Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 80

About how long will it take a perfume molecule to diffuse a distance of $5.00 \mathrm{m}\( in one direction in a room if the diffusion constant is \)1.00 \times 10^{-5} \mathrm{m}^{2} / \mathrm{s} ?$ Assume that the air is perfectly still-there are no air currents.

Short Answer

Expert verified
Answer: The time it takes for a perfume molecule to diffuse a distance of \(5.00 \,\mathrm{m}\) in one direction in a room, under the given conditions, is approximately \(1.25\times10^{6} \,\mathrm{s}\).

Step by step solution

01

Understand the problem and the given information

We are given the following information: - Diffusion distance: \(5.00 \,\mathrm{m}\) - Diffusion constant: \(1.00 \times 10^{-5} \,\mathrm{m}^{2}/\mathrm{s}\) Our goal is to determine the time it will take for a perfume molecule to diffuse a given distance in a room, assuming that the air is still.
02

Apply the formula for diffusion

We can use the formula for diffusive spread of particles, which involves the diffusion constant, distance, and time: $$x^{2} = 2Dt$$ Where: - \(x\) is the diffusion distance - \(D\) is the diffusion constant - \(t\) is the time it takes for the particle to diffuse In our case, we are looking for \(t\) and we know \(x = 5.00 \,\mathrm{m}\) and \(D = 1.00 \times 10^{-5} \,\mathrm{m}^{2}/\mathrm{s}\).
03

Solve the equation for time \(t\)

Let's isolate the variable \(t\) by dividing both sides by \(2D\): $$t = \frac{x^{2}}{2D}$$ Now, we plug in the given values for \(x\) and \(D\): $$t = \frac{(5.00 \,\mathrm{m})^{2}}{2(1.00\times10^{-5}\,\mathrm{m}^{2}/\mathrm{s})}$$
04

Calculate the time \(t\)

By performing the calculations, we get: $$t = \frac{25 \,\mathrm{m}^{2}}{2(1.00\times10^{-5}\,\mathrm{m}^{2}/\mathrm{s})}$$ $$t = \frac{25 \,\mathrm{m}^{2}}{2\times10^{-5}\,\mathrm{m}^{2}/\mathrm{s}}$$ $$t = \frac{25 \,\mathrm{m}^{2}}{2\times10^{-5}\,\mathrm{m}^{2}/\mathrm{s}} \times \frac{1\,\mathrm{s}}{2\times10^{-5}\,\mathrm{m}^{2}}$$ $$t = \frac{25}{2\times10^{-5}}$$ $$t = 1.25\times10^{6}\,\mathrm{s}$$ Therefore, it will take approximately \(1.25\times10^{6} \,\mathrm{s}\) for a perfume molecule to diffuse a distance of \(5.00 \,\mathrm{m}\) in one direction in a room, if the diffusion constant is \(1.00 \times 10^{-5} \,\mathrm{m}^{2}/\mathrm{s}\), and there are no air currents.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Ten students take a test and get the following scores: $83, 62,81,77,68,92,88,83,72,\( and \)75 .$ What are the average value, the rms value, and the most probable value, respectively, of these test scores?
A square brass plate, \(8.00 \mathrm{cm}\) on a side, has a hole cut into its center of area \(4.90874 \mathrm{cm}^{2}\) (at $\left.20.0^{\circ} \mathrm{C}\right) .$ The hole in the plate is to slide over a cylindrical steel shaft of cross-sectional area \(4.91000 \mathrm{cm}^{2}\) (also at \(20.0^{\circ} \mathrm{C}\) ). To what temperature must the brass plate be heated so that it can just slide over the steel cylinder (which remains at \(20.0^{\circ} \mathrm{C}\) )? [Hint: The steel cylinder is not heated so it does not expand; only the brass plate is heated. \(]\)
Steel railroad tracks of length \(18.30 \mathrm{m}\) are laid at $10.0^{\circ} \mathrm{C} .$ How much space should be left between the track sections if they are to just touch when the temperature is \(50.0^{\circ} \mathrm{C} ?\)
A cylinder with an interior cross-sectional area of \(70.0 \mathrm{cm}^{2}\) has a moveable piston of mass \(5.40 \mathrm{kg}\) at the top that can move up and down without friction. The cylinder contains $2.25 \times 10^{-3} \mathrm{mol}\( of an ideal gas at \)23.0^{\circ} \mathrm{C} .$ (a) What is the volume of the gas when the piston is in equilibrium? Assume the air pressure outside the cylinder is 1.00 atm. (b) By what factor does the volume change if the gas temperature is raised to \(223.0^{\circ} \mathrm{C}\) and the piston moves until it is again in equilibrium?
How many hydrogen atoms are present in \(684.6 \mathrm{g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) ?\)
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Measurements

Read Explanation

Fields in Physics

Read Explanation

Modern Physics

Read Explanation

Atoms and Radioactivity

Read Explanation
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.

Sign-up for free