How does the mass diffusivity of a gas mixture change with \((a)\) temperature and \((b)\) pressure?

Short Answer

Expert verified
Answer: As the temperature increases, the mass diffusivity of a gas mixture increases, which means gas molecules will diffuse faster at higher temperatures. On the other hand, as the pressure increases, the mass diffusivity of a gas mixture decreases, due to increased number of collisions between gas molecules.

Step by step solution

01

Understand mass diffusivity

Mass diffusivity (D) is a measure of how fast one gas diffuses into another gas. It is dependent on temperature (T), pressure (P), and the properties of the gases involved, such as their molecular weights and sizes. We will consider a binary gas mixture, which consists of two gases, A and B, diffusing into each other.
02

Formula for mass diffusivity

The mass diffusivity of a binary gas mixture can be calculated using the Chapman-Enskog theory, which gives the following formula: D = \frac{3}{16} \frac{(k_B T)^{3/2}}{(\pi \mu)^{1/2} \sigma^2 \Omega_D} where, - D is mass diffusivity - k_B is Boltzmann's constant - T is the temperature (in Kelvin) - \mu is the reduced mass of the gas pair, given by \frac{m_A m_B}{m_A+m_B} (m_A and m_B are the molecular masses of gases A and B) - \sigma is the collision diameter (a measure of the size of the gas molecules) - \Omega_D is a dimensionless quantity dependent on the gas properties and temperature, which accounts for the collision interactions between the two gases.
03

Effect of temperature on mass diffusivity

To study the effect of temperature on mass diffusivity, we need to look at how D changes with T, keeping pressure and other parameters constant. From the formula above, we can see that D is directly proportional to T^{3/2}. Therefore, as the temperature increases, the mass diffusivity also increases, which implies that gas molecules will diffuse faster at higher temperatures.
04

Effect of pressure on mass diffusivity

Pressure does not appear explicitly in the formula for mass diffusivity, so we need to analyze its effect indirectly. The collision diameter, \sigma, which is independent of pressure, can be considered constant. However, pressure affects the concentration (n/V) of the gas mixture, where n is the number of moles of the gas and V is the volume. As per the ideal gas equation, PV = nRT \Rightarrow \frac{n}{V}= \frac{P}{RT} As the pressure increases, the concentration increases which means that the number of collisions between gas molecules will also increase. An increased number of collisions will reduce the average distance traveled by the molecules between successive collisions, leading to a decrease in mass diffusivity. In summary, (a) As the temperature increases, the mass diffusivity of a gas mixture increases. (b) As the pressure increases, the mass diffusivity of a gas mixture decreases.

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 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

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.

Write down the relations for steady one-dimensional heat conduction and mass diffusion through a plane wall, and identify the quantities in the two equations that correspond to each other.

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 \(25^{\circ} \mathrm{C}\).

Liquid methanol is accidentally spilt on a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) laboratory bench and covered the entire bench surface. A fan is providing a \(20 \mathrm{~m} / \mathrm{s}\) air flow parallel over the bench surface. The air is maintained at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), and the concentration of methanol in the free stream is negligible. If the methanol vapor at the air-methanol interface has a pressure of \(4000 \mathrm{~Pa}\) and a temperature of \(25^{\circ} \mathrm{C}\), determine the evaporation rate of methanol in molar basis.

Air flows through a wet pipe at \(298 \mathrm{~K}\) and 1 atm, and the diffusion coefficient of water vapor in air is \(2.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). If the heat transfer coefficient is determined to be \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the mass transfer coefficient is (a) \(0.0326 \mathrm{~m} / \mathrm{s}\) (b) \(0.0387 \mathrm{~m} / \mathrm{s}\) (c) \(0.0517 \mathrm{~m} / \mathrm{s}\) (d) \(0.0583 \mathrm{~m} / \mathrm{s}\) (e) \(0.0707 \mathrm{~m} / \mathrm{s}\)

See all solutions

Recommended explanations on Physics 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.

Sign-up for free