What volume will a balloon occupy at 1.0 atm, if the balloon has a volume of 4.4 l at 2.2 atm?

The two states of the balloon are
State 1:
pressure, p₁ = 1 atm = 101325 Pa

State 2:
volume, V₂ = 4.4 L = 4.4*0.001 m³ = 0.0044 m³
pressure, p₂ = 2.2 atm = 2.2*101325 Pa = 222915 Pa

Let V₁ = the volume in state 1.
Assume that the air in the balloon is at the same temperature at the two given states.
Then according to Boyle's Law (pV = constant), obtain
(101325 Pa)*(V₁ m³) = (222915 Pa)*(0.0044 m³)
V₁ = 0.00968 m³
      = (0.00968 m³)/(0.001 m³/L) = 9.68 L

9.68 L 

The answer is 323 g.