Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 2 / A8 / d / i / Solution

Following the restricted model in the preceding part, its electronic factor for $N$ independent magnetic molecules is $Q_{\mathrm{mag}}=(2\cosh x{)}^N$. Its field-dependent Helmholtz free energy is thereforewhere $A_{\mathrm{ind}}$ denotes terms independent of the field. Equivalently, the change from zero field is $A(B)-A(0)=-N{k}_{\mathrm{B}}T\ln \cosh x$. Retaining the excited-state offset simply adds $N\Delta$ to the restricted free energy.

If $N$ instead counts all molecules in thermal equilibrium, the correct electronic expression isThe ordinary translational indistinguishability factor is field independent and does not alter either magnetic derivative.