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Chapter 5: Problem 6

Briefly explain the concept of steady state as it applies to diffusion.

Short Answer

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Answer: A steady-state in the context of diffusion is reached when the concentration gradient no longer changes with time, and the rate of diffusion in one direction is equal to the rate of diffusion in the opposite direction. At this point, there's no net movement of particles, meaning that the system is in a stable equilibrium. The concept of steady-state is important in various practical applications to understand how substances are transported within living organisms and predict the behavior of diffusion-based systems in physics, chemistry, and engineering. Knowing when a system achieves a steady state can help determine the optimal conditions and parameters for the desired outcome or application.

Step by step solution

01

Introduction to Diffusion

Diffusion is the process by which particles (such as molecules, atoms, or ions) move from an area of higher concentration to an area of lower concentration until a uniform distribution is achieved. This movement is driven by the random motion of the particles and aims to reduce the concentration gradient within the system.
02

Steady State in Diffusion

In the context of diffusion, a steady state is reached when the concentration gradient no longer changes with time, i.e., when the rate of the diffusion in one direction is equal to the rate of diffusion in the opposite direction. At this point, there's no net movement of particles, meaning that the system is in a stable equilibrium.
03

Importance of Steady State

The concept of the steady state is essential in various practical applications such as understanding how substances are transported within living organisms and predicting the behavior of diffusion-based systems in physics, chemistry, and engineering. In these contexts, knowing when a system achieves a steady state can help determine the optimal conditions and parameters for the desired outcome or application.
04

Example of Steady State in Diffusion

Consider a container divided into two halves by a permeable membrane, with the left half containing a higher concentration of dye molecules than the right half. Initially, dye molecules will diffuse from the left half to the right half, following the concentration gradient. Gradually, as dye molecules distribute more uniformly across the container, diffusion will continue but at a decreasing rate. Eventually, the concentration gradient will become negligible, and dye molecules will move between the two halves at equal rates in both directions, meaning there is no net movement of molecules. At this point, the system has reached a steady state.

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Most popular questions from this chapter

The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a \(6-\mathrm{mm}\) thick sheet of palladium having an area of \(0.25 \mathrm{~m}^{2}\) at \(600^{\circ} \mathrm{C}\). Assume a diffusion coefficient of \(1.7 \times 10^{-8} \mathrm{~m}^{2} / \mathrm{s}\), that the respective concentrations at the high- and low-pressure sides of the plate are \(2.0\) and \(0.4 \mathrm{~kg}\). of hydrogen per cubic meter of palladium, and that steady-state conditions have been attained.

Consider the diffusion of some hypothetical metal Y into another hypothetical metal \(Z\) at \(950^{\circ} \mathrm{C}\); after \(10 \mathrm{~h}\) the concentration at the \(0.5 \mathrm{~mm}\) position (in metal \(Z\) ) is \(2.0 \mathrm{wt} \% \mathrm{Y}\). At what position will the concentration also be \(2.0 \mathrm{wt} \% \mathrm{Y}\) after a \(17.5-\mathrm{h}\) heat treatment again at \(950^{\circ} \mathrm{C}\) ? Assume preexponential and activation energy values of \(4.3 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\) and \(180,000 \mathrm{~J} / \mathrm{mol}\), respectively, for this diffusion system.

Indium atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat treatments; the background concentration of In in this silicon material is known to be \(2 \times 10^{20}\) atoms \(/ \mathrm{m}^{3}\). The drive-in diffusion treatment is to be carried out at \(1175^{\circ} \mathrm{C}\) for a period of \(2.0 \mathrm{~h}\), whichgives a junction depth \(x_{j}\) of \(2.35 \mu \mathrm{m}\). Compute the predeposition diffusion time at \(925^{\circ} \mathrm{C}\) if the surface concentration is maintained at a constant level of \(2.5 \times 10^{26}\) atoms \(/ \mathrm{m}^{3}\). For the diffusion of In in Si. values of \(Q_{d}\) and \(D_{0}\) are \(3.63 \mathrm{eV} /\) atom and \(7.85 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively.

Nitrogen from a gaseous phase is to be diffused into pure iron at \(675^{\circ} \mathrm{C}\). If the surface concentration is maintained at \(0.2 \mathrm{wt} \% \mathrm{~N}\), what will be the concentration \(2 \mathrm{~mm}\) from the surface after \(25 \mathrm{~h}\) ? The diffusion coefficient for nitrogen in iron at \(675^{\circ} \mathrm{C}\) is \(2.8 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\).

A sheet of BCC iron \(2-\mathrm{mm}\) thick was exposed to a carburizing gas atmosphere on one side and a decarburizing atmosphere on the other side at \(675^{\circ} \mathrm{C}\). After reaching steady state, the iron was quickly cooled to room temperature. The carbon concentrations at the two surfaces of the sheet were determined to be \(0.015\) and \(0.0068\) wt \(\%\), respectively. Compute the diffusion coefficient ifthe diffusion flux is \(7.36 \times 10^{-9} \mathrm{~kg} / \mathrm{m}^{2}+\mathrm{s}\). Hint: Use Equation \(4.9\) to convert the concentrations from weight percent to kilograms of carbon per cubic meter of iron.
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