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.
Calculate the mass of air in an air filled tire.
Calculate the mas of helium in a helium-filled tire.
What is the mass difference between the two?

Answer 1

Answer:

The mass of air is 8.245 g

The mass of helium is 1.145 g

The difference is 7.1 g

Explanation:

Total mass (air and helium) = PVM/RT

P is total pressure in the tire = 120 Psi = 120/14.696 = 8.17 atm

V is volume of the tire = 860 mL = 860 cm^3

M is the total molar mass of air and helium = 28.8 + 4= 32.8 g/mol

R is gas constant = 82.057 cm^3.atm/mol.K

T is temperature = 26°C = 26+273 = 299 K

Total mass = 8.17×860×32.8/82.057×299 = 9.39 g

Mass of air = mass fraction of air × total mass = 28.8/32.8 × 9.39 = 8.245 g

Mass of helium = total mass - mass of air = 9.39 - 8.245 = 1.145 g

Difference = 8.245 - 1.145 = 7.1 g