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The diffusion coefficient of carbon in steel is given as $$ D_{A B}=2.67 \times 10^{-5} \exp (-17,400 / T) \quad\left(\mathrm{m}^{2} / \mathrm{s}\right) $$ where \(T\) is in \(\mathrm{K}\). Determine the diffusion coefficient from \(300 \mathrm{~K}\) to \(1500 \mathrm{~K}\) in \(100 \mathrm{~K}\) increments and plot the results.

Short Answer

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Answer: The steps involved in calculating the diffusion coefficient of carbon in steel at different temperatures are: 1. Identify the temperature range and increments (300K to 1500K, with 100K increments). 2. Define the diffusion coefficient formula: $$D_{A B} = 2.67 \times 10^{-5} \exp (-17,400 / T) \quad\left(\mathrm{m}^{2} / \mathrm{s}\right)$$ 3. Calculate the diffusion coefficient at each temperature by substituting the value of T in the formula. 4. Record the values in a table for easier analysis. 5. Plot the results to visualize the relationship between the diffusion coefficient and temperature.

Step by step solution

01

Identify temperature range and increments

We are given a temperature range of 300K to 1500K and have to find the diffusion coefficient at every 100K increment. We need to create a list of temperatures in this range with a step of 100K.
02

Define the diffusion coefficient formula

The formula for the diffusion coefficient is given by $$ D_{A B} = 2.67 \times 10^{-5} \exp (-17,400 / T) \quad\left(\mathrm{m}^{2} / \mathrm{s}\right) $$ where \(D_{AB}\) is the diffusion coefficient and \(T\) is the temperature in Kelvin.
03

Calculate the diffusion coefficient at each temperature

For each temperature in the list, we will substitute the value of \(T\) in the formula and calculate the corresponding diffusion coefficient \(D_{AB}\).
04

Record the values in a table

It's useful to record the calculated values of the diffusion coefficient for each temperature in a table. This will make it easier to plot and analyze the results.
05

Plot the results

Using the values of the diffusion coefficient obtained in the previous step, we will plot \(D_{AB}\) against the temperature \(T\). The plot will help us visually understand the relationship between the diffusion coefficient and temperature.

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Key Concepts

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

Carbon Diffusion in Steel
Carbon diffusion in steel is a vital process in metallurgy that greatly impacts the material's properties. Steel, an alloy primarily composed of iron and carbon, undergoes significant changes due to the movement of carbon atoms within its lattice structure. This movement, called diffusion, determines many of the mechanical properties of steel, such as hardness, strength, and toughness. Diffusion is the process where atoms or molecules move from a region of higher concentration to a region of lower concentration. In steel, carbon atoms migrate through the iron lattice, a structure made of tightly packed iron atoms.
  • Carbon diffusion affects processes like carburizing, a heat treatment used to harden the surface of steel.
  • The rate of carbon diffusion is crucial for achieving the desired characteristics in steel production.
Understanding how temperature and other factors affect carbon diffusion helps metallurgists plan the appropriate heat treatments for steel products.
Temperature Dependence of Diffusion
Temperature has a profound effect on the diffusion process in solids. As temperature increases, atoms gain energy and move more rapidly, which enhances the diffusion rate. For carbon diffusion in steel, the role of temperature is particularly significant. According to the Arrhenius equation, the diffusion coefficient typically depends exponentially on temperature. The equation explains how the probability of an atom overcoming the energy barrier increases with temperature. This dependence means that as the temperature of the steel heats up, carbon diffuses more quickly through the metal.
  • This principle is crucial in processes like annealing or tempering.
  • High temperatures often lead to more effective diffusion and improved mechanical properties of steel.
In summary, higher temperatures enable a faster diffusion rate, allowing carbon atoms to spread more effectively throughout the steel matrix.
Exponential Functions in Diffusion Coefficient
In the context of diffusion coefficients like that of carbon in steel, the Arrhenius equation often appears as an exponential function. The diffusion coefficient (\[D_{AB} = 2.67 \times 10^{-5} \exp (-17,400 / T)\]indicates how diffusion is temperature-dependent. In this equation, \(D_{AB}\) is the diffusion coefficient, and \(T\) is the temperature in Kelvin.
  • The term \(\exp (-17,400 / T)\) illustrates the exponential nature of diffusion in relation to temperature.
  • The negative exponent suggests that as \(T\) increases, the overall value of the exponent (and hence the diffusion coefficient) becomes less negative, leading to a larger diffusion coefficient.
This exponential relationship is key to understanding how small changes in temperature can result in significant variations in diffusion rates.
Plotting Diffusion Coefficient vs. Temperature
Plotting the diffusion coefficient against temperature provides a visual representation of the data, aiding in the deep comprehension of the relationship. By plotting \(D_{AB}\) against temperature \(T\), students and practitioners can observe how the diffusion coefficient changes over a range of temperatures from 300 to 1500 K.This plot typically shows an upward curve, indicative of the exponential growth in diffusion rate with temperature. This visual aid not only helps in understanding theoretical concepts but also in practical applications when adjusting processing temperatures for steel.
  • A clear plot helps in identifying the exact temperatures where diffusion rates substantially increase.
  • Graphical representation complements the mathematical formulation, offering another perspective on diffusion behavior.
By engaging with such plots, users can better predict and manage the diffusion processes in materials like steel.

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Most popular questions from this chapter

Exposure to high concentration of gaseous ammonia can cause lung damage. The acceptable shortterm ammonia exposure level set by the Occupational Safety and Health Administration (OSHA) is 35 ppm for 15 minutes. Consider a vessel filled with gaseous ammonia at \(30 \mathrm{~mol} / \mathrm{L}\), and a 10 -cm- diameter circular plastic plug with a thickness of \(2 \mathrm{~mm}\) is used to contain the ammonia inside the vessel. The ventilation system is capable of keeping the room safe with fresh air, provided that the rate of ammonia being released is below \(0.2 \mathrm{mg} / \mathrm{s}\). If the diffusion coefficient of ammonia through the plug is \(1.3 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\), determine whether or not the plug can safely contain the ammonia inside the vessel.

Consider a nickel wall separating hydrogen gas that is maintained on one side at \(5 \mathrm{~atm}\) and on the opposite at \(3 \mathrm{~atm}\). If the temperature is constant at \(85^{\circ} \mathrm{C}\), determine \((a)\) the mass densities of hydrogen gas in the nickel wall on both sides and \((b)\) the mass densities of hydrogen outside the nickel wall on both sides.

Nitrogen gas at high pressure and \(298 \mathrm{~K}\) is contained in a \(2-\mathrm{m} \times 2-\mathrm{m} \times 2-\mathrm{m}\) cubical container made of natural rubber whose walls are \(4 \mathrm{~cm}\) thick. The concentration of nitrogen in the rubber at the inner and outer surfaces are \(0.067 \mathrm{~kg} / \mathrm{m}^{3}\) and \(0.009 \mathrm{~kg} / \mathrm{m}^{3}\), respectively. The diffusion coefficient of nitrogen through rubber is \(1.5 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{s}\). The mass flow rate of nitrogen by diffusion through the cubical container is (a) \(8.24 \times 10^{-10} \mathrm{~kg} / \mathrm{s}\) (b) \(1.35 \times 10^{-10} \mathrm{~kg} / \mathrm{s}\) (c) \(5.22 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (d) \(9.71 \times 10^{-9} \mathrm{~kg} / \mathrm{s}\) (e) \(3.58 \times 10^{-8} \mathrm{~kg} / \mathrm{s}\)

A recent attempt to circumnavigate the world in a balloon used a helium-filled balloon whose volume was \(7240 \mathrm{~m}^{3}\) and surface area was \(1800 \mathrm{~m}^{2}\). The skin of this balloon is \(2 \mathrm{~mm}\) thick and is made of a material whose helium diffusion coefficient is \(1 \times 10^{-9} \mathrm{~m}^{2} / \mathrm{s}\). The molar concentration of the helium at the inner surface of the balloon skin is \(0.2 \mathrm{kmol} / \mathrm{m}^{3}\) and the molar concentration at the outer surface is extremely small. The rate at which helium is lost from this balloon is (a) \(0.26 \mathrm{~kg} / \mathrm{h}\) (b) \(1.5 \mathrm{~kg} / \mathrm{h}\) (c) \(2.6 \mathrm{~kg} / \mathrm{h}\) (d) \(3.8 \mathrm{~kg} / \mathrm{h}\) (e) \(5.2 \mathrm{~kg} / \mathrm{h}\)

Dry air whose molar analysis is \(78.1\) percent \(\mathrm{N}_{2}\), \(20.9\) percent \(\mathrm{O}_{2}\), and 1 percent Ar flows over a water body until it is saturated. If the pressure and temperature of air remain constant at \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) during the process, determine (a) the molar analysis of the saturated air and \((b)\) the density of air before and after the process. What do you conclude from your results?

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