In a deep sea station used for scientific observation, pressure is maintained at 20.0 atm. If you were to fill the station with 5.5x10^5 L air that was originally at STP, what would be the volume of the air in the deep sea station?
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This is Boyle's Law problem:
P1V1 = P2V2
(20.0 atm) (x) = (1.00 atm) (5.5x10^5 L)
x = 2.75 x 10^4 L
http://www.chemteam.info/GasLaw/KMT-Gas-…
The reason I ignored temp is because (while I do have the T in STP), I have no knowledge of how it changed, so I assume it stayed constant.
P1V1 = P2V2
(20.0 atm) (x) = (1.00 atm) (5.5x10^5 L)
x = 2.75 x 10^4 L
http://www.chemteam.info/GasLaw/KMT-Gas-…
The reason I ignored temp is because (while I do have the T in STP), I have no knowledge of how it changed, so I assume it stayed constant.
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PV=nRT. R is a gas constant and T is temp. in Kelvins. Find the number or mols of air in 5.5x10^5L. V=5.5x10^5L. P=1atm. R=constant T=standard temp in Kelvins. Solve for n. n=(PV)/(RT). Then put in your value for n and solve for V. V=(nRT)/P where P now equals your new pressure of 20 atm.