Answer
See the detailed answer below.
Work Step by Step
First of all, we need to find the number of moles of the helium gas.
$$n=\dfrac{m}{M}$$
where $m$ is the mass of the sample and $M$ is the atomic mass of the atom.
$$n=\dfrac{120\times 10^{-3}}{4}=\bf 0.03\;\rm mol$$
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a)
$\Rightarrow$ At point 1: the only unknown is the temperature, so
$$T_1=\dfrac{P_1V_1}{nR}=\dfrac{(1\times 1.013\times 10^5)(1000\times 10^{-6})}{(0.03)(8.31)}$$
$$T_1=\color{red}{\bf 406}\;\rm K$$
$\Rightarrow$ At point 2: the only unknown is the temperature as well, so
$$T_2=\dfrac{P_2V_2}{nR}=\dfrac{(5\times 1.013\times 10^5)(1000\times 10^{-6})}{(0.03)(8.31)}$$
$$T_2=\color{red}{\bf 2032}\;\rm K$$
$\Rightarrow$ At point 3: the only unknown is the volume. Noting that we know its temperature since the process from 2 to 3 is an isothermal process.
$$V_3=\dfrac{nRT_3}{P_3}=\dfrac{(0.03)(8.31)(2032)}{(1\times 1.013\times 10^5) }$$
$$V_3=\color{red}{\bf 5000}\;\rm cm^3$$
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b) The engine's thermal efficiency is given by
$$\eta=\dfrac{W_s}{Q_{in}}$$
Now we need to find the net work done during the whole three processes and the heat entered the system as well.
$$W_s=(W_s)_{1\rightarrow2}+(W_s)_{2\rightarrow3}+(W_s)_{3\rightarrow1}$$
From 1 to 2, it is an isochoric process, so the work done is zero.
$$W_s=0+\int_2^3PdV+P_1(\Delta V)_{3\rightarrow1}$$
$$W_s= nRT_2\ln\left[\dfrac{V_3}{V_2}\right]+P_1(V_1-V_3)$$
Plugging the known;
$$W_s= (0.03)(8.31)(2032)\ln\left[\dfrac{5000}{1000}\right]+(1.013\times 10^5)(1000-5000)\times 10^{-6}$$
$$W_s=\color{blue}{\bf 410}\;\rm J$$
Now we need to find $Q_{in}$
In the first process, from 1 to 2: [Isochoric]
$$Q_{1\rightarrow2}=nC_{\rm V}\Delta T=\frac{3}{2}nR(T_2-T_1)$$
$$Q_{1\rightarrow2}= \frac{3}{2}(0.03)(8.31)(2032-406)=\bf 608\;\rm J$$
In the second process, from 2 to 3:[Isothermal]
$$Q_{2\rightarrow3}-(W_s)_{2\rightarrow3}=0$$
$$Q_{2\rightarrow3}=(W_s)_{2\rightarrow3}= nRT_2\ln\left[\dfrac{V_3}{V_2}\right]$$
$$Q_{2\rightarrow3}=(0.03)(8.31)(2032)\ln\left[\dfrac{5000}{1000}\right] =\bf 815\;\rm J$$
In the third process, from 3 to 1:[Isobaric]
$$Q_{3\rightarrow1}=nC_{\rm P}\Delta T=\frac{5}{2}nR(T_1-T_3)$$
$$Q_{3\rightarrow1}= \frac{5}{2}(0.03)(8.31)(406-2032)=-1013\;\rm J$$
Hence,
$$Q_{in}=Q_{1\rightarrow2}+Q_{2\rightarrow3}=608+815$$
$$Q_{in}=\color{blue}{\bf 1423}\;\rm J$$
Therefore,
$$\eta=\dfrac{410}{1423}=\color{red}{\bf 0.29}$$
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c) The engine's maximum efficiency is given by
$$\eta_{max}=1-\dfrac{T_C}{T_H}=1-\dfrac{406}{2032}=\color{red}{\bf 0.8}$$
