Question

Assuming atmospheric pressure to be 1.01 x 10^5 Pa and the density of sea water to be 1025 kg/m^3, what is the absolute pressure at a depth of 15.0 m below the surface of the ocean?

Answer 1

The absolute pressure at a depth of 15.0m below the surface of the ocean is 150975 Pa.

Step-by-step explanation:

To calculate the absolute pressure at a depth of 15.0m below the surface of the ocean, we need to consider the pressure due to the weight of the water column above that point.

The pressure due to the weight of 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.

Substituting the given values, we have P = (1025 kg/m3) × (9.8 m/s2) × (15.0m) = 150975 Pa.

Therefore, the absolute pressure at a depth of 15.0m below the surface of the ocean is 150975 Pa.

Answer 2

To solve for absolute pressure, you will need this formula:


`P_(total) = P_(atm) + (rgh)`

Where:
`P_(total)` = absolute pressure

`P_(total)` = atmospheric pressure
r (rho) = density
g = acceleration due to gravity constant
`9.8 (m)/( s^(2) )`
h = depth (in this case)

rgh is the formula for pressure of fluids

So with your given, we just need to insert it into the formula:


`P_(total) = P_(atm) + (rgh)`

`P_(total)` = 1.01 x
`10^(5)` x (1,025
`(kg)/(m^3)` x 9.8
`\frac{m}{{s^2}}` x 15 m

`P_(total)` = 1.01 x
`10^(5)` + 150,675

`P_(total)` = 1.01 x
`10^(5)` + 1.51 x
`10^(5)`


`P_(total)` = 2.52 x
`10^(5)` This is your absolute pressure.