Question

A fish maintains its depth in fresh water by adjusting the air content of porous bone or air sacs to make its average density the same as that of the water. Suppose that with its air sacs collapsed, a fish has a density of 1.09 g/cm³. To what fraction of its expanded body volume must the fish inflate the air sacs to reduce its density to that of water?

Answer 1

Concepts and reason

The concepts used to solve this problem are density of a substance concept and the percentage fraction of a value concept.

Initially write the formula formulas for density of the water and for the density of the fish by using density of a substance concept.

Write the formula for the percentage fraction of expanded volume of air sacs of the fish the percentage fraction of a value concept.

Then calculate the percentage fraction of expanded volume of air sacs when it filled with air by using the percentage fraction of expanded volume formula.

Fundamentals

The density of a substance is defined by the ratio of the mass of the substance to the volume of the substance.

$\rho = \frac{m}{V}$

Here, $\rho$ is the density of a substance, m is the mass of the substance, and V is the volume of the substance.

The formula for the percentage fraction of a value is,

${c'_{{\rm{fraction}}}} = \left( {\frac{{c'}}{{c' + c}}} \right) \times 100\%$

Here, ${c'_{{\rm{fraction}}}}$ is the percentage fraction of a value $c'$, $c$ is the total value, and $c'$ is a certain value.

The formula for the density of the fish is,

${\rho _{{\rm{fish}}}} = \frac{{{m_{{\rm{fish}}}}}}{{{V_{{\rm{air sacs}}}}}}$

Here, ${\rho _{{\rm{fish}}}}$ is the density of the fish when its air sacs collapsed, ${m_{{\rm{fish}}}}$ is the mass of the fish, and ${V_{{\rm{air sacs}}}}$ is the volume of the air sacs unfilled with air sacs.

The formula for the density of the water when the air sacs of the fish filled with air is,

${\rho _{{\rm{water}}}} = \frac{{{m_{{\rm{fish}}}}}}{{{V_{{\rm{expanded}}}} + {V_{{\rm{air sacs}}}}}}$

Here, ${\rho _{{\rm{water}}}}$ is the density of the water and ${V_{{\rm{expanded}}}}$ is the expanded body volume of the fish.

The formula for the percentage fraction of expanded body volume of air sacs of the fish is,

${V_{{\rm{fraction}}}} = \left( {\frac{{{V_{{\rm{expanded}}}}}}{{{V_{{\rm{expanded}}}} + {V_{{\rm{air sacs}}}}}}} \right) \times 100\%$

Here, ${V_{{\rm{fraction}}}}$ is the fraction of the expanded body volume of the fish

Rewrite the equation ${\rho _{{\rm{fish}}}} = \frac{{{m_{{\rm{fish}}}}}}{{{V_{{\rm{air sacs}}}}}}$ for ${m_{{\rm{fish}}}}$.

${m_{{\rm{fish}}}} = \left( {{\rho _{{\rm{fish}}}}} \right)\left( {{V_{{\rm{air sacs}}}}} \right)$

Substitute $\left( {{\rho _{{\rm{fish}}}}} \right)\left( {{V_{{\rm{air sacs}}}}} \right)$ for ${m_{{\rm{fish}}}}$ in the equation ${\rho _{{\rm{water}}}} = \frac{{{m_{{\rm{fish}}}}}}{{{V_{{\rm{expanded}}}} + {V_{{\rm{air sacs}}}}}}$.

$\begin{array}{c}\\{\rho _{{\rm{water}}}} = \frac{{\left( {{\rho _{{\rm{fish}}}}} \right)\left( {{V_{{\rm{air sacs}}}}} \right)}}{{{V_{{\rm{expanded}}}} + {V_{{\rm{air sacs}}}}}}\\\\\frac{{{V_{{\rm{air sacs}}}}}}{{{V_{{\rm{expanded}}}} + {V_{{\rm{air sacs}}}}}} = \frac{{{\rho _{{\rm{water}}}}}}{{{\rho _{{\rm{fish}}}}}}\\\end{array}$

Multiply the above equation with minus (-) and add one (1) on both sides of the equation.

$\begin{array}{l}\\1 + \left( - \right)\frac{{{V_{{\rm{air sacs}}}}}}{{{V_{{\rm{expanded}}}} + {V_{{\rm{air sacs}}}}}} = 1 + \left( - \right)\frac{{{\rho _{{\rm{water}}}}}}{{{\rho _{{\rm{fish}}}}}}\\\\1 - \frac{{{V_{{\rm{air sacs}}}}}}{{{V_{{\rm{expanded}}}} + {V_{{\rm{air sacs}}}}}} = 1 - \frac{{{\rho _{{\rm{water}}}}}}{{{\rho _{{\rm{fish}}}}}}\\\\\frac{{{V_{{\rm{expanded}}}}}}{{{V_{{\rm{expanded}}}} + {V_{{\rm{air sacs}}}}}} = \frac{{{\rho _{{\rm{fish}}}} - {\rho _{{\rm{water}}}}}}{{{\rho _{{\rm{fish}}}}}}\\\end{array}$

Substitute $\frac{{{\rho _{{\rm{fish}}}} - {\rho _{{\rm{water}}}}}}{{{\rho _{{\rm{fish}}}}}}$ for $\frac{{{V_{{\rm{expanded}}}}}}{{{V_{{\rm{expanded}}}} + {V_{{\rm{air sacs}}}}}}$ in the equation ${V_{{\rm{fraction}}}} = \left( {\frac{{{V_{{\rm{expanded}}}}}}{{{V_{{\rm{expanded}}}} + {V_{{\rm{air sacs}}}}}}} \right) \times 100\%$ and solve for ${V_{{\rm{fraction}}}}$.

${V_{{\rm{fraction}}}} = \left( {\frac{{{\rho _{{\rm{fish}}}} - {\rho _{{\rm{water}}}}}}{{{\rho _{{\rm{fish}}}}}}} \right) \times 100\%$

The density of the water is $1.0\,{\rm{g/c}}{{\rm{m}}^{\rm{3}}}$ and the density of the fish when its air sacs collapsed is $1.09\,{\rm{g/c}}{{\rm{m}}^{\rm{3}}}$.

Substitute $1.09\,{\rm{g/c}}{{\rm{m}}^{\rm{3}}}$ for ${\rho _{{\rm{fish}}}}$ and $1.0\,{\rm{g/c}}{{\rm{m}}^{\rm{3}}}$ for ${\rho _{{\rm{water}}}}$ in the above equation and solve for ${V_{{\rm{fraction}}}}$.

$\begin{array}{c}\\{V_{{\rm{fraction}}}} = \left( {\frac{{1.09\,{\rm{g/c}}{{\rm{m}}^{\rm{3}}} - 1.0\,{\rm{g/c}}{{\rm{m}}^{\rm{3}}}}}{{1.09\,{\rm{g/c}}{{\rm{m}}^{\rm{3}}}}}} \right) \times 100\% \\\\ = 8.26\% \\\end{array}$

Ans:

Therefore, the fraction of the expanded body volume of the fish when the air sacs reduces its density to that of the water is 8.26%.