Black-body radiation is said to be the thermal electromagnetic radiation within, or surrounding, a body in thermodynamic equilibrium with its environment or emitted by a black body. For the first case above, would the same be applied to gases or do they behave in a different way?
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$\begingroup$ Relevant: physics.stackexchange.com/questions/290528/… $\endgroup$– MauricioCommented Jan 12, 2023 at 9:47
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If the gas in question can emit and absorb radiation, then it will eventually become black body radiation (i.e., radiation in thermal equilibrium). Thus, we will have equilibrium between the gas and the radiation (a photon gas.)
What is a bit tricky here is how the radiation becomes black body radiation: e.g., if gas consists of atoms with discrete energy spectrum, it will mainly absorb and emit on the frequencies characteristic of this spectrum. The thermal distribution of radiation spanning all frequencies is then established via higher-order processes, such as Raman scattering, multi-photon absorption, etc. This might be a rather slow process: as the answers in the duplicate thread suggest it might never happen in practical situations (since the gas will cool down via the radiative losses to the environment), but it might happen at high gas densities, where there direct interaction between gas atoms contributes to spreading radiation over all frequencies.
