Question

The absolute pressure below the surface of a freshwater lake is 3.51 x 10^5 Pa. At what depth does this pressure occur? Assume that atmospheric pressure is 1.01 x 10^5 Pa. and the density of the water is 1.00 x 10^3 kg/m^3

Answer 1

The general formula for absolute pressure is:

$P_{total} = P_{atm} + (rgh)$

Where: $P_{total}$ = Absolute pressure
             $P_{atm}$  = atmospheric pressure
             r = density
             g = $9.8 \frac{m}{{s^2}}$
             h = depth

We can use this formula to derive our formula for h:

$P_{total} = P_{atm} + (rgh)$      transpose atmospheric pressure
$P_{total} - P_{atm}$ =  $(rgh)$     transpose r and g

$P_{total}$ - $P_{atm}$
-------------------------      = h
         rg

Now let us input our given into our new formula:

3.52 x  $10^{5}$ - 1.01 x $10^{5}$
----------------------------------------              = h
1.00 x $10^{3}$ x 9.8 $\frac{m}{ s^{2} }$   

$\frac{2.51 x {10^5}}{9.8 x {10^3} \frac{m}{s^2}} = h$

$25.61 m = h$