The steady-state diffusion flux through a metal plate is \(7.8 \times 10^{-8} \mathrm{kg} / \mathrm{m}^{2}-\mathrm{s}\) at a tem- perature of \(1200^{\circ} \mathrm{C}(1473 \mathrm{K})\) and when the concentration gradient is \(-500 \mathrm{kg} / \mathrm{m}^{4}\). Calculate the diffusion flux at \(1000^{\circ} \mathrm{C}(1273 \mathrm{K})\) for the same concentration gradient and assuming an activation energy for diffusion of \(145,000 \mathrm{J} / \mathrm{mol}\).

Short Answer

Expert verified
Answer: The diffusion flux at 1000°C (1273 K) is approximately 2.04 × 10⁻⁸ kg/(m²·s).

Step by step solution

01

Understand Fick's First Law

Fick's first law relates the diffusion flux (J) to the concentration gradient (dC/dx) and the diffusion coefficient (D). The formula for Fick's first law is: \(J = -D \frac{dC}{dx}\) In this case, we are given the concentration gradient (dC/dx = -500 kg/m³), but we need to find the diffusion coefficient (D) at the two given temperatures.
02

Apply the Arrhenius equation

The Arrhenius equation can help us determine the diffusion coefficient as a function of temperature. This equation is: \(D = D_0 e^{-\frac{Q}{RT}}\) In this equation, \(D_0\) is the pre-exponential factor, \(Q\) is the activation energy given in J/mol, \(R\) is the gas constant (8.314 J/mol·K), and \(T\) is the temperature in Kelvin. Since we know the diffusion flux \(J_1\) and the concentration gradient \(\frac{dC}{dx}\) at the first temperature \(T_1\), we can find \(D_1\) for the first temperature: \(D_1 = -\frac{J_1}{\frac{dC}{dx}}\) We will then use \(D_1\) and the given activation energy to find \(D_0\) and finally to find \(D_2\), the diffusion coefficient at the second temperature \(T_2\).
03

Calculate \(D_1\) and \(D_0\)

First, calculate the diffusion coefficient at the first temperature: \(D_1 = -\frac{J_1}{\frac{dC}{dx}}\) \(D_1 = -\frac{7.8 \times 10^{-8} kg / m^2s}{-500 kg / m^3}\) \(D_1 = \frac{1.56 \times 10^{-10} m^2 / s}{1}\) Now, using the Arrhenius equation, we can calculate the pre-exponential factor (\(D_0\)): \(D_1 = D_0 e^{-\frac{Q}{R T_1}}\) \(D_0 = \frac{D_1}{e^{-\frac{Q}{R T_1}}}\) \(D_0 = \frac{1.56 \times 10^{-10} m^2/s}{e^{-\frac{145000J/mol}{(8.314 J/mol·K)(1473K)}}}\) \(D_0 \approx 5.81 \times 10^{-7} m^2/s\)
04

Calculate \(D_2\) and the diffusion flux at the second temperature

Next, use the pre-exponential factor (\(D_0\)) to find the diffusion coefficient at the second temperature (\(T_2\)): \(D_2 = D_0 e^{-\frac{Q}{R T_2}}\) \(D_2 = 5.81 \times 10^{-7} m^2/s \times e^{-\frac{145000J/mol}{(8.314 J/mol·K)(1273K)}}\) \(D_2 \approx 4.08 \times 10^{-11} m^2/s\) Finally, we can use this \(D_2\) value to calculate the diffusion flux at the second temperature: \(J_2 = -D_2 \frac{dC}{dx}\) \(J_2 = -(4.08 \times 10^{-11} m^2/s) \times (-500 kg/m^3)\) \(J_2 \approx 2.04 \times 10^{-8} kg/(m^2 \cdot s)\) The diffusion flux at \(1000^{\circ} \mathrm{C}(1273 \mathrm{K})\) for the same concentration gradient and with the activation energy given is approximately \(2.04 \times 10^{-8} \mathrm{kg} / \mathrm{m}^{2}-\mathrm{s}\).

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Key Concepts

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

Fick's first law
In the realm of materials science, Fick's first law is an essential principle that connects the idea of diffusion flux with how substances move in response to concentration differences. It states that the flux of material through an area is directly proportional to the concentration gradient across that area. In simpler terms, it explains how particles flow from regions of high concentration to regions of low concentration. This behavior is mathematically represented as

\[ J = -D \frac{dC}{dx} \]

where \(J\) stands for the diffusion flux, \(D\) is the diffusion coefficient indicative of how easily particles diffuse within the material, and \(\frac{dC}{dx}\) represents the concentration gradient, or how the concentration changes with distance. The negative sign signifies that diffusion occurs in the direction of decreasing concentration.
Arrhenius equation
The Arrhenius equation is a vital tool that describes how the diffusion coefficient \(D\) varies with temperature. It is particularly useful in materials science for predicting how temperature changes affect the rate at which atoms diffuse. The Arrhenius equation is given by:

\[ D = D_0 e^{-\frac{Q}{RT}} \]

where \(D_0\) is known as the pre-exponential factor (a constant that involves factors like vibration frequency), \(Q\) is the activation energy for diffusion, \(R\) is the universal gas constant (8.314 J/mol·K), and \(T\) is the absolute temperature in Kelvins. According to this equation, as the temperature increases, the exponential term decreases, allowing \(D\) to increase, which means the diffusion process accelerates.
Concentration gradient
The term concentration gradient plays a pivotal role in the study of diffusion. Simply put, it is the measure of how the concentration of a substance changes from one location to another. Mathematically, it's the change in concentration \(dC\) over a distance \(dx\), expressed as \(\frac{dC}{dx}\). A high concentration gradient implies a steep change in concentration over a short distance and typically results in a higher diffusion flux according to Fick's first law. Imagine a hill—the steeper the hill (or the higher the concentration gradient), the faster a ball (or diffusing particles) would roll down (or move across the gradient).
Activation energy
In the processes of diffusion and chemical reactions, the concept of activation energy, represented by \(Q\), is crucial. It is the minimum energy barrier that must be overcome for atoms or molecules to move and rearrange themselves. In the context of diffusion, it is the energy required for atoms to break away from their positions and jump to new ones, enabling the transport of matter. A high activation energy means that the substance will diffuse more slowly at a given temperature, and vice versa. The Arrhenius equation incorporates this activation energy into the exponential term, linking it to temperature and highlighting its influence on the rate of diffusion.

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Most popular questions from this chapter

A sheet of steel 2.5 \(\mathrm{mm}\) thick has nitrogen atmospheres on both sides at \(900^{\circ} \mathrm{C}\) and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is \(1.2 \times 10^{-10} \mathrm{m}^{2} / \mathrm{s},\) and the diffusion flux is found to be \(1.0 \times 10^{-7} \mathrm{kg} / \mathrm{m}^{2}\) -s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is \(2 \mathrm{kg} / \mathrm{m}^{3} .\) How far into the sheet from this highpressure side will the concentration be 0.5 \(\mathrm{kg} / \mathrm{m}^{3} ?\) Assume a linear concentration profile.

An FCC iron-carbon alloy initially containing 0.55 wt \(\%\) C is exposed to an oxygen-rich and virtually carbon-free atmosphere at \(1325 \mathrm{K}\) \(\left(1052^{\circ} \mathrm{C}\right) .\) Under these circumstances the carbon diffuses from the alloy and reacts at the surface with the oxygen in the atmosphere that is, the carbon concentration at the surface position is maintained essentially at 0 wt \(\%\) C. (This process of carbon depletion is termed decarburization.) At what position will the carbon concentration be 0.25 wt\% after a 10-h treatment? The value of \(D\) at \(1325 \mathrm{K}\) is \(4.3 \times 10^{-11} \mathrm{m}^{2} / \mathrm{s}\).

For a steel alloy it has been determined that a carburizing heat treatment of 15 h duration will raise the carbon concentration to 0.35 wt \(\%\) at a point \(2.0 \mathrm{mm}\) from the surface. Estimate the time necessary to achieve the same concentration at a 6.0 -mm position for an identical steel and at the same carburizing temperature.

When \(\alpha\) -iron is subjected to an atmosphere of nitrogen gas, the concentration of nitrogen in the iron, \(C_{\mathrm{N}}\) (in weight percent), is a function of hydrogen pressure, \(p_{\mathrm{N}_{2}}(\text { in } \mathrm{MPa}),\) and absolute temperature \((T)\) according to $$C_{\mathrm{N}}=4.90 \times 10^{-3} \sqrt{p_{\mathrm{N}_{2}}} \exp \left(-\frac{37.6 \mathrm{kJ} / \mathrm{mol}}{R T}\right)$$.Furthermore, the values of \(D_{0}\) and \(Q_{d}\) for this diffusion system are \(3.0 \times 10^{-7} \mathrm{m}^{2} / \mathrm{s}\) and \(76,150 \mathrm{J} / \mathrm{mol}\), respectively. Consider a thin iron membrane \(1.5 \mathrm{mm}\) thick that is at \(300^{\circ} \mathrm{C}\) Compute the diffusion flux through this membrane if the nitrogen pressure on one side of the membrane is \(0.10 \mathrm{MPa}(0.99 \mathrm{atm}),\) and on the other side \(5.0 \mathrm{MPa}(49.3 \mathrm{atm})\).

Self-diffusion involves the motion of atoms that are all of the same type; therefore it is not subject to observation by compositional changes, as with inter-diffusion. Suggest one way in which self-diffusion may be monitored.

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