# 38 The Kinetic-Molecular Theory

### Learning Objectives

By the end of this section, you will be able to:

• State the postulates of the kinetic-molecular theory
• Use this theory’s postulates to explain the gas laws

The gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm. Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships.

The kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term “molecule” will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)

1. Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.
2. The molecules composing the gas are negligibly small compared to the distances between them.
3. The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.
4. Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy).
5. The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.

The test of the KMT and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle’s, Charles’s, Amontons’s, Avogadro’s, and Dalton’s laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham’s law.

## The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I

Recalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows:

• Amontons’s law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure (Figure 1).
• Charles’s law. If the temperature of a gas is increased, a constant pressure may be maintained only if the volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. These conditions will decrease the both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which outweigh those of increased collision forces due to the greater kinetic energy at the higher temperature. The net result is a decrease in gas pressure.
• Boyle’s law. If the gas volume is decreased, the container wall area decreases and the molecule-wall collision frequency increases, both of which increase the pressure exerted by the gas (Figure 1).
• Avogadro’s law. At constant pressure and temperature, the frequency and force of molecule-wall collisions are constant. Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions (Figure 1).
• Dalton’s Law. Because of the large distances between them, the molecules of one gas in a mixture bombard the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases. Figure 1. (a) When gas temperature increases, gas pressure increases due to increased force and frequency of molecular collisions. (b) When volume decreases, gas pressure increases due to increased frequency of molecular collisions. (c) When the amount of gas increases at a constant pressure, volume increases to yield a constant number of collisions per unit wall area per unit time.

## Molecular Velocities and Kinetic Energy

The previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at velocities and kinetic energies of gas molecules, and the temperature of a gas sample.

In a gas sample, individual molecules have widely varying speeds; however, because of the vast number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed (Figure 2). Figure 2. The molecular speed distribution for oxygen gas at 300 K is shown here. Very few molecules move at either very low or very high speeds. The number of molecules with intermediate speeds increases rapidly up to a maximum, which is the most probable speed, then drops off rapidly. Note that the most probable speed, νp, is a little less than 400 m/s, while the root mean square speed, urms, is closer to 500 m/s.

The kinetic energy (KE) of a particle of mass (m) and speed (u) is given by:

$\text{KE}=\frac{1}{2}m{u}^{2}$

Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules (J = kg m2 s–2). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the root mean square velocity of a particle,urms, is defined as the square root of the average of the squares of the velocities with n = the number of particles:

${u}_{rms}=\sqrt{\overline{{u}^{2}}}=\sqrt{\frac{{u}_{1}^{2}+{u}_{2}^{2}+{u}_{3}^{2}+{u}_{4}^{2}+\dots }{n}}$

The average kinetic energy, KEavg, is then equal to:

${\text{KE}}_{\text{avg}}=\frac{1}{2}{mu}_{\text{rms}}^{2}$

The KEavg of a collection of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation:

${\text{KE}}_{\text{avg}}=\frac{3}{2}RT$

where R is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is 8.314 J/K (8.314 kg m2s–2K–1). These two separate equations for KEavg may be combined and rearranged to yield a relation between molecular speed and temperature:

$\frac{1}{2}{mu}_{\text{rms}}^{2}=\frac{3}{2}RT$ ${u}_{\text{rms}}=\sqrt{\frac{3RT}{m}}$

### Example 1: Calculation of urms

Calculate the root-mean-square velocity for a nitrogen molecule at 30 °C.

Convert the temperature into Kelvin: $30^{\circ}\text{C}+273=\text{303 K}$

Determine the mass of a nitrogen molecule in kilograms:

$\frac{28.0\cancel{\text{g}}}{\text{1 mol}}\times \frac{\text{1 kg}}{1000\cancel{\text{g}}}=0.028\text{kg/mol}$

Replace the variables and constants in the root-mean-square velocity equation, replacing Joules with the equivalent kg m2s–2:

${u}_{\text{rms}}=\sqrt{\frac{3RT}{m}}$ ${u}_{rms}=\sqrt{\frac{3\left(8.314\text{J/mol K}\right)\left(\text{303 K}\right)}{\left(0.028\text{kg/mol}\right)}}=\sqrt{2.70\times {10}^{5}{\text{m}}^{2}{\text{s}}^{-2}}=519\text{m/s}$

Calculate the root-mean-square velocity for an oxygen molecule at –23 °C.

441 m/s

If the temperature of a gas increases, its KEavg increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, KEavg decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in Figure 3.

At a given temperature, all gases have the same KEavg for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher urms, with a speed distribution that peaks at relatively higher velocities. Gases consisting of heavier molecules have more low-speed particles, a lower urms, and a speed distribution that peaks at relatively lower velocities. This trend is demonstrated by the data for a series of noble gases shown in Figure 4.

The PhET gas simulator may be used to examine the effect of temperature on molecular velocities. Examine the simulator’s “energy histograms” (molecular speed distributions) and “species information” (which gives average speed values) for molecules of different masses at various temperatures.

## The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II

According to Graham’s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates.

The rate of effusion of a gas depends directly on the (average) speed of its molecules:

$\text{effusion rate}\propto {u}_{\text{rms}}$

Using this relation, and the equation relating molecular speed to mass, Graham’s law may be easily derived as shown here:

${u}_{\text{rms}}=\sqrt{\frac{3RT}{m}}$ $m=\frac{3RT}{{u}_{rms}^{2}}=\frac{3RT}{{\overline{u}}^{2}}$ $\frac{\text{effusion rate A}}{\text{effusion rate B}}=\frac{{u}_{rms\text{A}}}{{u}_{rms\text{B}}}=\frac{\sqrt{\frac{3RT}{{m}_{\text{A}}}}}{\sqrt{\frac{3RT}{{m}_{\text{B}}}}}=\sqrt{\frac{{m}_{\text{B}}}{{m}_{\text{A}}}}$

The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham’s law.

### Key Concepts and Summary

The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average velocities determined by their absolute temperatures. The individual molecules of a gas exhibit a range of velocities, the distribution of these velocities being dependent on the temperature of the gas and the mass of its molecules.

#### Key Equations

• ${u}_{rms}=\sqrt{\overline{{u}^{2}}}=\sqrt{\frac{{u}_{1}^{2}+{u}_{2}^{2}+{u}_{3}^{2}+{u}_{4}^{2}+\dots }{n}}$
• ${\text{KE}}_{\text{avg}}=\frac{3}{2}R\text{T}$
• ${u}_{\text{rms}}=\sqrt{\frac{3RT}{m}}$

### Exercises

1. Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape.
2. Can the speed of a given molecule in a gas double at constant temperature? Explain your answer.
3. Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:
1. The pressure of the gas is increased by reducing the volume at constant temperature.
2. The pressure of the gas is increased by increasing the temperature at constant volume.
3. The average velocity of the molecules is increased by a factor of 2.
4. The distribution of molecular velocities in a sample of helium is shown in Figure 9.34. If the sample is cooled, will the distribution of velocities look more like that of H2 or of H2O? Explain your answer.
5. What is the ratio of the average kinetic energy of a SO2 molecule to that of an O2 molecule in a mixture of two gases? What is the ratio of the root mean square speeds, urms, of the two gases?
6. A 1-L sample of CO initially at STP is heated to 546 °C, and its volume is increased to 2 L.
1. What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?
2. What is the effect on the average kinetic energy of the molecules?
3. What is the effect on the root mean square speed of the molecules?
7. The root mean square speed of H2 molecules at 25 °C is about 1.6 km/s. What is the root mean square speed of a N2 molecule at 25 °C?
8. Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas 2, $\frac{{R}_{1}}{{R}_{2}},$ is the same at 0 °C and 100 °C.

2. Yes. At any given instant, there are a range of values of molecular speeds in a sample of gas. Any single molecule can speed up or slow down as it collides with other molecules. The average velocity of all the molecules is constant at constant temperature.

4. H2O. Cooling slows the velocities of the He atoms, causing them to behave as though they were heavier.

6. Both the temperature and the volume are doubled for this gas (n constant), so P remains constant.

1. The number of collisions per unit area of the container wall is constant.
2. The average kinetic energy doubles; it is proportional to temperature.
3. The root mean square speed increases to $\sqrt{2}$ times its initial value; urms is proportional to $\sqrt{{\text{KE}}_{\text{avg}}}.$

8. The rate at which a gas will diffuse, R, is proportional lo urms, the root mean square speed of its molecules. The square of this value, in turn, is proportional to the average kinetic energy. The average kinetic energy is:

${\overline{\text{KE}}}_{\text{avg}}=kT$

For two different gases, 1 and 2, the constant of proportionality can be represented as k1 and k2, respectively. Thus,

$\frac{{R}_{1}}{{R}_{2}}=\frac{{k}_{1}\sqrt{T}}{{k}_{2}\sqrt{T}}$

As a result of this relationship, no matter at which temperature diffusion occurs, the temperature term will cancel out of the equation and the ratio of rates will be the same.

### Exercises

1. Is the pressure of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
2. Is the density of the gas in the hot air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
3. At a pressure of 1 atm and a temperature of 20 °C, dry air has a density of 1.2256 g/L. What is the (average) molar mass of dry air?
4. The average temperature of the gas in a hot air balloon is 1.30 × 102 °F. Calculate its density, assuming the molar mass equals that of dry air.
5. The lifting capacity of a hot air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?
6. An average balloon has a diameter of 60 feet and a volume of 1.1 × 105 ft3. What is the lifting power of such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?
7. A balloon carries 40.0 gallons of liquid propane (density 0.5005 g/L). What volume of CO2 and H2O gas is produced by the combustion of this propane?
8. A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in kJ/min) from the hot air in the bag during the flight?
1. equal, because the balloon is free to expand until the pressures are equalized
2. less than the density outside
3. assume three-place accuracy throughout unless greater accuracy is stated:
$\text{molar mass}=\frac{DRT}{P}=1.2256\text{g}\cancel{{\text{L}}^{-\text{1}}}\times \frac{0.08206\cancel{\text{L}}\cancel{\text{atm}}{\text{mol}}^{-\text{1}}\cancel{{\text{K}}^{-\text{1}}}\times 293.15\cancel{\text{K}}}{1.00\cancel{\text{atm}}}=29.48{\text{g mol}}^{-\text{1}}$
4. convert the temperature to °C; then use the ideal gas law:
$^{\circ}\text{C}=\frac{5}{9}\left(\text{F}-32\right)=\frac{5}{9}\left(130-32\right)=54.44^{\circ}\text{C}=327.6\text{K}$ $D=\frac{\mathcal{M}P}{RT}=29.48\text{g}\cancel{{\text{mol}}^{-\text{1}}}\times \frac{1.00\cancel{\text{atm}}}{0.08206\text{L}\cancel{\text{atm}}\cancel{{\text{mol}}^{-\text{1}}}\cancel{{\text{K}}^{-\text{1}}}\times 327.6\cancel{\text{K}}}=1.0966{\text{g L}}^{-\text{1}}$
5. 1.2256 g/L – 1.09966 g/L = 0.129 g/L;
6. calculate the volume in liters, multiply the volume by the density difference to find the lifting capacity of the balloon, subtract the weight of the balloon after converting to pounds:
$1.1\times 105{\text{ft}}^{3}\times {\left(\frac{\text{12 in}}{\text{91 ft}}\right)}^{3}\times {\left(\frac{\text{2.54 cm}}{\text{in}}\right)}^{3}\times \frac{\text{1 L}}{1000{\text{cm}}^{3}}=3.11\times {10}^{6}\text{L}$
3.11 × 106 L × 0.129 g/L = 4.01 × 105 g
$\frac{4.01\times {10}^{5}\text{g}}{453.59{\text{g lb}}^{-\text{1}}}=884\text{lb;}\text{884 lb}-\text{500 lb}=\text{384 lb}$ net lifting capacity = 384 lb
7. First, find the mass of propane contained in 40.0 gal. Then calculate the moles of CO2(g) and H2O(g) produced from the balanced equation.
$40.0\cancel{\text{gal}}\times \frac{4\left(0.9463\text{L}\right)}{1\cancel{\text{gal}}}=151.4\text{L}$ $151.4\cancel{\text{L}}\times 0.5005\text{g}{\cancel{\text{L}}}^{\cancel{-1}}=75.8\text{g}$ Molar mass of propane = 3(12.011) + 8(1.00794) = 36.033 + 8.064 = 44.097 g mol–1 $\frac{75.8\cancel{\text{g}}}{44.097\cancel{\text{g}}{\text{mol}}^{-\text{1}}}=1.72\text{mol}$
The reaction is ${\text{C}}_{3}{\text{H}}_{8}\left(g\right)+5{\text{O}}_{2}\left(g\right)\rightarrow 3{\text{CO}}_{2}\left(g\right)+4{\text{H}}_{2}\text{O}\left(g\right)$
For each 1.72 mol propane, there are 3 × 1.72 mol = 5.15 mol of CO2 and 4 × 1.72 mol = 6.88 mol H2 O. The total volume at STP = 22.4 L × 12.04 = 270 L
8. The total heat released is determined from the heat of combustion of the propane. Using the equation in question 7,
$\begin{array}{ll}\Delta {H}_{\text{combustion}}^{\circ}\hfill & =3{\Delta H}_{{\text{CO}}_{2}\left(g\right)}^{\circ}+4{\Delta H}_{{\text{H}}_{2}\text{O}\left(g\right)}^{\circ}-{\Delta H}_{\text{propane}}^{\circ}\hfill \\ \hfill & =3\left(-393.51\right)+4\left(-241.82\right)-\left(-103.85\right)\hfill \\ \hfill & =-1180.52-967.28+103.85=-2043.96{\text{kJ mol}}^{-1}\hfill \end{array}$
Since there is 1.72 mol propane, 1.72 × 2043.96 kJ mol-1 = 3.52 × 103 kJ used for heating. This heat is used over 90 minutes, so $\frac{3.52\times {10}^{3}\text{kJ}}{\text{90 min}}=39.1{\text{kJ min}}^{-\text{1}}$ is released.

## Glossary

kinetic molecular theory: theory based on simple principles and assumptions that effectively explains ideal gas behavior

root mean square velocity (urms): measure of average velocity for a group of particles calculated as the square root of the average squared velocity 