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

We have the air molar percent composition and it's required to calculate air density assuming ideal gas behaviour under given conditions of pressure and temperature. The result is 1,04 g/L

Air (molar %):    78.1 % Nitrogen, 20.9 % oxygen, 0.934 % argon

Density = ?      

T = 22 °C  + 273 =  295 K  

P = 760 Torr  = 1 atm

Using the ideal gas equation:

P . V = n . R . T

n = moles

P = pressure

V = volume

T = Temperature

R = universal ideal gas constant

Seeing that :    n = m / MW

         m = mass (g)

        MW = molecular weight (g/mol)

Then:

P . V = ( m/ MW ) R .T

P = ( m / V. MW ) . R. T

m/ V = Density ( ρ )

P = ( ρ / MW ) . R. T   (*)

Molecular weight  (Air)

                  moles/1 mol air        MW (g/mol)       Mass (g)

O₂                       0.209                     32                 2.88

N₂                        0.781                     28                 21.87

Ar                        0,010                     40                  0.40

                          1 mol Air                                     25.15

Using the equation (*), we can calculate density:

P = ( ρ / MW ) . R. T   (*)

ρ = (P / R. T ) . MW

ρ = ( 1 atm  ) ×25.15 g/mol /(0.082 atm. L / K. mol) 295 K )

ρ =  1.04 g/ L