= Solution
For one molecule whose translational motion separates from its internal degrees of freedom, the classical one-particle <partition function> is
$$
z_1(T,V)=\frac{V}{\lambda^3}q_{\rm int}(T),
\qquad
\lambda=\sqrt{\frac{2\pi\hbar^2}{mk_BT}},
$$
where $q_{\rm int}$ contains rotational, vibrational, and other internal states. For $N$ indistinguishable noninteracting molecules, the <classical ideal-gas partition function> is $Z_N=z_1^N/N!$. The thermodynamic pressure therefore obeys
$$
P=k_BT\left(\frac{\partial\log Z_N}{\partial V}\right)_{T,N}
=\frac{Nk_BT}{V},
$$
so
$$
\boxed{PV=Nk_BT.}
$$
The internal structure changes the energy, entropy, and heat capacity through $q_{\rm int}(T)$, but it does not change the <ideal gas> law while the internal partition function is independent of volume and the molecules have negligible interactions.
For a dilute monatomic gas $q_{\rm int}=1$. Using the <Stirling formula> in $Z_N=(V/\lambda^3)^N/N!$ and differentiating the <Helmholtz free energy> $F=-k_BT\log Z_N$ gives the <Sackur-Tetrode equation>
$$
\boxed{S=Nk_B\left[
\log\left(\frac{V}{N}\left(\frac{2\pi mk_BT}{h^2}\right)^{3/2}\right)
+\frac52\right].}
$$
With $N$ fixed, an <isentropic process> therefore satisfies $VT^{3/2}=\text{constant}$, or $TV^{2/3}=\text{constant}$. Eliminating $T$ with the ideal-gas law yields the monatomic <reversible ideal-gas adiabat>
$$
\boxed{PV^{5/3}=\text{constant}.}
$$
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