Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

According to the kinetic theory of gases: (a) The pressure exerted by a gas is proportional to mean square velocity of the molecules (b) The pressure exerted by the gas is proportional to the root mean square velocity of the molecules (c) The root mean square velocity is inversely proportional to the temperature (d) The mean translational K.E. of the molecule is directly proportional to the absolute temperature.

Short Answer

Expert verified
Correct answers: (a) and (d).

Step by step solution

01

Understanding the Kinetic Theory of Gases

The kinetic theory of gases relates the macroscopic properties of gases to the microscopic actions of their molecules. According to this theory, the pressure of a gas is caused by collisions of the gas molecules with the walls of the container.
02

Relation of Pressure to Molecular Velocities

The pressure of a gas, as given by the kinetic theory, is proportional to the mean square speed (\(v^2_{mean}\)) of the gas molecules, not the root mean square speed (\(v_{rms}\)). This makes option (a) correct and (b) incorrect.
03

Root Mean Square Velocity and Temperature

The root mean square velocity (\(v_{rms}\)) of gas molecules is directly proportional to the square root of the temperature (in Kelvin). It is expressed mathematically as \(v_{rms} = \sqrt{\frac{3kT}{m}}\), where \(k\) is the Boltzmann constant and \(m\) is the mass of a molecule. Thus, (c) is incorrect.
04

Mean Translational Kinetic Energy

The mean translational kinetic energy of a molecule is given by \(\frac{3}{2}kT\) and is directly proportional to the absolute temperature, \(T\). This makes option (d) correct.

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!

Key Concepts

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

Pressure and Molecular Velocities
The pressure of a gas is fundamentally linked to the behavior of its molecules. In the kinetic theory of gases, pressure arises from the collisions of gas molecules with the walls of their container. Each impact exerts a force, contributing to the overall pressure experienced by the container walls.

According to this theory, the pressure is proportional to the mean square velocity of the molecules, denoted as \(v^2_{mean}\). This means that as the velocity increases, so does the pressure, provided all other conditions remain constant.

It's important to note that the pressure is not related to the root mean square velocity \(v_{rms}\), which is a common misconception.

In essence, the more energetic the molecules (higher mean square velocity), the greater the pressure they exert upon collision.
Root Mean Square Velocity
Root Mean Square Velocity (\(v_{rms}\)) is an important concept when discussing molecular velocities. It represents an average velocity of gas molecules, providing a measure for understanding how fast molecules are moving in a system.

Mathematically, it is defined as \(v_{rms} = \sqrt{\frac{3kT}{m}}\), where \(k\) is the Boltzmann constant, \(T\) is the absolute temperature in Kelvin, and \(m\) is the mass of a molecule.

This equation shows that \(v_{rms}\) is directly proportional to the square root of temperature. Therefore, as the temperature increases, molecules move faster, and their root mean square velocity increases.

Contrary to option (c) in the original exercise, the root mean square velocity is not inversely proportional to temperature but actually increases with it.
Translational Kinetic Energy and Temperature
The translational kinetic energy of a gas molecule is a measure of the energy due to its motion through space. According to the kinetic theory of gases, this energy is directly related to the temperature of the gas.

The average translational kinetic energy for a molecule is expressed by the formula \(\frac{3}{2}kT\). In this equation, \(k\) represents the Boltzmann constant, and \(T\) is the absolute temperature in Kelvin.

This relationship indicates that as the temperature increases, so does the average translational kinetic energy.

This principle supports option (d) from the exercise: the mean translational kinetic energy of a molecule increases proportionally with absolute temperature.

In conclusion, temperature not only affects the speed of molecular movement but also influences the energy associated with their motion.

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

Compressibility factor for 1 mol of a van der Waals gas at \(0^{\circ} \mathrm{C}\) and 100 atmospheric pressure is found to be \(0.5\), the volume of gas molecule is: (a) \(2.0224 \mathrm{~L}\) (b) \(1.4666 \mathrm{~L}\) (c) \(0.8542 \mathrm{~L}\) (d) \(0.1119 \mathrm{~L}\)

At \(100^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), if the density of liquid water is \(1.0 \mathrm{~g} \mathrm{~cm}^{-3}\) and that of water vapour is \(0.0006 \mathrm{~g} \mathrm{~cm}^{-3}\), then the volume occupied by water molecules in 1 litre of steam at that temperature: (a) \(6 \mathrm{~cm}^{3}\) (b) \(60 \mathrm{~cm}^{3}\) (c) \(0.6 \mathrm{~cm}^{3}\) (d) \(0.06 \mathrm{~cm}^{3}\)

At constant volume and temperature conditions, the rates of diffusion \(D_{A}\) and \(D_{B}\) of gases \(A\) and \(B\) having densities \(\rho_{A}\) and \(\rho_{B}\) are related by the expression: (a) \(\mathrm{D}_{\mathrm{A}}=\left[\mathrm{D}_{\mathrm{B}} \rho_{A} / \rho_{\mathrm{B}}\right]^{1 / 2}\) (b) \(\mathrm{D}_{\mathrm{A}}=\left[\mathrm{D}_{\mathrm{B}} \rho_{\mathrm{B}} / \rho_{\mathrm{A}}\right]^{1 / 2}\) (c) \(\mathrm{D}_{\mathrm{A}}=\mathrm{D}_{\mathrm{B}}\left[\rho_{A} / \rho_{\mathrm{B}}\right]^{1 / 2}\) (d) \(\mathrm{D}_{\mathrm{A}}=\mathrm{D}_{\mathrm{B}}\left[\rho_{\mathrm{B}} / \rho_{\mathrm{A}}\right]^{1 / 2}\)

Ideal gas obeying kinetic theory of gases can be liquefied if: (a) \(\mathrm{T}>\mathrm{T}_{\mathrm{c}}\) (b) \(\mathrm{P}>\mathrm{P}\) (c) \(\mathrm{P}>\mathrm{P}_{c}\) and \(\mathrm{T}<\mathrm{T}_{\mathrm{d}}\) (d) It cannot be liquefied at any value of \(\mathrm{P}\) and \(\mathrm{T}\).

A gas behaves most like an ideal gas under conditions of: (a) High pressure and low temperature (b) High temperature and high pressure (c) Low pressure and high temperature (d) Low pressure and low temperature

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.

Sign-up for free