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(Estimating glacial ages) The following oxygen- 18 data were obtained from soil samples taken at different depths in Ontario. Canada. Assuming that all the \(^{18} \mathrm{O}\) was laid down during the last glacial age and that the transport of \(^{18} \mathrm{C}\) to the surface takes place by molecular diffusion. estimate the number of years since the last glacial age from the following data. Independent measure. ments give the diffusivity of \(^{18} \mathrm{O}\) in soil as \(2.64 \times 10^{-10} \mathrm{m}^{2} / \mathrm{s}\) $$\begin{array}{l|cccccc} \text {Depth (m)} &{\text { (surface) }} \\ & 0 & 3 & 6 & 9 & 12 & 18 \\ \hline ^{18} \mathrm{O} \text {Conc. Ratio }\left(\mathrm{C} / \mathrm{C}_{0}\right) & 0 & 0.35 & 0.65 & 0.83 & 0.94 & 1.0 \end{array}$$

Short Answer

Expert verified
It is estimated that approximately 4040 years have passed since the last glacial age, using the given oxygen-18 data and the Gaussian distribution function as a solution to Fick's second law for molecular diffusion.

Step by step solution

01

Choose the correct diffusion model

Since the transport of oxygen-18 to the surface takes place by molecular diffusion, we can use Fick's second law to model the diffusion: \[ \frac{\partial C}{\partial t} = D \cdot \frac{\partial^2 C}{\partial x^2} \] where \(C\) is the concentration of oxygen-18 at position \(x\) and time \(t\), and \(D\) is the diffusivity of oxygen-18 in the soil.
02

Solve the diffusion model for concentration as a function of time and position

Due to the initial condition of all the oxygen-18 laid down during the last glacial age and the given concentration ratios (\(C/C_0\)), we take the Gaussian distribution function as the solution for the diffusion model (Fick's second law) problem: \[ C(x, t) = C_0 \cdot \exp \left[ -\frac{x^2}{4 \cdot D \cdot t} \right] \]
03

Calculate the time elapsed

Rearrange the Gaussian distribution function to solve for time: \[t = \frac{x^2}{4 \cdot D \cdot \ln{\frac{C_0}{C}}} \] Now our task is to estimate the time elapsed since the last glacial age. To do this, we can choose any data point from the given data. For instance let's use the data at depth of 6 m, where the concentration ratio \(C/C_0 = 0.65\): \[t = \frac{(6 \,\mathrm{m})^2}{4 \cdot (2.64 \times 10^{-10}\, \mathrm{m}^2 / \mathrm{s}) \cdot \ln{\frac{1}{0.65}}}\]
04

Convert the time to years

Calculate the time in seconds and then convert it to years: \[t = \frac{(6 \,\mathrm{m})^2}{4 \cdot (2.64 \times 10^{-10}\, \mathrm{m}^2 / \mathrm{s}) \cdot \ln{\frac{1}{0.65}}} = 1.274 \times 10^{11} \, \mathrm{s} \] Since there are \(3.1536 \times 10^7\) seconds in a year: \[t = \frac{1.274 \times 10^{11}\, \mathrm{s}}{3.1536 \times 10^7 \, \mathrm{s/yr}} = 4.04 \times 10^3 \, \mathrm{years} \] So, it is estimated that approximately 4040 years have passed since the last glacial age.

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Key Concepts

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

Molecular Diffusion
Molecular diffusion is a process where molecules spread from an area of high concentration to an area of low concentration. It is driven by the random motion of molecules, also known as Brownian motion. This phenomenon is essential in many natural and industrial processes. In the scenario of estimating glacial ages from soil samples, we assume the transport of oxygen-18 isotopes to the surface occurs via molecular diffusion.

Fick's second law describes how diffusion causes the concentration of a substance to change over time in a given space. The law is represented by the equation:\[\frac{\partial C}{\partial t} = D \cdot \frac{\partial^2 C}{\partial x^2}\]Here, \(C\) is the concentration level, \(t\) is time, \(x\) represents the position, and \(D\) is the diffusion coefficient. Understanding molecular diffusion is crucial to accurately modeling and predicting how substances like oxygen-18 move through soil or other materials.
Oxygen-18 Isotopes
Oxygen-18 is a stable isotope of oxygen, often used in paleoenvironmental and paleoclimate studies. It naturally occurs in the Earth's environment and can be found in water, ice, carbonates, and silicates. By studying the concentration of oxygen-18 in historical layers, researchers can gather valuable information about past climates and environmental conditions.

When soil samples are analyzed for estimating glacial ages, the concentration ratio of oxygen-18 is investigated. This ratio can indicate the period when certain amounts of oxygen-18 were embedded in soil layers. The concentration of this isotope is integral to understanding how different climatic events — such as glaciations — have occurred over time. Because oxygen-18 can tell us about rainfall, temperature, and other climate factors in the past, it plays a vital role in geological and climatological research.
Glacial Age Estimation
Estimating glacial ages involves determining when the last major ice ages occurred by analyzing geological and isotopic evidence. In the given exercise, this is done using oxygen-18 isotopes found in soil samples. By applying principles of molecular diffusion and isotopic concentrations, scientists can backtrack the changes in these concentrations to estimate the age of past glacial events.

The process requires an understanding of both the initial conditions (i.e., when the oxygen-18 was last deposited during a glaciation event) and the molecular diffusion that spreads these isotopes over time. By solving the diffusion model — specifically using Fick's second law — and examining the isotopic concentration at certain depths, scientists can estimate the number of years since the last glacial period. This calculation provides insights into how Earth's climate has shifted over millennia.
Diffusion Coefficient Calculation
The diffusion coefficient \(D\) is a measure of how quickly molecules or particles spread out due to diffusion. For oxygen-18 in the soil, this coefficient is a known value that aids in predicting how the isotope moves through the Earth's layers. In the given exercise, the diffusion coefficient is provided as \(2.64 \times 10^{-10} \ \text{m}^2/\text{s}\).

Given this coefficient and the initial positioning of the isotopes, we use Fick's second law to model their diffusion over time. Specifically, the Gaussian distribution function helps solve the concentration over time:\[ C(x, t) = C_0 \cdot \exp\left(-\frac{x^2}{4 \cdot D \cdot t}\right) \]By rearranging this formula, the time \(t\) it took for the isotopes to diffuse can be calculated, utilizing the known values of \(D\) and initial concentrations. This approach is crucial in determining the relative period since an event, such as the end of a glacial age. Understanding and calculating diffusion coefficients provide a foundation for numerous practical applications beyond geology, including chemical engineering and environmental science.

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Most popular questions from this chapter

A spherical particle is dissolving in a liquid. The rate of dissolution is first order in the solvent concentration, \(C\). Assuming that the solvent is in excess, show that the following conversion time relationships hold. $$\begin{array}{cl} \hline \begin{array}{c} \text {Rate} \text { -Limiting} \\ \text {Regime} \end{array} & \text {Conversion Time Relationship} \\ \hline \text { Surface reaction } & 1-(1-X)^{1 / 3}=\frac{\alpha t}{D_{i}} \\ \text { Mass transfer } & \frac{D_{i}}{2 D^{*}} 11-(1-X)^{2 n} 1=\frac{\alpha t}{D_{i}} \\ \text { Mixed } & 11-(1-X)^{1 / 3} 1+\frac{D_{i}}{2 D^{*}}\left[1-(1-X)^{2 / 3}\right]=\frac{\alpha t}{D_{i}} \\ \hline \end{array}$$

Carbon disulphide (A) is evaporating into air (B) at \(35^{\circ} \mathrm{C} P_{\mathrm{vp} \in}=510 \mathrm{mm} \mathrm{Hg}\) and 1 atm from the bottom of a \(1.0 \mathrm{cm}\) diameter vertical tube. The distance from the \(\mathrm{CS}_{2}\) surface to the open end is \(20.0 \mathrm{cm}\), and this is held constant by continuous addition of liquid \(C S_{2}\) from below. The experiment is arranged so that the vapor concentration of \(\mathrm{CS}_{2}\) at the open end is zero. (a) Calculate the molecular diffusivity of \(\mathrm{CS}_{2}\) in air \(\left(D_{\mathrm{ca}}\right)\) and its vapor pressure at \(35^{\circ} \mathrm{C}\). (Ans.. \(D_{\mathrm{AB}}=0.12 \mathrm{cm}^{2} / \mathrm{s}\).) (b) Find the molar and mass fluxes \(\left(W_{\mathrm{A}_{5}} \text { and } n_{c} \text { of } \mathrm{CS}_{2}\right)\) in the tube. (c) Calculate the following properties at \(0.0,5.0,10.0,15.0,18.0,\) and 20.0 cm from the \(C S_{2}\) surface. Arrange columns in the following order on one sheet of paper. (Additional columns may be included for computational purposes if desired.) On a separate sheet give the relations used to obtain each quantity. Try to put each relation into a form involving the minimum computation and the highest accuracy: (1) \(y_{\mathrm{A}}\) and \(y_{\mathrm{B}}\) (mole fractions), \(C_{\mathrm{A}}\) (2) \(V_{\mathrm{A}}, V_{\mathrm{B}}, \mathrm{V}^{*}, \mathrm{V}\) (mass velocity) (3) \(J_{A}, J_{B}\) (d) Plot each of the groups of quantities in \((c)(1),(2),\) and (3) on separate graphs. Name all variables and show units. Do not plot those parameters in parentheses. (e) What is the rate of evaporation of \(\mathrm{CS}_{2}\) in cm/day? (f) Discuss the physical meaning of the value of \(V_{A}\) and \(J_{A}\) at the open end of the tube. (g) Is molecular diffusion of air taking place?

The irreversible gas-phase reaction $$A \stackrel{\text { cealyst }}{\longrightarrow} \text { B }$$ is carried out adiabatically over a packed bed of solid catalyst particles. The reaction is first order in the concentration of \(A\) on the catalyst surface: $$-r_{\mathrm{A} s}^{\prime}=k^{\prime} C_{\mathrm{A}}$$ The feed consists of \(50 \%\) (mole) \(A\) and \(50 \%\) inerts and enters the bed at a temperature of \(300 \mathrm{K}\). The entering volumetric flow rate is \(10 \mathrm{dm}^{3} / \mathrm{s}\) (i.e... \(10,000 \mathrm{cm}^{3} / \mathrm{s}\) ). The relationship between the Sherwood number and the Reynolds number is $$S h=100 \mathrm{Re}^{1 / 2}$$ As a first approximation, one may neglect pressure drop. The entering concentration of \(A\) is \(1.0 \mathrm{M}\). Calculate the catalyst weight necessary to achieve \(60 \%\) conversion of \(\mathrm{A}\) for (a) Isothermal operation. (b) Adiabatic operation. (c) What generalizations can you make after comparing parts (a) and (b)? Additional information: Kinematic viscosity: \(\mu / \rho=0.02 \mathrm{cm}^{2} / \mathrm{s}\) Particle diameter: \(d_{p}=0.1 \mathrm{cm}\) Superficial velocity: \(U=10 \mathrm{cm} / \mathrm{s}\) Catalyst surface area/mass of catalyst bed: \(a=60 \mathrm{cm}^{2} / \mathrm{g}\) cat. Diffusivity of \(\mathrm{A}: D_{e}=10^{-2} \mathrm{cm}^{2} / \mathrm{s}\) Heat of reaction: \(\Delta H_{\mathrm{Rx}}=-10,000 \mathrm{cal} / \mathrm{g}\) mol \(\mathrm{A}\) Heat capacities: $$\begin{aligned} C_{p A}=C_{p B} &=25 \mathrm{cal} / \mathrm{g} \mathrm{mol} \cdot \mathrm{K} \\\ C_{p S}(\text { solvent }) &=75 \mathrm{cal} / \mathrm{g} \mathrm{mol} \cdot \mathrm{K} \end{aligned}$$ $$k^{\prime}(300 \mathrm{K})=0.01 \mathrm{cm}^{3 / \mathrm{s} \cdot \mathrm{g} \text { cat with } E}=4000 \mathrm{cal} / \mathrm{mol}$$

(Pills) An antibiotic drug is contained in a solid inner core and is surrounded by an outer coating that makes it palatable. The outer coating and the drug are dissolved at different rates in the stomach, owing to their differences in equilibrium solubilities. (a) If \(D_{2}=4 \mathrm{mm}\) and \(D_{1}=3 \mathrm{mm}\), calculate the time necessary for the pill to dissolve completely. (b) Assuming first-order kinetics \(\left(k_{\mathrm{A}}=10 \mathrm{h}^{-1}\right)\) for the absorption of the di solved drug (i.e... in solution in the stomach) into the bloodstream, pl the concentration in grams of the drug in the blood per gram of boc weight as a function of time when the following three pills are take simultaneously: $$\begin{aligned} &\text { Pill } 1: \quad D_{2}=5 \mathrm{mm} . \quad D_{1}=3 \mathrm{mm}\\\ &\text { Pill 2: } \quad D_{2}=4 \mathrm{mm}, \quad D_{1}=3 \mathrm{mm}\\\ &\text { Pill 3: } \quad D_{2}=3.5 \mathrm{mm} . \quad D_{1}=3 \mathrm{mm} \end{aligned}$$ (c) Discuss how you would maintain the drug level in the blood at a constat level using different-size pills? (d) How could you arrange a distribution of pill sizes so that the concentri tion in the blood was constant over a period (e.g.. 3 hr) of time? Additional information: Amount of drug in inner core \(=500 \mathrm{mg}\) Solubility of outer layer at stomach conditions \(=1.0 \mathrm{mg} / \mathrm{cm}^{3}\) Solubility of inner layer at stomach conditions \(=0.4 \mathrm{mg} / \mathrm{cm}^{3}\) Volume of fluid in stomach \(=1.2 \mathrm{L}\) Typical body weight \(=75 \mathrm{kg}\) \(\mathrm{Sh}=2 . . D_{\mathrm{AB}}=6 \times 10^{-4} \mathrm{cm}^{2} / \mathrm{min}\)

The journals listed at the erid of Chapter 1 may be useful for part (b). (a) Example \(1-1\). Consider the mass transfer-limited reaction $$A \longrightarrow 2 B$$ What would your concentration (mole fraction) profile look like? Using the same values for \(D_{\mathrm{A} \mathrm{h}},\) and so on. in Example \(11-1,\) what is the flux of \(\mathrm{A} ?\) (b) Example \(11-2 .\) How would your answers change if the temperature was increased by \(50^{\circ} \mathrm{C}\). the particle diameter was doubled. and fluid velocity was cut in half? Assume properties of water can be used for this system. (c) Example \(11-3 .\) How would your answers change if you had a \(50-50\) mixture of hydrazine and helium? If you increase \(d_{p}\) by a factor of \(5 ?\) (d) Example \(11-4 .\) What if you were asked for representative values for \(\mathrm{Re}\), Sc, Sh, and \(k_{\varepsilon}\) for both liquid-and gas-phase systems for a velocity of 10 \(\mathrm{cm} / \mathrm{s} \text { and a pipe diameter of } 5 \mathrm{cm} \text { (or a packed-bed diameter of } 0.2 \mathrm{cm}) ?\) What numbers would you give? (e) Example \(11-5 .\) How would your answers change if the reaction were carried out in the liquid phase where kinetic viscosity varied as $$v\left(T_{2}\right)=v\left(T_{1}\right) \exp \left[-\frac{4000 K}{T}\right] ?$$ to predict the drug delivery as a function of time. Compare this result with that where the drug in the patch is in a dissolving solid and a hydro-gel and therefore constant with time. Explore this problem using different models and parameter values. Additional information $$\begin{aligned} &H=0.1, D_{\mathrm{AB} 1}=10^{-6} \mathrm{cm}^{2} / \mathrm{s}, D_{\mathrm{AB} 2}=10^{-5} \mathrm{cm}^{2} / \mathrm{s}, A_{\mathrm{p}}=5 \mathrm{cm}^{2}, V=1 \mathrm{cm}^{3}\\\ &\text { and } C_{\mathrm{AP}}=10 \mathrm{mg} / \mathrm{dm}^{3} \end{aligned}$$

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