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.

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.

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures