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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Solution

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.

For equations (i) and (ii), we get:

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

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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?

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?