Ex. 4

$n$ moles of an ideal gas at temperature $T_0$ are originally confined to half of an insulated container by a partition. The partition is removed without doing any work. What is the final change in entropy?

First, we must resist the temptation to say ${}\raise0.44ex\hbox{\bf\symbol{'040}}\llap{d}Q=0$ (because process is adiabatic) hence ${\rm d}S=0$. In fact ${\rm d}S={}\raise0.44ex\hbox{\bf\symbol{'040}}\llap{d}Q^{\rm rev}/T$ and this is not a reversible process. To solve this problem, we need to find a reversible process linking the same two endpoints. We've already seen this; the process is a reversible isothermal expansion (see here.) For such an expansion $Q=-W=nRT_0 \ln(V_2/V_1)$ so $\Delta S=Q/T_0=nR \ln 2$ (details of the work calculation here).

This is rather subtle; make sure you understand the difference between the actual process (with $Q=0$ and $W=0$) and the reversible process for which we could calculate the entropy change (with $Q=-W\ne0$). Remember, $Q$ and $W$ are not functions of state - only the sum is.