= Solution
An <adiabatic process> exchanges no heat with its surroundings, so $\delta Q=0$. For a reversible process it is also <entropy>[isentropic]. The first law then gives
$$
C_V\,dT=dE=-p\,dV.
$$
Using $p=Nk_BT/V$ and the Mayer relation,
$$
\frac{dT}{T}
=-\frac{Nk_B}{C_V}\frac{dV}{V}
=-(\gamma-1)\frac{dV}{V}.
$$
Integration yields
$$
TV^{\gamma-1}=\text{constant}.
$$
Substituting $T=pV/(Nk_B)$ gives the <reversible ideal-gas adiabat>
$$
\boxed{pV^\gamma=\text{constant}}.
$$
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