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

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 $9.8 \frac{m}{ s^{2} }$
             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 $\frac{kg}{m^3}$ 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.