What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m^–3?

Given data,

Pressure at the given depth, p = 80.0 atm

⇒ Pressure = 80× 1.01 × 10⁵ Pa.

Density of water at the surface, ρ₁ = 1.03 × 10³ kgm^-3

Let the given depth be h.

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

⇒ V = m[(1/ρ₁)-(1/ρ₂)]

∴ Volumetric strain = ΔV/V₁

⇒ Volumteric Strain = m [(1/ρ₁)-(1/ρ₂)] × (ρ₁/m)

⇒ ΔV/V₁ = 1-(ρ₁/ρ₂) ………… (1)

Bulk modulus, B = pV₁/ ΔV

ΔV/V₁ = p/B

Compressibility of water = (1/B) = 45.8× 10^-11 Pa^-1

Compressibility is the fractional change in volume per unit increase in pressure. For each atmosphere increase in pressure, the volume of water would decrease 46.4 parts per million.

∴ ΔV/V₁ = 80× 1.013× 10⁵× 45.8× 10^-11 = 3.71 × 10^-3 ……. (2)

From equations 1 & 2, we get:

1-(ρ₁/ρ₂) = 3.71 × 10^-3

⇒ ρ₂ = 1.03× 10³ / [1-(3.71 × 10^-3)]

⇒ ρ₂ = 1.034× 10³ kgm^-3

Therefore, the density of water at the given depth (h) is 1.034× 10³ kgm^-3.