Quote:
|
Originally Posted by Matt Adams
10,000 points to the first person that can calculate the weight difference between an inflated and uninflated ball... (30" diameter, 30 PSI, assume 1/8" wall thickness) 
Matt |
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