- #1
haleyy89
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Suppose that a tracheae is 1.17 mm long with a cross-sectional area of 1.01 x 10-9m2. The concentration of oxygen in the air outside the insect is 0.659 kg/m3, and the diffusion constant is 1.99 x 10-5 m2/s. If the mass per second of oxygen is diffusing through a trachea is 1.50 x 10-12 kg/s, then find the oxygen concentration at the interior end of the tube.
m= (DA(deltaC))t / L
I re-arranged the equation to solve for the change in concentration, (delta)C.
deltaC = mL / DA. I know the given value for the outside concentration will be moved over to the other side. I think that the concentration given for the outside of the insect would be the same as initial concentration and the concentration I am solving for would be the final concentration but not exactly sure on this. So I added the concentration given to the other side of the equation to isolate the final concentration. I calculated this to be 8.73E8 kg/s, however this is incorrect. Any suggestions?
m= (DA(deltaC))t / L
I re-arranged the equation to solve for the change in concentration, (delta)C.
deltaC = mL / DA. I know the given value for the outside concentration will be moved over to the other side. I think that the concentration given for the outside of the insect would be the same as initial concentration and the concentration I am solving for would be the final concentration but not exactly sure on this. So I added the concentration given to the other side of the equation to isolate the final concentration. I calculated this to be 8.73E8 kg/s, however this is incorrect. Any suggestions?