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Find the density of seawater at a depth where the pressure is 950 atm if the density at the surface is 1100 kg/m3. Seawater has a bulk modulus of 2.3 × 10^9N/m2.
Answer in units of kg/m3.

Respuesta :

Histrionicus

GIVEN DATA

Pressure = 950 atm.

              since 1 atm = 101325 N/m² ; 950 atm = 9.626×10⁷  N/m²

density of surface (ρ) = 1100 kg/m³ ,

Bulk modulus of sea water (K) = 2.3×10⁹ N/m² ,

Determine the density of sea water (ρ)=?

From the concept of compressibility and bulk modulus, compressibility is the reciprocal of elasticity. The bulk modulus is defined as the ratio of compressive stress to the volumetric strain.

Volumetric strain is defined as it is the ratio of change in volume to the original volume.

Therefore, Volumetric strain = -(dV/V)          

                   Negative sign indicates that decrese in pressure with incresed volume.

                  In the limit δP tends to zero.

                 K = -V dP/dV

                 consider V = 1/ρ, and differentiate

                K = ρ.dP/dρ

                dρ = ρ.dP/K, use this eqution to find the density of seawater

                     = (1100 kg/m³ × 9.626×10⁷  N/m²) / (2.3×10⁹ N/m²)

                     = 46.037 kg/m³

Density of sea water  at 950 atm. pressure is 46.037 kg/m³.

The density of the seawater at 950 atm when the bulk modulus of seawater is 2.9x10⁹ N/m² is 1146.097 kg/m³.

What is Bulk modulus?

Bulk modulus is the inverse of elasticity. It shows how resist a material or substance is under compression.

[tex]k = -V\dfrac{dp}{dV} = \rho\dfrac{dp}{d\rho}[/tex]

Given to us

Pressure inside the seawater = 950 atm = 9.626 x 10⁷ N/m²

Density of seawater at the surface, rho = 1100 kg/m³

The bulk modulus of seawater, k = 2.9x10⁹ N/m²

We know that the bulk modulus is the reciprocal of elasticity, therefore,

[tex]k = -V\dfrac{dp}{dV}[/tex]

We can also right bulk modulus as,

[tex]k = \rho\dfrac{dp}{d\rho}[/tex]

Substitute the values,

[tex]k = \rho\dfrac{dp}{d\rho}[/tex]

[tex]2.9\times 10^9 = 1100\times \dfrac{9.626 \times 10^7}{d\rho}\\\\d\rho = 46.037\[/tex]

Thus, the change in the density of the seawater at the surface and at pressure 950 atm is 46.037.

We know the density at the surface of the seawater and also the difference in the density of the seawater, therefore,

The difference in the density

= Density at 950 atm - Density at the surface of the water

46.037 = ρ₁ - 1100

-ρ₁ = -1100 - 46.037

ρ₁ = 1146.037 kg/m³

Hence, the density of the seawater at 950 atm when the bulk modulus of seawater is 2.9x10⁹ N/m² is 1146.097 kg/m³.

Learn more about Bulk Modulus:

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