Weight of fully inflated ball

[briholton]

Originally Posted by SuperDanman

well....

volume of a sphere = 4/3 * pi * r^3
30 inches = 0.762 meters
4/3 * pi * (0.762 / 2)^3 = .232 cubic meters = 232 liters

Now, considering 30 pounds per square inch = 2.0413 atm, if you rearrange the ideal gas law pV = nRT to solve for n,
n = pV/RT = [(2.0413 atm)(232 l)]/[(8.2057 * 10^-2 l*atm/(mol*K))((273+25) K)] = 19.4 moles of air

According to
this lab, the molar mass of air is about 29 g/mol, soooo

19.4 moles * 29 g/mol = 560 grams, or 1.23 pounds

That means that the air accounts for about 36% of what seems to be the measured mass of the inflated ball.


NOTE: For those of you who have read this far, I sincerely appologize, but I have a chem midterm next week.

[Edit]
BONUS POINTS to the first person to calculate how relativistic effects change the weight

Yeah, but what was the name of that guy....hmmmm... archimedes or something ..... bouyant force..... "A body submerged in a fluid experiences a buoyant force equal to the weight of the displaced fluid." good thing your are studying for chemistry and not physics. Offhand, I think you need to reduce your calculation by about half since you've calculated on about 2 atmosphere and you need to take 1 off.

[DanL]

Originally Posted by briholton

Offhand, I think you need to reduce your calculation by about half since you've calculated on about 2 atmosphere and you need to take 1 off.

The ball is inflated at around 30 psi which is a pressure of about 2atm.

[Paul Marshall]

Yeah, you're right, it would weight more. I was thinking that the air inside wasn't presurized.