= Solution
In the <Carnot cycle>, let $1\to2$ be reversible <isothermal expansion> at $T_h$, $2\to3$ reversible <adiabatic expansion>, $3\to4$ <isothermal compression> at $T_c$, and $4\to1$ <adiabatic compression>. The <pressure>–<volume> curve is clockwise. The <entropy>–<temperature> figure below uses <temperature> on the horizontal axis and <entropy> on the vertical axis, literally an $S$–$T$ plot: isotherms are vertical and adiabats horizontal. The hot isotherm increases <entropy> by $\Delta S$ and absorbs $Q_h=T_h\Delta S$; the cold one rejects $Q_c=T_c\Delta S$. Thus $W=(T_h-T_c)\Delta S$ and $\eta_C=1-T_c/T_h$.
For the second engine, choose the working direction with <heat> addition during $A\to B$ at $p_1$ and rejection during $C\to D$ at $p_2$. A <reversible ideal-gas adiabat> obeys $T\propto p^{(\gamma-1)/\gamma}$. Define
$$
r=\left(\frac{p_1}{p_2}\right)^{(\gamma-1)/\gamma}>1,\qquad
\frac{T_B}{T_C}=\frac{T_A}{T_D}=r.
$$
The two constant-pressure <heat> transfers are $Q_{\rm in}=c_p(T_B-T_A)$ and $Q_{\rm out}=c_p(T_C-T_D)=Q_{\rm in}/r$, where rejected <heat> is quoted positive. Therefore the <Brayton cycle> efficiency is
$$
\eta=1-\frac1r,\qquad
\boxed{\eta_{\rm mono}=1-\left(\frac{p_2}{p_1}\right)^{2/5}}
$$
for $\gamma=5/3$. All heat-capacity amount conventions cancel in the ratio.
Since $T_B>T_A$ and $T_C=T_B/r$, $T_D=T_A/r$, the maximum is \b[at B] and the minimum \b[at D]. The <Brayton-Carnot efficiency comparison> is
$$
\eta_C-\eta
=\left(1-\frac{T_D}{T_B}\right)-\left(1-\frac1r\right)
=\frac1r\left(1-\frac{T_A}{T_B}\right)>0.
$$
The inequality is strict for a nondegenerate cycle with positive <heat> input. Although each process can be reversible, the isobaric <heat> transfer spans a range of <temperatures>; making the whole cycle reversible requires a succession of matched-temperature reservoirs, rather than just the hottest and coldest reservoirs.
For an ideal diatomic gas with active rotation and frozen vibration, $\gamma=7/5$ and
$$
\eta_{\rm diatomic}=1-\left(\frac{p_2}{p_1}\right)^{2/7}<1-\left(\frac{p_2}{p_1}\right)^{2/5}.
$$
Here $0<p_2/p_1<1$. This comparison holds at the same <pressure> ratio and assumes temperature-independent <heat capacities> in the stated mode regime. Frozen rotation or excited vibration would require their actual <heat capacities>.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2011/paper-1-engine-cycles.png]
{title=Carnot and Brayton cycles}
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