Question

The pressure of seawater increases by 1.0 atm for each 10m increase in depth. By what percentage is the density of water increased in the depth of the ocean of about 12m? The compressibility of water is given by (5.0 × 10^{-5}) mm in air.

Answer 1

To calculate the percentage increase in density of seawater at a depth of 12m, we can use the pressure increase and the compressibility of water. The pressure of seawater increases by 1.0 atm for every 10m increase in depth. By using the formula relating bulk modulus, pressure change, and volume change, we can calculate the fractional change in volume and density.

Step-by-step explanation:

To calculate the percentage increase in density of seawater at a depth of 12m, we need to determine the pressure increase and use the compressibility of water. The pressure of seawater increases by 1.0 atm for every 10m increase in depth. So, for a depth of 12m, the pressure increase would be (1.0 atm / 10m) x 12m = 1.2 atm.

The bulk modulus of water is used to determine the change in volume due to pressure. Since the compressibility of water is given as (5.0 x 10-5) mm in air, we can calculate the decrease in volume using the formula:

∆V/V = B x ∆P/P

Where ∆V/V is the fractional change in volume, B is the bulk modulus, ∆P is the change in pressure, and P is the initial pressure.

Finally, to calculate the percentage increase in density, we can use the relationship between density and volume:

∆ρ/ρ = - ∆V/V

Where ∆ρ/ρ is the fractional change in density.