Final answer:
To calculate the mass of air in an air-filled tire, you can use the ideal gas law equation PV = nRT. The mass of air in the tire is 142.57 g. To calculate the mass of helium in a helium-filled tire, use the same equation and the molar mass of helium. The mass of helium in the tire is 19.84 g. The mass difference between the two is -122.73 g, indicating that the helium-filled tire is lighter than the air-filled tire.
Step-by-step explanation:
To calculate the mass of air in an air-filled tire, we can use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, we need to convert the volume from mL to L and the temperature from Celsius to Kelvin. So, the volume is 0.860 L, and the temperature is 26 + 273 = 299 K.
Next, we rearrange the ideal gas law equation to solve for n, the number of moles: n = PV / RT. Substituting the values, we get n = (120 psi x 0.860 L) / (0.0821 L.atm/mol.K x 299 K) = 4.96 moles.
The molar mass of air is given as 28.8 g/mol, so the mass of air in the tire is 4.96 moles x 28.8 g/mol = 142.57 g.
To calculate the mass of helium in a helium-filled tire, we can follow the same steps as above, but use the molar mass of helium (4 g/mol) instead. Substituting the values, we get n = (120 psi x 0.860 L) / (0.0821 L.atm/mol.K x 299 K) = 4.96 moles.
So, the mass of helium in the tire is 4.96 moles x 4 g/mol = 19.84 g.
To find the mass difference between the two, we subtract the mass of air from the mass of helium: 19.84 g - 142.57 g = -122.73 g. The negative sign indicates that the helium-filled tire is lighter than the air-filled tire by 122.73 g.