Answer
See the detailed answer below.
Work Step by Step
From the given steps, it is obvious that the gas undergoes a closed cycle of 4 processes.
1) The first process is heating the gas under constant volume until its pressure increases from 1 atm to 3 atm. This means that this process is an isochoric process that starts from point 1 where
$$\color{blue}{V_1=50\;{\rm cm^3},~~~~P_1=1\;{\rm atm},~~~~T_1=20^\circ\rm C}$$
And ends at point 2 where
$$V_2=V_1=50\;{\rm cm^3},~~~~P_2=3\;{\rm atm},~~~~T_2=?^\circ\rm C$$
Finding $T_2$, $$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$$
Hence,
$$T_2=\dfrac{P_2}{P_1}\cdot T_1=\dfrac{3}{1}\cdot (20+273)=\bf 879\;\rm K=\bf 606^\circ \rm C$$
2) The second process is heating the gas under constant pressure until its volume increases to 100 cm$^3$.
It starts at point 2 where
$$\color{blue}{V_2=50\;{\rm cm^3},~~~~P_2=3\;{\rm atm},~~~~T_2=606^\circ\rm C}$$
And ends at point 3 where
$$V_3=100\;{\rm cm^3},~~~~P_3=P_2=3\;{\rm atm},~~~~T_3=?^\circ\rm C$$
Finding $T_3$, $$\dfrac{V_3}{T_3}=\dfrac{V_2}{T_2}$$
Hence,
$$T_3=\dfrac{V_3}{V_2}\cdot T_2=\dfrac{100}{50}\cdot (606+273)=\bf 1758\;\rm K=\bf 1485^\circ \rm C$$
3) The third process is cooling down the gas under constant volume until its pressure returns to its initial state of 1 atm.
It starts at point 3 where
$$\color{blue}{V_3=100\;{\rm cm^3},~~~~P_3 =3\;{\rm atm},~~~~T_3=1485^\circ\rm C}$$
And ends at point 4 where
$$V_4=V_3=100\;{\rm cm^3},~~~~P_4=1\;{\rm atm},~~~~T_4=?^\circ\rm C$$
Finding $T_4$, $$\dfrac{P_3}{T_3}=\dfrac{P_4}{T_4}$$
Hence,
$$T_4=\dfrac{P_4}{P_3}\cdot T_3=\dfrac{1}{3}\cdot (1485+273)=\bf 586\;\rm K=\bf 313^\circ \rm C$$
4) The fourth process is cooling down the gas under constant pressure until its volume reaches its initial volume of 50 cm$^3$ and since the same mass is added again so the final pressure is also as the initial pressure.
The gas starts from point 4 where
$$\color{blue}{V_4=100\;{\rm cm^3},~~~~P_4=1\;{\rm atm},~~~~T_4=313^\circ\rm C}$$
and ends at point 1 where
$$ V_1=50\;{\rm cm^3},~~~~P_1=1\;{\rm atm},~~~~T_1=20^\circ\rm C$$
Now we need to find the number of the moles of the gas
$$n=\dfrac{P_1V_1}{RT_1}=\dfrac{(1\times 1.013\times 10^5)(50\times 10^{-6})}{(8.31)(20+273)}$$
$$n=\bf 0.00208\;\rm mol$$
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a) $\bullet\Rightarrow$ From the 4 blue cases above, we can draw the graph of this closed cycle as shown below.
b) The work done by this engine is equal to the closed area under the $P-V$ graph.
$$W=(P_{max}-P_{min})(V_{max}-V_{min})$$
$$W=(3-1)\times 1.013\times 10^5(100-50)\times 10^{-6}$$
$$W=\color{red}{\bf 10.13}\;\rm J$$
c) The engine’s thermal efficiency is given by
$$\eta=\dfrac{W}{Q_{in}}$$
We found $W$ above, so we need to find $Q_{in}$.
We can see that the gas is heated in the first two processes.
$$Q_{in}=Q_{1\rightarrow2}+Q_{2\rightarrow3}$$
$$Q_{in}=nC_{\rm V}(T_2-T_1)+nC_{\rm P}(T_3-T_2)$$
$$Q_{in}=\frac{5}{2}nR (T_2-T_1)+\frac{7}{2}nR(T_3-T_2)$$
$$Q_{in}=\frac{nR}{2} \left[5(T_2-T_1)+7 (T_3-T_2)\right]$$
$$Q_{in}=\frac{(0.00208)(8.31)}{2} \left[5(606-20)+7 (1485-606)\right]$$
$$Q_{in}=78.5\;\rm J$$
Thus,
$$\eta=\dfrac{W}{Q_{in}}=\dfrac{10.13}{78.5}=\color{red}{\bf 0.13}$$
