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Chapter 5: Problem 1

Briefly explain the difference between selfdiffusion and interdiffusion.

Short Answer

Expert verified
Answer: The key difference between self-diffusion and inter-diffusion lies in the substances involved. In self-diffusion, particles move within a single substance, whereas in interdiffusion, particles move between two different substances. Furthermore, self-diffusion is caused by the random thermal motion of particles, while interdiffusion is driven by concentration gradients between the two substances.

Step by step solution

01

Definition of Self-diffusion

Self-diffusion is the movement of individual atoms, ions, or molecules within a single substance, without any net direction. This process occurs due to the random thermal motion of the particles within the substance. For example, in a block of pure metal, self-diffusion occurs as the atoms within the metal move randomly from one lattice site to another.
02

Definition of Interdiffusion

In contrast, interdiffusion refers to the movement of atoms, ions, or molecules between two different substances or phases, resulting in the mixing of their components. This process takes place at the interface between the two substances and is driven by the difference in concentration of the components. An example of interdiffusion is the spreading of two gases in contact with each other, leading to a homogeneous mixture.
03

Key Difference

The key difference between self-diffusion and inter-diffusion lies in the substances involved. In self-diffusion, particles move within a single substance, whereas in interdiffusion, particles move between two different substances. Furthermore, self-diffusion is caused by the random thermal motion of particles, while interdiffusion is driven by concentration gradients between the two substances.
04

Example of Self-diffusion

Considering a glass of water, self-diffusion occurs as the water molecules move randomly within the body of the water. Their motion does not result in any net change in the water, i.e., the water remains uniform and undisturbed.
05

Example of Interdiffusion

When a drop of ink is added to the glass of water, interdiffusion occurs as the ink molecules move and mix with the water molecules, resulting in a gradual spreading of the ink throughout the glass. This process eventually leads to a homogeneous mixture of ink and water.

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Most popular questions from this chapter

The preexponential and activation energy for the diffusion of iron in cobalt are \(1.1 \times 10^{-5}\) \(\mathrm{m}^{2} / \mathrm{s}\) and \(253,300 \mathrm{~J} / \mathrm{mol}\), respectively. At what temperature will the diffusion coefficient have a value of \(2.1 \times 10^{-14} \mathrm{~m}^{2} / \mathrm{s}\) ?

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).

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

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.
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