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Maximizing the Gibbs entropy under normalization alone gives a uniform distribution. Adding a fixed mean-energy constraint gives the Boltzmann distribution $p_n=Z^{-1}{e}^{-\beta E_n}$.

Here $p(n)$ is the probability of microstate $n$, including all occupation data. In the grand canonical ensemble, with energy $E_n$, particle number $N_n$, chemical potential $\mu$ and $T>0$,Substitution in the Gibbs entropy gives $S=k\log \mathcal{Z}+(U-\mu N)/T$, where $U=⟨E⟩$ and $N=⟨N_n⟩$. Differentiating at fixed $\mu ,V$ gives ${\partial }_T\log \mathcal{Z}=(U-\mu N)/(k{T}^2)$. Therefore

The chemical potential is the change in internal energy when one particle is added at fixed entropy and volume:Equivalently, $\mu =(\partial F/\partial N{)}_{T,V}$ for the Helmholtz free energy.

Let a microstate $s$ have energy $E_s$ and particle number $N_s$. Maximizing the Gibbs entropy subject to normalization and fixed mean values of $E$ and $N$ is the maximum-entropy derivation of equilibrium ensembles. With Lagrange multipliers $\alpha$, $\beta$, and $-\beta \mu$, varyThe stationarity equation is $-\log p_s-1-\alpha -\beta (E_s-\mu N_s)=0$. Normalization therefore gives the grand canonical ensembleHere $\mathcal{Z}$ is the grand canonical partition function and $k_B$ is the Boltzmann constant.

For one fermionic quantum state of energy $\epsilon$, the Pauli exclusion principle permits occupation numbers only $n=0,1$. Its two grand-canonical weights are $1$ and $e^{-\beta (\epsilon -\mu )}$, so its mean occupation is the Fermi-Dirac distribution

For a free nonrelativistic particle, $\epsilon ={\hslash }^2{k}^2/(2m)$. In a region of area $A$, the number of wave-vector states in the annulus from $k$ to $k+dk$, including the two spin angular momentum states, isSince $kdk=(m/{\hslash }^2)d\epsilon$, the two-dimensional free-electron density of states is constant:

We now use units in which $k_B=1$, as in the question. At zero temperature, the Fermi-Dirac distribution is a step function, and henceAt positive temperature the same fixed particle number satisfies the exact relationThusThis is the low-temperature particle-number cancellation for constant density of states. Therefore, for $T\ll {\epsilon }_F$, the mean number of particles in the energy interval $[\epsilon ,\epsilon +d\epsilon ]$ is

Finally, compare the internal energy with its zero-temperature value. The thermally excited particles above ${\epsilon }_F$ and holes below ${\epsilon }_F$ have equal leading particle numbers, so their terms proportional to ${\epsilon }_F$ cancel. With $x=|\epsilon -{\epsilon }_F|/T$,where the integral follows from the fermion-to-boson thermal integral ratio. Differentiation gives the linear low-temperature heat capacity of a two-dimensional Fermi gas:The requested power law is therefore linear in $T$, with exponent one.

For the Gibbs entropyfirst maximize subject only to ${\sum }_n{p}_n=1$ over $\Omega$ accessible microstates. A Lagrange multiplier givesso every $p_n$ is equal. Normalization yields the microcanonical ensemble

For the canonical ensemble, impose both normalization and the mean-energy constraint ${\sum }_n{p}_n{E}_n=E$. Stationarity ofgives $p_n=C{e}^{-\beta E_n}$. The canonical partition function fixes $C$, soThermodynamic consistency identifies $\beta =(k_{B}T{)}^{-1}$. This is the maximum-entropy derivation of equilibrium ensembles.

Substitution of the canonical probabilities into the Gibbs entropy givesHence