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.

The difference in the mass = 7.0869 g

Explanation:

Pressure = 120 psi

The conversion of P(psi) to P(atm) is shown below:

So,  

Pressure = 120 / 14.696 atm = 8.16552 atm

Temperature = 26 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (26 + 273.15) K = 299.15 K

Volume = 860 mL = 0.86 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

8.16552 atm × 0.86 L = n × 0.0821 L.atm/K.mol × 299.15 K  

⇒n = 0.2859 moles

Molar mass of air = 28.8 g/mol

Moles = Mass/Molar mass

Mass of air in the tire= Moles* Molar mass = 0.2859 moles  * 28.8 g/mol = 8.2339 g

Molar mass of helium = 4.0026 g/mol

Mass of helium in the tire = 0.2859 moles  * 4.0026 g/mol = 1.1443 g

The difference in the mass = 7.0869 g