a scuba diver 40 ft below the ocean surface inhales 50.0 mL of compressed air in a scuba tank at a pressure of 3.00 atm and a temperature of 8 C. What is the pressure of air in the lungs if the gas expands to 150.0 mL at a body temp of 37 C?
2007-03-08 07:06:25 · 4 answers · asked by Jacquline M 1 in Science & Mathematics ➔ Chemistry
PV = nRT
n and R are constants so:
PV/T = constant
P1V1/T1 = P2V2/T2
P1 = 3 atm
T1 = 8ºC +273 = 281 K
V1 = 0.05 liters
P2 = ?
T2 = 37ºC +273 = 310 K
V1 = 0.150 liters
solve for P2 = P1V1/T2 * T1/V2
=3atm*(0.05L)*281K/
(310K*0.150L) = 0.906 atm
2007-03-08 07:15:02 · answer #1 · answered by Dr Dave P 7 · 0⤊ 0⤋
Assuming this is from a chem teacher/professor who's never been scuba diving:
Rearranging the ideal gas law, we can determine:
P1V1/T1 = P2V2/T2
so. (3 atm)(50mL)/(281K) = (X atm)(150mL)/(310K)
X = 1.1 atm
However, this problem is not realistic. You have to account for the depth of the diver (absolute pressure) and no scuba regulator would function well at 3atm (45 psi). etc.etc.etc
2007-03-08 15:17:37 · answer #2 · answered by tickdhero 4 · 0⤊ 0⤋
To work this out, all that is needed is to know the first volume.
If the volume of air is 50mL before, then when it becomes 150mL, that pressure will be 3 times lower.
Since the pressure is 8atm, then it would become 2.67atm.
2007-03-08 15:10:55 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Doesn't stay in there long enough to change its' temperature completely.
and...
The outgoing air-breath will accommodate some of the expansion.
2007-03-08 15:19:52 · answer #4 · answered by occluderx 4 · 0⤊ 0⤋