A mixture of gases at s.t.p contains 65% of N2, 15% of CO2 and 20% of O2 by volume. What is the partial pressure of each gas in mmHg?

Answer 1

Assume 1 liter total volume:

65% N₂ = 0.65 L N₂

15% CO₂ = 0.15 L CO₂

20% O₂ = 0.20 L O₂

At STP, the pressure is 1 atm and partial pressure is mole fraction x total pressure.

Also, mole fraction is the same as % by volume because from PV = nRT, moles ∝ volume since R,T and P are constant. Therefore...

Partial pressure N2 = 0.65 atm

Partial pressure CO2 = 0.15 atm

Partial pressure O2 = 0.20 atm

If you want to look at this another way, we know that for an ideal gas at STP 1 mole = 22.4 L. We can now use the % composition to find moles:

0.65 L N₂ x 22.4 mol/L = 14.56 moles N₂

0.15 L CO₂ x 22.4 mol/L = 3.36 moles CO₂

0.20 L O₂ x 22.4 mol/L = 4.48 moles O₂

Total moles = 22.4 moles

According to Dalton's Law of Partial Pressures, mole fraction x total pressure = partial pressure

Partial pressure N₂ = 14.56/22.4 x 1 atm = 0.65 atm

Partial pressure CO₂ = 3.36/22.4 x 1 atm = 0.15 atm

Partial pressure O₂ = 4.48/22/4 x 1 atm = 0.20 atm