Loss of pressure during sound propagation
The acoustic intensity of sound from an isotropic point source reduces as a square of the distance. 

This is exactly the same as the power of a received radio wave in free space from an isotropic antenna so I think the question might be considered relevant (somewhat) to this site.

\$I = \dfrac{W}{4\pi r^2}\$

Where I is received acoustic intensity at distance r (watts per sq metre) and W is power emitted from the point source.

The equation describes the surface of a sphere of radius r - in other words all the power from the point source is evenly distributed on the surface of a sphere at distance r. If your receiving microphone (antenna) has a surface area of 1 sq metre and r is 10 metres then theoretically, the power received is 1.59 mW for each watt emitted at the source. For a microphone with an area of 1 sq cm then the received power is one-ten-thousandth of this power i.e. 159 nW.

It's exactly the same analysis as would be done for an isotropic antenna.

If the emitter is directional then this is the same in electromagnetics as having a transmit antenna with gain (and beam width) and ditto if the microphone is directional (normally the case of course).

See this for a more detailed explanation.

The above link goes on to explain that the relative sound pressure levels at two different points can also be calculated using this: -

\$L_{p1} - L_{p2} = 20 log_{10}(\dfrac{r_2}{r_1})\$

It's So similar to RF transmission that it's a little scary and, SPLs (\$L_{p1} or L_{p2}\$), translate exactly to E-field intensity or H-field intensity.