Answer
2.27 L
Procedure
The volume occupied by one mole of any gas at standard temperature(0°C) and pressure(1 atm) is 22.4 liters.
[tex]6.22\text{ dm}^3\frac{1\text{ L}}{1\text{ dm}^3}\frac{1\text{ mol}}{22.4\text{ L}}=0.2777\text{ mol}[/tex]
Then we will use the ideal gas formula to get the volume.
PV=nRT
Where R= 0.08206 L⋅atm⋅°K⁻¹⋅mol⁻¹
P= pressure
V=volume
n= moles
T= temperature in °K
Solving for V
[tex]V=\frac{nRT}{P}=\frac{0.2777\text{ mol }(0.08206)\text{ L.atm.}\degree\text{K}^{-1}\text{.mol}^{-1}(300.65)\text{ }\degree\text{K}}{3.0120898\text{ atm}}=2.274\text{ L}[/tex]
