Question

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?

Answer 1

Let the given depth be h.
Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 10 5 Pa
Density of water at the surface, ρ1 = 1.03 × 10 3 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

$= m (\frac{1 }{ ρ_1} - \frac{1 }{ ρ_2})$

∴ Volumetric strain = $\frac{ΔV }{ V_1 }$

= m (\frac{1 }{ ρ_1} - \frac{1 }{ ρ_2})$$× \frac{ρ_1 }{ m}

$∴ \frac{ΔV }{ V_1} = 1 - \frac{ρ_1 }{ ρ_2} ....(i)$

Bulk modulus, B = $\frac{PV_1 }{ ΔV}$

$\frac{ΔV }{ V_1} =\frac{ P }{ B}$

Compressibity of water =$\frac{1 }{ B}$ = 45.8 × 10 - 11 Pa- 1

∴ $\frac{ΔV }{ V_1}$ = 80 × 1.013 × 10 5 × 45.8 × 10 -11 = 3.71 × 10 - 3 ...(ii)

For equations (i) and (ii), we get : 1 - ρ1 / ρ2 = 3.71 × 10 - 3

ρ2 = $\frac{1.03 × 10^ 3 }{ 1- (3.71 × 10 ^- 3) }$

= 1.034 × 10 3 kg m-3

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