Question

The mass percentage composition of dry air at sea level is approximately N2: 75.5; O2: 23.2; Argon: 1.3. What is the partial pressure of each component when the total pressure is 1.60 atm?

Answer 1

Molar mass of N2 = 28.02 g/mol

Molar mass of O2 = 32 g/mol

Molar mass of Ar = 39.95 g/mol

n(N2) = mass/molar mass

= 75.5/28.02

= 2.6945

n(O2) = mass/molar mass

= 23.2/32.0

= 0.725

n(Ar) = mass/molar mass

= 1.3/39.95

= 0.0325

n(N2),n1 = 2.6945 mol

n(O2),n2 = 0.725 mol

n(Ar),n3 = 0.0325 mol

Total number of mol = n1+n2+n3

= 2.6945 + 0.725 + 0.0325

= 3.452 mol

Partial pressure of each components are

p(N2),p1 = (n1*Ptotal)/total mol

= (2.6945 * 1.6)/3.452

= 1.25 atm

p(O2),p2 = (n2*Ptotal)/total mol

= (0.725 * 1.6)/3.452

= 0.336 atm

p(Ar),p3 = (n3*Ptotal)/total mol

= (0.0325 * 1.6)/3.452

= 0.0151 atm