Question

A particular deep-sea fish has a swim bladder that can hold a maximum to 0.34 L of gas. It lives at 500m depth. If

Answer 1

Answer:

• The researcher's can safely raise the fish before reaching the 170 m depth.

Explanation:

Assuming constant temperature, you can use Boyle's law:

         $p_1V_1=p_2V_2$

Where:

• p₁ is the pressure at 500 m depth = 50.57 atm

• p₂ is the unknown pressure

• V₁ = 0.12 Liter

• V₂ is the maximum volume that the fish's swim bladder can hold = 0.34 liters

Then, determine p₂:

     $p_2=p_1\times \dfrac{V_1}{V_2}$

      $p_2=50.57atm\times \dfrac{0.12liter}{0.34liter}=17.85atm$

From the table, there is a linear relationship between pressure (atm) and water depth (m).

The slope of the linear function is about 0.099:

• (50 - 40)/(100-50) = 0.099

The y-intercept is 1.00 (the point 0, 1.00).

Then, the linear function is: Pressure = 1 + 0.099×depth

Now you can find the depth at wich the pressure is 17.85 atm

• 17.85atm = 1.00atm + 0.099×depth

• depth = (17.85 atm - 1.00atm) / 0.099 = 170.2 m ≈ 170 m

Hence, the fish's swimm bladder will ruputure at 170m.