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#7
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Re: Weight of fully inflated ball
Quote:
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 Last edited by DanL : 14-01-2004 at 20:49. |
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