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 x 103 kg m⁻³?

Answer 1

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) ]
∴ Volumetric strain = ΔV / V1
= m [ (1/ρ1) – (1/ρ2) ] × (ρ1 / m)
ΔV / V1 = 1 – (ρ1/ρ2)     ……(i)
Bulk modulus, B = pV1 / ΔV
ΔV / V1 = p / B
Compressibility of water = (1/B) = 45.8 × 10-11 Pa-1
∴ ΔV / V1 = 80 × 1.013 × 105 × 45.8 × 10-11  =  3.71 × 10-3    ….(ii)
For equations (i) and (ii), we get:
1 – (ρ1/ρ2)   =   3.71 × 10-3
ρ2 = 1.03 × 103 / [ 1 – (3.71 × 10-3) ]
= 1.034 × 103 kg m-3
Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.

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Answer 2

the density. of water is 997 kg/m³