135. Kinetic Theory: Atomic and Molecular Explanation of Pressure and TemperatureLearning Objectives
We have developed macroscopic definitions of pressure and temperature. Pressure is the force divided by the area on which the force is exerted, and temperature is measured with a thermometer. We gain a better understanding of pressure and temperature from the kinetic theory of gases, which assumes that atoms and molecules are in continuous random motion. Figure 1 shows an elastic collision of a gas molecule with the wall of a container, so that it exerts a force on the wall (by Newton’s third law). Because a huge number of molecules will collide with the wall in a short time, we observe an average force per unit area. These collisions are the source of pressure in a gas. As the number of molecules increases, the number of collisions and thus the pressure increase. Similarly, the gas pressure is higher if the average velocity of molecules is higher. The actual relationship is derived in the Things Great and Small feature below. The following relationship is found: $$\text{PV}=\frac{1}{3}\text{Nm}\overline{{v}^{2}},$$where $P$ is the pressure (average force per unit area), $V$ is the volume of gas in the container, $N$ is the number of molecules in the container, $m$ is the mass of a molecule, and $\overline{{v}^{2}}$ is the average of the molecular speed squared. What can we learn from this atomic and molecular version of the ideal gas law? We can derive a relationship between temperature and the average translational kinetic energy of molecules in a gas. Recall the previous expression of the ideal gas law: $$\text{PV}=\text{NkT}.$$Equating the righthand side of this equation with the righthand side of $\text{PV}=\frac{1}{3}\text{Nm}\overline{{v}^{2}}$ gives $$\frac{1}{3}\text{Nm}\overline{{v}^{2}}=\text{NkT}.$$Making Connections: Things Great and Small—Atomic and Molecular Origin of Pressure in a Gas:Figure 2 shows a box filled with a gas. We know from our previous discussions that putting more gas into the box produces greater pressure, and that increasing the temperature of the gas also produces a greater pressure. But why should increasing the temperature of the gas increase the pressure in the box? A look at the atomic and molecular scale gives us some answers, and an alternative expression for the ideal gas law. The figure shows an expanded view of an elastic collision of a gas molecule with the wall of a container. Calculating the average force exerted by such molecules will lead us to the ideal gas law, and to the connection between temperature and molecular kinetic energy. We assume that a molecule is small compared with the separation of molecules in the gas, and that its interaction with other molecules can be ignored. We also assume the wall is rigid and that the molecule’s direction changes, but that its speed remains constant (and hence its kinetic energy and the magnitude of its momentum remain constant as well). This assumption is not always valid, but the same result is obtained with a more detailed description of the molecule’s exchange of energy and momentum with the wall. If the molecule’s velocity changes in the $x$direction, its momentum changes from $\u2013{\text{mv}}_{x}$ to $+{\text{mv}}_{x}$. Thus, its change in momentum is $\text{\Delta}\text{mv}\phantom{\rule{0.25em}{0ex}}\text{= +}{\text{mv}}_{x}\u2013\left(\u2013{\text{mv}}_{x}\right)=2{\text{mv}}_{x}$. The force exerted on the molecule is given by $$F=\frac{\text{\Delta}p}{\text{\Delta}t}=\frac{2{\text{mv}}_{x}}{\text{\Delta}t}\text{.}$$There is no force between the wall and the molecule until the molecule hits the wall. During the short time of the collision, the force between the molecule and wall is relatively large. We are looking for an average force; we take $\text{\Delta}t$ to be the average time between collisions of the molecule with this wall. It is the time it would take the molecule to go across the box and back (a distance $2l)$ at a speed of ${v}_{x}$. Thus $\text{\Delta}t=2l/{v}_{x}$, and the expression for the force becomes $$F=\frac{2{\text{mv}}_{x}}{2l/{v}_{x}}=\frac{{\text{mv}}_{x}^{2}}{l}\text{.}$$This force is due to one molecule. We multiply by the number of molecules $N$ and use their average squared velocity to find the force $$F=N\frac{m\overline{{v}_{x}^{2}}}{l},$$where the bar over a quantity means its average value. We would like to have the force in terms of the speed $v$, rather than the $x$component of the velocity. We note that the total velocity squared is the sum of the squares of its components, so that $$\overline{{v}^{2}}=\overline{{v}_{x}^{2}}+\overline{{v}_{y}^{2}}+\overline{{v}_{z}^{2}}\text{.}$$Because the velocities are random, their average components in all directions are the same: $$\overline{{v}_{x}^{2}}=\overline{{v}_{y}^{2}}=\overline{{v}_{z}^{2}}\text{.}$$Thus, $$\overline{{v}^{2}}=3\overline{{v}_{x}^{2}},$$or $$\overline{{v}_{x}^{2}}=\frac{1}{3}\overline{{v}^{2}}.$$Substituting $\frac{1}{3}\overline{{v}^{2}}$ into the expression for $F$ gives $$F=N\frac{m\overline{{v}^{2}}}{3l}\text{.}$$The pressure is $F/A,$ so that we obtain $$P=\frac{F}{A}=N\frac{m\overline{{v}^{2}}}{3\text{Al}}=\frac{1}{3}\frac{\text{Nm}\overline{{v}^{2}}}{V},$$where we used $V=\text{Al}$ for the volume. This gives the important result. $$\text{PV}=\frac{1}{3}\text{Nm}\overline{{v}^{2}}$$This equation is another expression of the ideal gas law. We can get the average kinetic energy of a molecule, $\frac{1}{2}{\text{mv}}^{2}$, from the lefthand side of the equation by canceling $N$ and multiplying by 3/2. This calculation produces the result that the average kinetic energy of a molecule is directly related to absolute temperature. $$\overline{\text{KE}}=\frac{1}{2}m\overline{{v}^{2}}=\frac{3}{2}\text{kT}$$The average translational kinetic energy of a molecule, $\overline{\text{KE}}$, is called It is sometimes useful to rearrange $\overline{\text{KE}}=\frac{1}{2}m\overline{{v}^{2}}=\frac{3}{2}\text{kT}$, and solve for the average speed of molecules in a gas in terms of temperature, $$\sqrt{\overline{{v}^{2}}}={v}_{\text{rms}}=\sqrt{\frac{3\text{kT}}{m}},$$where ${v}_{\text{rms}}$ stands for rootmeansquare (rms) speed. Example 1: Calculating Kinetic Energy and Speed of a Gas Molecule(a) What is the average kinetic energy of a gas molecule at $\text{20}\text{.}0\text{\xba}\text{C}$ (room temperature)? (b) Find the rms speed of a nitrogen molecule $({\text{N}}_{2})$ at this temperature. Strategy for (a) The known in the equation for the average kinetic energy is the temperature. $$\overline{\text{KE}}=\frac{1}{2}m\overline{{v}^{2}}=\frac{3}{2}\text{kT}$$Before substituting values into this equation, we must convert the given temperature to kelvins. This conversion gives $T=(\text{20}\text{.}0+\text{273})\phantom{\rule{0.25em}{0ex}}\text{K = 293}\phantom{\rule{0.25em}{0ex}}\text{K}.$ Solution for (a) The temperature alone is sufficient to find the average translational kinetic energy. Substituting the temperature into the translational kinetic energy equation gives $$\overline{\text{KE}}=\frac{3}{2}\text{kT}=\frac{3}{2}\left(1\text{.}\text{38}\times {\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K}\right)\left(\text{293}\phantom{\rule{0.25em}{0ex}}\text{K}\right)=6\text{.}\text{07}\times {\text{10}}^{\text{21}}\phantom{\rule{0.25em}{0ex}}\text{J}\text{.}$$Strategy for (b) Finding the rms speed of a nitrogen molecule involves a straightforward calculation using the equation $$\sqrt{\overline{{v}^{2}}}={v}_{\text{rms}}=\sqrt{\frac{3\text{kT}}{m}},$$but we must first find the mass of a nitrogen molecule. Using the molecular mass of nitrogen ${\text{N}}_{2}$ from the periodic table, $$m=\frac{2\left(\text{14}\text{.}\text{0067}\right)\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{kg/mol}}{6\text{.}\text{02}\times {\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}{\text{mol}}^{1}}=4\text{.}\text{65}\times {\text{10}}^{\text{26}}\phantom{\rule{0.25em}{0ex}}\text{kg}\text{.}$$Solution for (b) Substituting this mass and the value for $k$ into the equation for ${v}_{\text{rms}}$ yields $${v}_{\text{rms}}=\sqrt{\frac{3\text{kT}}{m}}=\sqrt{\frac{3\left(1\text{.}\text{38}\times {\text{10}}^{\u2013\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K}\right)\left(\text{293 K}\right)}{4\text{.}\text{65}\times {\text{10}}^{\text{\u201326}}\phantom{\rule{0.25em}{0ex}}\text{kg}}}=\text{511}\phantom{\rule{0.25em}{0ex}}\text{m/s}\text{.}$$Discussion Note that the average kinetic energy of the molecule is independent of the type of molecule. The average translational kinetic energy depends only on absolute temperature. The kinetic energy is very small compared to macroscopic energies, so that we do not feel when an air molecule is hitting our skin. The rms velocity of the nitrogen molecule is surprisingly large. These large molecular velocities do not yield macroscopic movement of air, since the molecules move in all directions with equal likelihood. The mean free path (the distance a molecule can move on average between collisions) of molecules in air is very small, and so the molecules move rapidly but do not get very far in a second. The high value for rms speed is reflected in the speed of sound, however, which is about 340 m/s at room temperature. The faster the rms speed of air molecules, the faster that sound vibrations can be transferred through the air. The speed of sound increases with temperature and is greater in gases with small molecular masses, such as helium. (See Figure 3.) Making Connections: Historical Note—Kinetic Theory of Gases:The kinetic theory of gases was developed by Daniel Bernoulli (1700–1782), who is best known in physics for his work on fluid flow (hydrodynamics). Bernoulli’s work predates the atomistic view of matter established by Dalton. Distribution of Molecular SpeedsThe motion of molecules in a gas is random in magnitude and direction for individual molecules, but a gas of many molecules has a predictable distribution of molecular speeds. This distribution is called the MaxwellBoltzmann distribution, after its originators, who calculated it based on kinetic theory, and has since been confirmed experimentally. (See Figure 4.) The distribution has a long tail, because a few molecules may go several times the rms speed. The most probable speed ${v}_{\text{p}}$ is less than the rms speed ${v}_{\text{rms}}$. Figure 5 shows that the curve is shifted to higher speeds at higher temperatures, with a broader range of speeds. The distribution of thermal speeds depends strongly on temperature. As temperature increases, the speeds are shifted to higher values and the distribution is broadened. What is the implication of the change in distribution with temperature shown in Figure 5 for humans? All other things being equal, if a person has a fever, he or she is likely to lose more water molecules, particularly from linings along moist cavities such as the lungs and mouth, creating a dry sensation in the mouth. Example 2: Calculating Temperature: Escape Velocity of Helium AtomsIn order to escape Earth’s gravity, an object near the top of the atmosphere (at an altitude of 100 km) must travel away from Earth at 11.1 km/s. This speed is called the escape velocity. At what temperature would helium atoms have an rms speed equal to the escape velocity? Strategy Identify the knowns and unknowns and determine which equations to use to solve the problem. Solution 1. Identify the knowns: $v$ is the escape velocity, 11.1 km/s. 2. Identify the unknowns: We need to solve for temperature, $T$. We also need to solve for the mass $m$ of the helium atom. 3. Determine which equations are needed.
4. Plug the known values into the equations and solve for the unknowns. $$\begin{array}{lll}m& =& \frac{\text{molar mass}}{\text{number of atoms per mole}}=\frac{4\text{.}\text{0026}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{kg/mol}}{6\text{.}\text{02}\times {\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}\text{mol}}\\ & =& 6\text{.}\text{65}\times {\text{10}}^{\text{27}}\phantom{\rule{0.25em}{0ex}}\text{kg}\end{array}$$ $$T=\frac{\left(6\text{.}\text{65}\times {\text{10}}^{\text{27}}\phantom{\rule{0.25em}{0ex}}\text{kg}\right){\left(\text{11}\text{.}1\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)}^{2}}{3\left(1\text{.}\text{38}\times {\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K}\right)}=1\text{.}\text{98}\times {\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{K}$$Discussion This temperature is much higher than atmospheric temperature, which is approximately 250 K $(\u2013\text{25}\text{\xba}\text{C}$ or $\u2013\text{10}\text{\xba}\text{F})$ at high altitude. Very few helium atoms are left in the atmosphere, but there were many when the atmosphere was formed. The reason for the loss of helium atoms is that there are a small number of helium atoms with speeds higher than Earth’s escape velocity even at normal temperatures. The speed of a helium atom changes from one instant to the next, so that at any instant, there is a small, but nonzero chance that the speed is greater than the escape speed and the molecule escapes from Earth’s gravitational pull. Heavier molecules, such as oxygen, nitrogen, and water (very little of which reach a very high altitude), have smaller rms speeds, and so it is much less likely that any of them will have speeds greater than the escape velocity. In fact, so few have speeds above the escape velocity that billions of years are required to lose significant amounts of the atmosphere. Figure 6 shows the impact of a lack of an atmosphere on the Moon. Because the gravitational pull of the Moon is much weaker, it has lost almost its entire atmosphere. The comparison between Earth and the Moon is discussed in this chapter’s Problems and Exercises. Check Your UnderstandingIf you consider a very small object such as a grain of pollen, in a gas, then the number of atoms and molecules striking its surface would also be relatively small. Would the grain of pollen experience any fluctuations in pressure due to statistical fluctuations in the number of gas atoms and molecules striking it in a given amount of time? Show/Hide Solution SolutionYes. Such fluctuations actually occur for a body of any size in a gas, but since the numbers of atoms and molecules are immense for macroscopic bodies, the fluctuations are a tiny percentage of the number of collisions, and the averages spoken of in this section vary imperceptibly. Roughly speaking the fluctuations are proportional to the inverse square root of the number of collisions, so for small bodies they can become significant. This was actually observed in the 19th century for pollen grains in water, and is known as the Brownian effect. PhET Explorations: Gas Properties:Pump gas molecules into a box and see what happens as you change the volume, add or remove heat, change gravity, and more. Measure the temperature and pressure, and discover how the properties of the gas vary in relation to each other. Section Summary
Conceptual QuestionsExercise 1How is momentum related to the pressure exerted by a gas? Explain on the atomic and molecular level, considering the behavior of atoms and molecules. Problems & ExercisesExercise 1Some incandescent light bulbs are filled with argon gas. What is ${v}_{\text{rms}}$ for argon atoms near the filament, assuming their temperature is 2500 K? Show/Hide Solution Solution$1\text{.}\text{25}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ Exercise 2Average atomic and molecular speeds $({v}_{\text{rms}})$ are large, even at low temperatures. What is ${v}_{\text{rms}}$ for helium atoms at 5.00 K, just one degree above helium’s liquefaction temperature? Exercise 3(a) What is the average kinetic energy in joules of hydrogen atoms on the $\text{5500}\text{\xba}\text{C}$ surface of the Sun? (b) What is the average kinetic energy of helium atoms in a region of the solar corona where the temperature is $6\text{.}\text{00}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{K}$? Show/Hide Solution Solution(a) $1\text{.}\text{20}\times {\text{10}}^{\text{19}}\phantom{\rule{0.25em}{0ex}}\text{J}$ (b) $1\text{.}\text{24}\times {\text{10}}^{\text{17}}\phantom{\rule{0.25em}{0ex}}\text{J}$ Exercise 4The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would oxygen molecules (molecular mass is equal to 32.0 g/mol) have an average velocity ${v}_{\text{rms}}$ equal to Earth’s escape velocity of 11.1 km/s? Exercise 5The escape velocity from the Moon is much smaller than from Earth and is only 2.38 km/s. At what temperature would hydrogen molecules (molecular mass is equal to 2.016 g/mol) have an average velocity ${v}_{\text{rms}}$ equal to the Moon’s escape velocity? Show/Hide Solution Solution$\text{458}\phantom{\rule{0.25em}{0ex}}\text{K}$ Exercise 6Nuclear fusion, the energy source of the Sun, hydrogen bombs, and fusion reactors, occurs much more readily when the average kinetic energy of the atoms is high—that is, at high temperatures. Suppose you want the atoms in your fusion experiment to have average kinetic energies of $6\text{.}\text{40}\times {\text{10}}^{\u2013\text{14}}\phantom{\rule{0.25em}{0ex}}\text{J}$. What temperature is needed? Exercise 7Suppose that the average velocity $({v}_{\text{rms}})$ of carbon dioxide molecules (molecular mass is equal to 44.0 g/mol) in a flame is found to be $1\text{.}\text{05}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{m/s}$. What temperature does this represent? Show/Hide Solution Solution$1\text{.}\text{95}\times {\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{K}$ Exercise 8Hydrogen molecules (molecular mass is equal to 2.016 g/mol) have an average velocity ${v}_{\text{rms}}$ equal to 193 m/s. What is the temperature? Exercise 9Much of the gas near the Sun is atomic hydrogen. Its temperature would have to be $1\text{.}5\times {\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{K}$ for the average velocity ${v}_{\text{rms}}$ to equal the escape velocity from the Sun. What is that velocity? Show/Hide Solution Solution$6\text{.}\text{09}\times {\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{m/s}$ Exercise 10There are two important isotopes of uranium— ${}^{\text{235}}\text{U}$ and ${}^{\text{238}}\text{U}$; these isotopes are nearly identical chemically but have different atomic masses. Only ${}^{\text{235}}\text{U}$ is very useful in nuclear reactors. One of the techniques for separating them (gas diffusion) is based on the different average velocities ${v}_{\text{rms}}$ of uranium hexafluoride gas, ${\text{UF}}_{6}$. (a) The molecular masses for ${}^{\text{235}}\text{U}\phantom{\rule{0.25em}{0ex}}$${\text{UF}}_{6}$ and ${}^{\text{238}}\text{U}$$\phantom{\rule{0.25em}{0ex}}{\text{UF}}_{6}$ are 349.0 g/mol and 352.0 g/mol, respectively. What is the ratio of their average velocities? (b) At what temperature would their average velocities differ by 1.00 m/s? (c) Do your answers in this problem imply that this technique may be difficult?
