= Solution
This is the <Brayton cycle>. Assume fixed gas amount, constant <heat capacities>, and internally reversible <adiabatic processes>, as intended by the ideal engine model. Put $r_p=p_h/p_l=p_2/p_1>1$ and
$$
a=\frac{\gamma-1}{\gamma},\qquad k=r_p^a.
$$
Combining $pV^\gamma=\text{constant}$ with the <ideal gas law> gives $T\propto p^{(\gamma-1)/\gamma}$ along either <reversible ideal-gas adiabat>. Therefore
$$
\frac{T_2}{T_1}=k,\qquad \frac{T_3}{T_4}=k,
\qquad T_3-T_2=k(T_4-T_1).
$$
During a constant-pressure process the <heat> supplied equals the <enthalpy> increase. If $C_p$ is the molar <heat capacity at constant pressure>, the positive <heat> input and output are
$$
Q_{\rm in}=nC_p(T_3-T_2),\qquad
Q_{\rm out}=nC_p(T_4-T_1).
$$
The insulated branches exchange no <heat>, so the <first law of thermodynamics> gives $W=Q_{\rm in}-Q_{\rm out}$. Hence
$$
\boxed{\eta=1-\frac{Q_{\rm out}}{Q_{\rm in}}=1-\frac1k=1-\left(\frac{p_2}{p_1}\right)^{(1-\gamma)/\gamma}.}
$$
An <adiabatic process> alone need not be an <isentropic process>; the relations used here also require internal reversibility. Dissipative compression or expansion would require extra information and would not, in general, obey this efficiency formula.
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