= Solution
{disambiguate=cycle calculations}
Put $P_*=p_1V_1=nRT_1$. For the specified <ideal gas>, $C_V=3nR/2$, $C_p=5nR/2$ and $\gamma=5/3$. Isobaric expansion doubles <temperature>, so $T_2=2T_1$. Reversible adiabatic closure gives $TV^{\gamma-1}=\mathrm{constant}$, hence
$$
x\equiv\frac{T_3}{T_1}=2^{-2/3}=0.62996,
\qquad
\boxed{W_{31}^{\rm on}=\frac32(1-x)P_*=0.5551P_*.}
$$
There is no <heat> input on that compression. The other stages give
$$
Q_{12}=\frac52P_*,\quad W_{12}^{\rm by}=P_*,\quad
\boxed{Q_{23}=\frac32(x-2)P_*=-2.0551P_*.}
$$
Therefore $W_{\rm net}^{\rm by}=P_*-W_{31}^{\rm on}=0.44494P_*$ and
$$
\boxed{\eta=\frac{0.44494}{2.5}=0.17798\simeq17.8\%.}
$$
The hot reservoir changes <entropy> by $-Q_{12}/(2.5T_1)=-P_*/T_1$. The cold reservoir changes by $-Q_{23}/(0.5T_1)=4.11012P_*/T_1$. Since the gas completes a cycle,
$$
\boxed{\Delta S_{\rm universe}=3.11012\,\frac{p_1V_1}{T_1}>0.}
$$
Although the mechanical paths can be quasistatic, <heat> flows across finite <temperature> differences in both contacts. This <entropy> generation explains why the efficiency is well below the Carnot value $1-0.5/2.5=80\%$ between these reservoirs.
For an isothermal final compression at $T_1$, the preceding constant-volume cooling must instead end at $T_3=T_1$; keeping the old state 3 would not close the cycle. Compression then requires <work> $P_*\ln2$, while the <heat> input remains $5P_*/2$. Thus
$$
\boxed{\eta_{\rm iso}=\frac{1-\ln2}{5/2}=0.12274\simeq12.3\%,}
$$
which is \b[lower]: this isothermal compression occurs at a higher <temperature> throughout than the original adiabat and consumes more <work>. It also rejects <heat> during compression.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2014/paper-2-b2-cycle-balances.png]
{title=Energy balance and entropy production in each stage of the three-stage engine}
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