The Kinetic Molecular Theory and Real Gas Behavior

Warm Up

Assume the box shown here has a volume of 22.4 L and exists under standard temperature and pressure (STP) conditions. Draw a molecular representation of oxygen gas filling the box. List three ways in which your drawing fails to represent the actual situation.

A Model for Gases — The Kinetic Molecular Theory

Your first introduction to chemistry in elementary school may have been a study of the kinetic molecular theory of matter (KMT). This theory states that matter is composed of particles such as molecules, atoms, or ions in continuous motion (Figure 8.4.1). In the solid state, the particles are close together and only vibrate. In a liquid, the particles are farther apart and move around. In a gas, the particles are much farther apart and move around freely in a random fashion. A mole of gas occupies 22.4 L, while a mole of most liquids has a volume of only 18 mL. In this section, we will take a closer look at how this theory applies to gases.

Figure particles are farther apart and can move freely. A gas is virtually impossible to represent because the particles are no more than tiny points in space with less than 0.01% of the volume occupied by the actual gas molecules. There is a regular distribution of speeds for the molecules. Some are hardly moving while others have far more energy and are moving very quickly.

When we use KMT to explain the behavior of gases, we make a number of assumptions about the size of the molecules, the way they behave, and what does and does not occur during particle collisions. Such assumptions are included in the postulates of the kinetic molecular theory.

The Postulates of the Kinetic Molecular Theory

1. The volume of gas particles can be assumed to be negligible relative to the space between them.

2. Gas particles are in constant motion. Gas particles colliding with the walls of their container produce pressure.

3. Gas particles neither attract nor repel one another.

4. The average kinetic energy of a gas sample is directly proportional to the temperature of the gas in Kelvin degrees.

Molecular Size

The size of individual gas particles is so small compared to the distance between them that at room temperature and standard atmospheric pressure more that 99.99% of the volume of any gas is actually empty space. Because of all this empty space, it is easy to compress a gas so that it occupies a smaller volume just by applying an external force.

Continuous Motion of Gas Particles

Individual gas particles are in perpetual, random, straight-line motion. During the course of this motion, they frequently collide with one another and then rebound in a perfectly elastic manner. The continuous random motion of gas particles prevents them from collecting in one area and causes them to be uniformly distributed throughout any container they are placed in. This movement is called “diffusion” and is a property of all gases. Diffusion also allows gases of different types to be mixed together to form solutions of any proportion.

Kinetic Energy of Gas Particles

Kinetic energy is the energy resulting from movement. The value of the kinetic energy of an object depends on its mass and velocity. This may be expressed mathematically as kinetic energy = ½mv2 where m is the mass and v is the velocity.

Number of particles

Number of Particles vs. Energy

There is a normal distribution of energies in any sample of matter (Figure 8.4.2). Energy

Figure 8 .4 .2 Any gas sample contains a normal distribution of energies.

At any given temperature, however, the molecules in all gases have the same average kinetic energy. This average kinetic energy is directly proportional to the Kelvin temperature.

The Elasticity of Gas Particles

Thus, if we have a sample of hydrogen gas and oxygen gas at the same temperature, both gases will have the same average kinetic energy. It follows that the average velocity of an oxygen molecule must be less than that of a hydrogen molecule since the mass of the oxygen is 16 times greater. The relatively high velocities of molecules of hydrogen and atoms of helium account for the fact that Earth’s atmosphere contains very little of these two gases. They escaped from Earth’s gravitational attraction many millions of years ago.

In any sample of gas, the particles are moving very quickly. Gas particles at room temperature or 25°C move at speeds in the order of 200 to 2000 m/s, which convert to 450 to 4500 miles per hour. As explained on the previous page, the speed is related to the mass of the gas particles (Figure 8.4.3).

As a result of this tremendous speed, gas particles collide with one another very frequently. One of the assumptions of the KMT is that no energy is lost during these collisions. In other words, the collisions are perfectly elastic. If this were not the case, the particles would gradually lose their kinetic energy. In fact, particles do transfer kinetic energy from one to another during collisions, but the average kinetic energy of all the particles present in a sample remains constant. Collisions between the gas particles and the walls of their container produce pressure (Figure 8.4.3).

Quick Check

Low temperature High temperature

Figure 8 .4 .3 The average velocity of gas particles is directly related to the Kelvin temperature of the sample. There is a normal distribution of energies in any gas sample with some particles barely moving while others move extremely quickly. Collisions between the particles are perfectly elastic and collisions with the walls of the container produce pressure. The representations above show the molecules much larger than they actually are. In reality, less than 0.01% of the volume of the container is occupied by gas particles.

Use the postulates of the kinetic molecular theory to explain each of the following gas laws:

(a) Boyle’s law

(b) Charles’s law

(c) Gay-Lussac’s law

(d) Dalton’s law of partial pressures

Root Mean Square Velocity

The Kelvin temperature is a measure of the average kinetic energy (KE) of any sample of gas particles. For a mole of any gas:

KE avg = 3 2 R T (I) where R = 8.31 J/K mol

Also, KE avg = NA(½mμ2) (II) where NA = the number of particles in a mole (Avogadro’s number) m = the mass of the gas particle μ2 = average of the squares of the particle velocities

Thus, the average kinetic energy of one gas particle in a sample is equal to ½mμ2 .

The average velocity of a gas particle is called the root mean square velocity and is written as:

V rms = √ μ2

Rearranging equations (I) and (II) from above gives:

3

2 R T = ½mμ2

3 R T m

It is necessary to insert the molar mass in kg to allow proper unit analysis. One joule is equivalent to one kg•m2 s2 . The molar mass in kg cancels, leaving the square root of m2/s2, which gives an appropriate unit for velocity of m/s.

Sample Problem — Root Mean Square Velocity and Average Kinetic Energy

Calculate the root mean square velocity and the kinetic energy of carbon monoxide gas at 25°C.

What To Think About

1. The root mean square velocity equals the square root of 3RT/M, where R is 8.31 J/mol K, T is in Kelvin degrees and M is the molar mass in kg.

2. Be sure to divide before taking the square root.

3. Unit analysis requires the substitution of kg•m2 s2 for J and insertion of the molar mass in kg/mol.

4. The kinetic energy is simply 3/2RT where R = 8.31 J/ mol∙K and the temperature is in Kelvin degrees.

How To Do It

µrms = √ 3RT M µrms = √ 3(8.31 J/mol•K)(298 K) 0.0028 kg µrms = √ 3(8.31 kg•m2/s2)(298 K) 0.028 kg/mol mol•K = 515 m/s

KE avg = 3/2RT = 3/2 x 8.31 J/mol K x 298 K = 3710 J

Practice Problems — Root Mean Square Velocity & Average Kinetic Energy

1. Calculate the root mean square velocity and the average kinetic energy of atoms of Ar gas at 25°C.

2. Calculate the root mean square velocity and the average kinetic energy of ozone molecules, O3, high in the stratosphere where the temperature is –83°C.

Range of Velocities in a Gas Sample

In a real gas, there are large numbers of collisions between gas particles. When the lid is removed from a perfume bottle, it takes some time for the odor to move through the air. This is due to collisions occurring between the perfume vapor and the oxygen and nitrogen particles that make up most of the air. The average distance a particle travels between collisions in a sample of gas is called the mean free path. This distance is typically very small and in the range of 60 nanometers. The many collisions result in a wide range of velocities as particles collide and exchange kinetic energy.

Figure 8.4.4 and Figure 8.4.6 illustrate the velocity distributions for various gases at STP conditions and a particular gas under changing conditions of temperature. The peak of each curve indicates the most common particle velocity.

Effusion and Diffusion

Two important terms for describing gases are diffusion and effusion. The odor emitted from a perfume bottle takes several moments to permeate an entire room as the vapor bounces off oxygen and nitrogen molecules, zigzagging its way through the air in random straight-line motion. Diffusion is the term used to describe the mixing of one gas through another. The rate of diffusion is the rate of this mixing.

The rate of diffusion is calculated by directly relating the relative distance travelled by two types of molecules in the same period of time to their relative velocities. Both of these ratios are inversely related to the square root of the ratio of the molecules’ masses.

distance travelled by gas 1 = µrms for gas 1 distance travelled by gas 2 for gas 2 problem!

Effusion is the passage of a gas through a tiny orifice into an evacuated chamber. The rate of effusion measures the speed at which the gas is transferred into the chamber.

This might take a while...

Lighter gases move more quickly and effuse more rapidly than heavier ones, as predicted by the equation, KE

Figure

Scottish chemist Thomas Graham developed Graham’s law of effusion. It states the rate of effusion is inversely proportional to the square root of the mass of its particles.

Figure

The rate of effusion of a light gas such as helium is much faster than that of a more massive gas like ethylene oxide.

Since the effusion rate for a gas depends directly on the average velocity of its particles, the rates of effusion of two gases at the same temperature may be compared by an equation similar to the one used for diffusion above.

Sample Problem — Graham’s Law

Carbon monoxide effuses at a rate that is 2.165 times faster than that of an unknown noble gas. Calculate the molar mass of the unknown gas and determine its identity.

What to Think About

1. Graham’s law of effusion is required to solve this problem.

2. Identify the lighter gas as gas 1 so the effusion rate ratio is >1. This makes the algebra a little easier.

How To Do It

3. Substitute 2.165 for the effusion rate ratio. The molar mass of CO may be substituted as 28.0 g/mol. It is not necessary to convert the molar mass to kg as the kilo prefix will cancel in the ratio form of the equation.

4. Determine the identity of the noble gas by comparison of known molar masses of the noble gas family with the calculated molar mass.

Practice Problems — Graham’s Law

The molar mass matches that of xenon gas.

1. Calculate the ratio of the effusion rates of H2 and UF6. UF6 is a gas used to produce fuel for nuclear reactors.

2. A popular experiment involves the determination of the distances travelled by NH3(g) and HCl(g) from opposite ends of a glass tube. When the two gases, meet they form a visible white ring of NH4Cl.

NH3 17 g/mol NH4Cl HCl 36.5 g/mol distance 100 cm

(a) Which end of the tube will be closer to the white ring of ammonium chloride? Justify your answer with an explanation.