Answer
See the detailed answer below.
Work Step by Step
a)
We can see, from the given graph, that the pressure at point 1 is
$$P_1=\color{red}{\bf 100}\;\rm kPa$$
In the first process, the gas undergoes an adiabatic process, so
$$P_1V_1^\gamma=P_2V_2^\gamma$$
Hence,
$$V_1=V_2\left[\dfrac{P_2}{P_1}\right]^{1/\gamma}$$
where $\gamma$ for a diatomic gas is given by $\gamma=C_P/C_V=(7R/2)/(5R/2)=7/5=\bf 1.4$
$$V_1=(1000)\left[\dfrac{400}{100}\right]^{1/1.4}$$
$$V_1=\color{red}{\bf 2692}\;\rm cm^3$$
And for the adiabatic process,
$$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$$
Hence,
$$T_1=T_2\left[ \dfrac{V_2}{V_1} \right]^{\gamma-1}=400\left[ \dfrac{1000}{2692} \right]^{1.4-1}$$
$$T_1=\color{red}{\bf 269.2}\;\rm K$$
Now we can find the number of moles, using the ideal gas law and data of point 2.
$$n=\dfrac{P_2V_2}{RT_2}=\dfrac{(400\times 10^3)(1000\times 10^{-6})}{(8.31)(400)}$$
$$n=\color{blue}{\bf 0.1203}\;\rm mol$$
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b)
$$\textbf{The first process 1$\rightarrow$2: [Adiabtic] }$$
$$Q_{1\rightarrow 2}= \color{red}{\bf 0}\;\rm J$$
$$(\Delta E_{th})_{1\rightarrow 2}=\frac{5}{2}nR(T_2-T_1) $$
$$(\Delta E_{th})_{1\rightarrow 2}=\frac{5}{2}(0.1203)(8.31)(400-269.2) $$
$$(\Delta E_{th})_{1\rightarrow 2}=\color{red}{\bf 326.9}\;\rm J $$
In such a process,
$$$(\Delta E_{th})_{1\rightarrow 2}=-(W_s)_{1\rightarrow 2}$$
Hence,
$$ (W_s)_{1\rightarrow 2}=\color{red}{\bf -326.9}\;\rm J $$
$$\textbf{The second process 2$\rightarrow$3: [Isothermal] }$$
$$(Ws)_{2\rightarrow 3}=\int_2^3PdV=\int_2^3\dfrac{nRT}{V}dV=nRT\int_2^3\dfrac{dV}{V}$$
$$(Ws)_{2\rightarrow 3}= nRT\ln\left[\dfrac{V_3}{V_2}\right]= (0.1203)(8.31)(400)\ln\left[\dfrac{4000}{1000}\right]$$
$$(Ws)_{2\rightarrow 3}=\color{red}{\bf 554}\;\rm J$$
It is an isothermal process, so
$$(\Delta E_{th})_{2\rightarrow3}= \color{red}{\bf 0}\;\rm J $$
And hence,
$$Q_{2\rightarrow3}=(Ws)_{2\rightarrow 3}=\color{red}{\bf 554}\;\rm J$$
$$\textbf{The third process 2$\rightarrow$3: [Isobaric] }$$
$$(Ws)_{3\rightarrow 1}=P\Delta V=P_3(V_1-V_3)\\
(Ws)_{3\rightarrow 1}=(100\times 10^3)(2692-4000)\times 10^{-6}$$
$$(Ws)_{3\rightarrow 1}=\color{red}{\bf- 130.8}\;\rm J$$
$$(\Delta E_{th})_{3\rightarrow 1}=\frac{5}{2}nR(T_1-T_3) $$
$$(\Delta E_{th})_{3\rightarrow 1}=\frac{5}{2}(0.1203)(8.31)(269.2-400) $$
$$(\Delta E_{th})_{3\rightarrow 1}=\color{red}{\bf -326.9}\;\rm J $$
$$Q_{3\rightarrow 1}=nC_P\Delta T=\frac{7}{2}nR(T_1-T_3) $$
$$Q_{3\rightarrow 1}=\frac{7}{2}(0.1203)(8.31)(269.2-400) $$
$$Q_{3\rightarrow 1}=\color{red}{\bf -457.6}\;\rm J$$
\begin{array}{|c|c|c|c|}
\hline
& \Delta E_{th}\;{\rm (J)}& W_s\;{\rm (J)}&Q\;{\rm (J)}\\
\hline
1\rightarrow 2& 326.9& -326.9&0 \\
\hline
2\rightarrow 3 & 0& 554&554
\\
\hline
3\rightarrow1&-326.9 & -130.8&-457.7\\
\hline
{\rm Sum}& 0 & 96.3&Q_{in}=554\\
\hline
\end{array}
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c) The work done by this engine per cycle is 96.3 J, as we see in the table above.
$$W_{\rm cycle}=\color{red}{\bf 96.3}\;\rm J$$
The thermal efficiency of the engine is given by
$$\eta=\dfrac{W_{out}}{Q_{in}}=\dfrac{96.3}{554}$$
$$\eta=\color{red}{\bf 0.17}$$
