Ideal gas law practical application

[astrophysicsm]

Homework Statement

To measure how far below the ocean surface a bird dives to catch a fish, a scientist uses a method originated by Lord Kelvin. He dusts the interiors of plastic tubes with powdered sugar and then seals one end of each tube. He captures the bird at nighttime in its nest and attaches a tube to its back. He then catches the same bird the next night and removes the tube. In one trial, using a tube 7.40 cm long, water washes away the sugar over a distance of 2.46 cm from the open end of the tube. Find the greatest depth to which the bird dived, assuming the air in the tube stayed at constant temperature. (Assume the density of the ocean water is 1030 kg/m3.)

Homework Equations

Ideal gas law:
PV=nRT

Density = m/V

The Attempt at a Solution

I think this might have to do with the pressure of the air inside the tube and the pressure at the desired water depth. I can apply this given the change in volume, however, only length is provided and the volume of the tube is not known. I can replace the volume by mass/density in the equation:
P.(m/D) = nRT

But I'm stuck at this point.

Thanks for any help in advance!
astrophysicsm

 

[jack action]

Air in tube:

State 1 (the bird is flying):

V₁ = tube length * tube area
P₁ = P_atm

State 2 (the bird is diving):

V₂ = compressed length * tube area
P₂ = P_atm + P_water

Given: T₁ = T₂

To find P_water, use Bernoulli (assuming the bird is at rest when at deepest depth)

Compare state 1 with state 2 using the ideal gas law.

 

[astrophysicsm]

Thank you ! :)

I used the ideal gas equation to find final Pressure:
Pf = (Pi Li /Lf)

Using Bernoulli's equation:
Pf=Psurface+Dgh

I ended up with:
h= [(PiLi/Lf) – Pi]/Dg = 4.99m

Can anyone verify this is correct ?

 

[Ioliar]

What are you measuring Pressure in. Shouldn't it be in Pascals so 1 atm = 101325 Pa
So therefore, height = 20.16m?

 

[Skittler]

What are you measuring Pressure in. Shouldn't it be in Pascals so 1 atm = 101325 Pa
So therefore, height = 20.16m?

You made an error in your calculation. The final length Lf isn't 2.46 cm, it is (7.4-2.46=4.94)cm
If you use this value you will get astrophysicsm's answer, which is correct. (I made the same mistake you did).

 

[amiranda]

Hey guys,
Your post had helped me.

here is what I did on Maple, and you are right, Lf is the difference.
> h = ((101325*0.74e-1)/(0.74e-1-0.246e-1)-101325)/(1030*9.8);
h = 4.998750614
... and your 4.99 attempt was right.cheers