# Why pressure is a function of temprature for photon gas?

I am trying to understand formula $(45.17)$ from this Feynman's lecture.

We know that $U=3PV$ for photon gas. Then Feynman concludes

$$\Big( \frac{ \partial U }{ \partial V } \Big)_{T} = 3P$$

It would be obvious if we knew

\begin{equation} \label{*} \Big( \frac{ \partial P }{ \partial V } \Big)_{T} = 0 \tag{*} \end{equation}

Equivalently, $P$ depends only on $T$ but not on $V$.

Question Why should one expect $\eqref{*}$ holds?

The Internal energy of a photon gas is given by $$U=\frac{\pi^2k_B^4}{15 c^3 \hbar^3}V T^4.$$ If one rearranges the expression $U=3PV$ to get P one has: $$P=\frac{U}{3V}.$$ But from the first expression we know, that $U$ divided by $V$ is independent of $V$. So $$\left(\frac{\partial P}{\partial V}\right)_T=0.$$ EDIT: Since pressure is an intensive quanity it should not depend on an extensive quantity.
• There are several equivalent formulations of my question. Why $U \sim V$? Why pressure is intensive quantity? I do not know answer for either of them. I was lost when I tried to read the chapter about it. Maybe it is a bad idea to think naively about it. But if one make volume of ideal gas twice bigger (isothermally), then $U$ is not twice bigger because the concentration is twice smaller. What happens for photon gas? Should I think that when you make the volume twice larger (isothermally), the walls of the box radiates new photons? – quinque Aug 17 '16 at 12:49