Assume:
P0-pressure at 4500m above sea level.
P1-pressure at sea-level
P2- pressure at depth of 20.6m
P2=P1+dgh
= P1 + 1024*9.8*20.6
= P1+2.06x10^5 Pa .......(1)
P0=P1-dgh
=P1 - 1.3*9.8*4500
=P1 - 0.57x10^5 Pa .......(2)
[It is assumed density of air near earth's surface is 1.3Kg/m^3 and is uniform in the atmosphere]
Change in pressure = P2 -P0
= (2.06+0.57)x10^5 Pa
= 2.63x10^5 Pa
A scuba diver must decompress after a deep dive to allow excess nitrogen to exit safely from his bloodstream. The length of time required for decompression depends on the total change in pressure that the diver experienced. Find this total change in pressure for a diver who starts at a depth of d = 20.6 m in the ocean (density of seawater = 1024 kg/m3) and then travels aboard a small plane (with an unpressurized cabin) that rises to an altitude of h = 4500 m above sea level.
The answer is 2.49e+05 Pa. I'm having trouble figuring out how to get this answer. Any help is greatly appreciated.
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2 answers
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