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

(Compressibility of water=45.8×10^–11Pa^–1) 

 

Given,

$\mathrm{p}=80.0\mathrm{atm}=80.0\times 1.013\times {10}^5\mathrm{Pa}$

$\mathrm{compressibility}, \frac{1}{\mathrm{B}}=45.8\times {10}^{-11}{\mathrm{Pa}}^{-1}$

$\mathrm{\rho }=1.03\times {10}^3{\mathrm{kgm}}^{-3}$

Let ρ' represent the water's density at a particular depth. If volumes V and V' of a specific mass M of ocean water are present at the surface and at a particular depth, then

$\mathrm{V}=\frac{\mathrm{M}}{\mathrm{\rho }}$

$\mathrm{V}\text{'}=\frac{\mathrm{M}}{\mathrm{\rho }\text{'}}$

$$\begin{gathered} \mathrm{Change} \mathrm{in} \mathrm{volume}, \\ \mathrm{\Delta V}=\mathrm{V}-\mathrm{V}\text{'}=\mathrm{M}\left(\frac{1}{\mathrm{\rho }}-\frac{1}{\mathrm{\rho }\text{'}}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Volumetric} \mathrm{strain} \\ \frac{\mathrm{\Delta V}}{\mathrm{V}}=\mathrm{M}\left(\frac{1}{\mathrm{\rho }}-\frac{1}{\mathrm{\rho }\text{'}}\right)\times \frac{\mathrm{\rho }}{\mathrm{M}}=1-\frac{\mathrm{\rho }}{\mathrm{\rho }\text{'}} \end{gathered}$$

$\Rightarrow \frac{\mathrm{\Delta V}}{\mathrm{V}}=1-\frac{1.03\times {10}^3}{\mathrm{\rho }\text{'}}\mathrm{\_\_\_\_\_\_\_\_}\left(1\right)$

Bulk modulus,

$\mathrm{B}=\frac{\mathrm{pV}}{∆\mathrm{V}}\Rightarrow \frac{∆\mathrm{V}}{\mathrm{V}}=\frac{\mathrm{p}}{\mathrm{B}}=(80.0\times 1.013\times {10}^5)\times 45.8\times {10}^{-11}=3.712\times {10}^{-3}$

Substituting in equation 1,

$1-\frac{1.30\times {10}^3}{\mathrm{\rho }\text{'}}=3.712\times {10}^{-3}$

$\mathrm{\rho }\text{'}=\frac{1.30\times {10}^3}{1-3.712\times {10}^{-3}}=1.304\times {10}^3{\mathrm{kgm}}^{-3}$

Hence the correct answer is $1.304\times {10}^3{\mathrm{kgm}}^{-3}.$