CP The deepest point known in any of the earth’s oceans is in the Marianas Trench, 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 calculated value will be less because the density actually varies with depth. Using the compressibility of water and the actual pressure, find the density of the water at the bottom of the Marianas Trench. What is the percent change in the density of the water?

Solution 50P Problem (a) Step 1: 3 Depth of the ocean h = 10.92 km or 10.92 x 10 km Acceleration due to gravity g = 9.8 m/s 2 Density of seawater = 1030 kg/m 3 Atmospheric pressure p = 1.01 x 10 Pa 5 0 Step 2: To find the pressure at the depth if water is incompressible p = p 0hg p = 1.01 x 10 + 10.92 x 10 x 9.8 x 1030 p = 1.01 x 10 + 1.10 x 10 8 p = 1.101 x 10 Pa8 8 The pressure at the depth if water is incompressible is 1.101 x 10 Pa Problem (b) Step 1: The value found previously is not matched with the original value 1.16x10 Pa, because 8 the density under ocean varies with respect to change in volume. Compressibility of water plays the role in change of volume. -11 -1 Compressibility of water k = 45.8 x 10 Pa Density of seawater is 1030 or 1.03 x 10 kg/m 3 3 Mass of seawater m = 1030 kg Volume of water at the surface of sea V = 1.00 m 3 0 Step 2: Since the seawater is compressible the density is different and the volume also changes. To find the density under the ocean, the change in volume has to be found. It is calculated using the relationship between compressibility, volume strain and change in pressure. k = - 1 V -----(1) V 0 p To find the change in Volume under ocean V = -kV p 0 Where V 0 the volume of water at the surface of sea p = 1.101 x 10 Pa V = -45.8 x 10 x1.00*1.101 x 10 8 3 V = -0.050 m 3 Change in volume is -0.050 m Step 3: Therefore the actual Volume V = V + V 0 V = 1+( -0.050 ) 3 V = 0.9496 or 0.95m 3 Volume of seawater under the depth V = 0.95m