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Let $\Lambda =h/\sqrt{2\pi m{k}_{\mathrm{B}}T}$ be the thermal de Broglie wavelength. The two free momentum components of one adsorbed atom give the phase-space integralEach momentum integral is a Gaussian integral, soThe second expression is the classical ideal-gas partition function for mobile, indistinguishable surface atoms. Its validity requires a dilute nondegenerate gas, $N{\Lambda }^2/\sigma \ll 1$, and neglects interactions and excited motion normal to the surface. The $N!$ appears here because particles move between positions, unlike the fully occupied fixed-site model of A6.

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

For $N$ identical, noninteracting classical particles, the one-particle phase-space integrals factorize and indistinguishability supplies the Gibbs factor. Thus the classical ideal-gas partition function rule gives