Answer
a) $1096\;\rm W$
b) $0.0902$
Work Step by Step
We are given that the gas has a $\gamma$ of 1.25. This means that $$C_{\rm V}=4R~~~~~~~~~~~~~ {\rm and}~~~~~~~~ C_{\rm P}=5R$$
Recalling that $\gamma=C_{\rm P}/C_{\rm V}$ and that $C_{\rm P}=C_{\rm V}+R$
Now we need to find the number of moles of the gas, which is given by the ideal gas law, $PV=nRT$.
Using point (1);
$$n=\dfrac{P_1V_1}{RT_1}=\dfrac{(1\times 1.013\times 10^5)(600\times 10^{-6})}{(8.31)(300)}$$
$$n=\color{blue}{\bf 0.02438}\;\rm mol$$
Now we need to find $P_2$; noting that the process from 1 to 2 is an isothermal process, so $T_1=T_2=300$ K.
$$P_2=\dfrac{nRT_2}{V_2}=\dfrac{(0.02438)(8.31)(300)}{(200\times 10^{-6})}$$
$$P_2=303897\;\rm Pa=\color{blue}{\bf 3}\;\rm atm$$
Now we need to find $T_3$; noting that the process from 2 to 3 is an isobaric process, so $P_2=P_3=3$ atm.
$$T_3=\dfrac{P_3V_3}{nR}=\dfrac{(3\times 1.013\times 10^5)(600\times 10^{-6})}{(0.02438)(8.31)}$$
$$T_3 =\color{blue}{\bf 900}\;\rm K$$
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a) To find the output work we need to find the total work done by the system which is the sum of the work done in each process.
The work done in the first process [Isothermal]:
$$W_{1\rightarrow2}=\int_1^2 PdV=\int_1^2 \dfrac{nRT}{V}dV=nRT\int_1^2 \dfrac{dV}{V}$$
$$W_{1\rightarrow2}=\overbrace{nRT_1}^{=P_1V_1}\ln\left[\dfrac{V_2}{V_1}\right]=P_1V_1\ln\left[\dfrac{V_2}{V_1}\right]$$
$$W_{1\rightarrow2}= (1.013\times 10^5)(600\times 10^{-6})\ln\left[\dfrac{200}{600}\right]$$
$$W_{1\rightarrow2}=\color{green}{\bf -66.8}\;\rm J$$
The work done in the second process [Isobaric]:
$$W_{2\rightarrow3}=P_2\Delta V=P(V_3-V_2)$$
$$W_{2\rightarrow3}= (3\times 1.013\times 10^5)(600-200)\times 10^{-6}$$
$$W_{2\rightarrow3}=\color{green}{\bf 121.6}\;\rm J$$
The work done in the third process [Isochoric]:
$$W_{3\rightarrow1}=P_2\Delta V=P(0)$$
$$W_{3\rightarrow1}=\color{green}{\bf 0}\;\rm J$$
Therefore,
$$W_{\rm cycle}=W_{1\rightarrow2}+W_{2\rightarrow3}+W_{3\rightarrow1}$$
$$W_{\rm cycle}=-66.8+121.6+0=\bf 54.8\;\rm J$$
$${\rm Power\;output}=N W_{cycle}$$
where $N$ is the number of cycles per second,
$${\rm Power\;output}=(20)(54.8)=\color{red}{\bf 1096}\;\rm W$$
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b) The engine's thermal efficiency is given by
$$\eta=\dfrac{W_{out}}{Q_{in}}$$
we found $W_{out}$ in one cycle above, now we need to find $Q_{in}$.
In the first process [Isothermal]:
$$Q_{1\rightarrow2}=W_{1\rightarrow2}$$
where $\Delta E_{th}=0$
$$Q_{1\rightarrow2}=\color{green}{\bf -66.8}\;\rm J$$
In the second process [Isobaric]:
$$Q_{2\rightarrow3}=nC_{\rm P}\Delta T=n(5R)(T_3-T_2)$$
$$Q_{2\rightarrow3}= 5(0.02438)(8.31)(900-300)$$
$$Q_{2\rightarrow3}=\color{green}{\bf 607.8}\;\rm J$$
In the third process [Isochoric]:
$$Q_{3\rightarrow1}=nC_{\rm V}\Delta T=n(4R)(T_1-T_3)$$
$$Q_{3\rightarrow1}=4(0.02438)(8.31)(300-900)$$
$$Q_{3\rightarrow1}=\color{green}{\bf -486.2}\;\rm J$$
Therefore,
$$Q_{\rm in}= Q_{2\rightarrow3} $$
$$Q_{\rm in}=\bf 607.8\;\rm J$$
Therefore, the engine's thermal efficiency is given by
$$\eta=\dfrac{W_{out}}{Q_{in}}=\dfrac{54.8}{607.8}$$
$$\eta=\color{red}{\bf 0.0902}$$
