= Solution
For a single-particle state of energy $\epsilon$, put $z=e^{\mu/(k_BT)}$. The equilibrium mean occupancies are
$$
\boxed{\bar n_{\rm MB}=ze^{-\epsilon/(k_BT)},\qquad
\bar n_{\rm FD}=\frac1{z^{-1}e^{\epsilon/(k_BT)}+1},\qquad
\bar n_{\rm BE}=\frac1{z^{-1}e^{\epsilon/(k_BT)}-1}.}
$$
The <Maxwell-Boltzmann distribution> is the dilute classical limit. The <Fermi-Dirac distribution> applies to identical <fermions> with at most one particle in each complete single-particle state by the <Pauli exclusion principle>. The <Bose-Einstein distribution> applies to identical <bosons> with no such upper occupation bound. The <chemical potential> fixes total particle number; these are mean state occupancies, not three separately normalized single-particle probability densities. For a number-conserving ideal <Bose gas>, $\mu$ cannot exceed the lowest single-particle energy.
Take $\mathbf k$ to be a <wavevector> and count periodic-box states with cell volume $(2\pi)^3/V$ in <momentum space>, parameterized by $\mathbf k=\mathbf p/\hbar$. Write $g_s$ for internal degeneracy. The number of states in a spherical shell is $g_sV\,4\pi k^2dk/(2\pi)^3$. Since $\epsilon=\alpha k$,
$$
\boxed{g(\epsilon)=\frac{g_sV}{2\pi^2\alpha^3}\epsilon^2,\qquad \epsilon\ge0.}
$$
Thus the one-particle <canonical partition function> is
$$
Z_1=\int_0^\infty g(\epsilon)e^{-\beta\epsilon}d\epsilon
=\frac{g_sV}{2\pi^2\alpha^3}\frac2{\beta^3}
=\boxed{aVT^b,\qquad a=\frac{g_sk_B^3}{\pi^2\alpha^3},\quad b=3.}
$$
Here $\beta=1/(k_BT)$, $g_s=1$ for one nondegenerate branch, and $g_s=2$ for the spin-half case below. The integral uses $\int_0^\infty u^2e^{-u}du=2$. For classical indistinguishable particles, $Z_N=Z_1^N/N!$, giving
$$
\boxed{U=-\partial_\beta\ln Z_N=3Nk_BT,\qquad p=k_BT\left(\frac{\partial\ln Z_N}{\partial V}\right)_{T,N}=\frac{Nk_BT}{V}.}
$$
The factor three in the energy arises from the linear <dispersion relation> and three-dimensional state counting; the usual nonrelativistic quadratic-dispersion value $3Nk_BT/2$ does not apply.
For spin-half <fermions> at $T=0$, all states up to the <Fermi energy> are occupied. With $g_s=2$,
$$
N=\frac{V\epsilon_F^3}{3\pi^2\alpha^3},\qquad
\epsilon_F=\alpha(3\pi^2n)^{1/3},\qquad n=\frac NV.
$$
Integrating the occupied energy states gives
$$
\boxed{U_0=\frac34N\epsilon_F=\frac34\alpha(3\pi^2)^{1/3}N^{4/3}V^{-1/3},\qquad
p_0=-\left(\frac{\partial U_0}{\partial V}\right)_N=\frac14n\epsilon_F=\frac{U_0}{3V}.}
$$
These are the corresponding <ultrarelativistic ideal Fermi gas> formulae with general slope $\alpha$. The nonzero ground-state <internal energy> and <degeneracy pressure> reflect the <Pauli exclusion principle>: particles cannot all occupy the lowest-momentum state. Formally extrapolating the classical equations gives $U,p\to0$, but <Maxwell–Boltzmann statistics> is not valid at fixed positive $n$ as $T\to0$.
At large $T$, the number equation gives $z\simeq N/Z_1\ll1$, so expanding the <Fermi-Dirac distribution> recovers the <Maxwell-Boltzmann distribution>. A convenient crossover scale is the <Fermi temperature>,
$$
\boxed{T_0\sim T_F=\frac\alpha{k_B}(3\pi^2n)^{1/3}.}
$$
For $T\gg T_F$, both gases have $p\simeq nk_BT$. At fixed $n$, the Fermi curve begins at $p_0=n\epsilon_F/4$, has zero initial slope, lies above the classical line, and approaches it at high temperature. For example its low-temperature expansion is $p/(n\epsilon_F)=1/4+(\pi^2/6)(T/T_F)^2+\cdots$. The plot evaluates the finite-temperature occupation integrals at fixed number density.
\Image[/past-exam-of-the-thermal-physics-course-of-the-university-of-oxford/2016-linear-dispersion-gas-pressure.png]
{title=Pressure of classical and spin-half Fermi gases with linear dispersion}
{description=Original fixed-density calculation. The Fermi curve is obtained by solving the number equation for the chemical potential at each temperature; its positive zero-temperature pressure and the shared classical limit are marked.}
If the source's symbol $\mathbf k$ is interpreted literally as <momentum> rather than a <wavevector>, the state-space cell is $h^3/V$. In that convention replace $\alpha$ in the wavevector-based coefficients and energy scales above by $\hbar\alpha$. The dimensionless thermodynamics and the relation $p=U/(3V)$ are unchanged.
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