68.

Let us also consider a reversible adiabatic process. For this it is necessary not merely that the piston and the mantle but also that the bottom of the cylinder be assumed as completely reflecting, e.g., as white. Then the heat furnished on compression or expansion of the volume of radiation is $Q=0$ and the energy of radiation changes only by the value $pdV$ of the external work. To insure, however, that in a finite adiabatic process the radiation shall be perfectly stable at every instant, i.e., shall have the character of black radiation, we may assume that inside the evacuated cavity there is a carbon particle of minute size. This particle, which may be assumed to possess an absorbing power differing

from zero for all kinds of rays, serves merely to produce stable equilibrium of the radiation in the cavity ([sect:51.] Sec. 51 et seq.) and thereby to insure the reversibility of the process, while its heat contents may be taken as so small compared with the energy of radiation, $U$, that the addition of heat required for an appreciable temperature change of the particle is perfectly negligible. Then at every instant in the process there exists absolutely stable equilibrium of radiation and the radiation has the temperature of the particle in the cavity. The volume, energy, and entropy of the particle may be entirely neglected.

On a reversible adiabatic change, according to [eqn:(72)] (72), the entropy $S$ of the system remains constant. Hence from [eqn:(80)] (80) we have as a condition for such a process $T^{3}V=\text{const}.,$ or, according to [eqn:(77)] (77), $p{V}^{\frac{4}{3}}=\text{const}.,$ i.e., on an adiabatic compression the temperature and the pressure of the radiation increase in a manner that may be definitely stated. The energy of the radiation, $U$, in such a case varies according to the law $\frac{U}{T}=\frac{3}{4}S=\text{const}.,$ i.e., it increases in proportion to the absolute temperature, although the volume becomes smaller.