= Solution
For a reversible <adiabatic process> of an <ideal gas>, $\delta Q=0$ and $dU=-p\,dV$. At fixed composition and constant sample <heat capacity at constant volume> $C_V$,
$$
C_V\,dT=-\frac{n_{\rm mol}RT}{V}\,dV,
\qquad d\ln T=-\frac{n_{\rm mol}R}{C_V}d\ln V.
$$
<Mayer's relation> gives $n_{\rm mol}R=C_p-C_V$, so $TV^{\gamma-1}$ is constant and
$$
\boxed{pV^\gamma=\mathrm{constant},\qquad \gamma=\frac{C_p}{C_V}.}
$$
The finite-path power law requires approximately constant <heat capacities> as well as reversibility. When rotational or vibrational populations change appreciably, use $\int C_V(T)\,dT/T+n_{\rm mol}R\ln V=\mathrm{constant}$ instead; the local isentropic pressure exponent remains $\gamma(T)$.
For an intact heteronuclear <diatomic gas>, quantum excitation progressively activates molecular motion. With rotational and vibrational modes frozen, the three translational degrees of freedom give $C_{V,\rm m}=3R/2$ and $\gamma=5/3$. Two active rotations add $R$, giving $C_{V,\rm m}=5R/2$ and $\gamma=7/5$. One active harmonic vibration contributes a further $R$, through kinetic and potential terms, giving $C_{V,\rm m}=7R/2$ and $\gamma=9/7$.
For molecular <moment of inertia> $I=\mu_{\rm red}r_e^2$, the <rigid rotor> levels are $E_J=\hbar^2J(J+1)/(2I)$, so the rotational excitation scale is
$$
\boxed{\Theta_{\rm rot}=\frac{\hbar^2}{2Ik_B}.}
$$
The first gap corresponds to $2\Theta_{\rm rot}$, so the crossover occurs at temperatures of that order. For bond force constant $k_s$ and reduced molecular <mass> $\mu_{\rm red}$, the vibrational scale is
$$
\boxed{\Theta_{\rm vib}=\frac{\hbar\omega_{\rm vib}}{k_B}=\frac\hbar{k_B}\sqrt{\frac{k_s}{\mu_{\rm red}}}.}
$$
Typically $\Theta_{\rm vib}\gg\Theta_{\rm rot}$. Real gases may condense at the low-temperature end or dissociate and excite electronic states at the high-temperature end; the three plateau values describe the ideal intact-molecule model.
\Image[/past-exam-of-the-thermal-physics-course-of-the-university-of-oxford/2016-diatomic-heat-capacity-ratio.png]
{title=Translational, rotational and vibrational regimes of the heat capacity ratio}
{description=Original rigid-rotor and harmonic-oscillator calculation with an illustrative ratio of vibrational to rotational temperatures of 100. The three limiting <heat capacity ratios> are marked; detailed crossover shapes are not universal.}
For the piston apparatus, mechanical equilibrium requires $p_0A=p_{\rm air}A+mg$, hence $p_0=p_{\rm air}+mg/A$. Let $x$ be a small upward displacement, so $V=V_0+Ax$. The local <isentropic compressibility> gives
$$
p-p_0\simeq-\frac{\gamma p_0A}{V_0}x,
\qquad m\ddot x=A(p-p_0)=-\frac{\gamma p_0A^2}{V_0}x.
$$
Consequently the <adiabatic gas-spring piston period> is
$$
\boxed{\tau=2\pi\sqrt{\frac{B}{\gamma}},\qquad B=\frac{mV_0}{p_0A^2}.}
$$
This neglects friction, gas inertia, leakage and dissipation, with internal equilibration much faster than the oscillation and external thermal equilibration much slower. The small oscillation probes $\gamma$ locally; it need not keep $\gamma$ constant over a large temperature interval.
With $p_{\rm air}=101325\,\mathrm{Pa}$, the supplied dimensions give $p_0=101351.4\,\mathrm{Pa}$ and $B=2.20566\times10^{-5}\,\mathrm{s^2}$. Thus
$$
\boxed{\gamma_{\rm NO}=\frac{4\pi^2B}{(0.0251\,\mathrm s)^2}=1.382,\qquad
\gamma_{\rm He}=\frac{4\pi^2B}{(0.0229\,\mathrm s)^2}=1.660.}
$$
Helium is close to the monatomic <ideal gas> value $5/3$; NO is close to the <diatomic gas> value $7/5$, with rotations active and vibration largely frozen. The numbers are approximate experimental inferences, rather than exact rational values.
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