Answer:
24.44 atm
Explanation:
Considering that this gas mixture behaves like an ideal gass, and that all component gases are ideal gases, we can use:
PV=nRT
Then:
P=nRT/V
Where:
n= N° of moles
R= gas constant= 0.082 Lt*atm/K*mol
T= temperature (in Kelvin)
V = volume (in Lt)
Finally, statement says:
T = 25°C = 298 K
V = 5 Lt
n = 8 moles (for O₂)
P = [8 molx(0.082 Lt*atm/K*mol)x298 K]/5 Lt
P = 24.44 atm would be the pressure due to O₂ (partial pressure of the oxygen)