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 = 8.24 grams

b) the mass of helium = 1.14 grams

c) the mass difference = 7.10 grams

Explanation:

Step 1: Data given

Volume of the tire = 860 mL

Total pressure = 120 psi

Temperature = 26°C

molar mass of air = 28.8 g/mol

Step 2:  Convert psi to atm

(

120 psi) (1 atm / 14.7 psi) = 8.163

Step 4: Calculate moles

PV = nRT

 ⇒ with P = the pressure = 8.163 atm

⇒ with V = the volume = 860 mL = 0.860 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the universal gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 26 °C = 299 Kelvin

n = (8163*0.860)/(0.08206*299)

n = 0.2861 moles of gas

Step 5: Calculate the mass of air in an air-filled tire.

Mass = moles * molar mass

Mass = (0.2861 moles of gas) (28.8 g/mol)  

Mass = 8.24 grams

Step 6: Calculate the mass of helium in a helium- filled tire.

mass of helium = 0.2861 moles of gas * 4 g/mol)  

mass of helium  = 1.14 grams

Step 7: What is the mass difference between the two?

Δmass=  8.24 grams -  1.14 grams

Δmass= 7.10 grams

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.