Question

Insects obtain oxygen by diffusion through tracheae, tiny tubes connecting the outside air to the interior of the body. A typical such tube is about 2.3 mm long and has cross-sectional area of 2.2 × 10-9 m2. Assuming the concentration of oxygen inside is 48 % what it is outside in the atmosphere,

a. find the concentration of oxygen in the air at 24.9 ° C. Assume 21% is oxygen and the pressure is 1 atm.

--> using ideal gas law, a.) = 8.587 mol/m^3

b. Calculate the diffusion rate J of oxygen molecules in units of kg/s. Take the diffusion constant to be 1 × 10-5 m2/s.

*Diffusion constant is 1*10^-5

Answer 1

A) The ideal gas law is represented by the mathematical expression

PV=nRT --------(1)

where, P= pressure, V= volume, n=moles of gas, R=gas constant and T is temperature

Here we have given

P=1 atm,

V=21% or 0.21 m³ ,

R= 8.205 x 10^-5 m³.atm.K^-1mol^-1 and

T= 273.15 + 24.9 = 298.05K

n=?

After rearrangement equation (1), we will get

n=PV/RT

n= (1 x 0.21)/(8.205 x 10^-5 x 298.05

n=8.587 mol/m³

b)   Concentration of O₂ outside in Kg/m³ =8.587mol/m³ x 32 g/mol x 10^-3 Kg/gm = 0.275 Kg/ m³

Since the density of O₂ in inside is 48% what is outside in atmosphere

So, concentration of O₂ inside in Kg/m³ = 0.275 x 48/100 = 0.132 Kg/m³

now rate of diffusion can be calculate by using Fick's Law of diffusion which is mathematically expressed as

J= DA(C₂-C₁)/L

J= diffusion rate

where D= Diffusion constant

A= area of cross-section

C₂ and C₁ = concentration of solute

L= distance between two different solute concentration

So, here,

D= 1 × 10^-5 m²/s

C₂= 0.275 Kg/m³  

C₁ = 0.132 Kg/m³

A= 2.2 × 10^-9 m² and

L= 2.3 mm or 2.3 x10^-3m

So, J= (1 × 10^-5 x 2.2 × 10^-9 x (0.275-0.132))/(2.3 x10^-3)

J= 1.368 x10^-12 Kg/s