The Marianas Trench, near the Philippines, is the deepest part of Earth\'s ocean

Question

The Marianas Trench, near the Philippines, is the deepest part of Earth's oceans at nearly 11 km

(Part A) Calculate the pressure, in atmosheres, due to the ocean at the bottom of the Mariannas Trench, assuming the density of sea water is a constant 1025 kg/m3 all the way down to the bottom.

(Part B) Calculate the percent decrease in volume of sea water due to such pressure, assuming its bulk modulud is the same as that of water, 2.20 x 109N/m2, and is constant all the way to the bottom.

(Part C) What would be the percent increase in the density of the water?

Explanation / Answer

depth of the ocean, h=11km

density, rho=1025 kg/m^3

A)

pressure at depth,

P2-P1=rho*g*h

P2=Po+rho*g*h

P2-P0=rho*g*h

P2-1.032*10^5 =(1025*9.8*11*10^3)

P2-1.032*10^5 = 1.105*10^8

=1.106*10^8 pascal

B)

bulk modulus, k=2.2*10^9 N/m^2

now,

K=-v*dp/dv

===>

dv/v=-dP/k

=-1.105*10^8/(2.2*10^9)

=0.05023

=5.023 %

percentage decrease in volume=5.023 %

c)

density of water, rho=1000 kg/m^3

density of sea water, rho'=1025 kg/m^3

now,

pecentage increase in density of water, =(rho1-rho)/rho

=(1025-1000)/(1000)

=2.5 %

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