{k g}
$$ Similarly, $$
q_{\text {out }}=w_{3-4}=R T_3 \ln \frac{\boldsymbol{v}_4}{\boldsymbol{v}_3}=(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(298 \mathrm{~K}) \ln \left(\frac{1}{12}\right)=\mathbf{2 1 2 . 5}\ \mathrm{kJ} / \mathbf{k g}
$$ The net work is then $$
w_{\text {net }}=q_{\text {in }}-q_{\text {out }}=1275-212.5=\mathbf{1 0 6 3 ~k J / k g}
$$
