Just how much head pressure does a slug need behind it to be lethal

Okay so I know enough to know that in general this is a open ended question, that is to say to make a proper answer all variables would need to be filled in.

Caliber, Weight, Barrel length, smooth or rifled etc 

I "think" I understand for the most effective system the amount of psi/volume needs to be equal to the amount to fill the barrel at psi/volume with the projectile at the end of the barrel.

And so in very layman's terms and my admittedly limited understanding your tank would need to be at minimum 1 1/2 times the volume/psi of the amount needed to reach that goal.

(We are talking about a 1 shot wonder ie fill shoot n repeat) 

I have seen some examples of real math geniuses on this site so I hope someone can shed some light on this for me
 
I did some research on the “wind rifle” a few years back. They did fire a .46 cal. and shot as well. The .46 ball was a 210-215gr projectile and the max. range was estimated at 125 yds. However, the projectile was only being fired between 450-550fps. and if I remember correctly it took between 1300-1600 strokes with a hand pump to fill it up. I don’t remember ever reading that it was filled to 900psi. but I believe you. Really a cool design for the time. Killed Elk and Buffalo with it, probably at very close range.

Higher PSI’s and larger tanks mean your gun is capable of holding more air between fills. 
 
Hmm, challenge accepted. I think the better answer is "What is the MINIMUM pressure behind a slug to be lethal"



Lets see how low I can go...



20 mm Caliber...

275 gr projectile

28 inch barrel

50~ FPE / 290~ FPS @ 200 psi...



This shot would likely eject around 200~ cc's of air to accomplish the above...so yes a one shot wonder. There you have it.