More precisely, how small is the potency a listener hears, compared to the potency of the emitter.
I'd like to present a simple and yet reasonable approximation, to a high school audience (I am a teacher)
Intuitively, it should decrease with a square (of distance; because of the expansion of the area the wave covers) and exponentially (because of accumulation losses in the way) .Here, they mention both decreases: http://www.sfu.ca/sonic-studio/handbook/Sound_Propagation.html . The air absorption component is linear in dB (which, as far as I know, means exponential on energy)
I am just not sure of how to put them together, and if there are other ideas that a first approximation should cover.
Best Answer
It all depends how much you want to model the problem according to realities. Getting exact answers is almost an impossible task. I was working for a while on noise propagation calculation for traffic and industry sources as a business. Things are extremely complicated, you have dozens of formulas to calculate that. In fact very precise EU and state-dependent regulations exist for calculating sound/noise propagation. Eventually, everybody use special software for that and I guess only those who write software are fully aware of all details.
But as a bottom line: absorption due to air itself (which by the way largery depends on humidity and air temperature!) is much smaller than intensity fall due to geometric reasons (I am talking about earthly dimensions). Even when sound propagates through thick vegetation, the additive reduction according to EU-regulated modelling, is few dB per 100 meter. And bottom line, when doing real absorption calculation in air, exponential formula fails. I would gladly provide you exact formulas from calculation regulations, but they are no longer at my disposal.
As it happens, absorption due to presence of soft materials (like for example earth ground) is more important, and reflections are also a much more acute effect. Other effects are well described in the document you cite. So I think it would be fairly right to say for real problems, that intensity reduces $\frac{1}{r^2}$ for point sources (mostly industrial sources) and $\frac{1}{r}$ for linear sources (mostly traffic sources). If you only explain this difference (point and line sources) I think you shall do enough.
Note: The right derivation for point source (without theoretical absorption) is rather simple
$$A(r, t) = A_0\frac{\sin(k r-\omega t)}{r}.$$
$$I_0 = |A(1,t)|^2 = \frac{A_0^2}{2}$$
$$I = |A(r,t)|^2 = \frac{A_0^2}{2 r^2} = \frac{I_0}{r^2}$$
And here is for linear sources
$$A(\rho, t) = A_0\frac{\sin(k \rho-\omega t)}{\sqrt{\rho}}.$$
$$I_0 = |A(1,t)|^2 = \frac{A_0^2}{2}$$
$$I = |A(\rho,t)|^2 = \frac{A_0^2}{2 \rho} = \frac{I_0}{\rho}$$
There is even much simpler explanation for the formulas above. Intensity is power over area
$$I = \frac{P}{A}.$$
In case if there is no absorption, power remains constant, while area changes. In case of the point source, area increases as $A \propto r^2$, while in case of the linear source, area increases as $A \propto \rho$!
PS: In the document you cite it says "Under 'normal' circumstances, atmospheric absorption can be neglected except where long distances or very high frequencies are involved." This is in full agreement with general experience, as this absorption presents a serious effect in case you calculate distrubtion of noise from huge industrial transformers. Long distance effect is on the other hand completely diminished by other effects.