= Solution
For the same atoms in a three-dimensional volume $V$, the one-atom <translational partition function> is $q_{\rm trans}^{(3)}=V/\Lambda^3$. Comparing the same $N$ atoms at the same <temperature> cancels the common $N!$ in their <canonical partition functions>. Therefore the molar <adsorption confinement free energy>, with the sign defined as adsorbed minus free, is
$$
\boxed{\Delta F_m^{\rm trans}=-RT\ln\frac{q_{\rm trans}^{(2)}}{q_{\rm trans}^{(3)}}
=RT\ln\frac{V}{\sigma\Lambda}.}
$$
The logarithm is dimensionless: $V/\sigma$ and $\Lambda$ are both lengths. For a macroscopic gas depth $V/\sigma\gg\Lambda$, confinement \b[raises the translational <Helmholtz free energy>]. Equivalently, the translational state count is reduced by a factor $\sigma\Lambda/V$ per atom. The question supplies neither a volume nor an adsorption binding <energy>, so this answer is symbolic and describes the confinement cost, not a numerical net adsorption free <energy>.
For example, a binding <energy> $D>0$ per atom would supply a separate $-N_{\rm A}D$ to $\Delta F_m$, together with any normal-motion internal contribution retained by a more detailed model. Adsorption can then be favourable despite the loss of free translational phase space. Merely calling the change a loss must not reverse the sign of the explicitly defined $F_{\rm ads}-F_{\rm gas}$.
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