Density of water as a function of depth

[LBloom]

Homework Statement

Estimate the density of the water 5.3 km deep in the sea. (bulk modulus for water is B=2.0 *10^9

Homework Equations

\DeltaV/V=(-1/B)*\DeltaP

dP/dY=-\rhog

The Attempt at a Solution

I tried to use the equation:
P=P_o*e^^{(-\rho_o/P_o)*gh}
But i kept getting ridiculously large answers that i don't think could possibly be true.

I said that the initial pressure was 1atm or 1.013*10^5 Pa and that initial density was 1*10^3(kg/m^3).

I know that once i get the pressure under water, I can find the change in volume divided by volume (delta V over V) and from there figure out how what the density of a theoretical block of water would be if submerged 5300 meters. Its just finding out the pressure that trips me up. Am i using the wrong constants or must i derive a different equation for pressure as a function of height?

I tried to find volume as a function of pressure, but its finding the pressure with a variable density that keeps throwing me off. I'm not sure the third equation i used is correct for this kind of problem.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[lanedance]

as a sense check, first assume the density is constant, calculate the pressure at 5.3km (10m of water is ~1atm so will be around 530atm). now use this to estimate the density as the chaneg will probably be reasonably small... also gives you an idea of your error


now for a more accurate assumption, set up your de as follows..
\rho(V) = \frac{m}{V}
\frac{d}{dV} \rho(V) = -\frac{m}{V^2}
d \rho = -\frac{dV}{V} \rho

then from the bulk modulus
\frac{dV}{V} = -\frac{dP}{B}
d \rho= \frac{dP}{B} \rho
or
dP = B \frac{d \rho}{\rho}


then the pressure gradient with denisty you know
\frac{d}{dy} P(y) = \rho(y).g
re-arranging
dP(y) = \rho(y).g.dy

then combine the equations & integrate

 

[LBloom]

thank you for your help, and i certainly will look this this over to get a better understanding of the problem, but I actually made this problem more difficult than it had to be and i don't need differentials. thanks again though