To effectively suppress normal audible sound, how wide and how absolute would a vacuum space have to be? I don't know if this question is too specific or simple, but: to effectively suppress (say 99%) of normal audible sound (say 20-20kHz @ 100dB), how wide (mm?) and how absolute (torr?) would a vacuum space have to be? 
 A: This is really a nice question. After reading your question I have searched the web with the idea that sound is a pressure wave and in order to hear them it must transfer energy to your ears. I have found this article which is really nice. It says that (and so was my intuition) the energy density of the sound wave is directly proportional to the density of the medium. Hence if you decrease the density of the medium the energy stored in the medium decreases and less energy is transferred from that medium. Hence if you want to decrease the sound intensity by a factor of ~100 decrease the pressure to ~10 mbar.
Now if you have a vacuum curtain (i.e. suddenly the pressure of a portion of the medium is decreased) then the sound will not propagate. Now question is how thin we can make this curtain. I believe that the sound wave setup an oscillation in the gas molecules and the extent of this oscillation is of the order of wavelength. Hence if you want to effectively suppress the sound wave then the thickness of the vacuum curtain must be larger than the sound wavelength (or else the molecules will overshoot the curtain and transfer energy at other side). It may be noted that this analogy of free vacuum curtain is adopted to avoid any other type of effects i.e. sound propagation through the metal enclosures (I can hear sound of motors from inside the vacuum chambers) or scattering of sound waves from such enclosures.
