Answer
See the detailed answer below.
Work Step by Step
a) To find all the unknown, we first need to find the number of moles of Helium gas which is given by
$$n=\dfrac{m}{M_{He}}=\dfrac{120\times 10^{-3}\;\rm g}{4\;\rm g/mol}=\bf 0.03\;\rm mol$$
Now we need to find $P_1$, so we can use point 1 which we know the gas's temperature and volume.
$$P_1=\dfrac{nRT_1}{V_1}=\dfrac{(0.03)(8.31)(133+273)}{(1000\times 10^{-6})}$$
$$P_1=1.01\times 10^5\;\rm Pa=\bf 1\;\rm atm$$
Now we need to find $T_2$ and note that $T_2=T_3$ since it is an isothermal process from 2 to 3.
From 1 to 2, it is an isochoric process, so
$$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$$
$$T_2=\dfrac{P_2T_1}{P_1}= \dfrac{(5P_1) T_1}{P_1}=5T_1$$
Hence, $T_2$ in Celsius is given by
$$T_2 =5 \times (133+273)-273=\bf 1757^\circ$$
Now we need to find $V_3$, we can use the isothermal process.
$$P_2V_2=P_3V_3$$
$$V_3=\dfrac{P_2V_2}{P_3}$$
Noting that $P_3=P_1$, $P_2=5P_1$, and $V_2=V_1$
$$V_3=\dfrac{5P_1V_2}{P_1}=5V_2=\bf 5000\;\rm cm^3$$
\begin{array}{|c|c|c|}
\hline
{\rm Point}& P\;{\rm (atm)} & V\;{\rm (cm^3)}&T\;{\rm (^\circ C)}\\
\hline
1 & 1 & 1000 & 133 \\
\hline
2 & 5 & 1000 & 1757 \\
\hline
3 & 1 & 5000 & 1757 \\
\hline
\end{array}
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b)
$\Rightarrow$ From 1 to 2, it is an isochoric process, so
$$W_{1\rightarrow2}=-\int PdV=\color{red}{\bf 0}\;\rm J$$
$\Rightarrow$ From 2 to 3, it is an isothermal process, so
$$W_{2\rightarrow3}=-\int PdV= -\int_2^3\dfrac{nRT}{V}dV$$
where $T$ is constant.
$$W_{2\rightarrow3}=-nRT \int_2^3\dfrac{dV}{V}=-nRT\ln\left[\dfrac{V_3}{V_2}\right]$$
Plugging the known
$$W_{2\rightarrow3}=-(0.03)(8.31)(1757+273)\ln\left[\dfrac{5000}{1000}\right]$$
$$W_{2\rightarrow3}=\color{red}{\bf -815}\;\rm J$$
$\Rightarrow$ rom 3 to 1, it is an isobaric process, so
$$W_{3\rightarrow1}=-\int PdV=- P\Delta V $$
where $P$ is constant.
$$W_{3\rightarrow1} =- P(V_1-V_3)=-(1.01\times 10^5)(1000-5000)\times 10^{-6}$$
$$W_{3\rightarrow1} =\color{red}{\bf 404}\;\rm J$$
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c) The heat needed for each process:
$$Q_{1\rightarrow2}=nC_{\rm v}(T_2-T_1)$$
$$Q_{1\rightarrow2}=(0.03)(12.5)(1757-133)=\color{red}{\bf 609}\;\rm J$$
$$Q_{2\rightarrow3}=\Delta E_{th}-W_{2\rightarrow3}$$
$$Q_{2\rightarrow3}=0-(-815)$$
$$Q_{2\rightarrow3} =\color{red}{\bf 815}\;\rm J$$
$$Q_{3\rightarrow1}=nC_{\rm p}(T_1-T_3)$$
$$Q_{3\rightarrow1}=(0.03)(20.8)(133-1757 )=\color{red}{\bf -1013}\;\rm J$$
