= Solution
The <nuclear Fermi gas model> treats <protons> and <neutrons> as two species of <spin one-half> <fermions> moving approximately independently in a confining <nuclear mean field>. At a <temperature> small compared with their <Fermi temperature>, the <Pauli exclusion principle> fills the lowest available <momentum> states. It does not allow all <nucleons> to sit at zero <momentum>.
For species $i=p,n$, two <spin> states occupy each momentum-space cell $(2\pi\hbar)^3/V$. Counting a filled sphere gives
$$
N_i=\frac{2V}{(2\pi\hbar)^3}\frac{4\pi p_{F,i}^3}{3},\qquad
p_{F,i}=\hbar(3\pi^2 n_i)^{1/3},\qquad E_{F,i}=\frac{p_{F,i}^2}{2m_N},
$$
where $n_i=N_i/V$. The <Fermi energy> is measured from the mean-field well bottom; it need not equal the <energy> needed to remove a <nucleon> into vacuum. Occupied-state averaging gives
$$
E_{\mathrm{kin}}=\frac35(NE_{F,n}+ZE_{F,p}),\qquad P_i=\frac25 n_iE_{F,i}.
$$
Even at zero <temperature> there is <kinetic energy> and <degeneracy pressure>. As an explicitly illustrative input, total density $n=0.16\,\mathrm{fm}^{-3}$ and $N=Z$ give $p_Fc\simeq263\,\mathrm{MeV}$ and $E_F\simeq37\,\mathrm{MeV}$. These numbers illustrate the formula; they are not additional data supplied by the paper.
Separate <proton> and <neutron> seas explain a kinetic contribution to <nuclear symmetry energy>. At fixed $n$, put $\delta=(N-Z)/A$ and define $E_F$ at $n_p=n_n=n/2$. Then
$$
\frac{E_{\mathrm{kin}}}{A}
=\frac35E_F\frac{(1+\delta)^{5/3}+(1-\delta)^{5/3}}2
=\frac35E_F+\frac13E_F\delta^2+O(\delta^4).
$$
Increasing one species' population raises its <Fermi surface> and costs <energy>, favoring similar populations before electrostatic and interaction terms are included.
<pauli blocking> suppresses collisions deep in a filled sea because scattering needs unoccupied final states. An energetic external probe can knock a <nucleon> into an empty state and reveal its initial <Fermi motion>, making the model useful for broad inclusive scattering distributions. A finite surface can be approximated using a position-dependent <nuclear mean field>.
A free gas cannot produce nuclear binding or saturation: those require an attractive interaction and its density dependence. The smooth gas picture also omits discrete <nuclear shells>, <nuclear spin-orbit splitting>, <nuclear pairing>, short-distance correlations and deformation. It cannot assign all ground-state <spins> from a <momentum> sphere. \b[Its useful prediction is the scale and occupancy of <nucleon> motion, with a surrounding mean field supplied separately.]
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