= Solution
At fixed <temperature>, equilibrium <blackbody radiation> has fixed <energy density> $u$, with <radiation pressure> $p=u/3$. Doubling the <volume (thermodynamics)> therefore doubles its internal <energy>, so $\Delta U=U_0$. The <work> done by the radiation is
$$
W=\int_{V_0}^{2V_0}p\,dV=\frac u3V_0=\frac{U_0}{3}.
$$
The <first law of thermodynamics>, with positive <work> done by the radiation, gives
$$
\boxed{Q=\Delta U+W=\frac43U_0=1000\ \mu\mathrm J=1.00\ \mathrm{mJ}.}
$$
Thus \b[$1.00\ \mathrm{mJ}$ 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.
Back to article page
