Question

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?

Answer 1

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.

Step-by-step 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=(PV)/(RT)`

`n=(8.16552 atm* 0.860 L)/(0.0821 atm L/mol K* 299.15 K)`

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