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Chapter 13: Problem 102

In plants, water diffuses out through small openings known as stomatal pores. If \(D=2.4 \times 10^{-5} \mathrm{m}^{2} / \mathrm{s}\) for water vapor in air, and the length of the pores is \(2.5 \times 10^{-5} \mathrm{m},\) how long does it take for a water molecule to diffuse out through the pore?

Short Answer

Expert verified
Answer: The time taken for a water molecule to diffuse out through the stomatal pore is approximately 1.04 seconds.

Step by step solution

01

Identify the given values

We have the following given values: - Diffusion coefficient of water vapor in air (D): \(2.4 \times 10^{-5} \mathrm{m}^2/\mathrm{s}\) - Length of the stomatal pore (L): \(2.5 \times 10^{-5} \mathrm{m}\)
02

Use Fick's first law of diffusion

Fick's first law of diffusion states that the rate of diffusion (J) is proportional to the product of the diffusion coefficient (D) and the concentration gradient (Δc/Δx): $$J = -D\frac{\Delta c}{\Delta x}$$ In this case, we are not given the concentration gradient, but we know that it remains constant throughout the process. Therefore, we can use Fick's first law to find the time taken for a water molecule to diffuse out through the pore.
03

Find the time taken for diffusion

We can rearrange the Fick's first law equation to find the time taken for diffusion (t): $$t = \frac{\Delta x}{D}$$ Now, we can substitute the given values of L (the length of the pore, equivalent to Δx) and D (the diffusion coefficient) into the equation: $$t = \frac{2.5 \times 10^{-5}\mathrm{m}}{2.4 \times 10^{-5}\mathrm{m}^2/\mathrm{s}}$$ $$t = 1.04 \times 10^{0} \mathrm{s} \approx 1.04\,\mathrm{s}$$ So, it takes approximately 1.04 seconds for a water molecule to diffuse out through the stomatal pore.

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