Nice strawman - you build that yourself? The point is not that it somehow happens "instantaneously", but that the air is going to flow from high pressure to low pressure. In your tube, that means flowing down the tube from the open end to the closed end. Whether it takes 1 second, 10 seconds, ten minutes, or an hour to equalize the pressure with the outside atmosphere is neither here nor there - the point is that it's going to happen, and that flow is going to move down the tube in the opposite direction of your payload and meet it at some point.
Beg to differ. From beginning of depressurization to vehicle away would be less than maybe 1/8 of a second. No let's say I have 12 miles of tube. At mile marker one I place one baloon, at mile marker two I place two baloons, and so on, I pop all the balloons at once. The pressure readings are going to differ for some time after. No matter what scale you use this is the way it works. Been fun, but I gotta go coach soccer, I'll see what you have when I get back.
With my spelling being what it is, I should perhaps refrain from referencing grade levels ;)
Oops, I'm so smart I posted this to myself ;)
Now for you. In your model, how does the front of the wave "know" what the final equalized pressure will be?
In fact, as air enters the tunnel it creates a bit of a vacuum at the entry compared to the surrounding area. The air molecules at the front of the wave bounce around, vice proceed on a direct path, thus, some of that leading wave is always turning back. If the pressure behind them isn't as great, as it was at the beginning, a forgone conclusion, then not as many bounce back in the initial direction. As time progresses this leads to a diminishing pressure. This effect is constant trough out the air mass, and thus you have the effect of slowly (relatively speaking), or perhaps better put, lineally increasing pressure in the path of propagation.
Despite the subject, this isn't really rocket science, but rather freshman (in HS) chemistry.