# Oxygen concentration

• Archived

## Homework Statement

Insects do not have lungs as we do, nor do they breathe through their mouths. Instead, they have a system of tiny tubes, called tracheae, through which oxygen diffuses into their bodies. The tracheae begin at the surface of the insect's body and penetrate into the interior. Suppose that a trachea is 1.9 mm long with a cross-sectional area of 2.1 10-9 m2. The concentration of oxygen in the air outside the insect is 0.23 kg/m3, and the diffusion constant is 1.1 10-5 m2/s. If the mass per second of oxygen diffusing through a trachea is 1.7 10-12 kg/s, find the oxygen concentration at the interior end of the tube.

## Homework Equations

m/t = (DA(deltaC)) / L

## The Attempt at a Solution

m/t = (DA(deltaC)) / L
1.7 E-12 = ((1.1 E-5 m^2/s)(2.1 E-9 m^2)(change in C)) / .0019 m
change in c = .3698268
(inside concentration - outside concentration) = .3698268
(inside concentration - .23 kg/m^3) = .3698268
inside concentration = .5998 kg/m^3

Thats not the right answer... but that is what I keep getting. Someone at school told me that the equation m/t = (DA(deltaC)) / L was correct, so that means I must be messing up after that, but I don't know where and I can't figure it out!!! If anyone would help me, I would reeeally appreciate it!!! :-)

$$j = D \frac{(c_o - c_i)}{L}$$
Where $j = \frac{f}{A}$, and f is the mass flow rate of oxygen. I can derive the above equation if needed to. What we need is to find the concentration of oxygen inside the insect, so
$$c_i = c_o - \frac{fL}{DA}$$
$$c_i = 0.23 \frac{kg}{m^3} - \frac{\left(1.7 \cdot 10^{-12} \frac{kg}{s} \right) (1.9 \cdot 10^{-3} m)}{ \left( 1.1 \cdot 10^{-5} \frac{m^2}{s} \right) (2.1 \cdot 10^{-9} m^2)} = 0.09 \frac{kg}{m^3}$$