= Solution
For a diatomic <ideal gas> in the usual regime with three translational and two rotational quadratic <degrees of freedom>, but frozen vibrational excitation, the <equipartition theorem> gives
$$
C_V=\frac52R_g,\qquad C_p=\frac72R_g,\qquad \gamma=\frac75.
$$
The <Brayton cycle> efficiency is therefore
$$
\boxed{\eta=1-10^{-2/7}=0.4821\simeq48.2\%.}
$$
The constant-pressure expansion has $T_3/T_2=V_3/V_2=2$, while the compression has $T_2/T_1=10^{2/7}=1.93070$. The same relations give $T_4/T_1=2$. Thus, in units of the minimum <temperature>,
$$
(T_1,T_2,T_3,T_4)=T_1(1,1.93070,3.86140,2).
$$
The maximum and minimum <temperatures> are $T_3$ and $T_1$. A <Carnot cycle> between those extremes has
$$
\boxed{\eta_{\rm Carnot}=1-\frac{T_1}{T_3}=1-\frac1{2\,10^{2/7}}=0.7410\simeq74.1\%.}
$$
The difference is about $25.9$ percentage points. In the <Brayton cycle>, <heat> is supplied and rejected over ranges of <temperature>; the whole input is not supplied at $T_3$, nor is the whole rejection made at $T_1$. This explains the lower efficiency even for mechanically ideal stages. Absolute state <temperatures> are not supplied, so $\gamma=7/5$ is an explicit standard diatomic-gas approximation rather than a conclusion about all possible rotational and vibrational excitation regimes.
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