Olympic cyclists fill their tires with helium to make them lighter. assume that the volume of the tire is 860 ml , that it is filled to a total pressure of 120 psi , and that the temperature is 26 ∘c. also, assume an average molar mass for air of 28.8 g/mol.
a) calculate the mass of air in an air filled tire.
b) calculate the mas of helium in a helium-filled tire.
c) what is the mass difference between the two?

a) The mass of air in an air filled tire is 8.235 g.

b) The mass of helium in an air filled tire is 1.144 g.

c) 7.091 g is the mass difference between the two.

Explanation:

Volume of the tire,V =  860 mL = 0.860 L

Total pressure of the gases in tire,P = 120 psi = 8.16552 atm

1 psi = 0.068046 atm

Temperature of the gases in tire = T = 299.15 K

Total mole of gases = n

PV = nRT (Ideal gas equation)

n = 0.2859 mol

a) Mass of air =  molar mass of air × 0.2859 mol:

= 28.8 g/mol × 0.2859 mol = 8.235 g

b) Mass of helium =  molar mass of helium × 0.2859 mol:

= 4 g/mol × 0.2859 mol = 1.144 g

c) The mass difference between the two gases:

8.235 g - 1.144 g = 7.091 g

b is the correct