# 2.6 – Kinetic-Molecular Theory of Gases (Ideal Gas Behaviours)

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 lower pressures (*i.e.* less than 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.)

- 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.
- The molecules composing the gas are negligibly small compared to the distances between them.
- The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.
- 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). - The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.

**Figure 2.6.1. **Visualizing molecular motion. Molecules of a gas are in constant motion and collide with one another and with the container wall.

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**/Gay-Lussac’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 2.6.2(a)).*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 both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which balance the effect of increased collision forces due to the greater kinetic energy at the higher temperature.*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 2.6.2(b)).*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 2.6.2(c)).*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 2.6.2. **(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 2.6.3.).

**Figure 2.6.3. **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, *ν*_{p}, is a little less than 400 m/s, while the root mean square speed, *u*_{rms}, is closer to 500 m/s.

The kinetic energy (*E*_{k}) of a particle of mass (*m*) and speed (*u*) is given by:

**Equation 2.6.1 **Kinetic Energy to velocity and mass

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

**Equation 2.6.2 **Root mean square of a particle

Note that this value is mathematically different from the gas population’s arithmetic mean, and also does *not* correspond to the most probable molecular speed (*v*_{p}).

The *average* kinetic energy for a mole of particles, *E*_{k }_{avg}, is then equal to (where *M* is the average mass of the particles, expressed in kilograms):

**Equation 2.6.3 **Average kinetic energy for a mole of particles

The *E*_{k}_{ avg} of a mole of gas molecules is also directly proportional to the temperature of the gas and may be described by the equation (to extend your learning, check out the derivation for *E*_{k}_{ avg} here):

**Equation 2.6.4 **Average Kinetic Energy is directly proportional to temperature

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/mol⋅K (8.314 kg m^{2}s^{–2}mol^{–1}K^{–1}). These two separate equations for *E*_{k} may be combined and rearranged to yield a relation between the root-mean-square speed and temperature:

**Equation 2.6.5 **The Root-Mean-Square Speed vs Temperature

**Example 2.6.1 – Calculation of ***u*_{rms}

Calculate the root-mean-square speed for a nitrogen molecule (N_{2}) at 30 °C.

**Solution**

Convert the temperature into Kelvin:

30 °C + 273 = 303 K

Determine the molar mass of nitrogen in kilograms:

Replace the variables and constants in the root-mean-square velocity equation, replacing Joules with the equivalent kg m^{2}s^{–2}:

**Check Your Learning 2.6.1 – Calculation of u_{rms}**

Calculate the root-mean-square speed for a mole of oxygen molecules at –23 °C.

**Answer**

441 m/s

If the temperature of a gas increases, its *E*_{k}_{ avg} increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution spreads out toward higher speeds overall, that is, to the right (note that the curve always passes through the origin, meaning there are always some slow-moving particles). If temperature decreases, *E*_{k}_{ avg} 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 behaviour is illustrated for nitrogen gas in Figure 2.6.4.

**Figure 2.6.4.**The molecular speed distribution for nitrogen gas (N_{2}) shifts to the right and flattens as the temperature increases; it shifts to the left and heightens as the temperature decreases.

At a given temperature, all gases have the same *E*_{k}_{ avg} for their molecules. Gases composed of lighter molecules have more high-speed particles and thus a greater *u*_{rms}, with a speed distribution that peaks at relatively higher velocities. Gases consisting of heavier molecules have more low-speed particles, a lower *u*_{rms}, 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 2.6.5.

**Figure 2.6.5.** Molecular velocity is directly related to molecular mass. At a given temperature, lighter molecules move faster on average than heavier molecules.

The 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.

**Questions**

**★**** Questions**

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:

a. The pressure of the gas is increased by reducing the volume at constant temperature.

b. The pressure of the gas is increased by increasing the temperature at constant volume.

c. The average velocity of the molecules is increased by a factor of 2.

4. What is the ratio of the average kinetic energy of a SO_{2} molecule to that of an O_{2} molecule in a mixture of two gases? What is the ratio of the root mean square speeds, *u*_{rms}, of the two gases?

5. A 1-L sample of CO initially at STP is heated to 546 K, and its volume is increased to 2 L.

a. What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?

b. What is the effect on the average kinetic energy of the molecules?

c. What is the effect on the root mean square speed of the molecules?

6. The root mean square speed of H_{2} molecules at 25 °C is about 1.6 km/s. What is the root mean square speed of a N_{2} molecule at 25 °C?

**★★**** Questions**

7. Answer the following questions about a hot air balloon:

a. 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?

b. 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?

c. 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?

d. The average temperature of the gas in a hot air balloon is 1.30 × 10^{2} °F. Calculate its density, assuming the molar mass equals that of dry air.

e. 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)?

f. An average balloon has a diameter of 60 feet and a volume of 1.1 × 10^{5} ft^{3}. 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?

g. A balloon carries 40.0 gallons of liquid propane (density 0.5005 g/L). What volume of CO_{2} and H_{2}O gas is produced by the combustion of this propane?

h. 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?

8. Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas 2, ^{R1}/_{R2}, is the same at 0 °C and 100 °C.

**Answers**

1. Gases are composed of molecules that are in continuous motion, travelling in straight lines only deviating once colliding with other molecules. The pressure exerted by a gas in a container results from collisions between the gas molecules and the walls. The gas molecules are negligibly small in comparison to the distances between them.

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.

3. (a) Average kinetic energy remains unchanged, (b) *P* increase leads to average kinetic energy increase, (c) velocity doubled leads to a quadruple in kinetic energy

4. (a) The ratio of average kinetic energies is 1, (b) 0.71

5. (a) The number of collisions per unit area of the container wall is constant. (b) The average kinetic energy doubles. (c) The root mean square speed increases to sqrt(2) times its initial value; *u*_{rms} is proportional to sqrt(*E*_{k}_{ avg}).

6. (a) 1.92 km/s

7. (a) equal; (b) less than; (c) 29.48 g mol^{−1}; (d) 1.0966 g L^{−1}; (e) 0.129 g/L; (f) 4.01 × 10^{5} g; net lifting capacity = 384 lb; (g) 270 L; (h) 39.1 kJ min^{−1}

8. , assume a system of Gas 1 and Gas 2 with masses *m*_{1} and *m*_{2}, rate of diffusion is dependent on only mass and not temperature. Hence the rate of diffusion will remain constant at different temperatures.

Theory based on simple principles and assumptions that effectively explains ideal gas behavior by helping us understand gases at the molecular level and their physical properties

Measure of average velocity for a group of particles calculated as the square root of the average squared velocity