Power of thermal radiation in gases How much thermal radiation does a given volume of gas emit? Does it go according to Stefan-Boltzmann law, i.e. as
$P=A\varepsilon \sigma T^4$,
with $A$ as the cross-sectional area, $\varepsilon$ as emissivity, $\sigma$ as Stefan-Boltzmann constant? If so, what is the value of emissivity? My first guess is $\varepsilon=1$.
 A: You need to be clear what you mean by "thermal emission". The phrase means (to an astrophysicist) that the source function of the gas depends only on temperature. That is a necessary but not sufficient condition for the emitting object to be treated as a blackbody.
As such, it isn't clear at all that you can use the Stefan's law in any form whatsoever and certainly not by inserting a constant emissivity. A good counter-example would be thermal bremsstrahlung, which certainly is "thermal emission" that comes from gases (e.g. from the solar corona) and looks nothing like blackbody radiation.
That said, of course you can always say the power emitted is equal to $\epsilon A \sigma T^4$, where $\epsilon$ is a flux-weighted "emissivity" - basically a fudge factor that means the total power emitted is corrected for the fact your gas is not a blackbody. But $\epsilon$ is not an intrinsic property of the gas - it depends on composition, pressure, density and geometry. In such cases, it is also not a given that $\epsilon$ is temperature independent(!) because the amount of emission from a gas does not necessarily depend on $T^4$ (e.g. the bremsstrahlung example above).
However, the good news is that the total emission from an "optically thin" gas (one which is transparent to the radiation it is emitting) will be proportional to its volume. Or more accurately, it will be proportional to its emission measure, usually defined as $n_e^2 V$, where $n_e$ is the electron number density.
On the contrary, blackbody radiation does not necessarily tell you anything about the volume of the emitting object, only the emitting area visible to the observer. An example of a gas where this works would be the gas that makes up the Sun. The emission from the Sun approximates to a blackbody and can tell us what the emitting surface area of the Sun is, but not its volume without knowing or assuming something about its geometry.
If your gas is somewhere between optically thick and optically thin, then the solution is much more complicated and a proper radiative transfer approach is required. There is no simple equation and the link to the volume of the gas will be weak, but geometry-dependent.
A: I emailed a meteorologist, and he explained me that Stefan-Boltzmann law cannot be used for gases, as it assumes that the body can emit in all frequencies, whereas gases emit only in the very same frequencies where they absorb.
Consequently, I guess that the total power would need to be calculated by integrating Planck's law times the gas-specific spectrum, i.e.
as
$P(T)=\int B_{\nu}(\nu,T)S(\nu)d\nu$.
Well, this does not have units of power exactly, as $B_{\nu}(\nu,T)$ has units of $Wsr^{-1}m^{-2}Hz^{-1}$, and consequently $P(t)$ has units of $Wsr^{-1}m^{-2}$, but I guess one might call it a power density. 
