Calculate the moles of that can occupy the tire:

$\mathbf{PV}\mathbf{=}\mathbf{nRT}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{PV}}{\mathbf{RT}}\mathbf{=}\frac{\mathbf{n}\overline{)\mathbf{RT}}}{\overline{)\mathbf{RT}}}\phantom{\rule{0ex}{0ex}}\overline{)\mathbf{n}\mathbf{=}\frac{\mathbf{PV}}{\mathbf{RT}}}$

T = 25°C +273.15 = **298.15 K**

$\mathbf{V}\mathbf{=}\mathbf{860}\mathbf{}\overline{)\mathbf{mL}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{L}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mL}}}$

**V = 0.860 L**

$\mathbf{P}\mathbf{=}\mathbf{125}\mathbf{}\overline{)\mathbf{psi}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{atm}}{\mathbf{14}\mathbf{.}\mathbf{6959}\mathbf{}\overline{)\mathbf{psi}}}$

**P = 8.6058 atm**

**$\mathbf{n}\mathbf{=}\frac{\mathbf{(}\mathbf{8}\mathbf{.}\mathbf{6058}\mathbf{}\overline{)\mathbf{atm}}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{860}\mathbf{}\overline{)\mathbf{L}}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{08206}{\displaystyle \frac{\overline{)L}\xb7\overline{)\mathrm{atm}}}{\mathrm{mol}\xb7\overline{)K}}}\mathbf{)}\mathbf{(}\mathbf{298}\mathbf{.}\mathbf{15}\mathbf{}\overline{)\mathbf{K}}\mathbf{)}}$**

**n = 0.****299**** mol ≈ 0.30 mol**

Calculate the mass of helium (He) and air:

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.

What is the mass difference between the two?

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