Swim Bladder In order to stay at a constant depth, an aquatic organism needs to
ID: 1790775 • Letter: S
Question
Swim Bladder In order to stay at a constant depth, an aquatic organism needs to be neutrally buoyant, meaning that the buoyant force on it and the force of gravity on it have equal strengths. To achieve this, the animal must have an average density that is equal to that of water, so that the mass of its body is equal to the mass of the volume of water being displaced One of the complications to maintaining neutral buoyancy underwater is the fact that pressure changes with depth. Because air volume is inversely proportional to pressure, as we will see later in the course (and as you may already know if you have studied the ideal gas law), the volume of air in a fish's swim bladder or a mammal's lungs decreases with depth, changing the animal's overall volume and hence its overall density. The following problem explores some of the consequences of this. Consider a freshwater fish swimming in water of density 1,000 kg/m'. Assume that the average density of its tissues is 1,070 kg/m, and its swim bladder makes up 7% of its total volume at the surface. Calculate the average density of the fish at the surface a) b) Find the fish's average density when it is at a point that is 20 m below the surface. What would happen to the fish if it did nothing? What volume of air would have to be added to the swim bladder for the fish to remain neutrally buoyant at a depth of 20 m? c)Explanation / Answer
a. density of water, rho = 1,000 kg/m^3
rho' = 1070 kg/m^3 ( average density of tissues)
volume of swim bladder = 0.07 V ( where V is its total volume)
so swim bladder is filled with air of densioty rho' = 1.223 kg/m^3
hence average density of fish at surface = (1070*0.93V + 1.223*0.07V)/V = 995.18575 kg/m^3
b. at 20 m below the surface
density of air = rho"
now using ideal gas equation
Po/rho' = P/rho"
P = Po + rho*g*20 = 1.01*10^5 + 1000*20*9.81 = 297200 Pa
hence
1.01*10^5/1.223 = 297200/rho"
rho" = 3.598 kg/m^3
also, voluem of swim bladder reduces by 2.9377 times
new bladder volume = 0.0238V
total volume = V' = 0.93V + 0.0238V = 0.9538V
hence average density =(1070*0.93V + 3.598*0.0238V)/(0.9538V) = 1043.39 kg/m^3
c. let volume of air added be v
then
1000 = (1070*0.93V + 3.598*(0.0238(V)+ v))/(0.93V + (0.0238V + v))
930V + 23.8V + 1000v = 995.1856324V + 3.598v
v = 0.04153V ( where V is the volume of the fish on the surface of water)
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