Answer:
0.9307 moles have been introduced into the bag.
Explanation:
Pressure of the gas within the bag,P = 1.00 atm
Temperature of the gas remains at room temperature,T=20.0 °C = 293.15 K
Volume of the gas in the bag = V = 22.4 L
Number of moles of gas = n
Using an ideal gas equation:
[tex]PV=nRT[/tex]
[tex]1.00 atm\times 22.4 atm=n\times 0.0821 atm l/mol K\times 293.15 K[/tex]
n = 0.9307 moles
0.9307 moles have been introduced into the bag.