Question

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is $1.03\times {10}^{3}kg/m^3$?

Answer 1

Given, the pressure of the water at a depth is 80.0 atm and the density of water at the surface is $1.03\times {10}^{3}kg/m^3$.

Let $V_1$ and $V_2$ be the volume of water of mass $m$ at the surface and at a depth respectively. Let the density of water at the surface and at a depth be ${\rho }_1$ and ${\rho }_2$ respectively.

The change in the volume of the water is given by the equation,

$\Delta V=V_1-V_2=\frac{m}{{\rho }_1}-\frac{m}{{\rho }_2}$

The volumetric strain is given by the equation,

${\epsilon }_v=\frac{\Delta V}{V_1}$

Substituting the values in the above expression, we get:

${\epsilon }_v=\frac{\frac{m}{{\rho }_1}-\frac{m}{{\rho }_2}}{\frac{m}{{\rho }_1}}=1-\frac{{\rho }_1}{{\rho }_2}$

$\Rightarrow \frac{{\rho }_1}{{\rho }_2}=1-\frac{\Delta V}{V_1}$…… (1)

The compressibility of the water is 45.8× 10 −11   ( Pa ) −1

The compressibility is given by the equation,

$k=\frac{V_1}{\Delta p\Delta V}$

Compressibility of water =(1/B) = $45.8\times {10}^{-11}P{a}^{-1}$

Substituting the values in the above equation, we get:

$\frac{\Delta V}{V_1}=(80atm-1atm)(1.013\times {10}^{5}Pa)\times 45.8\times {10}^{-}11P{a}^{-1}=3.665\times {10}^{-5}$

Substituting the values in the equation (1), we get:

$\frac{1.03\times {10}^{3}kg/m^3}{{\rho }_2}=1-3.665\times {10}^{-5}$

${\rho }_2=1.034\times {10}^{3}kg/m^3$

Hence, the density of the water at a depth where pressure is 80.0 atm is $1.034\times {10}^{3}kg/m^3$.