Thermodynamic derivation of blackbody radiation pressure

For an equilibrium system with energy $E(T,V)$ and no conserved particle number, exactness of the entropy differential obtained from $dE=TdS-PdV$ givesIf $E=aV{T}^4$ and $P=o(T)$ as $T\to 0$, integration gives $P=a{T}^4/3=E/(3V)$.