Question

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

Answer 1

We know that change in volume is given by $\Delta V={{V}_{1}}-{{V}_{2}}$ and and relation between volume, density and mass can be given as, $V=\dfrac{m}{\rho }$. So, using these formulas and substituting the values in the formula of change in volume we will find the density of water at the given height.

Formula used:
$\Delta V={{V}_{1}}-{{V}_{2}}$, $V=\dfrac{m}{\rho }$

Complete answer:
Now, in question we are asked to find the density of water at a depth where pressure is $80.0\ atm$ and the density at the surface is given as $1.03\times {{10}^{3}}kg{{m}^{-3}}$, so now we know that change in volume can be given by, $\Delta V={{V}_{1}}-{{V}_{2}}$ and the relation between volume and density can be given as, $V=\dfrac{m}{\rho }$.
Where, V is volume, m is mass and $\rho$ is density of water.
Now, let the given depth be h,
density of water at surface be, ${{\rho }_{1}}=1.03\times {{10}^{-3}}$
${{\rho }_{2}}$ is the density of water at given depth.
Pressure at depth is $80.0\ atm$
Now, we know that $1\ atm=1.015\times {{10}^{5}}\ Pa$
So, $80.0\ atm=80\times 1.01\times {{10}^{5}}Pa$
Now, considering the change in volume and substituting values of volume in terms of density we will get,

$\Delta V={{V}_{1}}-{{V}_{2}}$
$V=\dfrac{m}{\rho }$
$\Rightarrow \Delta V=m\left( \dfrac{1}{{{\rho }_{1}}}-\dfrac{1}{{{\rho }_{2}}} \right)$ ………………….(i)

Now, as go in depth in water the volumetric strain of water increases which can be expressed mathematically as,
$\text{Volumetric}\ \text{strain}=\dfrac{\text{change in volume }}{\text{original volume}}=\dfrac{\Delta V}{{{V}_{1}}}$
Now, substituting the values of expression (i) in this equation we will get,

$\text{Volumetric}\ \text{strain}=\dfrac{\Delta V}{{{V}_{1}}}=m\left( \dfrac{1}{{{\rho }_{1}}}-\dfrac{1}{{{\rho }_{2}}} \right)\times \dfrac{{{\rho }_{1}}}{m}$
$\therefore \dfrac{\Delta V}{{{V}_{1}}}=1-\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}$ …………………….(ii)

Now, the bulk modulus can be given as, $B=\dfrac{p{{V}_{1}}}{\Delta V}$, where p is pressure at depth h.
$\Rightarrow \dfrac{\Delta V}{{{V}_{1}}}=\dfrac{p}{B}$ ……………………(iii)

Now, the inverse of bulk modulus is called compressibility and the value of compressibility of water is fixed which can be given as, $\text{compressibility}=\dfrac{1}{B}=45.8\times {{10}^{-11}}P{{a}^{-1}}$.
So, substituting this value in equation (iii) we will get,
$\dfrac{\Delta V}{{{V}_{1}}}=\dfrac{p}{B}=80\times 1.01\times 1{{0}^{3}}\times 45.8\times {{10}^{-11}}$
$\Rightarrow \dfrac{\Delta V}{{{V}_{1}}}=3.71\times {{10}^{-3}}$ …………..(iv)

Now, by equation expressions (ii) and (iv) we will get,
$1-\dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}=3.71\times {{10}^{-3}}$
$\Rightarrow \dfrac{{{\rho }_{1}}}{{{\rho }_{2}}}=\left( 1-3.71\times {{10}^{-3}} \right)$
$\Rightarrow {{\rho }_{2}}=\dfrac{1.03\times {{10}^{3}}}{\left( 1-3.71\times {{10}^{-3}} \right)}$
$\Rightarrow {{\rho }_{2}}=1.034\times {{10}^{3}}kg{{m}^{-3}}$
Hence, the density of water at a given height is $1.034\times {{10}^{3}}kg{{m}^{-3}}$.

Note:
Bulk modulus is used to show how much a substance can resist the compression acting on it due to change in pressure and volume on the object. It is one of the most important to be considered in Fluid mechanics and it is defined as a ratio of pressure increase to the resulting relative decrease of the volume. So, for such types of problems students must consider the Bulk modulus.