(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 represents the difference in concentration levels between two points in the medium. This gradient motivates particles to migrate from regions of higher concentration to lower concentration, leading to a constant diffusion rate until equilibrium is reached.

Step by step solution

01

(Step 1: Defining Driving Force)

A driving force can be defined as an external factor or influence that is responsible for causing a change in a given system. It provides the necessary energy to initiate a certain process or reaction, and establishes a gradient between the initial and final states of the system, thus driving the process toward its completion. In general, driving forces can include thermal, electrical, or concentration gradients.
02

(Step 2: Understanding Steady-State Diffusion)

Steady-state diffusion refers to the process where the rate of diffusion remains constant and does not change with time. In this type of diffusion, particles (such as atoms, ions, or molecules) move through a medium (solid or liquid) under the influence of a driving force. The movement of particles occurs from high-concentration regions to low-concentration regions, resulting in a smoother distribution of particles in the medium.
03

(Step 3: Identifying the Driving Force for Steady-State Diffusion)

The driving force for steady-state diffusion is the concentration gradient. This gradient represents the difference in concentration levels between two points in the medium. The concentration gradient acts as the motivating force for the particles to migrate from a region of higher concentration to a region of lower concentration. As long as the concentration gradient is maintained, the diffusion process will continue, with particles moving through the medium at a constant rate, until they reach an equilibrium state where the concentration becomes uniform throughout the medium.

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Most popular questions from this chapter

The steady-state diffusion flux through a metal plate is \(5.4 \times 10^{-10} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\) at a temperature of \(727^{\circ} \mathrm{C}(1000 \mathrm{~K})\) and when the concentration gradient is \(-350 \mathrm{~kg} / \mathrm{m}^{4} .\) Calculate the diffusion flux at \(1027^{\circ} \mathrm{C}(1300 \mathrm{~K})\) for the same concentration gradient and assuming an activation energy for diffusion of \(125,000 \mathrm{~J} / \mathrm{mol}\).

Consider a diffusion couple composed of two semi-infinite solids of the same metal, and that each side of the diffusion couple has a different concentration of the same elemental impurity; furthermore, assume each impurity level is constant throughout its side of the diffusion couple. For this situation, the solution to Fick's second law (assuming that the diffusion coefficient for the impurity is independent of concentration) is as follows: $$ C_{x}=\left(\frac{C_{1}+C_{2}}{2}\right)-\left(\frac{C_{1}-C_{2}}{2}\right) \operatorname{erf}\left(\frac{x}{2 \sqrt{D t}}\right) $$ In this expression, when the \(x=0\) position is taken as the initial diffusion couple interface, then \(C_{1}\) is the impurity concentration for \(x<0\); likewise, \(C_{2}\) is the impurity content for \(x>0\). A diffusion couple composed of two silver-gold alloys is formed; these alloys have compositions of \(98 \mathrm{wt} \% \mathrm{Ag}-2 \mathrm{wt} \% \mathrm{Au}\) and \(95 \mathrm{wt} \% \mathrm{Ag}-5 \mathrm{wt} \%\) Au. Determine the time this diffusion couple must be heated at \(750^{\circ} \mathrm{C}\) (1023 K) in order for the composition to be \(2.5 \mathrm{wt} \% \mathrm{Au}\) at the \(50 \mu \mathrm{m}\) position into the 2 wt \(\%\) Au side of the diffusion couple. Preexponential and activation energy values for Au diffusion in \(\mathrm{Ag}\) are \(8.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) and \(202,100 \mathrm{~J} / \mathrm{mol}\), respectively.

A sheet of steel \(1.5 \mathrm{~mm}\) thick has nitrogen atmospheres on both sides at \(1200^{\circ} \mathrm{C}\) and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is \(6 \times 10^{-11} \mathrm{~m}^{2} / \mathrm{s}\), and the diffusion flux is found to be \(1.2 \times\) \(10^{-7} \mathrm{~kg} / \mathrm{m}^{2} \cdot \mathrm{s}\). Also, it is known that the concentration of nitrogen in the steel at the highpressure surface is \(4 \mathrm{~kg} / \mathrm{m}^{3} .\) How far into the sheet from this high-pressure side will the concentration be \(2.0 \mathrm{~kg} / \mathrm{m}^{3}\) ? Assume a linear concentration profile.

For a steel alloy it has been determined that a carburizing heat treatment of 10 -h duration will raise the carbon concentration to \(0.45 \mathrm{wt} \%\) at a point \(2.5 \mathrm{~mm}\) from the surface. Estimate the time necessary to achieve the same concentration at a 5.0-mm position for an identical steel and at the same carburizing temperature.

When \(\alpha\)-iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, \(C_{\mathrm{H}}\) (in weight percent), is a function of hydrogen pressure, \(p_{\mathrm{H}_{2}}\) (in MPa), and absolute temperature \((T)\) according to $$ C_{\mathrm{H}}=1.34 \times 10^{-2} \sqrt{p_{\mathrm{H}}_{2}} \exp \left(-\frac{27.2 \mathrm{~kJ} / \mathrm{mol}}{R T}\right) $$ Furthermore, the values of \(D_{0}\) and \(Q_{d}\) for this diffusion system are \(1.4 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) and \(13,400 \mathrm{~J} / \mathrm{mol}\), respectively. Consider a thin iron membrane \(1 \mathrm{~mm}\) thick that is at \(250^{\circ} \mathrm{C}\). Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is \(0.15 \mathrm{MPa}\) (1.48 atm), and on the other side \(7.5 \mathrm{MPa}\) (74 atm).

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