Answer
See the detailed answer below.
Work Step by Step
First of all, we need to find the number of moles of the gas which is given by
$$PV=nRT$$
$$n=\dfrac{P_1V_1}{RT_1}=\dfrac{(0.5\times 1.013\times 10^5)(10\times 10^{-6})}{(8.31)(20+273)}$$
$$n=\color{green}{\bf 2.08\times 10^{-4}}\;\rm mol$$
Now we need to find $T_2$ and $T_3$;
$$T_2=\dfrac{P_2V_2}{nR}=\dfrac{(1.5\times 1.013\times 10^5)(40\times 10^{-6})}{(2.08\times 10^{-4})(8.31)}$$
$$T_2= \color{green}{\bf3516}\;\rm K$$
$$T_3=\dfrac{P_2V_2}{nR}=\dfrac{(0.5\times 1.013\times 10^5)(40\times 10^{-6})}{(2.08\times 10^{-4})(8.31)}$$
$$T_3= \color{green}{\bf 1172}\;\rm K$$
$\textbf{The first process 1$\rightarrow$2:}$
$$(W_s)_{1\rightarrow 2}=\rm \;Area\;under\;the \;({\it P-V})\;curve$$
$$(W_s)_{1\rightarrow 2}=\left[\frac{1}{2}(40-10)\times 10^{-6}(1.5-0.5)\times 1.013\times 10^5\right]+\left[(40-10)\times 10^{-6}(0.5\times 1.013\times 10^5)\right]$$
$$(W_s)_{1\rightarrow 2}=\color{red}{\bf 3.039}\;\rm J$$
$$(\Delta E_{th})_{1\rightarrow 2}=\frac{5}{2}nR(T_2-T_1) $$
$$(\Delta E_{th})_{1\rightarrow 2}=\frac{5}{2}(2.08\times 10^{-4})(8.31)(3516-[20+273]) $$
$$(\Delta E_{th})_{1\rightarrow 2}=\color{red}{\bf 13.93}\;\rm J$$
Hence,
$$Q_{1\rightarrow2}=(\Delta E_{th})_{1\rightarrow 2}+(W_s)_{1\rightarrow 2}$$
$$Q_{1\rightarrow2}=13.93+3.039$$
$$Q_{1\rightarrow 2}=\color{red}{\bf 16.97}\;\rm J$$
$\textbf{The second process 2$\rightarrow$3:[Isochoric]}$
$$(W_s)_{2\rightarrow 3}=P\Delta V=P(0)=0$$
$$(W_s)_{2\rightarrow 3}=\color{red}{\bf 0}\;\rm J$$
Hence,
$$(\Delta E_{th})_{2\rightarrow 3}=Q_{2\rightarrow3}=\frac{5}{2}nR(T_3-T_2) $$
$$(\Delta E_{th})_{2\rightarrow 3}=Q_{2\rightarrow3}=\frac{5}{2}(2.08\times 10^{-4})(8.31)(1172-3516) $$
$$(\Delta E_{th})_{2\rightarrow 3}=Q_{2\rightarrow3}=\color{red}{\bf -10.13}\;\rm J$$
$\textbf{The third process 3$\rightarrow$1:[Isobaric]}$
$$(W_s)_{3\rightarrow 1}=P\Delta V=P(V_1-V_3)$$
$$(W_s)_{3\rightarrow 1}=(0.5\times 1.013\times 10^5)(10-40)\times 10^{-6}$$
$$(W_s)_{3\rightarrow 1}=\color{red}{\bf -1.52}\;\rm J$$
$$(\Delta E_{th})_{3\rightarrow 1}=\frac{5}{2}nR(T_1-T_3) $$
$$(\Delta E_{th})_{3\rightarrow 1}=\frac{5}{2}(2.08\times 10^{-4})(8.31)([20+273]-1172) $$
$$(\Delta E_{th})_{3\rightarrow 1}=\color{red}{\bf-3.8}\;\rm J$$
$$Q_{3\rightarrow 1}=nC_{\rm P}(T_1-T_3)=\frac{7}{2}nR(T_1-T_3)$$
$$Q_{3\rightarrow 1}=\frac{7}{2}(2.08\times 10^{-4})(8.31)([20+273]-1172) $$
$$Q_{3\rightarrow 1}=\color{red}{\bf-5.32}\;\rm J$$
See the table below:
\begin{array}{|c|c|c|c|}
\hline
& \Delta E_{th}\;{\rm (J)}& W_s\;{\rm (J)}&Q\;{\rm (J)}\\
\hline
1\rightarrow 2&13.93 & 3.039&16.97\\
\hline
2\rightarrow 3 &-10.13&0 &-10.13 \\
\hline
3\rightarrow1& -3.8 & -1.52 & -5.32 \\
\hline
{\rm Sum}& & 1.52 & \\
\hline
\end{array}
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b) The thermal efficiency is given by
$$\eta=\dfrac{W}{Q_H}=\dfrac{1.52}{16.97}$$
$$\eta=\color{red}{\bf 0.089}$$
____________________________________________________________
c) The power out put is given by
$$P_{out}=\dfrac{500}{60}\times 1.52=\color{red}{\bf 12.67}\;\rm W$$
