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For one molecule whose translational motion separates from its internal degrees of freedom, the classical one-particle partition function iswhere $q_{\mathrm{int}}$ 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 obeyssoThe internal structure changes the energy, entropy, and heat capacity through $q_{\mathrm{int}}(T)$, 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 $q_{\mathrm{int}}=1$. Using the Stirling formula in $Z_N=(V/{\lambda }^3{)}^N/N!$ and differentiating the Helmholtz free energy $F=-k_{B}T\log Z_N$ gives the Sackur-Tetrode equationWith $N$ fixed, an isentropic process therefore satisfies $V{T}^{3/2}=\text{constant}$, or $T{V}^{2/3}=\text{constant}$. Eliminating $T$ with the ideal-gas law yields the monatomic reversible ideal-gas adiabat

An adiabatic process exchanges no heat with its surroundings, so $\delta Q=0$. For a reversible process it is also isentropic. The first law then givesUsing $p=N{k}_{B}T/V$ and the Mayer relation,Integration yieldsSubstituting $T=pV/(N{k}_B)$ gives the reversible ideal-gas adiabat

An adiabatic process has no heat transfer, $\delta Q=0$. With work done by the gas taken as positive, the first law of thermodynamics givesUsing $dE=C_{V}dT$ and $p=N{k}_{B}T/V$,Integration, followed by the Mayer relation $N{k}_B/C_V=\gamma -1$, yieldsEliminating $T$ with the equation of state gives the reversible ideal-gas adiabat