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Chapter 5: Problem 7

(a) Briefly explain the concept of a driving force. (b) What is the driving force for steady-state diffusion?

Short Answer

Expert verified
Answer: The driving force for steady-state diffusion is the concentration gradient, which stays constant during the process. It provides the necessary force for atoms or molecules to move from a region with higher concentration to a region with lower concentration.

Step by step solution

01

(Step 1: Define driving force)

A driving force can be defined as a factor that initiates and sustains a change or process in a system, causing it to move from one state to another. It is essential for understanding various material processes, such as diffusion. In general, the driving force can result from differences in concentration, temperature, or other physical properties across the system.
02

(Step 2: Discuss the concept of diffusion)

Diffusion is a process that occurs in materials where atoms or molecules move from a region with higher concentration to a region with lower concentration, as a result of random thermal motion. The driving force behind diffusion is the difference in concentration (or concentration gradient) between these regions. Diffusion tends to equalize the concentration throughout the material, thus minimizing the concentration differences and reducing the driving force.
03

(Step 3: Explain the concept of steady-state diffusion)

Steady-state diffusion, also known as stationary diffusion, is a specific case of diffusion where the concentration gradient remains constant over time. In steady-state diffusion, the rate at which the particles are moving through the material remains constant. This type of diffusion is often observed when the materials are maintained at a constant temperature and the external conditions remain consistent.
04

(Step 4: Identify the driving force for steady-state diffusion)

The driving force for steady-state diffusion is the concentration gradient, which stays constant during the process. The concentration gradient is the difference in concentration between the regions and provides the necessary force for atoms or molecules to move from the region with higher concentration to the region with lower concentration. In summary, the driving force is a factor that initiates and sustains changes in a system. In the case of steady-state diffusion, the driving force is the concentration gradient, which remains constant during the process.

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Most popular questions from this chapter

The preexponential and activation energy for the diffusion of chromium in nickel are \(1.1 \times 10^{-4}\) \(\mathrm{m}^{2} / \mathrm{s}\) and \(272,000 \mathrm{~J} / \mathrm{mol}\), respectively. At what temperature will the diffusion coefficient have a value of \(12 \times 10^{-14} \mathrm{~m}^{2} / \mathrm{s} 2\)

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.

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.

Carbon is allowed to diffuse through a steel plate 10 -mm thick. The concentrations of carbon at the two faces are \(0.85\) and \(0.40 \mathrm{~kg} \mathrm{C} / \mathrm{cm}^{3} \mathrm{Fe}\), which are maintained constant. If the preexponential and activation energy are \(5.0 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) and 77,000 \(\mathrm{J} / \mathrm{mol}\), respectively, compute the temperature at which the diffusion flux is \(6.3 \times 10^{-10} \mathrm{~kg} / \mathrm{m}^{2} \mathrm{~s}\).

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