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?
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Answer 1

$\large {\tt{EXPLAINATION : }}$

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.