= Solution
For a reversible adiabatic change, $dS=0$, equivalently $C_VdT=-p\,dV$. Substitute $p=RT/(V-b)$:
$$
C_V\frac{dT}{T}+R\frac{dV}{V-b}=0.
$$
Integration, with fixed positive reference values $T_*,W_*$, gives
$$
C_V\ln\frac{T}{T_*}+R\ln\frac{V-b}{W_*}=\mathrm{constant},
\qquad T(V-b)^{R/C_V}=\mathrm{constant}.
$$
Since $T=p(V-b)/R$ and $\gamma=1+R/C_V$,
$$
\boxed{p(V-b)^\gamma=\mathrm{constant}.}
$$
The relation applies to the quasistatic reversible <adiabat>. Adiabatic <heat> isolation alone does not enforce it for an irreversible process which generates <entropy>. The shifted <volume (thermodynamics)> $V-b$, rather than $V$ itself, is the appropriate coordinate in this model.
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