We\'ve seen that fish can control their buoyancy through the use of a swim bladd

Question

We've seen that fish can control their buoyancy through the use of a swim bladder, a gas-filled organ inside the body. You can assume that the gas pressure inside the swim bladder is roughly equal to the external water pressure. A fish swimming at a particular depth adjusts the volume of its swim bladder to give it neutral buoyancy. If the fish swims upward or downward, the changing water pressure causes the bladder to expand or contract. Consequently, the fish must adjust the quantity of gas to restore the original volume and thus reestablish neutral buoyancy. Consider a large, 7.0 kgstriped bass with a volume of 7.0 L. When neutrally buoyant, 7.0% of the fish's volume is taken up by the swim bladder. Assume a body temperature of 15C.

A. How many moles of air are in the swim bladder when the fish is at a depth of 100 ft ? Answer: 8.4x10^-2 mol

B. What will the volume of the swim bladder be if the fish ascends to a 50 ft depth without changing the quantity of gas? Assume that the density of the salt water is 1030 kg/m3. Answer: 7.9x10^-4 m^3

C. To return the swim bladder to its original size, how many moles of gas must be removed??? Please help with this last part that i'm stuck on! thank you!!

Explanation / Answer

We know that

PV = n R T

n = PV/RT

P = Patm + P gauge = p(atm) + h rho g

P = 1.01 x 10^5 + 1000 x 30.48 x 9.81 = 1.01 x 10^5 + 3 x 10^5 = 4.01 x 10^5 Pa = 4.04 atm

V = 7 x 7/100 = 0.49

T = 15 + 273 = 288

n = 4.04 x 0.49/0.082 x 288 = 8.4 x 10^-2 mol

Hence, n = 8.4 x 10^-2 mol

b)Pressure at 50 ft

P = 1.01 x 10^5 + 1000 x 15.24 x 9.8 = 2.5 x 10^5 atm = 2.48 atm

we know for an ideal gas:

P1V1 = P2V2

V2 = P1V1/P2 = 4.04 x 0.49/2.48 = 0.79 L

Hence, V = 7.9 x 10^-4 m^3 = 0.79 L

c)n' = P2V2/RT

n' = 2.48 x 0.79/0.082 x 288 = 0.083 mol

Hence, 0.083 mol should be removed.

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