= Solution
For an <ultrarelativistic particle>, $\varepsilon=pc$. Including the two <spin angular momentum>[spin] states of an electron, counting wavevectors in a sphere gives the <density of states>
$$
g(\varepsilon)
=\frac{V\varepsilon^2}{\pi^2\hbar^3c^3}.
$$
The <grand canonical partition function> of the ideal <Fermi gas> is therefore
$$
\log\mathcal Z
=\int_0^\infty g(\varepsilon)
\log\!\left(1+e^{-\beta(\varepsilon-\mu)}\right)d\varepsilon.
$$
Since the <grand potential> satisfies $pV=k_BT\log\mathcal Z$, <integration by parts> gives
$$
pV
=\frac13\int_0^\infty
\varepsilon g(\varepsilon)
\frac{d\varepsilon}{e^{\beta(\varepsilon-\mu)}+1}
=\frac E3.
$$
Thus
$$
\boxed{pV=AE,\qquad A=\frac13}.
$$
This is the <equation of state of an ideal ultrarelativistic gas>.
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