Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 11

At a given temperature and pressure, do you think the mass diffusivity of air in water vapor will be equal to the mass diffusivity of water vapor in air? Explain.

Short Answer

Expert verified
Answer: No, the mass diffusivity of air in water vapor will not be equal to the mass diffusivity of water vapor in air, even when the temperature and pressure are constant, due to the difference in molecular weights of the two substances.

Step by step solution

01

Understanding Mass Diffusivity

Mass diffusivity (D) is a measure of how quickly one substance can diffuse through another. It depends on factors such as temperature, pressure, and the properties of the substances involved. In this exercise, we are dealing with air and water vapor, so we will focus on those two substances.
02

Factors Affecting Mass Diffusivity

Some factors that affect mass diffusivity are temperature, pressure, and molecular weight of the substances involved. In this case, we have two scenarios: 1. The mass diffusivity of air in water vapor. 2. The mass diffusivity of water vapor in air. Even though temperature and pressure are given to be constant, the molecular weights of air and water vapor are different.
03

Molecular Weight and Mass Diffusivity

The molecular weight of a substance, denoted as M, plays a role in determining the mass diffusivity. The larger the difference in molecular weights between the two substances, the lower the mass diffusivity. In this case, the molecular weights of air and water vapor are not equal, with air having a molecular weight of approximately 29 g/mol and water vapor having a molecular weight of approximately 18 g/mol.
04

Conclusion

Since the molecular weights of air and water vapor are not equal, the mass diffusivity of air in water vapor will not be equal to the mass diffusivity of water vapor in air, even when the temperature and pressure are constant. This is because the molecular weights of the substances have a direct impact on the mass diffusivity, regardless of the temperature and pressure.

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

Does a mass transfer process have to involve heat transfer? Describe a process that involves both heat and mass transfer.

Consider a 15-cm-internal-diameter, 10-m-long circular duct whose interior surface is wet. The duct is to be dried by forcing dry air at \(1 \mathrm{~atm}\) and \(15^{\circ} \mathrm{C}\) through it at an average velocity of \(3 \mathrm{~m} / \mathrm{s}\). The duct passes through a chilled room, and it remains at an average temperature of \(15^{\circ} \mathrm{C}\) at all times. Determine the mass transfer coefficient in the duct.

Moisture migration in the walls, floors, and ceilings of buildings is controlled by vapor barriers or vapor retarders. Explain the difference between the two, and discuss which is more suitable for use in the walls of residential buildings.

You probably have noticed that balloons inflated with helium gas rise in the air the first day during a party but they fall down the next day and act like ordinary balloons filled with air. This is because the helium in the balloon slowly leaks out through the wall while air leaks in by diffusion. Consider a balloon that is made of \(0.1\)-mm-thick soft rubber and has a diameter of \(15 \mathrm{~cm}\) when inflated. The pressure and temperature inside the balloon are initially \(110 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). The permeability of rubber to helium, oxygen, and nitrogen at \(25^{\circ} \mathrm{C}\) are \(9.4 \times 10^{-13}, 7.05 \times 10^{-13}\), and \(2.6 \times 10^{-13} \mathrm{kmol} / \mathrm{m} \cdot \mathrm{s} \cdot\) bar, respectively. Determine the initial rates of diffusion of helium, oxygen, and nitrogen through the balloon wall and the mass fraction of helium that escapes the balloon during the first \(5 \mathrm{~h}\) assuming the helium pressure inside the balloon remains nearly constant. Assume air to be 21 percent oxygen and 79 percent nitrogen by mole numbers and take the room conditions to be \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\).

Benzene-free air at \(25^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) enters a 5 -cm-diameter tube at an average velocity of \(5 \mathrm{~m} / \mathrm{s}\). The inner surface of the \(6-m\)-long tube is coated with a thin film of pure benzene at \(25^{\circ} \mathrm{C}\). The vapor pressure of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) at \(25^{\circ} \mathrm{C}\) is \(13 \mathrm{kPa}\), and the solubility of air in benezene is assumed to be negligible. Calculate \((a)\) the average mass transfer coefficient in \(\mathrm{m} / \mathrm{s},(b)\) the molar concentration of benzene in the outlet air, and \((c)\) the evaporation rate of benzene in \(\mathrm{kg} / \mathrm{h}\).
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Wave Optics

Read Explanation

Electricity and Magnetism

Read Explanation

Nuclear Physics

Read Explanation

Mechanics and Materials

Read Explanation

Modern Physics

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