How Does Pressure Affect Seawater Density at Depth?

[Beginner@Phys]

Homework Statement

Part A
What is the pressure at a depth of 5000 in the ocean? Given rho_water=1,030 kg/m^3
pTotal=p_0 + rho*g*h=5.06×107


Part B
What is the fractional volume change of seawater at this pressure? Given B=0.2*10^10
dV/v=p/-B =−2.53×10−2



Part C
What is the density of seawater at this pressure? (This is the question that I don't understand)

Homework Equations

(1)pTotal=p_0 + rho*g*h
(2)dV/v=p/-B
(3)p=F/A

The Attempt at a Solution

Part C: The liquid is incompressible. So. I guessed that the density of the liquid is still 1,030 kg/m^3 at a depth of 5,000m. And, I don't know why my answer was considered wrong.

 

[LowlyPion]

You determined the volume change from the bulk modulus.

But what is density? weight/volume?

So what was occupying a m³ is now occupying less volume
... so isn't the density now greater since you can now squeeze more into a m³?

 

[thomate1]

As a scientist, it is important to provide an accurate and evidence-based response. In this case, the density of seawater at a depth of 5000m may not necessarily be 1,030 kg/m^3. This is because, at such high pressures, seawater can undergo changes in density due to the compression of its molecules. Therefore, it is important to take into account the bulk modulus (B) of seawater, which is a measure of its resistance to compression. Using the equation (2) provided in the homework, we can calculate the change in volume (dV) of seawater at this pressure, which is equal to -2.53×10−2. This means that the density of seawater at this pressure will be slightly higher than 1,030 kg/m^3, as there is a decrease in volume. To accurately calculate the density, we would also need to consider the initial volume and the change in volume. Overall, as a scientist, it is important to consider all relevant factors and equations in order to provide a comprehensive and accurate response.