...as you've assumed, because the rate of expansion is far too rapid to allow the cooling of expansion to be balanced by absorption of heat from the barrel walls, etc. This makes the rate of pressure drop much faster than just V/V'.
The actual math is P' = P(V/V')^y, where y = 1.3 for CO2
So it only takes a V/V' ratio of about 26:1 to drop 70bar down to 1bar. Not 70:1. So nearly 3x as much gas is required to fill a barrel than your arithmetic suggests.
But that's just the beginning.