= Solution
For relativistic one-particle energy
$\epsilon(p)=\sqrt{p^2c^2+m^2c^4}$, replace $q_1$ by
$$
q_1^{\rm rel}=\frac{V}{h^3}\int_{\mathbb R^3}
e^{-\beta\epsilon(p)}\,d^3p.
$$
Classical particle counting still gives
$$
\mathcal Z=\exp(e^{\beta\mu}q_1^{\rm rel}),\qquad
\langle N\rangle=e^{\beta\mu}q_1^{\rm rel},
\qquad
(\Delta N)^2=\langle N\rangle.
$$
Since $q_1^{\rm rel}$ is proportional to $V$, $\Phi=-k_BT\langle N\rangle=-pV$, and therefore
$$
\boxed{pV=\langle N\rangle k_BT,\qquad
\frac{\Delta N}{\langle N\rangle}
=\frac1{\sqrt{\langle N\rangle}}}.
$$
The classical ideal-gas equation of state is independent of the dispersion relation.
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