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9.3The Boltzmann equation

The molecules of a gas move about randomly with a range of speeds. The graph in Figure 9.21 shows the distribution of speeds for oxygen molecules kept at two different temperatures: the blue curve is at 100 K and the red curve at 300 K. The vertical axis shows the fraction of molecules having a given speed v. You will not be examined on this graph but knowing a few of its features helps a lot in understanding how gases behave.

Figure 9.21: The distribution of speeds at two different temperatures.

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We see that there is a speed that corresponds to the peak of the curve. For the blue curve this is about 200 m s−1 and for the red curve it is 400 m s−1. The speed at the peak represents the most probable speed that would be found if you picked a molecule at random. One other speed is important:

The root mean square (rms) speed c, is the square root of the average of the squares of the speeds of the molecules; that is:

c=v 1 2+v 2 2+v 3 2+⋯+v N 2 N

Why do we bother to work with an rms speed? Consider the average kinetic energy for the N molecules, which is given by:

E¯K=1 2 m v 1 2+1 2 m v 2 2+1 2 m v 3 2+⋯+1 2 m v N 2 N=1 2 m(v 1 2+v 2 2+v 3 2+⋯+v N 2 N)=1 2 m c 2

So we see that the average kinetic energy involves the rms speed. These two speeds (most probable and rms speed as well as the average speed) are all different but numerically close to each other. So, even though it is not technically correct, we may assume that all three speeds mean the same thing and we will use the symbol c for all of them.

Worked example 9.12

A flask contains a gas at a temperature of 300 K. An identical flask is aboard a fast-moving aircraft. The temperature inside the aircraft is 300 K. Suggest how the rms speeds of the two gases compare.

Answer

The temperature of the gas depends on the random motion of the molecules and not on any additional uniform motion imposed on the gas as a result of the motion of the container. The speeds are the same.

We will now show that the pressure of a gas can be written as P=1 3 ρ c 2, where c stands for the rms speed and ρ is the density of the gas. You do not need to recall the proof of this equation for exam purposes.

Figure 9.22 shows a molecule of mass m moving with horizontal velocity v x. The molecule collides with the walls elastically.

Figure 9.22: A molecule moves back and forth between the walls exerting a force at every collision.

The magnitude of the impulse during a collision with the right wall is:

J=m v x−(−m v x)=2 m v x

The molecule will hit the right wall again after a time:

T=2 L v

We need to find the average force over a time interval long compared to T, say a time interval NT, where N is large. In this interval the total impulse is NJ and so the average force is N J N T i.e. J T.

J T=2 m v x 2 L v=m v x 2 L

The pressure from one molecule is then (we assume the container is a cube):

P 1=m v x 2 L L 2=m v x 2 L 3=m v x 2 V

There are N molecules in the container so the total pressure becomes:

P=m v 1 x 2 V+m v 2 x 2 V+⋯+m v N x 2 V=m V(v 1 x 2+v 2 x 2+⋯+v N x 2).

The average of the squares of the x components of velocity is:

v x 2―=v 1 x 2+v 2 x 2+⋯+v N x 2 N

and so P=m N V v x 2―.

By symmetry, v x 2―=v y 2―=v z 2― (all directions are equivalent)

And since v x 2―+v y 2―+v z 2―=c 2, it follows that v x 2―=c 2 3.

Hence the pressure is:

P=1 3 N m c 2 V

But Nm is the total mass M of the gas and so finally we get (ρ=M V):

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P=1 3 ρ c 2

Pressure is related to the rms speed and the density of the gas.

Links

Notice how the ideas we studied in Chapter 4 are also heavily used here but in a different context.

We get a very interesting result if we combine this equation with the ideal gas law. In what follows, N stands for the number of molecules and m for the mass of one molecule.

Start with the ideal gas law:

P V=N k B T

replace pressure with P=1 3 ρ c 2 to get:

(1 3 ρ c 2)V=N k B T

replace density with mass volume to get:

1 3(N m)c 2 V V=N k B T

multiply both sides by 3 2 to get:

1 2 m c 2=3 2 k B T

(Recall that the Boltzmann constant is k B=R N A=8.31 6.02×10 23=1.38×10−23 J K−1.)

The last equation states that the average random kinetic energy of the particles is directly proportional to the kelvin temperature:

E¯K=3 2 k B T

This exceptional equation relates a microscopic quantity, the average kinetic energy of molecules, to a macroscopic quantity, temperature. It also says that at T = 0, all molecular motion stops.

Check yourself 4

When the volume of a fixed quantity of an ideal gas is decreased at constant temperature, the pressure will increase. Three reasons are proposed for this increase:

athe molecules collide with each other more frequently

bthe molecules collide with the walls more frequently

cthe molecules are moving faster on average

Which is/are the correct reason(s) for the pressure increase?

Using this equation we can find an expression for the internal energy of an ideal gas. Remember that the internal energy of an ideal gas consists only of the random kinetic energy of its molecules and zero potential energy. Suppose that the gas has N molecules. Then, since the average kinetic energy is 3 2 k B T, the total random kinetic energy, i.e. the internal energy U, is:

U=3 2 N k B T

But recall that k B=R N A, so that another expression is:

U=3 2 n R T

Yet another expression comes from using the ideal gas law, PV = nRT, which gives:

U=3 2 P V

Exam tip

You must be able to obtain an expression for the internal energy of an ideal gas.

Check yourself 5

Which has a greater internal energy: 40 g of helium (molar mass 4 g mol−1) or 40 g of argon (molar mass 40 g mol−1)? Both gases are at the same temperature.

Worked example 9.13

The kelvin temperature of a gas is doubled. By what factor does the average speed increase?

Answer

From 1 2 m c 2=3 2 k B T we find that when T is doubled then c 2 will double, so c itself will increase by a factor of 2.

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Worked example 9.14

Calculate the ratio of the average speed of helium (molar mass 4.0 g mol−1) to argon (molar mass 40 g mol−1) atoms when both gases are at the same temperature.

Answer

Since the temperature is the same for both gases, using 1 2 m c 2=3 2 k B T we find that:

1 2 m He c He 2=1 2 m Ar c Ar 2 and so c He 2 c Ar 2=m Ar m He

So, we need to find the ratio of the masses of the molecules.

One mole of helium has a mass of 4 g so one molecule has a mass (in grams) of 4 N A.

Similarly, the mass in grams of an argon molecule is 40 N A.

So, c He 2 c Ar 2=40 N A 4 N A=10⇒c He c Ar=10≈3.2

Worked example 9.15

Calculate the average speed of helium (2 4 He) atoms at a temperature of −15 °C.

Answer

We use 1 2 m c 2=3 2 k B T.

First we need to find the mass m of a helium atom. One mole of helium has a mass of 4.0 g so the mass of one molecule is given by:

m=4.0 N A=4.0 6.02×10 23=6.64×10−24 g=6.64×10−27 kg

Now remember to convert the temperature into kelvin: 273 − 15 = 258 K.

So we have:

1 2×6.64×10−27×c 2=3 2×1.38×10−23×258

This gives c 2 = 1.61 × 106 and so c = 1.3 × 103 m s−1.

NATURE OF SCIENCE

Models must be correct but also simple

Boyle thought that a gas consists of particles joined by springs. Newton thought that a gas consists of particles that exert repulsive forces on each other. Bernoulli thought that a gas is a collection of a very large number of particles that exert forces on each other only when they collide. All three could explain why a gas exerts a pressure on its container, but Bernoulli’s picture is the simplest. We assume that the ordinary laws of mechanics apply to the individual particles making up the gas. Even though the laws apply to each individual particle, we cannot observe or analyse each particle individually since there are so many of them. By concentrating on average behaviours of the whole gas and using probability and statistics, physicists developed a new field of physics known as statistical mechanics. This has had enormous success in advancing our understanding of gases and other systems, including where the approximation to an ideal gas breaks down.

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TEST YOUR UNDERSTANDING

23Helium has molar mass 4 g mol−1 and neon has molar mass 20 g mol−1. The helium and neon are kept in the same container. What is the ratio of:

athe average kinetic energies of

molecules E¯He E¯Ne

bthe average rms speeds of molecules c He C Ne?

24The density of an ideal gas is 1.35 kg m−3. The temperature in kelvin and the pressure are both doubled. What happens to the density of the gas?

25A gas is kept at pressure 2.8 × 105 Pa. The density of the gas is 1.2 kg m−3. What is the rms speed of the molecules?

26The pressure and the volume of a fixed mass of an ideal gas are both doubled.

aWhat happens to the rms speed of the molecules?

bWhat happens to the internal energy?

27A fixed mass of an ideal gas is compressed to half its volume at constant pressure.

aWhat happens to the rms speed of the molecules?

bWhat happens to the internal energy?

28A fixed mass of an ideal gas is compressed to half its volume at constant temperature. Explain what happens to the rms speed of the molecules by using:

a P=1 3 ρ c 2

b E¯K=3 2 k B T.

29Calculate the rms speed of krypton atoms at a temperature of 350 K. The molar mass of krypton is 84 g mol−1.

30Show that the rms speed of molecules of a gas of molar mass M (expressed in kg mol−1) kept at a temperature T is given by c=3 R T M.

31Krypton (molar mass 84 g mol−1) is kept at 300 K. At what temperature would xenon (molar mass 132 g mol−1) have the same rms speed as krypton?

32Calculate the internal energy of a gas kept at pressure 1.3 × 105 Pa and volume 0.25 m3.