Question

What is the density of ocean water at a depth where the pressure is 80.0 atmosphere . Given that its density at the surface = 1.03?10^3 kg/m^3. Compressiblity of water = 45.8 ?10^-11 Pa^-1

Answer 1

Dear Student,

Let the given depth be h.

Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 10⁵ Pa

Density of water at the surface, ρ₁ = 1.03 × 10³ kg m^–3

Let ρ₂ be the density of water at the depth h.

Let V₁ be the volume of water of mass m at the surface.

Let V₂ be the volume of water of mass m at the depth h.

Let ΔV be the change in volume.

ΔV = V₁ - V₂

=m (1/ ρ₁ – 1/ ρ₂)

Therefore, Volumetric strain = ΔV/ V₁

=m (1/ ρ₁ – 1/ ρ₂) x ρ₁/m

Therefore, ΔV/ V₁ = (1- ρ₁/ ρ₂) … (i)

Bulk modulus, B = p V₁/ ΔV

ΔV/ V₁ = p/B

Compressibility of water =1/B =45.8x10^-11 Pa^-1

Therefore, ΔV/ V₁= 80x1.013x10⁵x45.8x10^-11 = 3.71x10^-3 … (ii)

For equations (i) and (ii), we get:

1- ρ₁ / ρ₂ = 3.71 x10^-3

ρ₂ = 1.03x10³/(1-(3.71x10^-3)

=1.034x10³kgm^-3

Therefore, the density of water at the given depth (h) is 1.034 × 10³ kg m^–3.

Regards