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Chapter 9: Problem 73

Sucrose is allowed to diffuse along a \(10-\mathrm{cm}\) length of tubing filled with water. The tube is \(6.0 \mathrm{~cm}^{2}\) in crosssectional area. The diffusion coefficient is equal to \(5.0 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\), and \(8.0 \times 10^{-14} \mathrm{~kg}\) is transported along the tube in \(15 \mathrm{~s}\). What is the difference in the concentraton levels of sucrose at the two ends of the tube?

Short Answer

Expert verified
Using the provided values and steps, calculate the difference in concentration levels of sucrose at the two ends of the tube.

Step by step solution

01

Calculate the Molar Flux

The molar flux, \(J\), can be calculated from the given mass of sucrose that is being transported and the time it takes. \nGiven that the mass of sucrose transported \(m = 8.0 \times 10^{-14}~ \mathrm{kg}\), time \(t = 15~\mathrm{s}\), and the cross-sectional area of the tube \(A = 6.0~\mathrm{cm}^2\). \nWe convert the area from cm虏 to m虏: \(A = 6.0 \times 10^{-4}~\mathrm{m}^2\) (since 1m虏 = 10鈦碿m虏).\nSo, \(J = \frac{m}{A \cdot t}\). Calculate this value.
02

Convert the diffusion coefficient to the correct units

The given diffusion coefficient, \(D\), is \(5.0 \times 10^{-10}~\mathrm{m}^2/\mathrm{s}\). This is already in the appropriate units, so no conversion is necessary.
03

Use Fick's Law to find the concentration difference

Fick's first law of diffusion, \(J = -D \frac{\Delta \phi}{\Delta x}\), can be rearranged to find the concentration difference: \(\Delta \phi = -\frac{J \cdot \Delta x}{D}\).\nGiven that the length of the tube, or \(\Delta x = 10~\mathrm{cm} = 0.1~\mathrm{m}\) (since 1m = 100cm). Insert the calculated molar flux from step 1, the given diffusion coefficient from step 2, and the length of the tube into the equation to find the concentration difference, \(\Delta \phi\).

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Key Concepts

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

Molar Flux
Molar flux is a measure of the amount of substance that passes through a given surface per unit time. In the context of diffusion, it's often symbolized as \( J \). It indicates how fast a substance is moving from an area of high concentration to an area of low concentration.

So, when we calculate the molar flux in our exercise, we're finding out the rate at which sucrose is diffusing through the water in the tubing. The calculation involves the mass of sucrose transported, the area of the cross-section through which it diffuses, and the time it takes for this transport. The basic unit of molar flux is moles per second per square meter \( (mol/s \cdot m^2) \).

To provide a deeper understanding, imagine molar flux as the number of molecules of sucrose 'walking' across a 'bridge' (the tubing's cross-sectional area) in a second. The larger the crowd (mass) or the narrower the bridge (smaller area), or the faster the crowd moves (less time), the greater the molar flux.
Concentration Gradient
The concentration gradient in diffusion is essentially the driving force that causes particles to move. It is the difference in the concentration of a substance between two regions, and particles will naturally move from the region of higher concentration to the region of lower concentration, down their concentration gradient.

Plotting the changing concentration of sucrose along the length of the tubing in our exercise would show a slope or gradient. This gradient is what induces the sucrose molecules to distribute themselves more evenly through the tubing. In mathematical terms, a steeper gradient means a larger concentration difference, leading to a higher rate of diffusion.

In the world of small molecules like sucrose, a concentration gradient is like a hill. Molecules 'roll down' this hill from where there are many (high concentration) to where there are few (low concentration), reaching a state known as equilibrium, where no net movement occurs because the 'hill' becomes level (no gradient).
Diffusion Coefficient
The diffusion coefficient, symbolized as \( D \), is a factor that impacts the rate of diffusion and is specific to the substance and the medium through which it diffuses. It reflects how easily particles can move and has units of area per unit time, such as \(m^2/s\).

In our exercise, the diffusion coefficient for sucrose in water is given and doesn't require conversion, since it's already in appropriate units. Different substances have different diffusion coefficients in the same medium due to their varying sizes, shapes, and interactions with the solvent.

One can compare the diffusion coefficient to the ease with which a person can walk through different mediums. Walking on a clear path (high diffusion coefficient) is faster and easier compared to walking in a dense forest (low diffusion coefficient). This coefficient determines how quickly equilibrium is reached, or, going back to our earlier metaphor, how quickly the molecules 'fill in' the valleys to level out the 'hill' of the concentration gradient.

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Most popular questions from this chapter

Water is pumped through a pipe of diameter \(15.0 \mathrm{~cm}\) from the Colorado River up to Grand Canyon Village, on the rim of the canyon. The river is at \(564 \mathrm{~m}\) elevation and the village is at \(2096 \mathrm{~m}\). (a) At what minimum pressure must the water be pumped to arrive at the village? (b) If \(4500 \mathrm{~m}^{3}\) are pumped per day, what is the speed of the water in the pipe? (c) What additional pressure is necessary to deliver this flow? Note: You may assume the free-fall acceleration and the density of air are constant over the given range of elevations.

A bathysphere used for deep sea exploration has a radius of \(1.50 \mathrm{~m}\) and a mass of \(1.20 \times 10^{4} \mathrm{~kg}\). In order to dive, the sphere takes on mass in the form of sea water. Determine the mass the bathysphere must take on so that it can descend at a constant speed of \(1.20 \mathrm{~m} / \mathrm{s}\) when the resistive force on it is \(1100 \mathrm{~N}\) upward. The density of sea water is \(1.03 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\)

Take the density of blood to be \(\rho\) and the distance between the feet and the heart to be \(h_{H}\). Ignore the flow of blood, (a) Show that the difference in blood pressure between the feet and the heart is given by \(P_{F}-P_{H}=\rho g h_{j F}\) (b) Take the density of blood to be \(1.05 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3}\) and the distance between the heart and the feet to be \(1.20 \mathrm{~m}\). Find the difference in blood pressure between these two points. This problem indicates that pumping blood from the extremities is very difficult for the heart. The veins in the legs have valves in them that open when blood is pumped toward the heart and close when blood flows away from the heart. Also, pumping action produced by physical activities such as walking and breathing assists the heart.

For safety in climbing, a mountaineer uses a nylon rope that is \(50 \mathrm{~m}\) long and \(1.0 \mathrm{~cm}\) in diameter. When supporting a \(90-\mathrm{kg}\) climber, the rope elongates \(1.6 \mathrm{~m}\). Find its Young's modulus.

Figure \(\mathrm{P} 9.27\) shows the essential parts of a hydraulic brake system. The area of the piston in the master cylinder is \(1.8 \mathrm{~cm}^{2}\) and that of the piston in the brake cylinder is \(6.4 \mathrm{~cm}^{2}\). The coefficient of friction between shoe and wheel drum is \(0.50 .\) If the wheel has a radius of \(34 \mathrm{~cm}\), determine the frictional torque about the axle when a force of \(44 \mathrm{~N}\) is exerted on the brake pedal.
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