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.