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.

The

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

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first step to solving this problem is to determine the pressure at each depth. The pressure at a specific depth in a fluid is given by the equation:

P = ρgh

Where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.

At the surface of the ocean (depth = 0), the pressure is simply atmospheric pressure, which is 101325 Pa.

To find the pressure at a depth of 20.6 m, substitute the given values into the equation:

P = (1024 kg/m3) * (9.8 m/s2) * (20.6 m)
P = 212267.2 Pa (approximately)

Next, we need to find the pressure at an altitude of 4500 m above sea level. The pressure at a specific altitude can be calculated using the equation:

P = P0 * (1 - (L * h / T0)) ^ (g * M / (R * L))

Where P is the pressure, P0 is the pressure at sea level, L is the temperature lapse rate (0.0065 K/m), h is the altitude, T0 is the standard temperature at sea level (288 K), g is the acceleration due to gravity, M is the molar mass of air (0.0289644 kg/mol), and R is the universal gas constant (8.314 J/(mol·K)).

Substituting the given values into the equation, we have:

P = 101325 Pa * (1 - (0.0065 K/m * 4500 m) / 288 K)) ^ ((9.8 m/s2 * 0.0289644 kg/mol) / (8.314 J/(mol·K) * 0.0065 K/m))
P ≈ 75500.195 Pa

Now we have the pressures at the initial depth of 20.6 m and the altitude of 4500 m. To find the total change in pressure, we subtract the pressure at the altitude from the pressure at the initial depth:

Total change in pressure = 212267.2 Pa - 75500.195 Pa
Total change in pressure ≈ 136767 Pa

However, it's important to note that this is the change in pressure, not the total pressure. To convert the change in pressure to the total pressure, add the atmospheric pressure:

Total pressure = 136767 Pa + 101325 Pa
Total pressure ≈ 238092 Pa

Therefore, the total change in pressure for the diver is approximately 238092 Pa or 2.38 x 105 Pa.