Answer
See the detailed answer below.
Work Step by Step
a) The volume at point 2 is on the given graph,
$$V_2=\color{red}{\bf 1000}\;\rm cm^3$$
From 2 to 3, the gas undergoes an adiabatic process, so
$$P_2V_2^\gamma=P_3V_3^\gamma$$
Hence,
$$P_2=P_3\left[ \dfrac{V_3}{V_2}\right]^\gamma$$
For diatomic gas, we know that, $C_V=5R/2$ and $C_P=7R/2$, and hence, $\gamma=C_P/C_V=7/5$
$$P_2=(100)\left[ \dfrac{4000}{1000}\right]^\frac{7}{5}$$
$$P_2=\color{red}{\bf696.4}\;\rm kPa$$
From 1 to 2, the gas undergoes an isochoric process, so
$$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$$
Hence,
$$T_2=\dfrac{P_2T_1}{P_1}=\dfrac{(696.4)(300)}{400}$$
$$T_2=\color{red}{\bf 522.3}\;\rm K$$
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b)
First of all, we need to find the number of moles of the gas which is given by
$$PV=nRT$$
$$n=\dfrac{P_1V_1}{RT_1}=\dfrac{(400\times 10^3)(1000\times 10^{-6})}{(8.31)(300)}$$
$$n=\color{green}{\bf 0.16045}\;\rm mol$$
$\bullet$ $\textbf{The first process 1$\rightarrow$2: [Isochoric] }$
$$W_{1\rightarrow 2}=P\Delta V=P(0)$$
$$W_{1\rightarrow 2}=\color{red}{\bf 0}\;\rm J$$
$$(\Delta E_{th})_{1\rightarrow 2}=\frac{5}{2}nR(T_2-T_1) $$
$$(\Delta E_{th})_{1\rightarrow 2}=\frac{5}{2}(0.16045)(8.31)(522.3-300) $$
$$(\Delta E_{th})_{1\rightarrow 2}=\color{red}{\bf 741}\;\rm J$$
$$Q_{1\rightarrow 2}=(\Delta E_{th})_{1\rightarrow 2}=\color{red}{\bf 741}\;\rm J$$
$\bullet \bullet$ $\textbf{The second process 2$\rightarrow$3: [Adiabtic] }$
$$Q_{2\rightarrow3}=\color{red}{\bf 0}\;\rm J$$
$$(\Delta E_{th})_{2\rightarrow3}=\frac{5}{2}nR(T_3-T_2) $$
$$(\Delta E_{th})_{2\rightarrow3}=\frac{5}{2}(0.16045)(8.31)(300-522.3) $$
$$(\Delta E_{th})_{2\rightarrow3}=\color{red}{\bf- 741}\;\rm J$$
And hence,
$$(W_s)_{2\rightarrow3}=-(\Delta E_{th})_{2\rightarrow3}=-(-741)$$
$$(W_s)_{2\rightarrow3}=\color{red}{\bf 741}\;\rm J$$
$\bullet\bullet\bullet$ $\textbf{The third process 3$\rightarrow$1: [Isothermal] }$
$$(\Delta E_{th})_{3\rightarrow1}=\color{red}{\bf0}\;\rm J$$
$$(W_s)_{3\rightarrow1}=\int_3^1PdV=\int_3^1\dfrac{nRT}{V}dV\\
=nRT\int_3^1\dfrac{dV}{V}=nRT\ln\left[\dfrac{V_1}{V_3}\right]$$
$$(W_s)_{3\rightarrow1} =(0.16045)(8.31)(300)\ln\left[\dfrac{1000}{4000}\right]$$
$$(W_s)_{3\rightarrow1}=\color{red}{\bf -554.5}\;\rm J$$
And hence,
$$Q_{3\rightarrow1}-(W_{s})_{3\rightarrow1}=0$$
$$Q_{3\rightarrow1}=(W_s)_{3\rightarrow1}=\color{red}{\bf -554.5}\;\rm J$$
\begin{array}{|c|c|c|c|}
\hline
& \Delta E_{th}\;{\rm (J)}& W_s\;{\rm (J)}&Q\;{\rm (J)}\\
\hline
1\rightarrow 2& 741& 0&741\\
\hline
2\rightarrow 3 & -741& 741&0\\
\hline
3\rightarrow1& 0& -554.5&-554.5 \\
\hline
{\rm Sum}& 0 & \color{red}{186.5}&\\
\hline
\end{array}
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c) The work done by this engine per cycle is 186.5 J, as we see in the table above.
$$W_{\rm cycle}=\color{red}{\bf 186.5}\;\rm J$$
The thermal efficiency of the engine is given by
$$\eta=\dfrac{W_{out}}{Q_{in}}$$
Plugging from the table above;
$$\eta=\dfrac{186.5}{741}=\color{red}{\bf 0.25}$$
