 Solution

At fixed temperature, equilibrium blackbody radiation has fixed energy density $u$, with radiation pressure $p=u/3$. Doubling the volume therefore doubles its internal energy, so $\Delta U=U_0$. The work done by the radiation isThe first law of thermodynamics, with positive work done by the radiation, givesThus $1.00\mathrm{mJ}$ is absorbed from the walls. The walls must replace the energy used to do work as well as supply the extra stored energy; using only $\Delta U=750\mu \mathrm{J}$ would omit that work. This is a reversible isothermal expansion of a photon gas, not an adiabatic expansion.