Contents

Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature

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.

A green vector v, representing a molecule colliding with a wall, is pointing at the surface of a wall at an angle. A second vector v primed starts at the point of impact and travels away from the wall at an angle. A dotted line perpendicular to the wall through the point of impact represents the component of the molecule’s momentum that is perpendicular to the wall. A red vector F is pointing into the wall from the point of impact, representing the force of the molecule hitting the wall.

Figure 13.20 When a molecule collides with a rigid wall, the component of its momentum perpendicular to the wall is reversed. A force is thus exerted on the wall, creating pressure.

Figure 13.20 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:

PV = \frac{1}{3} Nm \bar{v^{2}}, PV = \frac{1}{3} Nm \bar{v^{2}}, size 12{ ital "PV"= {  {1}  over  {3} }  ital "Nm" {overline  {v rSup { size 8{2} } }} ,} {}

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where $P P size 12{P} {}$ is the pressure (average force per unit area), $V V size 12{V} {}$ is the volume of gas in the container, $N N size 12{N} {}$ is the number of molecules in the container, $m m size 12{m} {}$ is the mass of a molecule, and $\bar{v^{2}} \bar{v^{2}} size 12{ {overline {v rSup { size 8{2} } }} } {}$ 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:

PV = NkT . PV = NkT . size 12{ ital "PV"= ital "NkT"} {}

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Equating the right-hand side of this equation with the right-hand side of $PV = \frac{1}{3} Nm \bar{v^{2}} PV = \frac{1}{3} Nm \bar{v^{2}} size 12{ ital "PV"= { {1} over {3} } ital "Nm" {overline {v rSup { size 8{2} } }} } {}$ gives

\frac{1}{3} Nm \bar{v^{2}} = NkT . \frac{1}{3} Nm \bar{v^{2}} = NkT . size 12{ {  {1}  over  {3} }  ital "Nm" {overline  {v rSup { size 8{2} } }} = ital "NkT"} {}

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Making Connections: Things Great and Small—Atomic and Molecular Origin of Pressure in a Gas

Figure 13.21 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.

Diagram representing the pressures that a gas exerts on the walls of a box in a three-dimensional coordinate system with x, y, and z components.

Figure 13.21 Gas in a box exerts an outward pressure on its walls. A molecule colliding with a rigid wall has the direction of its velocity and momentum in the

x x size 12{x} {}

-direction reversed. This direction is perpendicular to the wall. The components of its velocity momentum in the

y y size 12{y} {}

- and

z z size 12{z} {}

-directions are not changed, which means there is no force parallel to the wall.

If the molecule’s velocity changes in the $x x size 12{x} {}$ -direction, its momentum changes from $- {mv}_{x} - {mv}_{x} size 12{– ital "mv" rSub { size 8{x} } } {}$ to $+ {mv}_{x} + {mv}_{x} size 12{+ ital "mv" rSub { size 8{x} } } {}$ . Thus, its change in momentum is $Δ mv = + {mv}_{x} - (- {mv}_{x}) = 2 {mv}_{x} Δ mv size 12{Δ ital "mv""=+" ital "mv" rSub { size 8{x} } – left (– ital "mv" rSub { size 8{x} } right )=2 ital "mv" rSub { size 8{x} } } {}$ . The force exerted on the molecule is given by

F = \frac{Δ p}{Δ t} = \frac{2 {mv}_{x}}{Δ t} . F = \frac{Δ p}{Δ t} = \frac{2 {mv}_{x}}{Δ t} . size 12{F= {  {Δp}  over  {Δt} } = {  {2 ital "mv" rSub { size 8{x} } }  over  {Δt} }  "." } {}

13.45

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 $Δ t Δ t size 12{Dt} {}$ 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 $2 l) 2 l) size 12{2l \) } {}$ at a speed of $v_{x} v_{x} size 12{v rSub { size 8{x} } } {}$ . Thus $Δ t = 2 l / v_{x} Δ t = 2 l / v_{x} size 12{Δt=2l/v rSub { size 8{x} } } {}$ , and the expression for the force becomes

F = \frac{2 {mv}_{x}}{2 l / v_{x}} = \frac{{mv}_{x}^{2}}{l} . F = \frac{2 {mv}_{x}}{2 l / v_{x}} = \frac{{mv}_{x}^{2}}{l} . size 12{F= {  {2 ital "mv" rSub { size 8{x} } }  over  { {2l} slash {v rSub { size 8{x} } } } } = {  { ital "mv" rSub { size 8{x} }  rSup { size 8{2} } }  over  {l} }  "." } {}

13.46

This force is due to one molecule. We multiply by the number of molecules $N N size 12{N} {}$ and use their average squared velocity to find the force

F = N \frac{m \bar{v_{x}^{2}}}{l}, F = N \frac{m \bar{v_{x}^{2}}}{l}, size 12{F=N {  {m {overline  {v rSub { size 8{x} }  rSup { size 8{2} } }} }  over  {l} } ,} {}

13.47

where the bar over a quantity means its average value. We would like to have the force in terms of the speed $v v size 12{v} {}$ , rather than the $x x size 12{x} {}$ -component of the velocity. We note that the total velocity squared is the sum of the squares of its components, so that

\bar{v^{2}} = \bar{v_{x}^{2}} + \bar{v_{y}^{2}} + \bar{v_{z}^{2}} . \bar{v^{2}} = \bar{v_{x}^{2}} + \bar{v_{y}^{2}} + \bar{v_{z}^{2}} . size 12{ {overline  {v rSup { size 8{2} } }} = {overline  {v rSub { size 8{x} }  rSup { size 8{2} } }} + {overline  {v rSub { size 8{y} }  rSup { size 8{2} } }} + {overline  {v rSub { size 8{z} }  rSup { size 8{2} } }}  "." } {}

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Because the velocities are random, their average components in all directions are the same:

\bar{v_{x}^{2}} = \bar{v_{y}^{2}} = \bar{v_{z}^{2}} . \bar{v_{x}^{2}} = \bar{v_{y}^{2}} = \bar{v_{z}^{2}} . size 12{ {overline  {v rSub { size 8{x} }  rSup { size 8{2} } }} = {overline  {v rSub { size 8{y} }  rSup { size 8{2} } }} = {overline  {v rSub { size 8{z} }  rSup { size 8{2} } }}  "." } {}

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Thus,

\bar{v^{2}} = 3 \bar{v_{x}^{2}}, \bar{v^{2}} = 3 \bar{v_{x}^{2}}, size 12{ {overline  {v rSup { size 8{2} } }} =3 {overline  {v rSub { size 8{x} }  rSup { size 8{2} } }} ,} {}

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\bar{v_{x}^{2}} = \frac{1}{3} \bar{v^{2}} . \bar{v_{x}^{2}} = \frac{1}{3} \bar{v^{2}} . size 12{ {overline  {v rSub { size 8{x} }  rSup { size 8{2} } }} = {  {1}  over  {3} }  {overline  {v rSup { size 8{2} } }} } {}

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Substituting $\frac{1}{3} \bar{v^{2}} \frac{1}{3} \bar{v^{2}} size 12{ { {1} over {3} } {overline {v rSup { size 8{2} } }} } {}$ into the expression for $F F size 12{F} {}$ gives

F = N \frac{m \bar{v^{2}}}{3 l} . F = N \frac{m \bar{v^{2}}}{3 l} . size 12{F=N {  {m {overline  {v rSup { size 8{2} } }} }  over  {3l} }  "." } {}

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The pressure is $F / A, F / A, size 12{F/A,} {}$ so that we obtain

P = \frac{F}{A} = N \frac{m \bar{v^{2}}}{3 Al} = \frac{1}{3} \frac{Nm \bar{v^{2}}}{V}, P = \frac{F}{A} = N \frac{m \bar{v^{2}}}{3 Al} = \frac{1}{3} \frac{Nm \bar{v^{2}}}{V}, size 12{P= {  {F}  over  {A} } =N {  {m {overline  {v rSup { size 8{2} } }} }  over  {3 ital "Al"} } = {  {1}  over  {3} }  {  { ital "Nm" {overline  {v rSup { size 8{2} } }} }  over  {V} } ,} {}

13.53

where we used $V = Al V = Al size 12{V= ital "Al"} {}$ for the volume. This gives the important result.

PV = \frac{1}{3} Nm \bar{v^{2}} PV = \frac{1}{3} Nm \bar{v^{2}} size 12{ ital "PV"= {  {1}  over  {3} }  ital "Nm" {overline  {v rSup { size 8{2} } }} } {}

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This equation is another expression of the ideal gas law.

We can get the average kinetic energy of a molecule, $\frac{1}{2} {mv}^{2} \frac{1}{2} {mv}^{2} size 12{ { { size 8{1} } over { size 8{2} } } ital "mv" rSup { size 8{2} } } {}$ , from the right-hand side of the equation by canceling $N N size 12{N} {}$ and multiplying by 3/2. This calculation produces the result that the average kinetic energy of a molecule is directly related to absolute temperature.

\bar{KE} = \frac{1}{2} m \bar{v^{2}} = \frac{3}{2} kT \bar{KE} = \frac{1}{2} m \bar{v^{2}} = \frac{3}{2} kT size 12{ {overline  {"KE"}} = {  {1}  over  {2} } m {overline  {v rSup { size 8{2} } }} = {  {3}  over  {2} }  ital "kT"} {}

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The average translational kinetic energy of a molecule, $\bar{KE} \bar{KE} size 12{ {overline {"KE"}} } {}$ , is called thermal energy. The equation $\bar{KE} = \frac{1}{2} m \bar{v^{2}} = \frac{3}{2} kT \bar{KE} = \frac{1}{2} m \bar{v^{2}} = \frac{3}{2} kT size 12{ {overline { size 11{"KE"}}} = { {1} over {2} } m {overline { size 11{v rSup { size 8{2} } }}} = { {3} over {2} } ital "kT"} {}$ is a molecular interpretation of temperature, and it has been found to be valid for gases and reasonably accurate in liquids and solids. It is another definition of temperature based on an expression of the molecular energy.

It is sometimes useful to rearrange $\bar{KE} = \frac{1}{2} m \bar{v^{2}} = \frac{3}{2} kT \bar{KE} = \frac{1}{2} m \bar{v^{2}} = \frac{3}{2} kT size 12{ {overline { size 11{"KE"}}} = { {1} over {2} } m {overline { size 11{v rSup { size 8{2} } }}} = { {3} over {2} } ital "kT"} {}$ , and solve for the average speed of molecules in a gas in terms of temperature,

\sqrt{\bar{v^{2}}} = v_{rms} = \sqrt{\frac{3 kT}{m}}, \sqrt{\bar{v^{2}}} = v_{rms} = \sqrt{\frac{3 kT}{m}}, size 12{ sqrt { {overline  {v rSup { size 8{2} } }} } =v rSub { size 8{"rms"} } = sqrt { {  {3 ital "kT"}  over  {m} } } ,} {}

13.56

where $v_{rms} v_{rms} size 12{v rSub { size 8{"rms"} } } {}$ stands for root-mean-square (rms) speed.

Example 13.10

Calculating Kinetic Energy and Speed of a Gas Molecule

(a) What is the average kinetic energy of a gas molecule at $20.0 º C 20.0 º C size 12{"20" "." 0°C} {}$ (room temperature)? (b) Find the rms speed of a nitrogen molecule $(N_{2}) (N_{2}) size 12{ $ N rSub { size 8{2} } $ } {}$ at this temperature.

Strategy for (a)

The known in the equation for the average kinetic energy is the temperature.

\bar{KE} = \frac{1}{2} m \bar{v^{2}} = \frac{3}{2} kT \bar{KE} = \frac{1}{2} m \bar{v^{2}} = \frac{3}{2} kT size 12{ {overline  {"KE"}} = {  {1}  over  {2} } m {overline  {v rSup { size 8{2} } }} = {  {3}  over  {2} }  ital "kT"} {}

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Before substituting values into this equation, we must convert the given temperature to kelvins. This conversion gives $T = (20 . 0 + 273) K = 293 K . T = (20 . 0 + 273) size 12{T= $ "20" "." 0+"273" $ " K=293 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

\bar{KE} = \frac{3}{2} kT = \frac{3}{2} (1 . 38 \times 10^{- 23} J/K) (293 K) = 6 . 07 \times 10^{- 21} J . \bar{KE} = \frac{3}{2} kT = \frac{3}{2} (1 . 38 \times 10^{- 23}) (293) = 6 . 07 \times 10^{- 21} size 12{ {overline  {"KE"}} = {  {3}  over  {2} }  ital "kT"= {  {3}  over  {2} }  left (1 "." "38" times "10" rSup { size 8{ - "23"} } " J/K" right ) left ("293"" K" right )=6 "." "07" times "10" rSup { size 8{ - "21"} } `J "." } {}

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Strategy for (b)

Finding the rms speed of a nitrogen molecule involves a straightforward calculation using the equation

\sqrt{\bar{v^{2}}} = v_{rms} = \sqrt{\frac{3 kT}{m}}, \sqrt{\bar{v^{2}}} = v_{rms} = \sqrt{\frac{3 kT}{m}}, size 12{ sqrt { {overline  {v rSup { size 8{2} } }} } =v rSub { size 8{"rms"} } = sqrt { {  {3 ital "kT"}  over  {m} } } ,} {}

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but we must first find the mass of a nitrogen molecule. Using the molecular mass of nitrogen $N_{2} N_{2} size 12{N rSub { size 8{2} } } {}$ from the periodic table,

m = \frac{2 (14.0067) \times 10^{- 3} kg/mol}{6 . 02 \times 10^{23} {mol}^{- 1}} = 4 . 65 \times 10^{- 26} kg . m = \frac{2 (14.0067) \times 10^{- 3}}{6 . 02 \times 10^{23}} = 4 . 65 \times 10^{- 26} size 12{m= {  {2 left ("14" "." "0067" right ) times "10" rSup { size 8{ - 3} } `"kg/mol"}  over  {6 "." "02" times "10" rSup { size 8{"23"} } `"mol" rSup { size 8{ - 1} } } } =4 "." "65" times "10" rSup { size 8{ - "26"} } `"kg" "." } {}

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Solution for (b)

Substituting this mass and the value for $k k size 12{k} {}$ into the equation for $v_{rms} v_{rms} size 12{v rSub { size 8{"rms"} } } {}$ yields

v_{rms} = \sqrt{\frac{3 kT}{m}} = \sqrt{\frac{3 (1 . 38 \times 10^{- 23} J/K) (293 K)}{4 . 65 \times 10^{–26} kg}} = 511 m/s . v_{rms} = \sqrt{\frac{3 kT}{m}} = \sqrt{\frac{3 (1 . 38 \times 10^{- 23}) (293 K)}{4 . 65 \times 10^{–26}}} = 511 size 12{v rSub { size 8{"rms"} } = sqrt { {  {3 ital "kT"}  over  {m} } } = sqrt { {  {3 left (1 "." "38" times "10" rSup { size 8{–"23"} } " J/K" right ) left ("293 K" right )}  over  {4 "." "65" times "10" rSup { size 8{"–26"} } " kg"} } } ="511"" m/s" "." } {}

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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 13.22.)

In part a of the figure, circles represent molecules distributed in a gas. Attached to each circle is a vector representing velocity. The circles have a random arrangement, while the vector arrows have random orientations and lengths. In part b of the figure, an arc represents a sound wave as it passes through a gas. The velocity of each molecule along the peak of the wave is roughly oriented parallel to the transmission direction of the wave.

Figure 13.22 (a) There are many molecules moving so fast in an ordinary gas that they collide a billion times every second. (b) Individual molecules do not move very far in a small amount of time, but disturbances like sound waves are transmitted at speeds related to the molecular speeds.

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 Speeds

The 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 Maxwell-Boltzmann distribution, after its originators, who calculated it based on kinetic theory, and has since been confirmed experimentally. (See Figure 13.23.) The distribution has a long tail, because a few molecules may go several times the rms speed. The most probable speed $v_{p} v_{p} size 12{v rSub { size 8{p} } } {}$ is less than the rms speed $v_{rms} v_{rms} size 12{v rSub { size 8{"rms"} } } {}$ . Figure 13.24 shows that the curve is shifted to higher speeds at higher temperatures, with a broader range of speeds.

A line graph of probability versus velocity in meters per second of oxygen gas at 300 kelvin. The graph is skewed to the right, with a peak probability just under 400 meters per second and a root-mean-square probability of about 500 meters per second.

Figure 13.23 The Maxwell-Boltzmann distribution of molecular speeds in an ideal gas. The most likely speed

v_{p} v_{p} size 12{v rSub { size 8{p} } } {}

is less than the rms speed

v_{rms} v_{rms} size 12{v rSub { size 8{"rms"} } } {}

. Although very high speeds are possible, only a tiny fraction of the molecules have speeds that are an order of magnitude greater than

v_{rms} v_{rms} size 12{v rSub { size 8{"rms"} } } {}

The distribution of thermal speeds depends strongly on temperature. As temperature increases, the speeds are shifted to higher values and the distribution is broadened.

Two distributions of probability versus velocity at two different temperatures plotted on the same graph. Temperature two is greater than Temperature one. The distribution for Temperature two has a peak with a lower probability, but a higher velocity than the distribution for Temperature one. The T sub two graph has a more normal distribution and is broader while the T sub one graph is more narrow and has a tail extending to the right.

Figure 13.24 The Maxwell-Boltzmann distribution is shifted to higher speeds and is broadened at higher temperatures.

What is the implication of the change in distribution with temperature shown in Figure 13.24 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 13.11

Calculating Temperature: Escape Velocity of Helium Atoms

In 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: $v v size 12{v} {}$ is the escape velocity, 11.1 km/s.

2. Identify the unknowns: We need to solve for temperature, $T T size 12{T} {}$ . We also need to solve for the mass $m m size 12{m} {}$ of the helium atom.

3. Determine which equations are needed.

To solve for mass $m m size 12{m} {}$ of the helium atom, we can use information from the periodic table:
$m = \frac{molar mass}{number of atoms per mole} . m = \frac{molar mass}{number of atoms per mole} . size 12{m= { { size 11{"molar mass"}} over { size 11{"number of atoms per mole"}} } } {}$
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To solve for temperature $T T size 12{T} {}$ , we can rearrange either
$\bar{KE} = \frac{1}{2} m \bar{v^{2}} = \frac{3}{2} kT \bar{KE} = \frac{1}{2} m \bar{v^{2}} = \frac{3}{2} kT size 12{ {overline {"KE"}} = { {1} over {2} } m {overline {v rSup { size 8{2} } }} = { {3} over {2} } ital "kT"} {}$
13.63

or

$\sqrt{\bar{v^{2}}} = v_{rms} = \sqrt{\frac{3 kT}{m}} \sqrt{\bar{v^{2}}} = v_{rms} = \sqrt{\frac{3 kT}{m}} size 12{ sqrt { {overline {v rSup { size 8{2} } }} } =v rSub { size 8{"rms"} } = sqrt { { {3 ital "kT"} over {m} } } } {}$
13.64

to yield

$T = \frac{m \bar{v^{2}}}{3 k}, T = \frac{m \bar{v^{2}}}{3 k}, size 12{T= { {m {overline {v rSup { size 8{2} } }} } over {3k} } ,} {}$
13.65

where $k k size 12{k} {}$ is the Boltzmann constant and $m m size 12{m} {}$ is the mass of a helium atom.

4. Plug the known values into the equations and solve for the unknowns.

m = \frac{molar mass}{number of atoms per mole} = \frac{4 . 0026 \times 10^{- 3} kg/mol}{6 . 02 \times 10^{23} mol} = 6 . 65 \times 10^{- 27} kg m = \frac{molar mass}{number of atoms per mole} = \frac{4 . 0026 \times 10^{- 3}}{6 . 02 \times 10^{23}} = 6 . 65 \times 10^{- 27} size 12{m= {  { size 11{"molar mass"}}  over  { size 11{"number of atoms per mole"}} } = {  { size 11{4 "." "0026" times "10" rSup { size 8{ - 3} } " kg/mol"}}  over  { size 12{6 "." "02" times "10" rSup { size 8{"23"} } " mol"} } } =6 "." "65" times "10" rSup { size 8{ - "27"} } " kg"} {}

13.66

T = \frac{(6 . 65 \times 10^{- 27} kg) {(11 . 1 \times 10^{3} m/s)}^{2}}{3 (1 . 38 \times 10^{- 23} J/K)} = 1 . 98 \times 10^{4} K T = \frac{(6 . 65 \times 10^{- 27}) {(11 . 1 \times 10^{3})}^{2}}{3 (1 . 38 \times 10^{- 23})} = 1 . 98 \times 10^{4} size 12{T= {  { left (6 "." "65" times "10" rSup { size 8{ - "27"} } `"kg" right ) left ("11" "." 1 times "10" rSup { size 8{3} } `"m/s" right ) rSup { size 8{2} } }  over  {3 left (1 "." "38" times "10" rSup { size 8{ - "23"} } `"J/K" right )} } =1 "." "98" times "10" rSup { size 8{4} } `K} {}

13.67

Discussion

This temperature is much higher than atmospheric temperature, which is approximately 250 K $(- 25 º C (- 25 º C size 12{ $ –"25"°C} {}$ or $- 10 º F) - 10 º F) size 12{–"10"°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 13.25 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.

Photograph of the lunar rover on the Moon. The photo looks like it was taken at night with a powerful spotlight shining on the rover from the left: light reflects off the rover, the astronaut, and the Moon’s surface, but the sky is black. The shadow of the rover is very sharp.

Figure 13.25 This photograph of Apollo 17 Commander Eugene Cernan driving the lunar rover on the Moon in 1972 looks as though it was taken at night with a large spotlight. In fact, the light is coming from the Sun. Because the acceleration due to gravity on the Moon is so low (about 1/6 that of Earth), the Moon’s escape velocity is much smaller. As a result, gas molecules escape very easily from the Moon, leaving it with virtually no atmosphere. Even during the daytime, the sky is black because there is no gas to scatter sunlight. (credit: Harrison H. Schmitt/NASA)

Check Your Understanding

If 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 solution

Solution

Yes. 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.

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