/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 An infinitely sharp zone of solu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An infinitely sharp zone of solute is placed at the center of a column at time \(t=0\). After diffusion for time \(t_{1}\), the standard deviation of the Gaussian band is \(1.0 \mathrm{~mm}\). After \(20 \mathrm{~min}\) more, at time \(t_{2}\), the standard deviation is \(2.0 \mathrm{~mm}\). What will be the width after another \(20 \mathrm{~min}\), at time \(t_{3}\) ?

Short Answer

Expert verified
The width after another 20 minutes is approximately 3.0 mm.

Step by step solution

01

Understand the Problem

We are given an initial Gaussian distribution of a solute at the center of a column. We know the standard deviation after certain time intervals: - After time \( t_1 \), \( \sigma(t_1) = 1.0 \text{ mm} \).- After an additional 20 minutes, \( \sigma(t_2) = 2.0 \text{ mm} \).We need to find \( \sigma(t_3) \) after another 20 minutes.
02

Use the Diffusion Equation

The standard deviation of a Gaussian process increases over time with diffusion. The relationship for the standard deviation \( \sigma \) at a given time related to the diffusion time can be expressed as:\[ \sigma(t) = \sqrt{2Dt} \]Where \( D \) is the diffusion coefficient.
03

Establish Equations for Given Times

From the information provided:- At time \( t_1 \), \( \sigma_1 = 1.0 \text{ mm} \). This implies \( \sqrt{2Dt_1} = 1.0 \).- At time \( t_2 = t_1 + 20 \text{ minutes} \), \( \sigma_2 = 2.0 \text{ mm} \). This implies \( \sqrt{2Dt_2} = 2.0 \).
04

Solve for the Diffusion Coefficient

From \( \sigma_1 = 1.0 \), we have:\[ 1.0 = \sqrt{2Dt_1} \Rightarrow t_1 = \frac{1.0^2}{2D} \]From \( \sigma_2 = 2.0 \), we have:\[ 2.0 = \sqrt{2D(t_1 + 20)}\Rightarrow t_1 + 20 = \frac{4.0}{2D}\]Solve these two equations to find \( D \).
05

Calculate Width after Additional 20 Minutes

Now, using the value of \( D \), we can calculate \( \sigma(t_3) \) after another 20 minutes, i.e., \( t_3 = t_1 + 40 \text{ minutes} \):\[ \sigma(t_3) = \sqrt{2D(t_1 + 40)} \]
06

Compute Width

Since \( 2D = \frac{2.0^2}{60} = \frac{4.0}{60} \) from step 4: - Find \( D = \frac{0.067}{2} \approx 0.0335 \). - Then, \( t_1 \approx 7.5 \) min, so \( t_3 \approx 47.5 \) min.Final width calculation:\[ \sigma(t_3) = \sqrt{2 \times 0.0335 \times 47.5} \approx \text{3.0 mm}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Diffusion Equation
The diffusion equation describes how substances distribute themselves over time. When we talk about diffusion, we often refer to how particles spread from an area of high concentration to low concentration. This behavior can be mathematically expressed using the diffusion equation:

\[ \frac{\partial C}{\partial t} = D \frac{\partial^2 C}{\partial x^2} \]

where \(C\) is the concentration, \(D\) is the diffusion coefficient, \(x\) is the position, and \(t\) is time. This partial differential equation suggests that the rate of change of concentration with respect to time is proportional to the second derivative of concentration with respect to space.

- **Key Components**
- **Concentration (\(C\))**: Describes how much of a solute is present in a given amount of space.
- **Diffusion Coefficient (\(D\))**: A constant that quantifies how quickly diffusion occurs.

In simple terms, the diffusion equation is a foundational principle explaining how substances move and mix, driven by the natural tendency to reach equilibrium.
Standard Deviation
In the context of diffusion and Gaussian distribution, standard deviation is a measure of how spread out the particles are in a distribution. Specifically, it tells us the average distance of the particles from the mean position in the distribution.

- **Key Functions**
- A small standard deviation means particles are closely packed around the mean.
- A larger standard deviation means particles are more spread out.

For a Gaussian distribution, the standard deviation \(\sigma\) evolves over time as particles move away from an initial sharp-point source, like in the exercise with the solute in a column. Mathematically for a diffusion process, the standard deviation can be described as:

\[ \sigma(t) = \sqrt{2Dt} \]

Here, \(t\) is time and \(D\) is the diffusion coefficient. This equation shows how the spread of particles (the width of the Gaussian curve) increases as time progresses.
Diffusion Coefficient
The diffusion coefficient, denoted by \(D\), is a parameter that defines the rate at which particles disperse over time. It's essential for understanding how fast or slow the diffusion process is happening.

- **Influences on \(D\)**
- **Temperature**: Generally, higher temperatures increase \(D\), leading to faster diffusion.
- **Medium**: Diffusion through gases is usually faster than liquids because molecular collisions occur less frequently.
- **Particle Size**: Smaller particles tend to diffuse faster than larger ones in the same medium.

In the exercise, we applied the diffusion coefficient to calculate how the standard deviation of a solute evolves over specific time intervals. By understanding \(D\), we can predict changes in particle distribution and effectively solve real-world diffusion problems in different contexts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the extraction of \(\mathrm{M}^{w+}\) from aqueous solution into organic solution by reaction with protonated ligand, HL: $$ \begin{aligned} \mathrm{M}^{n+}(a q)+& n \text { HL(org) } \Longrightarrow \mathrm{ML}_{n}(\text { org })+n \mathrm{H}^{+}(a q) \\ & K_{\text {extraction }}=\frac{\left[\mathrm{ML}_{n}\right]_{\text {arg }}\left[\mathrm{H}^{+}\right]_{\text {aq }}^{m}}{\left[\mathrm{M}^{w+}\right]_{\text {aq }}[\mathrm{HL}]_{\text {org }}^{w}} \end{aligned} $$ Rewrite Equation 22-13 in terms of \(K_{\text {evtraction }}\) and express \(K_{\text {entraction }}\) in terms of the constants in Equation 22-13. Give a physical reason why each constant increases or decreases \(K_{\text {exiraction- }}\)

A chromatogram with ideal Gaussian bands has \(t_{r}=9.0 \mathrm{~min}\) and \(w_{1 / 2}=2.0 \mathrm{~min}\). (a) How many theoretical plates are present? (b) Find the plate height if the column is \(10 \mathrm{~cm}\) long.

A chromatographic procedure separates \(4.0 \mathrm{mg}\) of unknown mixture on a column with a length of \(40 \mathrm{~cm}\) and a diameter of \(0.85 \mathrm{~cm}\). (a) What size column would you use to separate \(100 \mathrm{mg}\) of the same mixture? (b) If the flow is \(0.22 \mathrm{~mL} / \mathrm{min}\) on the small column, what volume flow rate should be used on the large column? (c) If mobile phase occupies \(35 \%\) of the column volume, calculate the linear flow rate for the small column and the large column.

Match the terms in the first list with the characteristics in the second list. 1\. adsorption chromatography 2\. partition chromatography 3\. ion-exchange chromatography 4\. molecular exclusion chromatography 5\. affinity chromatography A. Ions in mobile phase are attracted to counterions cowalently attached to stationary phase. B. Solute in mobile phase is attracted to specific groups covalently attached to stationary phase. C. Solute equilibrates between mobile phase and surface of stationary phase. D. Solute equilibrates between mobile phase and film of liquid attached to stationary phase. E. Different-sized solutes penetrate voids in stationary phase to different extents. Largest solutes are cluted first.

An open tubular column has an inner diameter of \(207 \mu \mathrm{m}\) and the thickness of the stationary phase on the inner wall is \(0.50 \mu \mathrm{m}\). Unretained solute passes through in \(63 \mathrm{~s}\) and a particular solute emerges in \(433 \mathrm{~s}\). Find the partition coefficient for this solute and find the fraction of time spent in the stationary phase.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.