Calculate the density of water near the bottom of the ocean where the pressure is about 500 atm and density of water is 1025 $k g / m^{3}$ approximately. Take bulk modulus of water as for water as $2.0 \times 10^{9} N / m^{2} .$

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Solution

The density of water at the oceans surface where the pressure is 1 atm is about $1025 k g m^{3}$ . Thus, 1 m3 of water at the surface has a mass of 1025kg. Let us calculate the change in volume $Δ V$ of this water when it is at a pressure of 500 atm.
$Δ V = \frac{(Δ p) V}{K}$
The pressure change $Δ p i s 500 a t m - 1 a t m = 499 a t m,$ which is equivalent to $499 \times 10^{5} N m^{- 2} .$ Thus, the volume change $Δ V i s$
$∴ Δ V = \frac{(499 \times 10^{5} N / m^{2}) (1 m^{3})}{2.0 \times 10^{9} N / m^{2}} = 0.025 m^{3}$

The density of the water is then
$ρ = \frac{m}{V - Δ V} = \frac{1025 k g}{1.00 m^{3} - 0.025 m^{3}} = \frac{1025 k g}{0.9775 m^{3}} = 1051 k g m^{- 3}$

The water is compressed from only 2% to 3% at the bottom of the ocean where the pressure is several hundred atmosphere. Liquids and solids are not easily compressed.

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Calculate the density of water near the bottom of the ocean where the pressure is about 500 atm and density of water is 1025 kg/m3 approximately. Take bulk modulus of water as for water as 2.0×109N/m2.

Calculate the density of water near the bottom of the ocean where the pressure is about 500 atm and density of water is 1025 $k g / m^{3}$ approximately. Take bulk modulus of water as for water as $2.0 \times 10^{9} N / m^{2} .$