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What is the density of ocean water at a depth where the pressure is $80 \mathrm{~atm}$, given that its density at the surface is $1.03 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$ ? Compressibility of water $=45.8$ $\times 10^{11} / \mathrm{Pa}$. Given: $1 \mathrm{~atm}=1.013 \times 10^5 \mathrm{~Pa}$.
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$P=80 \mathrm{~atm}=80 \times 1.013 \times 10^5 \mathrm{~Pa}$
Compressibility $=\frac{1}{B}=45.8 \times 10^{-11} / \mathrm{Pa}$; Density of water at the surface $=\rho=1.03 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$
Let $\rho^{\prime}=$ density of water at a given depth;
$V^{\prime}=$ Vol. of water of mass $M$ at given depth
$V=$ Vol. of same mass of water at the surface
$\therefore \quad V=\frac{M}{\rho}$ and $V^{\prime}=\frac{M}{\rho^{\prime}}$
Change in Vol. $=\Delta V=V-V=M\left(\frac{1}{\rho}-\frac{1}{\rho^{\prime}}\right)$
Vol. strain $=\frac{\Delta V}{V}=\frac{M\left(\frac{1}{\rho}-\frac{1}{\rho^{\prime}}\right)}{\frac{M}{\rho}}=1-\frac{\rho}{\rho^{\prime}}$
$\Rightarrow \frac{\Delta V}{V}=1-\frac{1.03 \times 10^3}{\rho^{\prime}}$
Also $B=\frac{P}{\Delta V / V} \quad \therefore \frac{\Delta V}{V}=\frac{P}{B}$
$=80 \times 1.013 \times 10^5 \times 45.8 \times 10^{-11}=3.712 \times 10^{-3}$
$\therefore \quad 1-\frac{1.03 \times 10^3}{\rho^{\prime}}=3.712 \times 10^{-3}$
$\Rightarrow \rho^{\prime}=\frac{1.03 \times 10^3}{1-3.712 \times 10^{-3}}=1.034 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$
Compressibility $=\frac{1}{B}=45.8 \times 10^{-11} / \mathrm{Pa}$; Density of water at the surface $=\rho=1.03 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$
Let $\rho^{\prime}=$ density of water at a given depth;
$V^{\prime}=$ Vol. of water of mass $M$ at given depth
$V=$ Vol. of same mass of water at the surface
$\therefore \quad V=\frac{M}{\rho}$ and $V^{\prime}=\frac{M}{\rho^{\prime}}$
Change in Vol. $=\Delta V=V-V=M\left(\frac{1}{\rho}-\frac{1}{\rho^{\prime}}\right)$
Vol. strain $=\frac{\Delta V}{V}=\frac{M\left(\frac{1}{\rho}-\frac{1}{\rho^{\prime}}\right)}{\frac{M}{\rho}}=1-\frac{\rho}{\rho^{\prime}}$
$\Rightarrow \frac{\Delta V}{V}=1-\frac{1.03 \times 10^3}{\rho^{\prime}}$
Also $B=\frac{P}{\Delta V / V} \quad \therefore \frac{\Delta V}{V}=\frac{P}{B}$
$=80 \times 1.013 \times 10^5 \times 45.8 \times 10^{-11}=3.712 \times 10^{-3}$
$\therefore \quad 1-\frac{1.03 \times 10^3}{\rho^{\prime}}=3.712 \times 10^{-3}$
$\Rightarrow \rho^{\prime}=\frac{1.03 \times 10^3}{1-3.712 \times 10^{-3}}=1.034 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$
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