Past exam of the physics B course of the University of Cambridge / 2017 / Paper 2 / Section C / C11 / c / Solution

For a reversible adiabatic change, $dS=0$, equivalently $C_{V}dT=-pdV$. Substitute $p=RT/(V-b)$:Integration, with fixed positive reference values $T_{\ast },W_{\ast }$, givesSince $T=p(V-b)/R$ and $\gamma =1+R/C_V$,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 $V-b$, rather than $V$ itself, is the appropriate coordinate in this model.