We are asked to calculate the mass of air in an air-filled tire.

Given:

$\mathbf{V}\mathbf{}\mathbf{=}\mathbf{}\mathbf{860}\mathbf{}\overline{)\mathbf{mL}}\mathbf{}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{L}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}\mathbf{}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{86}\mathbf{}\mathbf{L}\phantom{\rule{0ex}{0ex}}\mathbf{P}\mathbf{}\mathbf{=}\mathbf{}\mathbf{125}\mathbf{}\overline{)\mathbf{psi}}\mathbf{}\mathbf{\times}\mathbf{}\frac{\mathbf{1}\mathbf{}\mathbf{atm}}{\mathbf{14}\mathbf{.}\mathbf{7}\overline{)\mathbf{psi}}\mathbf{}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{8}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{atm}\phantom{\rule{0ex}{0ex}}\mathbf{T}\mathbf{}\mathbf{=}\mathbf{}\mathbf{25}\mathbf{}\mathbf{\xb0}\mathbf{C}\mathbf{}\mathbf{+}\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{=}\mathbf{}\mathbf{298}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{K}$

MM = 28.8 g/mol

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 125 psi , and that the temperature is 25 ^{o}C. Also, assume an average molar mass for air of 28.8 g/mol.

Calculate the mass of air in an air-filled tire.

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