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Chapter 34: Problem 4

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

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?

What is the flow of blood through a small vein (diameter \(=3.00 \mathrm{mm}\) ) that is \(1.00 \mathrm{cm}\) long? The drop in blood pressure over this length is 40.0 Torr, and the viscosity of blood is approximately \(4.00 \mathrm{cP}\) at \(310 \mathrm{K}\)

a. The diffusion coefficient of the protein lysozyme \((\mathrm{MW}=14.1 \mathrm{kg} / \mathrm{mol})\) is \(0.104 \times 10^{-5} \mathrm{cm}^{2} \mathrm{s}^{-1} .\) How long will it take this protein to diffuse an rms distance of \(1 \mu \mathrm{m} ?\) Model the diffusion as a three-dimensional process. b. You are about to perform a microscopy experiment in which you will monitor the fluorescence from a single lysozyme molecule. The spatial resolution of the microscope is \(1 \mu \mathrm{m} .\) You intend to monitor the diffusion using a camera that is capable of one image every 60 s. Is the imaging rate of the camera sufficient to detect the diffusion of a single lysozyme protein over a length of \(1 \mu \mathrm{m} ?\) c. Assume that in the microscopy experiment of part (b) you use a thin layer of water such that diffusion is constrained to two dimensions. How long will it take a protein to diffuse an rms distance of \(1 \mu \mathrm{m}\) under these conditions?

The Reynolds' number (Re) is defined as \(\mathrm{Re}=\rho\left\langle\mathrm{v}_{x}\right\rangle d / \eta,\) where \(\rho\) and \(\eta\) are the fluid density and viscosity, respectively; \(d\) is the diameter of the tube through which the fluid is flowing; and \(\left\langle\mathrm{v}_{x}\right\rangle\) is the average velocity. Laminar flow occurs when \(\operatorname{Re}<2000\), the limit in which the equations for gas viscosity were derived in this chapter. Turbulent flow occurs when \(\mathrm{Re}>2000 .\) For the following species, determine the maximum value of \(\left\langle\mathrm{v}_{x}\right\rangle\) for which laminar flow will occur: a. \(\mathrm{Ne}\) at \(293 \mathrm{K}(\eta=313 \mu \mathrm{P}, \rho=\) that of an ideal gas) through a 2.00 -mm-diameter pipe. b. Liquid water at \(293 \mathrm{K}\left(\eta=0.891 \mathrm{cP}, \rho=0.998 \mathrm{g} \mathrm{mL}^{-1}\right)\) through a 2.00 -mm-diameter pipe.

You are interested in purifying a sample containing the protein alcohol dehydrogenase obtained from horse liver; however, the sample also contains a second protein, catalase. These two proteins have the following transport properties at \(298 \mathrm{K}\). $$\begin{array}{lll} &{\text { Catalase }} & \text { Alcohol Dehydrogenase } \\ \hline \overline{\mathrm{s}}(\mathrm{s}) & 11.3 \times 10^{-13} & 4.88 \times 10^{-13} \\ D\left(\mathrm{m}^{2} \mathrm{s}^{-1}\right) & 4.1 \times 10^{-11} & 6.5 \times 10^{-11} \\ \bar{V}\left(\mathrm{cm}^{3} \mathrm{g}^{-1}\right) & 0.715 & 0.751 \end{array}$$ a. Determine the molecular weight of catalase and alcohol dehydrogenase. b. You have access to a centrifuge that can provide angular velocities up to 35,000 rpm. For the species you expect to travel the greatest distance in the centrifuge tube, determine the time it will take to centrifuge until a \(3 \mathrm{cm}\) displacement of the boundary layer occurs relative to the initial \(5 \mathrm{cm}\) location of the boundary layer relative to the centrifuge axis. c. To separate the proteins, you need a separation of at least \(1.5 \mathrm{cm}\) between the boundary layers associated with each protein. Using your answer to part (b), will it be possible to separate the proteins by centrifugation?
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