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?
To solve for absolute pressure, you will need this formula:
[tex] P_{total} = P_{atm} + (rgh)[/tex]
Where: [tex] P_{total}[/tex] = absolute pressure [tex] P_{total}[/tex] = atmospheric pressure r (rho) = density g = acceleration due to gravity constant [tex]9.8 \frac{m}{ s^{2} } [/tex] 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:
[tex] P_{total} = P_{atm} + (rgh)[/tex] [tex] P_{total}[/tex] = 1.01 x [tex] 10^{5}[/tex] x (1,025 [tex]\frac{kg}{m^3} [/tex] x 9.8 [tex] \frac{m}{{s^2}} [/tex] x 15 m [tex] P_{total}[/tex] = 1.01 x [tex] 10^{5}[/tex] + 150,675 [tex] P_{total}[/tex] = 1.01 x [tex] 10^{5}[/tex] + 1.51 x [tex] 10^{5}[/tex]
[tex] P_{total}[/tex] = 2.52 x [tex] 10^{5} [/tex] This is your absolute pressure.
