69.

Let us finally, as a further example, consider a simple case of an irreversible process. Let the cavity of volume $V$, which is everywhere enclosed by absolutely reflecting walls, be uniformly filled with black radiation. Now let us make a small hole through any part of the walls, e.g., by opening a stopcock, so that the radiation may escape into another completely evacuated space, which may also be surrounded by rigid, absolutely reflecting walls. The radiation will at first be of a very irregular character; after some time, however, it will assume a stationary condition and will fill both communicating spaces uniformly, its total volume being, say, $V^{\prime }$. The presence of a carbon particle will cause all conditions of black radiation to be satisfied in the new

state. Then, since there is neither external work nor addition of heat from the outside, the energy of the new state is, according to the first principle, equal to that of the original one, or $U^{\prime }=U$ and hence from [eqn:(78)] (78)

$$\begin{gathered} T^{\prime 4}{V}^{\prime }=T^{4}V \\ \frac{T^{\prime }}{T}=\sqrt[4]{\frac{V}{V^{\prime }}} \end{gathered}$$

which defines completely the new state of equilibrium. Since $V^{\prime }>V$ the temperature of the radiation has been lowered by the process.

According to the second principle of thermodynamics the entropy of the system must have increased, since no external changes have occurred; in fact we have from [eqn:(80)] (80) $\frac{S^{\prime }}{S}=\frac{T^{\prime 3}{V}^{\prime }}{T^{3}V}=\sqrt[4]{\frac{V^{\prime }}{V}}>1.\mathrm{\backslash Label}[eqn](82)\text{upshape (82)}$