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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 sâ1 and for the red curve it is 400 m sâ1. 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=v21+v22+v23+â¯+v2NNââââââââââââââââââââ
Why do we bother to work with an rms speed? Consider the average kinetic energy for the N molecules, which is given by:
EÂ¯K=12mv21+12mv22+12mv23+â¯+12mv2NN=12m(v21+v22+v23+â¯+v2NN)=12mc2
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=13Ïc2, where c stands for the rms speed and Ï 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 vx. 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=mvxâ(âmvx)=2mvx
The molecule will hit the right wall again after a time:
T=2Lv
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 NJNT i.e. JT.
JT=2mvx2Lv=mv2xL
The pressure from one molecule is then (we assume the container is a cube):
P1=mv2xLL2=mv2xL3=mv2xV
There are N molecules in the container so the total pressure becomes:
P=mv21xV+mv22xV+â¯+mv2NxV=mV(v21x+v22x+â¯+v2Nx).
The average of the squares of the x components of velocity is:
v2xââ=v21x+v22x+â¯+v2NxN
and so P=mNVv2xââ.
By symmetry, v2xââ=v2yââ=v2zââ (all directions are equivalent)
And since v2xââ+v2yââ+v2zââ=c2, it follows that v2xââ=c23.
Hence the pressure is:
P=13Nmc2V
But Nm is the total mass M of the gas and so finally we get (Ï=MV):
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P=13Ïc2
Pressure is related to the rms speed and the density of the gas.
Links
Notice how the ideas we studied in ChapterÂ 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:
PV=NkBT
replace pressure with P=13Ïc2 to get:
(13Ïc2)V=NkBT
replace density with massÂ Â volumeÂ  to get:
13(Nm)c2VV=NkBT
multiply both sides by 32 to get:
12mc2=32kBT
(Recall that the Boltzmann constant is kB=RNA=8.316.02Ã1023=1.38Ã10â23Â JÂ Kâ1.)
The last equation states that the average random kinetic energy of the particles is directly proportional to the kelvin temperature:
EÂ¯K=32kBT
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 32kBT, the total random kinetic energy, i.e. the internal energy U, is:
U=32NkBT
But recall that kB=RNA, so that another expression is:
U=32nRT
Yet another expression comes from using the ideal gas law, PV = nRT, which gives:
U=32PV
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 molâ1) or 40 g of argon (molar mass 40 g molâ1)? 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 12mc2=32kBT we find that when T is doubled then c2 will double, so c itself will increase by a factor of 2â.
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WORKED EXAMPLE 9.14
Calculate the ratio of the average speed of helium (molar mass 4.0 g molâ1) to argon (molar mass 40 g molâ1) atoms when both gases are at the same temperature.
Answer
Since the temperature is the same for both gases, using 12mc2=32kBT we find that:
12mHec2He=12mArc2Ar and so c2Hec2Ar=mArmHe
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 4NA.
Similarly, the mass in grams of an argon molecule is 40NA.
So, c2Hec2Ar=40NA4NA=10âcHecAr=10ââââ3.2
WORKED EXAMPLE 9.15
Calculate the average speed of helium Â (42He) atoms at a temperature of â15 Â°C.
Answer
We use 12mc2=32kBT.
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=4.0NA=4.06.02Ã1023=6.64Ã10â24Â g=6.64Ã10â27Â kg
Now remember to convert the temperature into kelvin: 273 â 15 = 258 K.
So we have:
12Ã6.64Ã10â27Ãc2=32Ã1.38Ã10â23Ã258
This gives c2 = 1.61 Ã 106 and so c = 1.3 Ã 103 m sâ1.
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 Bernoulliâs 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 molâ1 and neon has molar mass 20 g molâ1. The helium and neon are kept in the same container. What is the ratio of:
athe average kinetic energies of
molecules EÂ¯HeEÂ¯Ne
bthe average rms speeds of molecules cHeCNe?
24The density of an ideal gas is 1.35 kg mâ3. 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 Ã 105 Pa. The density of the gas is 1.2 kg mâ3. 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:
aP=13Ïc2
bEÂ¯K=32kBT.
29Calculate the rms speed of krypton atoms at a temperature of 350 K. The molar mass of krypton is 84 g molâ1.
30Show that the rms speed of molecules of a gas of molar mass M (expressed in kg molâ1) kept at a temperature T is given by c=3RTMââââââ.
31Krypton (molar mass 84 g molâ1) is kept at 300 K. At what temperature would xenon (molar mass 132 g molâ1) have the same rms speed as krypton?
32Calculate the internal energy of a gas kept at pressure 1.3 Ã 105 Pa and volume 0.25 m3.
