Thermal radiation of a nitrogen sphere Let's imagine a :


*

*sphere of 100% pure nitrogen (N2), 

*(edit: 1 m diameter) with a constant volume (edit: using a kind of "magic forcefield")

*(edit : at 1 bar)

*in the void

*far from any light source (edit: background is at 0 K)

*at 300K


At this temperature, we should except any solid material to emit IR and to cool down ; it may be approximated by the dark body using an emissivity coefficient. But what happens to our N2 sphere ?
Will it cool down ? At which rate ?
(edit : Bonus question : if the sphere is at 1000K, will it glow as red iron ?)
Thanks in advance for your answer
 A: At 300K, radiation from your ball of Nitrogen should be pretty well approximated by a black body spectrum.  The power density emitted by a black body is given by the Stefan-Boltzmann law:
$$j = \sigma T^4$$
Where $j$ is the power radiated per unit surface area.  Therefore, we'll need to know the size of your sphere.  Let's keep it general, and say it has a radius $R$, and therefore an area $A = 4 \pi R^2$. I will also assume for simplicity that this sphere of gas is confined by some kind of magic forcefield, and not worry about volume changes.  The total power radiating is therefore
$$P = A j$$
$$P = 4 \pi R^2 j$$
$$P = 4 \pi R^2 \sigma T^4$$
For a diatomic gas, the internal energy is given by $U = \frac{5}{2} N k T$.  You haven't specified how much gas is in this sphere, or the pressure, so we will again retain generality by keeping the number of particles (not moles) of gas at a constant $N$.
Now we are close to the goal of finding out how fast the sphere cools:
$$P = -\frac{dU}{dt}$$
$$4 \pi R^2 \sigma T^4 = -\frac{d}{dt} (\frac{5}{2} N k T)$$
$$4 \pi R^2 \sigma T^4 =  -\frac{5}{2} N k \frac{dT}{dt}$$
$$\frac{dT}{dt} = -\frac{8 \pi R^2 \sigma}{5 N k} T^4 $$
$$\frac{dT}{T^4} = -\frac{8 \pi R^2 \sigma}{5 N k} dt $$
Integrating...
$$- \frac{1}{3} (T^{-3} - T_0^{-3}) = -\frac{8 \pi R^2 \sigma}{5 N k} t$$
$$T^{-3} = \frac{24 \pi R^2 \sigma}{5 N k} t + T_0^{-3}$$
Finally, the temperature as a function of time for a sphere of nitrogen gas due to radiative losses is approximated by
$$T(t) = \left (\frac{1}{ \frac{24 \pi R^2 \sigma}{5 N k} t + T_0^{-3}}\right)^{\frac{1}{3}}$$
Where $T_0$ is the initial temperature, $R$ is the radius of the sphere, and $N$ is the number of particles (not moles) of gas in the sphere.
Here is a Desmos graph of the temperature as a function of time.
Here is a screenshot of the graph in case you can't use Desmos.

Caveats:


*

*This cooling function will hold until the gas becomes cool enough that rotational degrees of freedom are frozen out; then the internal energy will have a different relation to temperature.  The cooling function will have the same form, just with different constants.

*Nitrogen gas is not a perfect blackbody - there are rotational, vibrational, and atomic resonances, but it's probably pretty close, especially at a temperature as low as 300°K.

*This model assumes the thermal energy is lost from the entire volume uniformly, when in fact the nitrogen on the surface will lose the most.  This will result in non-uniform temperature, as well as complicated heat movement via conduction and convection that are difficult to analyze.  These effects will be smaller, and this result more accurate, if the sphere radius $R$ is small.
A: The body will lose heat according to the Stefan-Boltman law, that is the heat loss per unit surface area goes as
$$\epsilon \sigma T^4$$
where $\epsilon$ is the emissivity (which is a function of wavelength). It will be hard to judge how quickly the sphere will lose heat - but it is absolutely certain that it will lose heat, given that the surroundings are cooler and therefore heat can be lost by radiation, but not absorbed.
The fact that this is a gas and not a solid doesn't come into it: that just changes how closely the gas molecules are associated with each other; and molecules inside will radiate but they have equal probability of receiving radiation "from outside" so the next flux is still determined by the surface, that is the number of atoms "on the outside".
On further reflection it is possible that the heat loss per unit time scales with the density of the material - but then so does the heat capacity, so I think that if you do the calculation for "solid nitrogen" and scale it, you will get the right answer. Assuming you can estimate the emissivity accurately.
