The amount of energy in a thermos like that is high enough that I'm not sure that would be safe.

Quick approximation, assuming adiabatic expansion:

Gamma for CO2 is ~1.29.

    Pf*Vf^y = Pi*Vi^y
    Vf = (Pi / Pf) ^ (1/y) * Vi
    Vf = (100atm / 1atm) ^ (1/1.29) * 1L
    Vf = 35.51L
    
    W = (PfVf - PiVi) / (y - 1)
    W = (1atm*35.51L - 100atm*1L ) / (1.29 - 1)
    W = 22.31 KJ

For comparison, a .50 cal round has ~15kJ of muzzle energy. Now, a lot of that energy won't be focused (if nothing else, the final temperature with adiabatic expansion is such that a large chunk should sublimate again), but still.