# Homework Help: Density of sea water

1. Nov 17, 2009

### drwma

What the density of sea water at depth of 1000 M where the water pressure is about 1.0*10^7 pascal the density of sea water at the surface is 1.03*10^3 kilograms per cubic M?

2. Relevant equations
BULK MODULUS=dP/dV/V
p=pressure
v=volume
B of water=,21*10^10 n/m^2
it is approx to the bulk of sea water

3. The attempt at a solution
Ihave used the (1000m)depth as aforce at the bottom then A=f/p=1000/1.0*10^7=10^-4 M^2
dV=A*h=10^-4*1000=10^-1 m^3
from bulk modulus=dp/dv/v=,21*10^10=10^7/dv/v then dv/v=4,76*10^-3 then v(surface)= x then density surface=mass/v then mass = x then v depth = dv+v surface(because as account to the bulk modulus the volume smaller at the depth) then the density of the depth= mass/v depth

it right to assume that the depth is aforce??????

Last edited: Nov 17, 2009
2. Nov 17, 2009

### lanedance

no, depth is not a force, depth is a distance

a supported column of fluid in a gravitational field leads to a hydrostatic pressure (which has units of Force/Area)

the hydrostatic pressure in a constant gravitational field, is given by
$$P = \rho g h$$
where h is the depth

3. Nov 17, 2009

### drwma

thank you for help but g is the acceleration due gravity ?what p
can you help me mor is there another method ?but I have to use the bulk modulus in the solution I will be very grateful for your help

thank you alot

Last edited: Nov 17, 2009
4. Nov 17, 2009

### lanedance

g is the accleration due to gravity

so, first calculate the pressure at 1000m assuming the density is constant

then assume the bulk modulus is constant with pressure & calculate the corresponding volume & density change. (similar to what you attempted previously)

This will be a pretty good approximatino to the densety change, to do any better you would have to set up an integral