Answered: Consider a large 10.0 kg striped bass…

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Question

Consider a large 10.0 kg striped bass swimming in fresh water. When neutrally buoyant, 10% of the total volume of the fish is taken up by air im the swim bladder. Assume a constant bodyy temperature of 20 degrees C. and neglect the mass of the air in your calculations.

a) What is the volume of the swim bladder, in m^3, when the fish is neutrally bouyant?

B) What is the water pressure at a depth of 25m?

C) How many moles of air ar in the swim bladder when the fish is neutrally bouyant at a depth of 25m?

D) What will the volume of the swim bladder be if the fish ascends to a depth of 15m without changing the quantity of gas?

Expert Solution

Step 1

Introduction:

When the density of the body is equal to the density of the fluid in which it is present it neither sinks and does not float and it said to be neutrally buoyant.

Calculation:

Write the expression for the density of the fish.

$ρ_{f} = \frac{m_{f}}{V_{f}}$

Here $ρ_{f}$ is the density of fish, $m_{f}$ is the mass of the fish, and $V_{f}$ is the volume of the fish.

Equate the density of water to the density of fish.

$ρ_{w} = ρ_{f}$

Here $ρ_{w}$ is the density of water.

Substitute $\frac{m_{f}}{V_{f}}$ for $ρ_{f}$ in the above expression.

$ρ_{w} = \frac{m_{f}}{v_{f}}$

Substitute $1000 \frac{kg}{m^{3}}$ for $ρ_{w}$ , and $10 kg$ for $m_{f}$ in the above expression.

$\begin{array}{rcl} 1000 \frac{kg}{m^{3}} & = & \frac{10 kg}{V_{f}} \\ V_{f} & = & \frac{10 kg}{1000 \frac{kg}{m^{3}}} \\ = & 0.01 m^{3} \end{array}$

The volume of swim bladder is $10 %$ of the total volume of fish.

Therefore, write the expression for the volume of swim bladder.

$V_{b} = \frac{10}{100} V_{f}$

Here, $V_{b}$ is the volume of the swim bladder. Substitute $0.01 m^{3}$ for $V_{f}$ in the above expression.

$\begin{array}{rcl} V_{b} & = & \frac{10}{100} (0.01 m^{3}) \\ = & 0.001 m^{3} \\ = & 0.001 m^{3} \times \frac{10^{6} {cm}^{3}}{1 m^{3}} \\ = & 1000 {cm}^{3} \end{array}$

Thus, the volume of the swim bladder is $1000 {cm}^{3}$ .

Step 2

b) The water pressure at the depth $25 m$ per meter square area will due to the weight of the water column of height $25 m$ and surface area of $1 m^{2}$ .

Write the expression for the water pressure at the depth of $25 m$ .

$P = ρ_{w} g h$

Here, $P$ is the pressure, $g$ is the acceleration due to gravity, and $h$ is the height.

Substitute $1000 \frac{kg}{m^{3}}$ for $ρ_{w}$ , $9.8 \frac{m}{s^{2}}$ for $g$ and $25 m$ for $h$ in the above expression.

$\begin{array}{rcl} P & = & (1000 \frac{kg}{m^{3}}) (9.8 \frac{m}{s^{2}}) (25 m) \\ = & 2.45 \times 10^{5} Pa \end{array}$

Thus, the water pressure at the depth of $25 m$ is $\begin{array}{rcl} 2 \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} 45 \end{array} \begin{array}{rcl} \times \end{array} \begin{array}{rcl} 10 \end{array} \begin{array}{rcl} \end{array}$ ⁵ Pa.

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