Question

What the density of water at a depth where pressure is 80 atm given that it density as a surface is zero 3 multi out and give 3 kg metre cube given that girls model of water is equal 2.2.19 pascal?

Answer 1

Please explain your question properly. It just seems to irrelevant.

Answer 2

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Given:

Pressure of water at a certain depth = 80 atm

At the surface of water, density of water (d) = $\begin{lgathered}1.013 \times {10}^{13} \: \: \frac{kg}{ {m}^{3} } \\\end{lgathered}$

To Find:

Density of water at a certain depth

$\huge{\red{\star}}$${\blue{\underline{\huge{\textsf{Solution}}}}} \huge{\red{\star}}$

Let Volume at surface of water = $\begin{lgathered}\frac{m}{ d_{1} } \\\end{lgathered}$

Let volume at a certain depth = $\begin{lgathered}\frac{m}{ d_{2} } \\\end{lgathered}$

Change in volume =

$\begin{lgathered}v_{2} - v_{1} \\ \\ = \frac{m}{ d_{2} } - \frac{m}{ d_{1} } \\ \\ = m( \frac{1}{ d_{2} } - \frac{1}{ d_{1} } ) \\ \\\end{lgathered}$

We know Bulk modulus (B) =

$\begin{lgathered}\frac{change \: \: in \ \: pressure \: \times \: volume}{change \: \: in \: \: volume} \\\end{lgathered}$

Compressibility of water = $\begin{lgathered}\frac{1} {b} \\ \\ = \frac{change \: \: in \: \: volume \: \: }{change \: \: in \: \: pressure \: \times \: volume} \\ \\ = 45.8 \times {10}^{ - 11 \:} pascals\end{lgathered}$

Let change in pressure = dp

Now,

$\begin{lgathered}45.8 \times {10}^{ - 11} = \frac{m( \frac{1}{ d_{2} } - \frac{1}{ d_{1}} ) }{dp \times \frac{m}{ d_{1} } } \\ \\ 45.8 \times {10}^{ - 11} = \frac{ d_{1} }{dp}( \frac{1}{ d_{2} } - \frac{1}{ d_{1} } ) \\ \\ 45.8 \times {10}^{ - 11} = \frac{1}{dp} (1 - \frac{ d_{1} }{d _{2} } ) \\ \\\end{lgathered}$

Substitute all the given values,

$\begin{lgathered}45.8 \times {10}^{ - 11} = 1 - (\frac{1.03 \: \times \: {10}^{3} }{ d_{2} } ) \: \times \: \frac{1}{80 \times 1.013 \: \times \: {10}^{5} } \\ \\ by \: \: solving \\ \\ d_{2} = 1.034 \times {10}^{3} \: \: \frac {kg}{ {m}^{3} }\end{lgathered}$

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