what is the relationship of sound volume to atmospheric pressure? what is the relationship of sound volume to atmospheric pressure? if I was in a plane with a cabin pressure equal to 8000 feet would the volume drop be noticable?
 A: While sound intensity is usually measured in decibels (dB), the physical effect of sound can be measured as the amplitude of pressure in the medium in which it is propagating (in this case, Earth's atmosphere).  Remember that decibels are a logarithmic scale, but pressure is measured in SI units as pascals (Pa).
The accepted scientific standard for the threshold of human hearing is defined as $\text{0 dB}$ sound pressure level (SPL) on the logarithmic decibel scale, and the equivalent pressure is $\textrm{20 }\mu \text{Pa}$.
Standard atmospheric pressure at sea level is $\text{101325 Pa}$, and this converts to $\text{194.1 dB}$ SPL.
\begin{equation*}
\text{20} \cdot \log_{10}\dfrac{101325\text{ Pa}}{0.00002\text{ Pa}} = 194.1 \text{ dB}.
\end{equation*}
WARNING:  Do NOT attempt to reproduce, much less listen to sounds of this intensity.  Sounds above $\text{120 dB}$ SPL can permanently damage your hearing and also cause bodily injury.  Interestingly, $\text{120 dB}$ SPL is "only" $\text{20 Pa}$.
As you climb higher in the atmosphere, the pressure does indeed decrease, and at $\text{8000 ft}$ the standard pressure is $\text{75263 Pa}$.  Converting this to dB yields $\text{191.5 dB}$.  An interesting correlation is that the difference in theoretical max SPL between two altitudes also indicates the SPL attenuation of ordinary sounds between the two heights (some examples listed below).
Being just $\text{2.6 dB}$ "softer" than at sea level, I don't think anyone would notice a difference.
Here are some maximum SPLs (and attenuation relative to sea level) calculated based on the pressure at various altitudes of the ISA:


*

*At $\text{8848 m} / \text{29029 ft}$ (Mount Everest):  $183.9 \text{ dB}$ max SPL, $10.2\text{ dB}$ attenuation.  Just noticeably quieter than at low altitudes.

*At $\text{30000 m} / \text{98400 ft}:  155.4 \text{ dB}$ max SPL, $38.7\text{ dB}$ attenuation.  About the same as most industrial-grade hearing protection.

*At $\text{50000 m} / \text{164000 ft}:  131.6 \text{ dB}$ max SPL, $62.5\text{ dB}$ attenuation.  A human voice a few feet away would not be audible.

*At $\text{100000 m} / \text{62.1 mi}$ (The Kármán Line):  $62.2 \text{ dB}$ max SPL, $131.9\text{ dB}$ attenuation.  Sounds loud enough to permanently damage hearing and cause injury at sea level are inaudible at this height.

*At $\text{400000 ft}$ (NASA "entry interface" for manned spacecraft re-entry):  $33.7 \text{ dB}$ max SPL, $160.4 \text{ dB}$ attenuation.  Imagine the volume of sound standing right next to the Space Shuttle at launch--even this would be inaudible at that height.

*At $\text{200 km}: 3.69 \text{ dB}$ max SPL.  The loudest noise theoretically possible at this height would barely be audible.  The pressure calculated at this height is merely theoretical; actual pressure would be highly variable.


NOTE: I haven't factored in the maximum frequency heard at altitude--the maximum frequency decreases as height increases, due to the mean free path getting larger.
EDIT Corrected minor math errors in attenuation values. Added Mount Everest bullet.
ADDENDUM As a practical example of how sound attenuation "happens" as you increase altitude, check out this video of the STS-115 launch from the external tank (ET) camera.  Notice that the "rumble" of the SRBs and SSMEs tends to diminish to nearly inaudible (the ET camera records audio, too) as the vehicle climbs higher.
A: There is not going to be any generic answer to your question.
Suppose I drop a book on the floor, and the book hits the floor with 1 joule of kinetic energy. Let's assume that there is very little internal friction in the book or the floor, so that none of the energy can be dissipated into heat, and that the floor is very stiff, so that no energy can be transmitted as vibrations in the floor. Then by conservation of energy, the amount of sound energy is going to be exactly 1 joule. This argument is independent of whether this is the floor of my house, near sea level, or the deck of an airplane. It only depends on the assumption that the mechanisms for dissipating energy into heat and vibration are much less efficient than the mechanism for dissipating it into sound waves in the air.
On the other hand, let's say for the sake of argument that in a hand clap at sea level, 1/3 of the energy goes into sound, 1/3 into vibrations transmitted down through the arms, and 1/3 into heating in the palms. In an airplane, these 1:1:1 proportions will be altered, because the conversion of energy into sound waves in air will be less efficient. Because the competition among the three dissipative processes is fairly equal, there will be a big reduction in loudness when the pressure is low.
So the answer is that it depends completely on the details of the process of production of the sound waves.
In the extreme case, suppose you do these experiments in outer space. Outer space is not a perfect vacuum, so it can support sound waves. However, the efficiency of the coupling to sound waves is extremely low, so almost any other mechanism for dissipating the energy will be many orders of magnitude more efficient.
