= Solution
The <first law of thermodynamics> is
$$
dE=T\,dS-P\,dV,
$$
so, regarding $S$ as a function of $(T,V)$,
$$
\left(\frac{\partial S}{\partial T}\right)_V
=\frac1T\left(\frac{\partial E}{\partial T}\right)_V,
\qquad
\left(\frac{\partial S}{\partial V}\right)_T
=\frac1T\left[
\left(\frac{\partial E}{\partial V}\right)_T+P
\right].
$$
Equality of mixed <partial derivative>[partial derivatives] gives the <thermodynamic derivation of blackbody radiation pressure>
$$
\boxed{
\left(\frac{\partial E}{\partial V}\right)_T
=T\left(\frac{\partial P}{\partial T}\right)_V-P}.
$$
Writing $a=\pi^2/(15\hbar^3c^3)$, we have $E=aVT^4$, and hence
$$
T\frac{dP}{dT}-P=aT^4.
$$
Equivalently, $d(P/T)/dT=aT^2$, so
$$
P=\frac a3T^4+CT.
$$
The condition $P=o(T)$ as $T\to0$ forces $C=0$. Therefore the <radiation pressure> is
$$
\boxed{P=\frac{\pi^2T^4}{45\hbar^3c^3}
=\frac{E}{3V}}.
$$
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