14.3

Kinetic Theory of Gases

CONCEPTS AT A GLANCE As useful as it is, the ideal gas law provides no insight as to how pressure and temperature are related to properties of the molecules themselves, such as their masses and speeds. To show how such microscopic properties are related to the pressure and temperature of an ideal gas, this section examines the dynamics of molecular motion. The pressure that a gas exerts on the walls of a container is due to the force exerted by the gas molecules when they collide with the walls. Therefore, as the Concepts-at-a-Glance chart in Figure 14-8 illustrates, we will begin by combining the notion of collisional forces exerted by a fluid (Section 11.2) with Newton’s second and third laws of motion (Sections 4.3 and 4.5). These concepts will allow us to obtain an expression for the pressure in terms of microscopic properties. We will then combine this with the ideal gas law to show that the average translational kinetic energy of a particle in an ideal gas is , where k is Boltzmann’s constant and T is the Kelvin temperature. In the process, we will also see that the internal energy U of a monatomic ideal gas is , where n is the number of moles and R is the universal gas constant.

Figure 14-8 CONCEPTS AT A GLANCE To formulate the kinetic theory of gases, Newton’s second and third laws are brought together with the ideal gas law. One of the predictions of the theory is that the internal energy of an ideal gas is proportional to the Kelvin temperature of the gas.

| The Distribution of Molecular Speeds

A macroscopic container filled with a gas at standard temperature and pressure contains a large number of particles (atoms or molecules). These particles are in constant, random motion, colliding with each other and with the walls of the container. In the course of one second, a particle undergoes many collisions, and each one changes the particle’s speed and direction of motion. As a result, the atoms or molecules have different speeds. It is possible, however, to speak about an average particle speed. At any given instant, some particles have speeds less than, some near, and some greater than the average. For conditions of low gas density, the distribution of speeds within a large collection of molecules at a constant temperature was calculated by the Scottish physicist James Clerk Maxwell (1831–1879). Figure 14-9 displays the Maxwell speed distribution curves for O2 gas at two different temperatures. When the temperature is 300 K, the maximum in the curve indicates that the most probable speed is about 400 m /s. At a temperature of 1200 K, the distribution curve shifts to the right, and the most probable speed increases to about 800 m /s. One particularly useful type of average speed, known as the rms speed and written as vrms, is also shown in the drawing. When the temperature of the oxygen gas is 300 K the rms speed is 480 m/s, and it increases to 970 m/s when the temperature rises to 1200 K. The reason why the rms speed is so important will be discussed shortly.

Figure 14-9 The Maxwell distribution curves for molecular speeds in oxygen gas at temperatures of 300 and 1200 K.

| Kinetic Theory

If a ball is thrown against a wall, it exerts a force on the wall. As Figure 14-10 suggests, gas particles do the same thing, except that their masses are smaller and their speeds are greater. The number of particles is so great and they strike the wall so often that the effect of their individual impacts appears as a continuous force. Dividing the magnitude of this force by the area of the wall gives the pressure exerted by the gas.

Figure 14-10 The pressure that a gas exerts is caused by the collisions of its molecules with the walls of the container.

To calculate the force, consider an ideal gas composed of N identical particles in a cubical container whose sides have length L. Except for elastic* collisions, these particles do not interact. Figure 14-11 focuses attention on one particle of mass m as it strikes the right wall perpendicularly and rebounds elastically. While approaching the wall, the particle has a velocity and linear momentum (see Section 7.1 for a review of linear momentum). The particle rebounds with a velocity and momentum , travels to the left wall, rebounds again, and heads back toward the right. The time t between collisions with the right wall is the round-trip distance 2L divided by the speed of the particle; that is, . According to Newton’s second law of motion, in the form of the impulse–momentum theorem, the average force exerted on the particle by the wall is given by the change in the particle’s momentum per unit time:

| (7.4)

According to Newton’s law of action–reaction, the force applied to the wall by the particle is equal in magnitude to this value, but oppositely directed (i.e., ). The magnitude F of the total force exerted on the right wall is equal to the number of particles that collide with the wall during the time t multiplied by the average force exerted by each particle. Since the N particles move randomly in three dimensions, one-third of them on the average strike the right wall during the time t. Therefore, the total force is

In this result v2 has been replaced by , the average value of the squared speed. The collection of particles possesses a Maxwell distribution of speeds, so an average value for v2 must be used, rather than a value for any individual particle. The square root of the quantity is called the root-mean-square speed, or, for short, the rms speed; With this substitution, the total force becomes

Pressure is force per unit area, so the pressure P acting on a wall of area L2 is

Since the volume of the box is , the equation above can be written as

| (14.5)

Figure 14-11 A gas particle is shown colliding elastically with the right wall of the container and rebounding from it.

Equation 14.5 relates the macroscopic properties of the gas, its pressure and volume, to the microscopic properties of the constituent particles, their mass and speed. Since the term is the average translational kinetic energy of an individual particle, it follows that

This result is similar to the ideal gas law, (Equation 14.2). Both equations have identical terms on the left, so the terms on the right must be equal: . Therefore,

| (14.6)

Equation 14.6 is significant, because it allows us to interpret temperature in terms of the motion of gas particles. This equation indicates that the Kelvin temperature is directly proportional to the average translational kinetic energy per particle in an ideal gas, no matter what the pressure and volume are. On the average, the particles have greater kinetic energies when the gas is hotter than when it is cooler. Conceptual Example 5 discusses a common misconception about the relation between kinetic energy and temperature.

| Conceptual Example 5 | | | Does a Single Particle Have a Temperature?

Each particle in a gas has kinetic energy. Furthermore, the equation establishes the relationship between the average kinetic energy per particle and the temperature of an ideal gas. Is it valid, then, to conclude that a single particle has a temperature?
Reasoning and Solution We know that a gas contains an enormous number of particles that are traveling with a distribution of speeds, such as those indicated by the graphs in Figure 14-9. Therefore, the particles do not all have the same kinetic energy, but possess a distribution of kinetic energies ranging from very nearly zero to extremely large values. If each particle had a temperature that was associated with its kinetic energy, there would be a whole range of different temperatures within the gas. This is not so, for a gas at thermal equilibrium has only one temperature (see Section 15.2), a temperature that would be registered by a thermometer placed in the gas. Thus, temperature is a property that characterizes the gas as a whole, a fact that is inherent in the relation . The term vrms is a kind of average particle speed. Therefore, is the average kinetic energy per particle and is characteristic of the gas as a whole. Since the Kelvin temperature is proportional to , it is also a characteristic of the gas as a whole and cannot be ascribed to each gas particle individually. Thus, a single gas particle does not have a temperature.

Check Your Understanding 3

The pressure of a monatomic ideal gas is doubled, while its volume is reduced by a factor of four. What is the ratio of the new rms speed of the atoms to the initial rms speed? (The answer is given at the end of the book.) Background: The ideal gas law and the kinetic theory of gases are required to answer this question.
For similar questions (including calculational counterparts), consult Self-Assessment Test 14.1. This test is described at the end of Section 14.4.

If two ideal gases have the same temperature, the relation indicates that the average kinetic energy of each kind of gas particle is the same. In general, however, the rms speeds of the different particles are not the same, for the masses may be different. The next example illustrates these facts and shows how rapidly gas particles move at normal temperatures.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 6 | | | The Speed of Molecules in Air

Air is primarily a mixture of nitrogen N2 and oxygen O2 . Assume that each behaves like an ideal gas and determine the rms speed of the nitrogen and oxygen molecules when the air temperature is 293 K. Reasoning As Figure 14-9 illustrates, the same type of molecules (e.g., O2 molecules) within a gas have different speeds, even though the gas itself has a constant temperature. The rms speed vrms of the molecules is a kind of average speed and is related to the kinetic energy (also an average) according to Equation 6.2. Thus, the average translational kinetic energy of a molecule is , where m is the mass of a molecule. We know that the average kinetic energy depends on the Kelvin temperature T of the gas through the relation (Equation 14.6). We can find the rms speed of each type of molecule, then, from a knowledge of its mass and the temperature.
Knowns and Unknowns The data for this problem are:

Description | Symbol | Value
Molecular mass of nitrogen (N2) | – | 28.0 u
Molecular mass of oxygen (O2) | – | 32.0 u
Air temperature | T | 293 K
Unknown Variable |  |
Rms speed of nitrogen and oxygen molecules | vrms | ?

Modeling the Problem

| Step 1 Rms Speed and Average Kinetic Energy
| The rms speed vrms of a molecule in a gas is related to the average kinetic energy by

| (6.2)

where m is its mass. Solving for vrms gives Equation 1 at the right. The average kinetic energy will be dealt with in Step 2.

| (1)

| Step 2 Average Kinetic Energy and Temperature
| Since the gas is assumed to be an ideal gas, the average kinetic energy of a molecule is directly proportional to the Kelvin temperature T of the gas according to Equation 14.6:

where k is Boltzmann’s constant. By substituting this relation into Equation 1 at the right, we can obtain an expression for the rms speed.

Solution Algebraically combining the results of the two steps, we have
Since the molecular masses of nitrogen and oxygen are given in atomic mass units (28.0 u and 32.0 u, respectively), we must convert them to kilograms by using the conversion factor (see Section 14.1). Thus,

The rms speed for each type of molecule is

Note that the nitrogen and oxygen molecules have the same average kinetic energy, since the temperature is the same for both. The fact that nitrogen has the greater rms speed is due to its smaller mass.
Related Homework: Problems 28, 32

Problem solving insight The average translational kinetic energy is the same for all ideal-gas molecules at the same temperature, regardless of their masses. The rms translational speed of the molecules is not the same, however, but depends on the mass.

The equation has also been applied to particles much larger than atoms or molecules. The English botanist Robert Brown (1773–1858) observed through a microscope that pollen grains suspended in water move on very irregular, zigzag paths. This Brownian motion can also be observed with other particle suspensions, such as fine smoke particles in air. In 1905, Albert Einstein (1879–1955) showed that Brownian motion could be explained as a response of the large suspended particles to impacts from the moving molecules of the fluid medium (e.g., water or air). As a result of the impacts, the suspended particles have the same average translational kinetic energy as the fluid molecules—namely, . But unlike the molecules, the particles are large enough to be seen through a microscope and, because of their relatively large mass, have a comparatively small average speed.

| The Internal Energy of a Monatomic Ideal Gas

Chapter 15 deals with the science of thermodynamics, in which the concept of internal energy plays an important role. Using the results just developed for the average translational kinetic energy, we conclude this section by expressing the internal energy of a monatomic ideal gas in a form that is suitable for use later on.
The internal energy of a substance is the sum of the various kinds of energy that the atoms or molecules of the substance possess. A monatomic ideal gas is composed of single atoms. These atoms are assumed to be so small that the mass is concentrated at a point, with the result that the moment of inertia I about the center of mass is negligible. Thus, the rotational kinetic energy is also negligible. Vibrational kinetic and potential energies are absent, because the atoms are not connected by chemical bonds and, except for elastic collisions, do not interact. As a result, the internal energy U is the total translational kinetic energy of the N atoms that constitute the gas: . Since according to Equation 14.6, the internal energy can be written in terms of the Kelvin temperature as

Usually, U is expressed in terms of the number of moles n, rather than the number of atoms N. Using the fact that Boltzmann’s constant is , where R is the universal gas constant and NA is Avogadro’s number, and realizing that , we find that

| (14.7)

Thus, the internal energy depends on the number of moles and Kelvin temperature of the gas. In fact, it can be shown that the internal energy is proportional to the Kelvin temperature for any type of ideal gas (e.g., monatomic, diatomic, etc.). For example, when hotair balloonists turn on the burner, they increase the temperature, and hence the internal energy per mole, of the air inside the balloon (see Figure 14-12).

Figure 14-12 Since air behaves approximately as an ideal gas, the internal energy per mole inside a hot-air balloon increases as the temperature rises. (Erik Lam/Alamy Images)

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