Pressure and incompressible

The deepest point known in any of earth's oceans is in the Marianas Tench, 10.92 km deep.

a) Assuming water is incompressible, what is the pressure at this depth? Use the density of seawater.

b) The actual pressure is 1.16 X 108 Pa your calculation will be less because the density actually varies with depth. Using the compressibility of water and the actualpressure, find the density of the water at the bottom of the Marianas Trench. What is the percent change in the density of the water?

The answer to a is 1.0712 x 10^8 Pa.
What is b?

Solutions

Expert Solution

(1) We know that
pressure= pgh
where p is the density , g is the acceleration due to gravity and h is the height or depth
Pressure (P) = 1025*9.81*10.92*10³ = 1.098*10⁸ Pa
(2) We know that
compressibility = volumetric strain /Pressure = (dV/V) /P
dV = V_S - V_D
where V_S is the volume at surface and V_D is the volume at depth
Compressiblity = (V_S - V_D) /V_S*P = (1-V_D/V_S) /P
(1-V_D/V_S) = Compressibility*Pressure
(1-p_s /p_D) = 45.8*10^-11*(1.16*10⁸ Pa)
p_s = 1025 kg/m³
p_D = 1082.511 kg/m³
(3)
percentage change in the density = (p_D - p_s / p_s )*100 = ((1082.511 - 1025) / 1025)*100 = 5.61%

genius_generous answered 2 years ago

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