 Solution

For an isolated system with equally probable accessible microstates, Boltzmann entropy is $S=k_B\ln \Omega$. The second law states that an isolated system's total entropy cannot decrease; it remains constant in a reversible process and increases in an irreversible one.

The insulated mixture exchanges no heat with its surroundings. At constant pressure, with only pressure-volume work, its enthalpy is conserved. Constant and equal specific heat capacities therefore give $T_f=(T_1+T_2)/2$. Calculate each entropy change along a hypothetical reversible temperature path:The surroundings gain no entropy through heat exchange; ideal mechanical work carries no entropy. The arithmetic-geometric mean inequality gives $\Delta S\ge 0$, with equality only for $T_1=T_2$. Mixing unequal temperatures is irreversible even though total energy is conserved.