= Solution
An <adiabatic process> has no heat transfer, $\delta Q=0$. With work done by the gas taken as positive, the <first law of thermodynamics> gives
$$
dE=\delta Q-p\,dV=-p\,dV.
$$
Using $dE=C_VdT$ and $p=Nk_BT/V$,
$$
C_V\frac{dT}{T}=-Nk_B\frac{dV}{V}.
$$
Integration, followed by the <Mayer relation> $Nk_B/C_V=\gamma-1$, yields
$$
TV^{\gamma-1}=\text{constant}.
$$
Eliminating $T$ with the equation of state gives the <reversible ideal-gas adiabat>
$$
\boxed{pV^\gamma=\text{constant}},
\qquad \gamma=\frac{C_p}{C_V}.
$$
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