I feel almost as foolish answering this as you should feel for asking the question! If you remember nothing else in life remember GOOGLE IS YOUR FRIEND!! The quote below is what I found using GOOGLE. GH
" I'm delivering this answer off the top of my head, since I don't have any hard information on the dimensions of a typical automobile tire, but I am using the following estimates:
Total weight of tire + wheel + air ~60 lbs = ~30 kg
Inside diameter of tire 15" = ~40 cm = 4 dm
Outside diameter of tire (actually the space inside the tire) 21" = ~50 cm = 5 dm
Width of wheel 6" = ~15 cm = 1.5 dm
Pressure inside the tire 30 psi = ~ 2 bar (that is, twice "standard" pressure)
Temperature 25° C ("standard" temperature)
(I'm using dm = decimeters for length because I want to measure volume in liters, that is, cubic decimeters.)
Air, at "standard temperature and pressure" (STP), has a molar volume of 22.4 liters and an average molar mass of 29 grams, that is, one mole of air masses 29 grams and takes up 22.4 liters at STP. Since we're estimating a pressure of twice "standard," the volume of 29 grams of air will be only 11.2 liters.
To get the mass of air, we need the volume inside the tire. For that we use the volume difference between two cylinders, one representing the wheel and the other representing the tire mounted on the wheel. The volume of a cylinder is given by
V = p × radius<SUP>2</SUP> × height
Remember that radius is diameter ¸ 2. For the wheel, the volume is given by
p × (4 dm ¸ 2)<SUP>2</SUP> × 1.5 dm = 19 liters (cubic decimeters).
For the tire mounted on the wheel, the total volume is given by
p × (5 dm ¸ 2)<SUP>2</SUP> × 1.5 dm = 29 liters (cubic decimeters).
The volume difference is just 10 liters, which means that the air in the tire will mass about 26 grams, less than 0.1% of the total weight (30 kg = 30,000 g) we estimated for the tire + the wheel.
Because of buoyancy effects, the air will only weigh 13 grams, less than 0.05%! Also, the thickness of the tire is negligible for our purposes. If the tire is being shipped alone, though, the air in the package weighs <SMALL>NOTHING</SMALL>--because it's at the same pressure as the surrounding air. This is a buoyancy effect."
<TABLE cellSpacing=5 cellPadding=0 align=right border=0><TBODY><TR><TD align=left> (bergerd@bluffton.edu)</TD></TR><TR><TD align=left> (http://www.bluffton.edu/)</TD></TR><TR><TD align=left> (http://www.bluffton.edu/~bergerd)</TD></TR></TBODY></TABLE>
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