Question

(II) Oxygen diffuses from the surface of insects to the interior through tiny tubes called tracheae. An average trachea is about 2 mm long and has cross-sectional area of \(2 \times {10^{ - 9}}\;{{\rm{m}}^{\rm{2}}}\). Assuming the concentration of oxygen inside is half what it is outside in the atmosphere, (a) show that the concentration of oxygen in the air (assume 21% is oxygen) at 20°C is about \(8.7\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}}\), then (b) calculate the diffusion rate J, and (c) estimate the average time for a molecule to diffuse in. Assume the diffusion constant is \(1 \times {10^{ - 5}}\;{{\rm{m}}^{\rm{2}}}{\rm{/s}}\)

Short answer

(a) It has been proved that the concentration of oxygen in the air is \(8.7\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}}\).

(b) The diffusion rate is \(4 \times {10^{ - 11}}\;{\rm{mol/s}}\).

(c) The time required for the diffusion is 0.6 s.

Step-by-step solution

Given data

The length of the trachea is \(\Delta x = 2\;{\rm{mm}} = 2 \times {10^{ - 3}}\;{\rm{m}}\).

The cross-sectional area is \(A = 2 \times {10^{ - 9}}\;{{\rm{m}}^{\rm{2}}}\).

The oxygen in the air is 21%.

The diffusion constant is \(D = 1 \times {10^{ - 5}}\;{{\rm{m}}^{\rm{2}}}{\rm{/s}}\).

The atmospheric pressure is \(P = 1\;{\rm{atm}} = 1.013 \times {10^5}\;{\rm{Pa}}\).

Concepts

The ideal gas equation is\(PV = nRT\).

The time for the diffusion is\(t = \frac{{\bar C}}{{\Delta C}}\frac{{{{\left( {\Delta x} \right)}^2}}}{D}\)and the diffusion constant is\(J = DA\frac{{{C_1} - {C_2}}}{{\Delta x}}\).

Calculation of part (a)

Part (a)

The pressure of the oxygen is \({P_o} = 0.21P\) as there is 21% oxygen in the air.

Now, using the ideal gas equation, you get:

\(\begin{aligned}{c}{P_o}V &= nRT\\\frac{n}{V} &= \frac{{{P_o}}}{{RT}}\\\frac{n}{V} &= \frac{{0.21P}}{{RT}}\end{aligned}\)

Substituting the values in the above equation, you get:

\(\begin{aligned}{c}\frac{n}{V} &= \frac{{0.21 \times \left( {1.013 \times {{10}^5}\;{\rm{Pa}}} \right)}}{{\left( {8.314\;{\rm{J/mol}} \cdot {\rm{K}}} \right) \times \left( {293\;{\rm{K}}} \right)}}\\ &= 8.7\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}}\end{aligned}\)

Hence, it has been proved that the concentration of oxygen in the air is \(8.7\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}}\).

Calculation of part (b)

Part (b)

The concentration of oxygen outside the insects is \({C_1} = 8.7\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}}\).

The concentration of oxygen inside the insects is \({C_2} = \frac{1}{2}{C_1}\).

Now, the diffusion rate is:

\(\begin{aligned}{c}J &= DA\frac{{{C_1} - {C_2}}}{{\Delta x}}\\ &= DA\frac{{{C_1} - \frac{1}{2}{C_1}}}{{\Delta x}}\\ &= DA{C_1}\frac{{0.5}}{{\Delta x}}\end{aligned}\)

Substituting the values in the above equation, you get:

\(\begin{aligned}{c}J &= \left( {1 \times {{10}^{ - 5}}\;{{\rm{m}}^{\rm{2}}}{\rm{/s}}} \right) \times \left( {2 \times {{10}^{ - 9}}\;{{\rm{m}}^{\rm{2}}}} \right) \times \left( {8.7\;{\rm{mol/}}{{\rm{m}}^{\rm{3}}}} \right) \times \frac{{0.5}}{{2 \times {{10}^{ - 3}}\;{\rm{m}}}}\\ &= 4 \times {10^{ - 11}}\;{\rm{mol/s}}\end{aligned}\)

Hence, the diffusion rate is \(4 \times {10^{ - 11}}\;{\rm{mol/s}}\).

Calculation of part (c)

Part (c)

The average concentration is:

\(\begin{aligned}{c}\bar C &= \frac{1}{2}\left( {{C_1} + {C_2}} \right)\\ &= \frac{1}{2}\left( {{C_1} + \frac{1}{2}{C_1}} \right)\\ &= \frac{1}{2} \times \frac{3}{2}{C_1}\\ &= \frac{3}{4}{C_1}\end{aligned}\)

The difference between the concertations is:

\(\begin{aligned}{c}{C_1} - {C_2} &= {C_1} - \frac{1}{2}{C_1}\\ &= \frac{1}{2}{C_1}\end{aligned}\)

The average diffusion time is:

\(\begin{aligned}{c}t &= \frac{{\bar C}}{{\Delta C}}\frac{{{{\left( {\Delta x} \right)}^2}}}{D}\\t &= \frac{{\frac{3}{4}{C_1}}}{{\frac{1}{2}{C_1}}}\frac{{{{\left( {\Delta x} \right)}^2}}}{D}\\t &= \frac{3}{2}\frac{{{{\left( {2 \times {{10}^{ - 3}}\;{\rm{m}}} \right)}^2}}}{{1 \times {{10}^{ - 5}}\;{{\rm{m}}^{\rm{2}}}{\rm{/s}}}}\\t &= 0.6\;{\rm{s}}\end{aligned}\)

Hence, the time required for the diffusion is 0.6 s.