A person accidentally swallows a drop of liquid oxygen, 02 (l), which has density of 1.149 g/mL. Assuming the drop has a volume of 0.050 mL, what volume of gas will be produced in the person's stomach at a body temperature of 98.6 degrees F and a pressure of 1.0 atm?
2007-11-30 13:36:30 · 3 answers · asked by confused44 1 in Science & Mathematics ➔ Chemistry
98.6 F = 37 C = 310 K
moles of O2 = ( 0.05 ml)(1.149 g / ml)( 1.00 mole / 18g) = 0.00319 mole
PV = nRT
V = nRT / P = (0.00319 mole)(.0821 l atm/mole K)(310 K) / 1.0 atm
V = 0.0812 liter = 81.2 ml
This poor person will have a lot less difficulty from the "burp" or oxygen gas than he will from the scarring of his esophagus by the liquid oxygen. I seriously doubt that the drop of liquid oxygen would ever make it to his stomach. It would unquestionably vaporize long before it ever got there.
2007-11-30 13:57:42 · answer #1 · answered by Dennis M 6 · 0⤊ 0⤋
use the density and the volume of the drop to get grams of oxygen. Divide by 32 to get moles.
Now use that in the ideal gas equation. But the temp. must be in Kelvin. You have the pressure and moles, use 0.081 for R so solve for V in liters
2007-11-30 13:56:00 · answer #2 · answered by reb1240 7 · 0⤊ 0⤋
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2007-12-02 00:16:07 · answer #3 · answered by Test I 1 · 0⤊ 0⤋