(II) Oxygen diffuses from the surface of insects to the...

(II) Oxygen diffuses from the surface of insects to the interior through tiny tubes called tracheae. An average trachea is about 2 $\mathrm{mm}$ long and has cross-sectional area of $2 \times 10^{-9} \mathrm{m}^{2} .$ 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 $20^{\circ} \mathrm{C}$ is about $8.7 \mathrm{mol} / \mathrm{m}^{3},$ then $(b)$ calculate the diffusion rate $J,$ and $(c)$ estimate the average time for a molecule to diffuse in. Assume the diffusion constant is $1 \times 10^{-5} \mathrm{m}^{2} / \mathrm{s}.$

Key Concepts

Diffusion Time Estimation

Diffusion time estimation involves calculating the average time required for a molecule to traverse a distance via diffusion. This is typically done using the characteristic length scale and the diffusion constant, often applying the relationship that the time scales with the square of the distance divided by the diffusion coefficient.

Molecular Diffusion

Molecular diffusion is the process by which molecules spread from areas of higher concentration to areas of lower concentration due to random thermal motion. This concept underpins many phenomena in physics, chemistry, and biology, including the transport of oxygen in various systems.

Ideal Gas Law

The Ideal Gas Law, expressed as PV = nRT, is a fundamental relationship linking the pressure, volume, temperature, and the amount (in moles) of a gas. It is commonly used to calculate the concentration of gases in a given volume at a specified temperature and pressure, which is essential for estimating the molar density of oxygen in air.

Fick’s First Law of Diffusion

Fick’s First Law states that the diffusive flux of a substance is proportional to the concentration gradient across space. It is used to determine the rate at which molecules (such as oxygen) diffuse from regions of higher concentration to regions of lower concentration, thereby quantifying the diffusion rate.

Transcript

00:01 Okay, so we're doing chapter 18, problem 55 here.

00:04 So this problem says oxygen diffuses from the surface of insects to the interior through tiny tubes called trache.

00:13 Treki.

00:15 An average trache is about two millimeters long and has a cross -sectional area of two times 10 to negative 9 square meters.

00:24 Assuming the concentration of oxygen is half what it is outside in the atmosphere.

00:31 A says show that the concentration of oxygen in the air assume 21 % oxygen at 20 degrees celsius.

00:42 21 % at 20 degrees celsius is 8 .7 moles per cubic meters.

00:52 Okay, so we want to use the ideal gas law first.

00:56 So if we assume a pressure of one atmosphere, then what that means is the pressure caused by the oxygen is the percent of oxygen times the pressure.

01:17 So this is 21 percent times p or 0 .21 atmospheres.

01:24 If we're just talking in terms of oxygen here.

01:27 So from this, we know that ideal gas law can use this.

01:32 So let's relate this to n over v or moles per volume.

01:38 This is a concentration.

01:41 So n over v then becomes p over r t.

01:45 So 0 .21 atmospheres times the conversion to si units times 10 in the fifth pascal's per atmosphere.

01:57 So those will cancel.

01:58 And then 8 .315 times 293.

02:07 If we plug that in, we should see this comes out to roughly 8 .732 moles per meter cubed, which is exactly what we were trying to show.

02:18 So awesome.

02:18 Check us out.

02:20 Part b then.

02:23 Calculate the diffusion rate, j.

02:27 Okay.

02:28 So let's do that.

02:29 So if we look at equation 1811, it gives us j equaling d times a times the change and concentration with respect to x.

02:40 And we can approximate this as d -a -c -1 minus c2 over denta -x.

02:50 Cool.

02:52 So that means the diffusion rate now is given approximately diffusion constant here, which is 1 times 10 to the negative 5 square meters per second, times the cross -sectional area, which we have 2 times 10 to the negative 9 square meters...

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