Question

Air is 78.1% nitrogen, 20.9% oxygen, and 0.934% argon by moles. What is the density of air at 22 °C and 760 torr? Assume ideal behavior.

Answer 1

Step 1: Explanation

We know the ideal gas equation PV=nRT

where, R=universal gas constant, T=temperature , P=pressure, V=volume, n=moles.

Step 2: Calculation of  the total volume

Total moles = ( 78.1+20.9+0.934 ) moles = 99.934 mol

Pressure = 760 torr = 1 atm and Temperature = ( 22 +273.15)K = 295.15 K

By using the ideal gas equation

V = nRT / P

V = (99.934 mol) x ( 0.08206 L.atm / K.mol) x ( 295.15 K) / ( 1 atm) = 2420.4 L

Step 3: Calculate the total mass

We know, Mass = moles x molar mass

Mass of N₂= (78.1 mol ) x (28.01344 g/mol) = 2187.85 g N2
Mass of O₂ = (20.9 mol ) x (31.99886 g/mol) = 668.78 g O2
Mass of Argon =  (0.934 mol ) x (39.9481 g/mol) = 37.31 g Ar

Total mass of Air = ( 2187.85 g + 668.78 g + 37.31 g) = 2893.94 g

Step 4: Calculate the density

We know, density = mass / volume

we got mass = 2893.94 g and Volume = 2420.4 L

Density = (2893.94 g) / (2420.4 L ) = 1.196 g/L ≈1.2 g/L