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a. The diffusion coefficient of sucrose in water at \(298 \mathrm{K}\) is \(0.522 \times 10^{-9} \mathrm{m}^{2} \mathrm{s}^{-1} .\) Determine the time it will take a sucrose molecule on average to diffuse an rms distance of \(1 \mathrm{mm}\) b. If the molecular diameter of sucrose is taken to be \(0.8 \mathrm{nm}\) what is the time per random walk step?

Short Answer

Expert verified
The time for a sucrose molecule to diffuse an rms distance of 1 mm is 958 seconds, and the time per random walk step is approximately \(6.13 \times 10^{-10}\) seconds.

Step by step solution

01

Use the given diffusion coefficient and rms distance formula

First, we have to use the diffusion coefficient and the relationship between rms distance and time. The rms distance formula is: \(x_{rms} = \sqrt{2D_{s}t}\) Where: - \(x_{rms}\) is the root-mean-square displacement - \(D_{s}\) is the diffusion coefficient - \(t\) is the time We have \(x_{rms} = 1 \times 10^{-3} \mathrm{m}\), and \(D_s = 0.522 \times 10^{-9} \mathrm{m}^2\mathrm{s}^{-1}\). To find the time \(t\), we must rearrange the formula and solve for it.
02

Rearrange the formula and find the time

Rearrange the rms distance formula and solve for time: \(t = \frac{x_{rms}^2}{2 D_s}\) Plug in the given values: \(t = \frac{(1 \times 10^{-3} \mathrm{m})^2}{2 \times (0.522 \times 10^{-9} \mathrm{m}^2\mathrm{s}^{-1})}\)
03

Calculate the time it takes to diffuse 1 mm

Perform the calculations: \(t = \frac{1 \times 10^{-6} \mathrm{m}^2}{1.044 \times 10^{-9} \mathrm{m}^2\mathrm{s}^{-1}}\) \(t = 958 \mathrm{s}\) Now we have the time it takes for a sucrose molecule to diffuse an rms distance of 1 mm, which is 958 seconds.
04

Calculate time per random walk step

To find the time per random walk step, we will use the molecular diameter of sucrose (\(d_{s}\)) and the rms distance formula: For one random walk step, \(x_{rms}\) = \(d_{s}\) \(t_{step} = \frac{d_{s}^2}{2 D_s}\) Plug in the values: \(t_{step} = \frac{(0.8 \times 10^{-9} \mathrm{m})^2}{2 \times (0.522 \times 10^{-9} \mathrm{m}^2\mathrm{s}^{-1})}\)
05

Calculate time per random walk step

Perform the calculations: \(t_{step} = \frac{6.4 \times 10^{-19} \mathrm{m}^2}{1.044 \times 10^{-9} \mathrm{m}^2\mathrm{s}^{-1}}\) \(t_{step} = 6.13 \times 10^{-10} \mathrm{s}\) The time per random walk step is approximately \(6.13 \times 10^{-10}\) seconds. In summary, the time for a sucrose molecule to diffuse an rms distance of 1 mm is 958 seconds, and the time per random walk step is approximately \(6.13 \times 10^{-10}\) seconds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Root-Mean-Square Displacement
The concept of root-mean-square (rms) displacement is crucial for understanding diffusion processes. In the context of molecular movement through a medium like water, rms displacement quantifies the average distance a molecule will move over a specific time period. It's derived from the motion of many particles and represents a statistical measure of the spatial extent of random fluctuations. To ease into this, imagine you're taking steps in random directions; the rms displacement will be the distance you've traveled after a given number of steps.

Using the rms displacement formula \(x_{rms} = \sqrt{2D_{s}t}\), where \(D_{s}\) is the diffusion coefficient and \(t\) is time, you can calculate how long it will take for a particle to travel a certain distance. For our sucrose molecule in water, we see that calculating this requires an understanding of the relationship between the molecule's diffusion and time. This relationship reveals that the further the distance you are curious about, the longer the time it will take on average for a particle to cover that distance due to diffusion.
The Random Walk Step Explained
The random walk step is a term used to describe the individual movements that a particle, like a sucrose molecule, makes as it diffuses through a substance such as water. Each of these tiny, seemingly erratic movements is a 'step' in the particle's path. This path is not straight but rather follows a random, zigzag trajectory, resembling the steps a person might take if blindfolded.

The size of each step in the random walk is related to the molecular diameter, impacting how quickly the substance diffuses. By understanding the time per random walk step, which is a reflection of both the diffusion coefficient and the molecular diameter, scientists and engineers can predict how substances move through environments, which is important in fields like pharmaceuticals, material science, and environmental engineering. The calculated time per step helps us to infer the dynamics of molecular interactions and the efficiency of the diffusion process.
Significance of Molecular Diameter in Diffusion
Molecular diameter is another significant factor in the discussion of diffusion. It's effectively the 'size' of a molecule, which influences how easily it can move through the spaces between other molecules in a solvent like water. In our exercise, the molecular diameter of sucrose dictates the scale of a 'single step' in the random walk model.

When you know the molecular diameter, you can start to visualize how these microscopic particles jostle and bump their way through a liquid. Larger diameters may mean slower diffusion rates due to increased interactions with surrounding molecules and a greater likelihood of finding obstructions in its path. This aspect is integral to understanding how substances mix at the molecular level and can affect everything from the speed of a chemical reaction to the taste of your morning coffee.

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Most popular questions from this chapter

The collisional cross section of \(\mathrm{N}_{2}\) is \(0.43 \mathrm{nm}^{2}\). What is the diffusion coefficient of \(\mathrm{N}_{2}\) at a pressure of 1 atm and a temperature of \(298 \mathrm{K} ?\)

a. The viscosity of \(\mathrm{Cl}_{2}\) at \(293 \mathrm{K}\) and 1 atm is \(132 \mu \mathrm{P}\) Determine the collisional cross section of this molecule based on the viscosity. b. Given your answer in part (a), estimate the thermal conductivity of \(\mathrm{Cl}_{2}\) under the same pressure and temperature conditions.

a. Derive the general relationship between the diffusion coefficient and viscosity for a gas. b. Given that the viscosity of Ar is \(223 \mu \mathrm{P}\) at \(293 \mathrm{K}\) and \(1 \mathrm{atm}\) what is the diffusion coefficient? c. What is the thermal conductivity of Ne under these same conditions (the collisional cross section of Ar is 1.5 times that of \(\mathrm{Ne}\) )?

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A current of 2.00 A is applied to a metal wire for 30. s. How many electrons pass through a given point in the wire during this time?

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