# 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 ?)

• According to you, the Nitrogen is at 300K, yeah? What is the temperature of the surroundings you've kept in? Is it absolute zero? – Hritik Narayan Feb 7 '15 at 13:10
• The sphere is in the void, and let's say that the sorrounding "observing" universe is at absolute zero. – Hyades Feb 7 '15 at 13:14
• Unless the sphere has a surface the N2 gas is gonna expand and diffuse into space – pentane Feb 7 '15 at 16:29
• @pentane. Yes, you're right, but this is a theoretical case, for studying radiative thermal behaviour of a gas. As written by Brionius, there is a magical forcefield that keeps its shape and volume constant. – Hyades Feb 7 '15 at 17:30
• Nitrogen at 1 bar in a 1m sphere at 300K is nothing like a blackbody. It is optically thin to visible and IR radiation. You need the emission coefficient integrated over all wavelengths (which will not be $\sigma T^4$) multiplied by the (small) optical depth. – Rob Jeffries Oct 29 '15 at 23:15

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 screenshot of the graph in case you can't use Desmos. Caveats:

1. 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.

2. 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.

3. 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.

• I've tested the expression for a 1 m sphere at 1 bar for 1 s, and it gives 1.4e-5 K. Is it an error from my calculation or do you have the same result ? – Hyades Feb 7 '15 at 15:20
• Do we need to include an emissivity for the pure N2 in the SB law ? It's difficult to find any value for N2 regarding this quantity, do you know any reference website where I may find it ? – Hyades Feb 7 '15 at 15:22
• Just a precision : I used the perfect gas law to find N=168 mol – Hyades Feb 7 '15 at 15:23
• Yes, I agree that the blackbody assumption breaks down when the distance between the particles approaches the dimensions of the sphere. But for a large number of particles, it would radiate the same amount of power as a sphere of iron at the same temperature. Note power, not energy. The amount of energy radiated would be much less, and the temperature would drop considerably faster. – Brionius Feb 7 '15 at 20:12
• I think it's a stretch to say the surface emission is the same as it would be for a solid. Atoms deeper inside the gas can participate in the emission since their radiation has a good probability of escape; this will affect the results in a way that I find difficult to calculate. I think you are making unsubstantiated assumptions - how the emission will vary with density is an interesting question which I don't think your answer addresses. – Floris Feb 7 '15 at 20:41

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.

• Is it sure that the black-body law can be used in the case of a gas ? Will the pure N2 sphere emit a black-body spectrum ? I mean that I find it hard to link the black-body assumptions (cavity) with the case of N2 sphere. – Hyades Feb 7 '15 at 14:27
• I've read on numerous website that the light from a flame comes from soot, and not from the heated atmosphere. If the N2 sphere above is at 1000 K, will it glow as red iron, as it would be if a black-body – Hyades Feb 7 '15 at 17:49
• @Hyades yes the soot particles, being more dense and more black, make good radiators. I did say that you need to take emissivity of the molecules into account - this may be quite low. – Floris Feb 7 '15 at 20:43