= Solution
Use <heat> into the <gas> as positive. The two adiabatic stages have $Q_{12}=Q_{34}=0$. During constant-pressure heating and constant-volume cooling,
$$
\boxed{Q_{23}=C_p(T_3-T_2)>0,\qquad Q_{41}=C_V(T_1-T_4)<0.}
$$
The first law over a complete cycle gives $W_{\rm out}=Q_{23}+Q_{41}$, since the <internal energy> returns to its initial value. Therefore
$$
\eta=1-\frac{C_V(T_4-T_1)}{C_p(T_3-T_2)}.
$$
Substitute the <temperatures>:
$$
\frac{T_4-T_1}{T_3-T_2}
=\frac{(V_3/V_2)^\gamma-1}
{(V_1/V_2)^{\gamma-1}(V_3/V_2-1)}
=\frac{V_3^\gamma-V_2^\gamma}{V_1^{\gamma-1}(V_3-V_2)}.
$$
Hence
$$
\boxed{\eta=1-\frac1\gamma
\frac{(V_3/V_1)^\gamma-(V_2/V_1)^\gamma}
{V_3/V_1-V_2/V_1}.}
$$
Equivalently, define compression ratio $r=V_1/V_2$ and cutoff ratio $\rho=V_3/V_2$:
$$
\boxed{\eta=1-\frac{\rho^\gamma-1}{\gamma r^{\gamma-1}(\rho-1)}.}
$$
These formulas use the ideal constant-heat-capacity model and reversible adiabatic strokes. The heat-addition and heat-rejection strokes occur over changing <gas> <temperatures>, unlike the isothermal <heat> exchanges of a Carnot engine.
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