12.1 Introduction
Chapter 12 — Kinetic Theory
In 1827, a Scottish botanist named Robert Brown was peering through his microscope at tiny pollen grains floating in water. He expected them to sit still. Instead, each grain trembled and darted about in an endless, jerky dance, as though it were alive. Brown suspected he had stumbled upon some hidden “life force” inside the pollen. But when he repeated the experiment with fine dust and ground-up rock — things that were certainly not living — they jittered in exactly the same restless way.
The motion was real, but its cause stayed a mystery for nearly eighty years. Then, in 1905, Albert Einstein offered an explanation: the visible grains were being knocked about by countless invisible water molecules striking them from every side, millions of times each second. The dance Brown had seen was the fingerprint of molecules in perpetual motion — motion far too small to watch directly. A few years later, careful experiments by Jean Perrin confirmed the idea and convinced the last doubters that matter really is built from atoms and molecules.
This ceaseless molecular motion is the very heart of the kinetic theory of gases. By picturing a gas as a swarm of tiny particles in constant random motion, and applying simple mechanics to them, we can explain everyday properties like pressure and temperature — and even predict how much heat a gas needs to warm up.
Figure to come
Fig. 12.0 – A microscope view of a large pollen grain suspended in water, its zig-zag path traced by a dotted line, with many small water molecules shown colliding with it from all directions.
By the end of this chapter, you will be able to answer each of these questions using the kinetic theory of gases.
Long before anyone could see a single molecule, scientists began to suspect that the behaviour of gases held a clue to the hidden structure of matter. One of the earliest breakthroughs came in 1661, when Robert Boyle discovered the gas law that now carries his name — a simple relationship between the pressure and volume of a gas. Boyle, Isaac Newton, and several others tried to explain such behaviour by imagining that a gas was made of tiny particles.
However, the real atomic theory of matter took much longer to become firmly established — more than 150 years passed before it was accepted on solid scientific ground. What finally tied everything together was the kinetic theory of gases.
The word “kinetic” comes from a Greek word meaning motion, and that is the central idea of this whole chapter. Kinetic theory explains the behaviour of gases by treating a gas as a collection of a very large number of atoms or molecules that are always moving rapidly and randomly.
Why can we get away with such a simple picture for a gas? The reason lies in the forces between particles. Atoms and molecules attract or repel each other through inter-atomic forces, but these are short-range forces — they matter only when particles are very close together. In solids and liquids the particles are packed close, so these forces are important. In a gas, the particles are usually far apart, so most of the time these forces can be neglected. This is exactly why a gas is so much simpler to describe than a solid or a liquid.
The kinetic theory was developed in the nineteenth century, mainly by James Clerk Maxwell, Ludwig Boltzmann, and others. It turned out to be remarkably successful, and it still forms the foundation of how we understand gases today.
The power of the theory shows up in how much it can explain from a single simple idea. Kinetic theory gives a molecular interpretation of two familiar large-scale properties of a gas — its pressure (the force the gas exerts per unit area on the walls of its container) and its temperature (a measure of how hot the gas is). It shows that these everyday quantities are really the combined effect of countless molecules in motion.
The theory is also fully consistent with the gas laws you have already met, such as Boyle’s law, and with Avogadro’s hypothesis about equal volumes of gases. On top of this, it correctly predicts the specific heat capacities of many gases — the amount of heat needed to raise their temperature.
Finally, kinetic theory does something even more impressive: it links measurable bulk properties of gases — such as viscosity (internal friction in a flowing gas), thermal conduction, and diffusion — to microscopic quantities like the size and mass of individual molecules. In this way, kinetic theory allowed scientists to estimate how big and how heavy molecules are, long before instruments could image them directly.
This chapter gives an introduction to kinetic theory. We will start from the idea of moving molecules and build up, step by step, to explanations of pressure, temperature, specific heat, and how far a molecule can travel before it collides with another.
12.2 Molecular Nature of Matter
Some ideas in science are so basic that everything else is built on top of them. Richard Feynman, one of the greatest physicists of the twentieth century, believed that the single most important discovery of all was simply this: matter is made up of atoms.
Feynman put it in a striking way. Suppose some disaster wiped out all of human knowledge, and only one sentence could be passed on to the next generation of intelligent creatures. Feynman said that sentence should be the atomic hypothesis — the idea that everything is made of tiny particles.
Notice how much is packed into that one statement. Atoms are always moving. They pull on each other when slightly apart, and push each other away when forced too close. Almost all of this chapter is an unpacking of these few words.
The suspicion that matter is not continuous — that it cannot be divided forever — is actually very old. It appeared in many cultures. In India, a thinker named Kanada, and in Greece, a philosopher named Democritus, both suggested that matter is made of indivisible building blocks. These were remarkable guesses made thousands of years before any experiment could test them.
The scientific atomic theory, as we know it today, is usually credited to John Dalton, about two hundred years ago. He proposed the atomic theory to explain two experimental laws that elements obey when they combine to form compounds.
The first is the law of definite proportions, and the second is the law of multiple proportions.
To explain these laws, Dalton suggested that the smallest constituents of an element are its atoms. Atoms of the same element are identical to each other, but differ from the atoms of other elements. A small number of atoms of different elements combine to form a molecule of a compound. This neatly accounts for why the mass ratios in a compound are always fixed and simple.
Around the early nineteenth century, another important rule was found — Gay Lussac’s law.
Soon after came a key insight from Avogadro.
Avogadro’s law, when combined with Dalton’s atomic theory, successfully explains Gay Lussac’s law of combining volumes. And because elements very often exist as molecules rather than single atoms, Dalton’s atomic theory is also called the molecular theory of matter.
Today this theory is fully accepted by scientists. But it is worth remembering that atoms were not always taken for granted — even at the end of the nineteenth century, there were still famous scientists who did not believe in the existence of atoms.
That doubt has now completely vanished. From many observations we know that matter is made of molecules, which are themselves made of one or more atoms. Modern instruments such as the electron microscope and the scanning tunnelling microscope let us do more than infer atoms indirectly — they let us actually image individual atoms.
The size of an atom is about one angstrom. An angstrom is simply a convenient small unit of length equal to \(10^{-10}\,\text{m}\), written as \(1\,\text{Å} = 10^{-10}\,\text{m}\). Because atoms are this small, their spacing tells us a great deal about whether a substance behaves as a solid, a liquid, or a gas.
In solids, atoms are tightly packed, spaced only a few angstroms apart — about \(2\,\text{Å}\). In liquids, the separation between atoms is roughly the same as in solids, so liquids are almost as dense. The difference is that in a liquid the atoms are not locked rigidly in place; they can move around one another, and this is exactly what allows a liquid to flow.
Figure to come
Fig. 12.2 – Three panels showing atomic arrangement: (a) solid with atoms in a rigid regular lattice spaced ~2 Å, (b) liquid with atoms close but disordered and able to slide past each other, (c) gas with widely separated atoms in random positions.
Gases are very different. Here the distances between molecules are much larger — of the order of tens of angstroms. Because the molecules are so far apart, a gas molecule can travel a long way before it bumps into another one. The average distance a molecule travels between two collisions is called the mean free path.
In gases the mean free path is of the order of thousands of angstroms — far larger than the atomic size. This means the molecules of a gas are much freer than those in a solid or liquid, and can travel long distances without colliding. It also explains a familiar fact: if a gas is not kept enclosed in a container, it simply disperses and spreads away into its surroundings. This freedom of movement, and the large gaps between molecules, is the deep reason a gas fills whatever container it is placed in, while a solid keeps its shape.
In solids and liquids, because the atoms sit so close together, the force between atoms becomes very important. This interatomic force has two parts: a long-range attraction that pulls atoms together when they are a few angstroms apart, and a short-range repulsion that pushes them apart when they are squeezed too close. So atoms attract at moderate distances but strongly repel when brought very near — exactly the behaviour described in Feynman’s atomic hypothesis.
Figure to come
Fig. 12.3 – A graph of interatomic force versus separation, showing attraction (force pulling atoms together) at larger separations of a few angstroms and strong repulsion (force pushing apart) at very small separations.
One warning about gases: their calm, unchanging appearance is misleading. A gas sitting quietly in a jar is actually full of frantic activity. Its molecules are constantly rushing about and colliding, and the equilibrium of the gas is a dynamic equilibrium, not a state of rest.
In dynamic equilibrium, molecules collide and change their speeds during each collision, but only the average properties of the whole gas remain steady. No single molecule stays at the same speed for long.
Finally, it is worth noting that atomic theory is not the end of the story of matter, but the beginning. We now know that atoms are themselves not indivisible or elementary. Each atom consists of a central nucleus surrounded by electrons. The nucleus is made of protons and neutrons, and these in turn are made of still smaller particles called quarks. Even quarks may not be the final layer — nature may hold even more fundamental, possibly string-like, entities. The search continues, and each answer opens new questions.
For the rest of this chapter, however, we deliberately keep things simple. We will limit ourselves to understanding the behaviour of gases (and a little about solids) by picturing them as a collection of a huge number of molecules in incessant, restless motion.
12.3 Behaviour of Gases
Of the three states of matter, gases are actually the easiest to understand. The reason goes back to what we saw in the last section: in a gas the molecules are far apart, so their mutual forces are negligible almost all the time. These forces matter only during the brief instant when two molecules collide. Between collisions, each molecule moves freely, almost as if the others were not there.
This simplicity means a gas obeys neat, general rules — rules that hold for all gases, regardless of which gas it is, as long as conditions are right.
Experiments show that gases at low pressures and high temperatures — that is, well above the temperature at which they would liquefy or solidify — approximately obey a simple relation between their pressure, volume, and temperature (you have already met this idea in an earlier chapter):
\[PV = KT \qquad (12.1)\]
Here \(P\) is the pressure of the gas (SI unit: pascal, Pa), \(V\) is its volume (m³), and \(T\) is the temperature measured on the kelvin (absolute) scale (K). We must always use the kelvin scale here, not celsius, because \(T\) appears directly in the equation and must never be zero or negative for a real sample.
In this relation \(K\) is a constant for a given sample of gas, but it is not universal — its value changes if we take a different amount (a different volume) of gas.
Now let us bring in the idea of molecules. If a sample contains \(N\) molecules, then the constant \(K\) turns out to be proportional to \(N\). So we may write \(K = Nk\), where \(k\) is a new constant.
Remarkably, experiments show that this \(k\) has the same value for every gas. Because of this universal character, \(k\) is given a special name — the Boltzmann constant — and its symbol is \(k_B\).
Writing the relation for two different samples of gas, we get:
\[\frac{P_1 V_1}{N_1 T_1} = \frac{P_2 V_2}{N_2 T_2} = \text{constant} = k_B \qquad (12.2)\]
This equation carries a powerful consequence. If two gases have the same \(P\), \(V\), and \(T\), then they must also have the same \(N\) — the same number of molecules. This is exactly Avogadro’s hypothesis, which we met earlier: the number of molecules per unit volume is the same for all gases at a fixed temperature and pressure.
It is a measured fact that the number of molecules in 22.4 litres of any gas is \(6.02 \times 10^{23}\). This important number is called the Avogadro number, written \(N_A\).
The mass of 22.4 litres of any gas equals its molecular weight expressed in grams, measured at STP (standard temperature \(273\ \text{K}\) and standard pressure \(1\ \text{atm}\)). This particular amount of a substance — the amount containing \(N_A\) particles — is called one mole.
Historically, Avogadro guessed that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, purely from patterns he saw in chemical reactions. He had no way to prove it. Kinetic theory, developed later, finally justified this bold hypothesis on solid physical grounds.
Using the idea of the mole, the relation for a gas can be written in its most familiar form — the perfect gas equation (also called the ideal gas equation):
\[PV = \mu RT \qquad (12.3)\]
Here \(\mu\) is the number of moles of gas, and \(R = N_A k_B\) is called the universal gas constant — universal because, like \(k_B\), it is the same for every gas. On the kelvin scale its value is:
\[R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}\]
The number of moles \(\mu\) can be found in two equivalent ways:
\[\mu = \frac{M}{M_0} = \frac{N}{N_A} \qquad (12.4)\]
where \(M\) is the mass of the gas containing \(N\) molecules, \(M_0\) is the molar mass (the mass of one mole), and \(N_A\) is Avogadro’s number.
Substituting \(\mu = N/N_A\) into Eq. (12.3) and using \(R = N_A k_B\), the perfect gas equation can also be written in terms of the actual number of molecules:
\[PV = k_B NT \qquad \text{or} \qquad P = k_B nT\]
Here \(n = N/V\) is the number density of the gas.
There is yet another useful form of Eq. (12.3). Writing the mass density of the gas as \(\rho\), we can show:
\[P = \frac{\rho RT}{M_0} \qquad (12.5)\]
where \(\rho\) is the mass density (kg m⁻³) and \(M_0\) is the molar mass. This version is handy when we know the density of a gas rather than its number of moles.
A gas that obeys Eq. (12.3) exactly at all pressures and all temperatures is defined to be an ideal gas.
Real gases only approach this behaviour; they never obey it perfectly. Fig. 12.1 shows how a real gas departs from ideal behaviour at three different temperatures, by plotting the quantity \(PV/\mu T\) against pressure \(P\). Notice that all the curves bend away from the flat ideal-gas line at high pressure, but they all approach the ideal-gas value at low pressures and high temperatures.
Figure to come
Fig. 12.1 – Graph of \(PV/\mu T\) versus \(P\) for a real gas at three temperatures, showing all curves approaching the horizontal ideal-gas line at low pressure and high temperature.
The reason is exactly the point we started with. At low pressures or high temperatures the molecules are far apart, so their mutual interactions are negligible. And a gas whose molecules do not interact behaves like an ideal gas.
We can now recover the familiar gas laws as special cases of the ideal gas equation.
Boyle’s law. If we keep the amount of gas \(\mu\) and the temperature \(T\) fixed in Eq. (12.3), the right-hand side becomes constant, so:
\[PV = \text{constant} \qquad (12.6)\]
That is, at constant temperature, the pressure of a fixed mass of gas varies inversely with its volume — squeeze it into half the volume and its pressure doubles.
Fig. 12.2 compares the experimental pressure–volume curves for steam (solid lines) with the curves predicted by Boyle’s law (dotted lines). Once again, the agreement is best at high temperatures and low pressures.
Figure to come
Fig. 12.2 – Experimental P–V curves (solid lines) for steam at three temperatures, compared with Boyle’s law predictions (dotted lines); P in units of 22 atm, V in units of 0.09 litres.
Charles’ law. Instead, if we keep the pressure \(P\) fixed, Eq. (12.1) tells us that \(V \propto T\). That is, at constant pressure, the volume of a gas is directly proportional to its absolute temperature \(T\). This is Charles’ law, illustrated in Fig. 12.3.
Fig. 12.3 shows experimental temperature–volume curves for carbon dioxide (solid lines) at three pressures, compared with the straight-line prediction of Charles’ law (dotted lines).
Figure to come
Fig. 12.3 – Experimental T–V curves (solid lines) for CO₂ at three pressures, compared with Charles’ law predictions (dotted lines); T in units of 300 K, V in units of 0.13 litres.
Mixtures of gases. Finally, consider a mixture of non-interacting ideal gases sealed together — say \(\mu_1\) moles of gas 1, \(\mu_2\) moles of gas 2, and so on — all in a vessel of volume \(V\) at temperature \(T\) and total pressure \(P\). Experiment shows that the equation of state of the mixture is:
\[PV = (\mu_1 + \mu_2 + \dots)\, RT \qquad (12.7)\]
We can rearrange this to write the pressure as a sum:
\[P = \mu_1 \frac{RT}{V} + \mu_2 \frac{RT}{V} + \dots \qquad (12.8)\]
\[= P_1 + P_2 + \dots \qquad (12.9)\]
Here each term \(P_1 = \mu_1 RT/V\) is the pressure that gas 1 alone would exert if it occupied the same vessel at the same temperature with no other gas present. This is called the partial pressure of gas 1.
Equation (12.9) then states a simple, powerful rule.
We now look at a few worked examples. These use the ideas above to estimate something we cannot measure with a ruler — the volume actually occupied by the molecules, and the size of a single molecule.
12.4 Kinetic Theory of an Ideal Gas
We are now ready to build the kinetic theory of gases from the ground up. The whole theory rests on the molecular picture of matter: a gas is nothing more than a very large collection of molecules — typically of the order of Avogadro’s number — all in incessant, random motion.
Let us recall the scale of things. At ordinary pressure and temperature, the average distance between molecules in a gas is about ten times (or more) the size of a single molecule, which is roughly \(2\ \text{Å}\). Because the molecules are so far apart compared with their own size, the force between them is negligible for most of the time.
This has an important consequence. Since the molecules barely interact, we can assume that between encounters they move freely in straight lines, obeying Newton’s first law — an object with no net force on it keeps moving in a straight line at constant speed. So most of the time a gas molecule simply coasts along undisturbed.
Only occasionally do two molecules come close enough to feel each other’s intermolecular forces (the short-range attraction and repulsion introduced earlier). When that happens, their velocities suddenly change. These brief interactions are called collisions.
The molecules of a gas collide constantly, either with one another or with the walls of the container, and each collision alters their velocities. To make the theory workable, we assume all these collisions are elastic.
You met elastic collisions earlier while studying work, energy, and power. The key facts we carry forward are simple: in an elastic collision both the total kinetic energy and the total momentum are conserved.
With this molecular picture in place — huge numbers of molecules, moving freely in straight lines, colliding elastically, conserving total kinetic energy and momentum — we can now do something remarkable. In the next part we will use ordinary mechanics on these molecules to derive an expression for the pressure of a gas, and then connect it to temperature.
12.4.1 Pressure of an Ideal Gas
We now come to the first great success of kinetic theory: deriving the pressure of a gas purely from the mechanics of its molecules. Pressure, remember, is force per unit area — and here that force comes from countless molecules hammering on the walls of the container.
Consider a gas enclosed in a cube of side \(l\). Choose the coordinate axes parallel to the sides of the cube, as shown in Fig. 12.4. Focus on one molecule moving with velocity \((v_x, v_y, v_z)\) that is about to strike the wall lying parallel to the yz-plane. This wall has area \(A = l^2\).
Figure to come
Fig. 12.4 – A cube with axes x, y, z; a single molecule approaching the yz-wall with velocity \((v_x, v_y, v_z)\) and rebounding with velocity \((-v_x, v_y, v_z)\), the x-component reversed.
Because the collision is elastic, the molecule bounces back with the same speed. Its \(y\) and \(z\) velocity components are unchanged, but the \(x\)-component simply reverses direction. So the velocity after the collision is \((-v_x, v_y, v_z)\).
Let us track the molecule’s momentum in the \(x\)-direction. Before the hit it is \(mv_x\); after the hit it is \(-mv_x\). The change in the molecule’s momentum is therefore
\[-mv_x - (mv_x) = -2mv_x.\]
By the principle of conservation of momentum, whatever momentum the molecule loses, the wall must gain. So the momentum imparted to the wall in one collision is \(2mv_x\).
To get the pressure, we need the momentum delivered to the wall per unit time, because force is the rate of change of momentum. So we must count how many molecules strike the wall in a small time interval \(\Delta t\).
A molecule with \(x\)-velocity \(v_x\) can reach the wall in time \(\Delta t\) only if it starts within a distance \(v_x \Delta t\) of it. In other words, only molecules lying inside a thin slab of volume \(A\, v_x \Delta t\) next to the wall are close enough to hit it during \(\Delta t\), as illustrated in Fig. 12.4a.
Figure to come
Fig. 12.4a – A thin slab of gas of area A and thickness \(v_x\Delta t\) against the wall; molecules inside it moving toward the wall are the ones that strike it within time \(\Delta t\).
But there is a subtlety. Inside that slab, on average only half the molecules are moving toward the wall; the other half are moving away from it. So the number of molecules that actually hit the wall in time \(\Delta t\) is
\[\tfrac{1}{2}\, A\, v_x \Delta t\, n,\]
where \(n\) is the number of molecules per unit volume (the number density introduced earlier).
The total momentum transferred to the wall in time \(\Delta t\) is the momentum per collision times the number of collisions:
\[Q = (2mv_x)\left(\tfrac{1}{2}\, n\, A\, v_x \Delta t\right) \qquad (12.10)\]
The force on the wall is the rate of momentum transfer, \(Q/\Delta t\), and pressure is force per unit area. Dividing by \(A\Delta t\):
\[P = \frac{Q}{A\Delta t} = n\, m\, v_x^2 \qquad (12.11)\]
Notice something satisfying: both \(A\) and \(\Delta t\) have cancelled out. The pressure does not depend on the size of the patch of wall we chose or the time interval — as it should not.
So far we pretended every molecule has the same \(v_x\). In reality, molecules move with a whole range of speeds — there is a distribution of velocities. So Eq. (12.11) really gives the pressure due only to the group of molecules that happen to have a particular \(v_x\), with \(n\) being the number density of that one group.
To get the total pressure, we add up the contributions from all such groups. This amounts to replacing \(v_x^2\) by its average value:
\[P = n\, m\, \overline{v_x^2} \qquad (12.12)\]
Here \(\overline{v_x^2}\) is the average of \(v_x^2\) taken over all the molecules.
Now we use a key physical fact about a gas: it is isotropic. That is, no direction is special — the molecules are just as likely to move along \(x\) as along \(y\) or \(z\).
Because of this symmetry, the average of the squared velocity component must be the same for all three directions:
\[\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2}\]
\[= \tfrac{1}{3}\left[\overline{v_x^2} + \overline{v_y^2} + \overline{v_z^2}\right] = \tfrac{1}{3}\,\overline{v^2} \qquad (12.13)\]
where \(v\) is the speed of a molecule and \(\overline{v^2}\) is the mean of the squared speed. The step above simply says: since the three direction-averages are equal, each one equals one-third of their sum, and their sum is the mean square speed \(\overline{v^2}\) (because \(v^2 = v_x^2 + v_y^2 + v_z^2\)).
Substituting Eq. (12.13) into Eq. (12.12) gives the central result of this section:
\[P = \tfrac{1}{3}\, n\, m\, \overline{v^2} \qquad (12.14)\]
This is a beautiful equation. On the left is pressure — something we can measure with a gauge. On the right are microscopic quantities: the number density \(n\), the mass \(m\) of one molecule, and the mean square speed \(\overline{v^2}\) of the molecules. Kinetic theory has connected the visible world to the invisible one.
Here \(P\) is pressure (Pa), \(n\) is number density (m⁻³), \(m\) is the mass of a single molecule (kg), and \(\overline{v^2}\) is the mean square speed (m²s⁻²).
Before moving on, a few honest remarks about this derivation are worth making.
First, the shape of the container does not matter. We chose a cube only for convenience. For a vessel of any shape, we can always pick a tiny flat area on its surface and carry out exactly the same argument. This is reassuring because it fits with Pascal’s law, which you studied earlier with fluids: in a fluid (or gas) in equilibrium, the pressure is the same throughout, so any small patch of wall would give the same result.
Second, we quietly ignored collisions between molecules during the derivation. Justifying this rigorously is hard, but we can see qualitatively why it does no harm. We found the number of molecules hitting the wall in time \(\Delta t\) to be \(\tfrac{1}{2}\,n\,A\,v_x\Delta t\). Now, the collisions are random and the gas is in a steady state.
So if one molecule with velocity \((v_x, v_y, v_z)\) gets knocked into a different velocity by a collision, then somewhere else another molecule with a different starting velocity gets knocked into \((v_x, v_y, v_z)\). The two swaps balance out. If this balance did not hold, the distribution of velocities would not stay steady — but it does.
In any case, the quantity we actually use is the average \(\overline{v_x^2}\), and averages are not disturbed by such swaps. So as long as collisions are not too frequent and each collision lasts a negligibly short time compared with the free time between collisions, molecular collisions do not affect the pressure result above.
12.4.2 Kinetic Interpretation of Temperature
We have just found that pressure is tied to the motion of molecules. The next step is one of the deepest ideas in all of physics: connecting temperature to that same molecular motion. What is temperature, really, at the level of the molecules?
Let us start from the pressure result of the last section, Eq. (12.14), and rewrite it. Multiplying both sides by the volume \(V\):
\[PV = \tfrac{1}{3}\, nV\, m\, \overline{v^2} \qquad (12.15\text{a})\]
Since \(n\) is the number of molecules per unit volume, the product \(nV\) is simply \(N\), the total number of molecules in the sample. Rearranging slightly to bring out a familiar quantity:
\[PV = \tfrac{2}{3}\, N \times \tfrac{1}{2} m\, \overline{v^2} \qquad (12.15\text{b})\]
where \(N (= nV)\) is the total number of molecules.
Look closely at the quantity in the bracket, \(\tfrac{1}{2}m\overline{v^2}\). This is just the average translational kinetic energy of a molecule — the energy a molecule has because it is moving bodily from one place to another.
For an ideal gas, the molecules are so far apart that we ignore forces between them. This means the gas has no potential energy stored between molecules — its entire internal energy \(E\) is just the total kinetic energy of all its molecules:
\[E = N \times \tfrac{1}{2} m\, \overline{v^2} \qquad (12.16)\]
Comparing Eq. (12.16) with Eq. (12.15b), we can replace the bracket by \(E/N\), which gives a clean link between pressure–volume and internal energy:
\[PV = \tfrac{2}{3}\, E \qquad (12.17)\]
Now comes the crucial move. In Section 12.3 we saw that the ideal gas equation can be written as \(PV = k_B NT\). Setting this equal to Eq. (12.17):
\[\tfrac{2}{3} E = k_B N T \quad\Rightarrow\quad E = \tfrac{3}{2}\, k_B N T \qquad (12.18)\]
Dividing by \(N\) to get the energy per molecule:
\[\frac{E}{N} = \tfrac{1}{2} m\, \overline{v^2} = \tfrac{3}{2}\, k_B T \qquad (12.19)\]
This single equation is one of the great results of physics. It says that the average kinetic energy of a molecule is directly proportional to the absolute temperature of the gas — and to nothing else. It does not depend on the pressure, the volume, or even which gas it is.
This is exactly what temperature is, seen through molecular eyes: a measure of the average kinetic energy of the molecules. A hotter gas is simply one whose molecules are, on average, moving faster and carrying more kinetic energy.
Figure to come
Fig. 12.4b – A straight-line graph of average translational kinetic energy per molecule versus absolute temperature \(T\), passing through the origin, illustrating \(\tfrac{1}{2}m\overline{v^2} \propto T\).
Here \(E\) is the internal (translational) energy in joules, \(N\) is the number of molecules, \(k_B = 1.38\times10^{-23}\ \text{J K}^{-1}\) is the Boltzmann constant, \(T\) is the absolute temperature in kelvin, \(m\) is the mass of one molecule in kilograms, and \(\overline{v^2}\) is the mean square speed in m²s⁻².
This is a profound bridge. On one side is temperature, a large-scale quantity we read off a thermometer (what physicists call a macroscopic or thermodynamic variable). On the other side is the average kinetic energy of a single molecule, a microscopic quantity we can never see directly. The Boltzmann constant \(k_B\) is exactly the bridge that connects these two worlds.
Notice also what Eq. (12.18) is quietly telling us: the internal energy of an ideal gas, \(E = \tfrac{3}{2}k_B NT\), depends only on temperature — not on pressure or volume. Squeeze the gas or let it expand, and as long as the temperature is unchanged, its internal energy stays the same. With this interpretation of temperature, kinetic theory is now fully consistent with the ideal gas equation and all the gas laws built on it.
Mixtures of gases, revisited. For a mixture of non-reactive ideal gases, each gas contributes to the total pressure. Extending Eq. (12.14) to several gases:
\[P = \tfrac{1}{3}\left[n_1 m_1 \overline{v_1^2} + n_2 m_2 \overline{v_2^2} + \dots\right] \qquad (12.20)\]
In thermal equilibrium, every gas in the mixture is at the same temperature, so by Eq. (12.19) the average kinetic energy per molecule is the same for all of them:
\[\tfrac{1}{2} m_1 \overline{v_1^2} = \tfrac{1}{2} m_2 \overline{v_2^2} = \tfrac{3}{2} k_B T\]
Substituting this back into Eq. (12.20), each term \(\tfrac{1}{3}n_i m_i \overline{v_i^2}\) becomes \(n_i k_B T\), giving:
\[P = (n_1 + n_2 + \dots)\, k_B T \qquad (12.21)\]
This is once again Dalton’s law of partial pressures — but now we have derived it from the kinetic theory, rather than just stating it as an experimental fact.
How fast do molecules actually move? Equation (12.19) lets us estimate the typical molecular speed. Take nitrogen gas at \(T = 300\ \text{K}\). The mass of one nitrogen molecule is
\[m = \frac{M_{N_2}}{N_A} = \frac{28}{6.02\times10^{26}} = 4.65\times10^{-26}\ \text{kg}.\]
(Here \(28\) is the molar mass of \(N_2\) in grams, and the factor \(10^{26}\) takes care of converting to the mass of a single molecule in kilograms.) Then from \(\overline{v^2} = 3k_B T/m\):
\[\overline{v^2} = \frac{3 k_B T}{m} = (516)^2\ \text{m}^2\text{s}^{-2}.\]
The square root of the mean square speed is called the root mean square (rms) speed, written \(v_{rms}\).
For nitrogen at 300 K, this gives
\[v_{rms} = 516\ \text{m s}^{-1}.\]
That is remarkable: the molecules are travelling at about the speed of sound in air. And from Eq. (12.19), since the average kinetic energy is fixed by temperature alone, at the same temperature, lighter molecules must move faster than heavier ones (smaller \(m\) means larger \(v_{rms}\)).
12.5 Law of Equipartition of Energy
In the last section we learned that the average translational kinetic energy of a molecule is \(\tfrac{3}{2}k_B T\). But a molecule can do more than just fly in a straight line — it can also spin and, sometimes, vibrate. How is a molecule’s energy shared among all these different kinds of motion? The answer is a beautifully simple rule called the law of equipartition of energy.
Let us begin with translation. The kinetic energy of a single molecule moving in three dimensions is the sum of its energy along the three axes:
\[\varepsilon_t = \tfrac{1}{2}m v_x^2 + \tfrac{1}{2}m v_y^2 + \tfrac{1}{2}m v_z^2 \qquad (12.22)\]
Here \(\varepsilon_t\) is the translational kinetic energy of the molecule, \(m\) is its mass (kg), and \(v_x, v_y, v_z\) are its velocity components along the three axes (m s⁻¹).
For a gas in thermal equilibrium at temperature \(T\), we take the average of this energy over all molecules. Writing the average with angle brackets \(\langle\ \rangle\):
\[\langle \varepsilon_t \rangle = \left\langle \tfrac{1}{2}m v_x^2 \right\rangle + \left\langle \tfrac{1}{2}m v_y^2 \right\rangle + \left\langle \tfrac{1}{2}m v_z^2 \right\rangle = \tfrac{3}{2} k_B T \qquad (12.23)\]
The last step uses the result from Section 12.4.2. Now, because a gas has no preferred direction (it is isotropic), the three terms on the right must be equal to one another. Since they add up to \(\tfrac{3}{2}k_B T\), each one must be exactly one-third of that:
\[\left\langle \tfrac{1}{2}m v_x^2 \right\rangle = \tfrac{1}{2}k_B T, \quad \left\langle \tfrac{1}{2}m v_y^2 \right\rangle = \tfrac{1}{2}k_B T, \quad \left\langle \tfrac{1}{2}m v_z^2 \right\rangle = \tfrac{1}{2}k_B T \qquad (12.24)\]
This is the seed of the whole idea: each independent way of moving carries, on average, the same share of energy — namely \(\tfrac{1}{2}k_B T\).
To make this precise, we need the idea of degrees of freedom. Think about how many numbers you need to pin down a molecule’s position. A molecule free to move anywhere in space needs three coordinates \((x, y, z)\). If it were confined to a plane, it would need two; if confined to a line, just one.
We express this by saying the molecule has that many degrees of freedom for its motion: one for motion along a line, two for motion in a plane, and three for motion in space.
Motion of a body as a whole from one point to another is called translation. So a molecule free to move in space has three translational degrees of freedom.
Notice a pattern: each translational degree of freedom contributes a term to the energy that contains the square of a motion variable — for example \(\tfrac{1}{2}m v_x^2\), and similar terms in \(v_y\) and \(v_z\). Equation (12.24) tells us that in thermal equilibrium, the average of each such squared term is \(\tfrac{1}{2}k_B T\).
The molecules of a monatomic gas such as argon or helium are single atoms. They can only translate, so they have only these three translational degrees of freedom — nothing more.
Now consider a diatomic gas such as oxygen (\(O_2\)) or nitrogen (\(N_2\)). A molecule of \(O_2\) still has three translational degrees of freedom. But because it is a dumbbell of two atoms, it can also rotate about its centre of mass.
Figure 12.6 shows the two independent axes about which a diatomic molecule can spin — both perpendicular to the line joining the two atoms. So a diatomic molecule has two rotational degrees of freedom in addition to its three translational ones.
Figure to come
Fig. 12.6 – A dumbbell-shaped diatomic molecule with the two atoms joined by a bond, showing two independent rotation axes (1 and 2), both perpendicular to the bond axis.
Each rotational degree of freedom adds its own energy term. The total energy is now the translational energy \(\varepsilon_t\) plus the rotational energy \(\varepsilon_r\):
\[\varepsilon_t + \varepsilon_r = \tfrac{1}{2}m v_x^2 + \tfrac{1}{2}m v_y^2 + \tfrac{1}{2}m v_z^2 + \tfrac{1}{2}I_1\omega_1^2 + \tfrac{1}{2}I_2\omega_2^2 \qquad (12.25)\]
Here \(\omega_1\) and \(\omega_2\) are the angular speeds about axes 1 and 2 (rad s⁻¹), and \(I_1, I_2\) are the corresponding moments of inertia (kg m²) — quantities you met while studying rotational motion. Notice that each rotational degree of freedom, just like each translational one, contributes a term containing the square of a motion variable (here an angular speed).
So far we treated the \(O_2\) molecule as a rigid rotator — a dumbbell whose bond length never changes, so the atoms do not vibrate.
This rigid-rotator picture works well for \(O_2\) at moderate temperatures, but it is not always valid. Some molecules, such as carbon monoxide (CO), vibrate even at moderate temperatures: the two atoms oscillate back and forth along the bond, like two masses joined by a spring — a one-dimensional oscillator, as shown in Fig. 12.6a.
Figure to come
Fig. 12.6a – A diatomic molecule drawn as two atoms connected by a spring, with arrows showing the atoms oscillating toward and away from each other along the bond axis.
Such a vibrational mode adds a further energy term \(\varepsilon_v\) to the total. This vibrational energy has two parts — a kinetic part from the atoms’ motion and a potential part from the stretched/compressed bond:
\[\varepsilon_v = \tfrac{1}{2}m\left(\frac{dy}{dt}\right)^2 + \tfrac{1}{2}k y^2\]
\[\varepsilon = \varepsilon_t + \varepsilon_r + \varepsilon_v \qquad (12.26)\]
Here \(k\) is the force constant of the oscillator (the stiffness of the bond, N m⁻¹) and \(y\) is the vibrational coordinate — how far the bond is stretched or compressed from its normal length. The first term \(\tfrac{1}{2}m(dy/dt)^2\) is the vibrational kinetic energy and the second term \(\tfrac{1}{2}ky^2\) is the vibrational potential energy, exactly as in simple harmonic motion.
Now notice a very important difference in Eq. (12.26). Each translational and each rotational degree of freedom contributed just one squared term. But a single vibrational mode contributes two squared terms — one kinetic (\(\tfrac{1}{2}m(dy/dt)^2\)) and one potential (\(\tfrac{1}{2}ky^2\)).
Each squared (quadratic) term in the energy expression is a separate way for the molecule to absorb and store energy — what we call a mode of energy.
We already know that for each translational mode, the average energy in thermal equilibrium is \(\tfrac{1}{2}k_B T\). The elegant discovery of classical statistical mechanics — first proved by Maxwell — is that this same share applies to every mode, whether translational, rotational, or vibrational.
Applying this rule: - Each translational degree of freedom contributes \(\tfrac{1}{2}k_B T\). - Each rotational degree of freedom contributes \(\tfrac{1}{2}k_B T\). - Each vibrational mode contributes \(2 \times \tfrac{1}{2}k_B T = k_B T\), because it has two squared terms (kinetic and potential).
That last point is the one students most often miss, so it is worth stating plainly: a vibrational mode is worth \(k_B T\), not \(\tfrac{1}{2}k_B T\).
The full proof of the law of equipartition of energy is beyond the scope of this book. What we will do next is put the law to work: in the following section we use it to predict the specific heat capacities of gases, and later of solids — and then compare these predictions with experiment.
12.6 Specific Heat Capacity
We have built up a molecular picture of energy in a gas. Now we cash it in for something directly measurable: the specific heat capacity — how much heat a gas needs to warm up. This is where kinetic theory makes sharp, testable predictions that we can check against experiment.
You have met specific heat before: it is the heat required to raise the temperature of a given amount of a substance by one degree. For gases it is convenient to work per mole, giving the molar specific heat. And for a gas there is an important twist — the answer depends on how we heat it: at constant volume, or at constant pressure.
The two differ because of what happens to the added heat. At constant volume, the gas cannot expand, so all the heat goes into raising the internal energy (and hence temperature). At constant pressure, the gas expands as it warms, so some of the heat is spent doing work pushing back the surroundings — meaning more heat is needed for the same temperature rise. That is why \(C_p\) is always larger than \(C_v\).
We will now use the law of equipartition of energy to predict these specific heats, starting with the simplest case.
12.6.1 Monatomic Gases
A monatomic gas — like helium, neon, or argon — is made of single atoms. As we saw in the previous section, such an atom has only three translational degrees of freedom and nothing else.
By the law of equipartition, each degree of freedom carries an average energy of \(\tfrac{1}{2}k_B T\), so the average energy of one molecule at temperature \(T\) is \(\tfrac{3}{2}k_B T\).
The total internal energy of one mole of the gas is this per-molecule energy multiplied by Avogadro’s number \(N_A\):
\[U = \tfrac{3}{2}k_B T \times N_A = \tfrac{3}{2}RT \qquad (12.27)\]
where we used \(N_A k_B = R\). Here \(U\) is the internal energy of one mole (J), \(R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}\) is the universal gas constant, and \(T\) is the absolute temperature (K).
The molar specific heat at constant volume is the rate at which this internal energy rises with temperature, \(C_v = dU/dT\). Differentiating Eq. (12.27):
\[C_v \text{ (monatomic gas)} = \frac{dU}{dT} = \tfrac{3}{2}R \qquad (12.28)\]
Now for the constant-pressure case. For any ideal gas, the two molar specific heats are related by a simple and important rule:
\[C_p - C_v = R \qquad (12.29)\]
This relation captures exactly the physical point made above — the extra \(R\) is the work the gas does while expanding at constant pressure.
Using \(C_v = \tfrac{3}{2}R\) in Eq. (12.29):
\[C_p = C_v + R = \tfrac{3}{2}R + R = \tfrac{5}{2}R \qquad (12.30)\]
Finally, the ratio of the two specific heats, written \(\gamma\) (gamma), is a very useful quantity:
\[\gamma = \frac{C_p}{C_v} = \frac{\tfrac{5}{2}R}{\tfrac{3}{2}R} = \frac{5}{3} \qquad (12.31)\]
These are not just theoretical numbers. When the molar specific heats of the noble gases helium, neon, and argon are measured, they come out extremely close to the predicted \(\tfrac{3}{2}R\) and \(\tfrac{5}{2}R\) — an early and striking confirmation of both kinetic theory and the law of equipartition.
12.6.2 Diatomic Gases
A diatomic gas — such as oxygen (\(O_2\)), nitrogen (\(N_2\)), or hydrogen (\(H_2\)) — is made of two-atom molecules shaped like tiny dumbbells. As we saw in Section 12.5, if we treat such a molecule as a rigid rotator (a dumbbell whose bond length does not change), it has 5 degrees of freedom: 3 translational and 2 rotational.
Applying the law of equipartition — \(\tfrac{1}{2}k_B T\) per degree of freedom — the average energy of one molecule is \(\tfrac{5}{2}k_B T\). So the total internal energy of one mole is:
\[U = \tfrac{5}{2}k_B T \times N_A = \tfrac{5}{2}RT \qquad (12.32)\]
where \(U\) is the internal energy per mole (J), \(R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}\), and \(T\) is the absolute temperature (K).
The molar specific heats follow at once. Differentiating, \(C_v = dU/dT = \tfrac{5}{2}R\), and using \(C_p - C_v = R\):
\[C_v \text{ (rigid diatomic)} = \tfrac{5}{2}R, \qquad C_p = \tfrac{7}{2}R \qquad (12.33)\]
\[\gamma \text{ (rigid diatomic)} = \frac{C_p}{C_v} = \frac{7}{5} \qquad (12.34)\]
Now compare this with the monatomic case. A monatomic gas had \(C_v = \tfrac{3}{2}R\); a diatomic gas has \(C_v = \tfrac{5}{2}R\). The diatomic molecule needs more heat for the same temperature rise, simply because it has more ways to store energy — its two extra rotational degrees of freedom soak up part of the added heat, leaving less to speed up translation. This is exactly why a monatomic gas like helium and a diatomic gas like oxygen require different amounts of heat to warm up by the same amount.
Figure to come
Fig. 12.6b – A bar chart comparing molar specific heat \(C_v\) for a monatomic gas (\(\tfrac{3}{2}R\)), a rigid diatomic gas (\(\tfrac{5}{2}R\)), and a vibrating diatomic gas (\(\tfrac{7}{2}R\)), showing \(C_v\) rising as the number of energy modes increases.
So far we assumed the diatomic molecule is rigid. But if the molecule is not rigid and also has a vibrational mode, that mode adds a further \(k_B T\) of energy per molecule (recall a vibration counts as two squared terms, hence \(k_B T\), not \(\tfrac{1}{2}k_B T\)). The internal energy per mole becomes:
\[U = \left(\tfrac{5}{2}k_B T + k_B T\right)N_A = \tfrac{7}{2}RT\]
and the specific heats become:
\[C_v = \tfrac{7}{2}R, \qquad C_p = \tfrac{9}{2}R, \qquad \gamma = \frac{9}{7} \qquad (12.35)\]
12.6.3 Polyatomic Gases
Finally we come to the most complex case — a polyatomic gas, whose molecules contain more than two atoms (for example water vapour \(H_2O\), carbon dioxide \(CO_2\), or methane \(CH_4\)).
Such a molecule can do everything a simpler one can, and more. In general it has 3 translational degrees of freedom, 3 rotational degrees of freedom (it can spin about three independent axes), and some number \(f\) of vibrational modes — the many ways its atoms can oscillate relative to one another.
Here \(f\) simply counts how many independent vibrational modes the particular molecule has; different polyatomic molecules have different values of \(f\).
Applying the law of equipartition — \(\tfrac{1}{2}k_B T\) for each translational and rotational degree of freedom, and \(k_B T\) for each vibrational mode — the internal energy of one mole is:
\[U = \left(\tfrac{3}{2}k_B T + \tfrac{3}{2}k_B T + f\, k_B T\right)N_A\]
Carrying out the differentiation \(C_v = dU/dT\) and using \(C_p = C_v + R\), we get:
\[C_v = (3 + f)R, \qquad C_p = (4 + f)R, \qquad \gamma = \frac{(4+f)}{(3+f)} \qquad (12.36)\]
Here \(C_v\) and \(C_p\) are the molar specific heats (J mol⁻¹K⁻¹), \(R\) is the universal gas constant, \(f\) is the number of vibrational modes, and \(\gamma\) is the ratio of specific heats.
It is worth pausing on one general fact that has held throughout: for any ideal gas — monatomic, diatomic, or polyatomic — the relation
\[C_p - C_v = R\]
is always true. You can check it in every case above: the difference is always one \(R\).
Comparing theory with experiment. Table 12.1 summarises the specific heats predicted by kinetic theory when vibrational modes are ignored (so a monatomic gas has 3 degrees of freedom, a diatomic 5, and a nonlinear triatomic 6).
Table 12.1 — Predicted values of specific heat capacities of gases (ignoring vibrational modes):
| Nature of Gas | \(C_v\) (J mol⁻¹K⁻¹) | \(C_p\) (J mol⁻¹K⁻¹) | \(C_p - C_v\) (J mol⁻¹K⁻¹) | \(\gamma\) |
|---|---|---|---|---|
| Monatomic | 12.5 | 20.8 | 8.31 | 1.67 |
| Diatomic | 20.8 | 29.1 | 8.31 | 1.40 |
| Triatomic | 24.93 | 33.24 | 8.31 | 1.33 |
These predictions match the measured values of several real gases remarkably well, as shown in Table 12.2.
Table 12.2 — Measured values of specific heat capacities of some gases:
| Nature of Gas | Gas | \(C_v\) (J mol⁻¹K⁻¹) | \(C_p\) (J mol⁻¹K⁻¹) | \(C_p - C_v\) (J mol⁻¹K⁻¹) | \(\gamma\) |
|---|---|---|---|---|---|
| Monatomic | He | 12.5 | 20.8 | 8.30 | 1.66 |
| Monatomic | Ne | 12.7 | 20.8 | 8.12 | 1.64 |
| Monatomic | Ar | 12.5 | 20.8 | 8.30 | 1.67 |
| Diatomic | H₂ | 20.4 | 28.8 | 8.45 | 1.41 |
| Diatomic | O₂ | 21.0 | 29.3 | 8.32 | 1.40 |
| Diatomic | N₂ | 20.8 | 29.1 | 8.32 | 1.40 |
| Triatomic | H₂O | 27.0 | 35.4 | 8.35 | 1.31 |
| Polyatomic | CH₄ | 27.1 | 35.4 | 8.36 | 1.31 |
The agreement between the two tables is excellent for these gases, which is a real triumph of the law of equipartition.
Of course, the agreement is not perfect for every gas. For several other gases not listed — such as chlorine (\(Cl_2\)) and ethane (\(C_2H_6\)) — the measured specific heats are noticeably higher than the simple predictions.
This is actually reassuring rather than troubling. The fact that including vibrational modes brings theory and experiment closer together shows that the law of equipartition of energy is well verified experimentally at ordinary temperatures.
Before turning from gases to solids, here is one consolidating example that ties together everything about the specific heats of gases.
12.6.4 Specific Heat Capacity of Solids
The law of equipartition is not limited to gases — it also predicts the specific heat of solids, and does so with surprising success.
Picture a solid as a regular arrangement of \(N\) atoms, each one held in place but able to vibrate about its mean position, as if connected to its neighbours by tiny springs (Fig. 12.6c).
Figure to come
Fig. 12.6c – A solid modelled as a 3D lattice of atoms, each atom connected to its neighbours by springs, vibrating about its fixed mean position along all three axes.
Consider one atom’s vibration along a single direction. A one-dimensional oscillation has two squared energy terms — kinetic and potential — so by equipartition its average energy is
\[2 \times \tfrac{1}{2}k_B T = k_B T.\]
Since an atom in a solid can vibrate along all three directions (\(x\), \(y\), \(z\)), its total average energy is three times this:
\[3 \times k_B T = 3k_B T.\]
For one mole of the solid, \(N = N_A\), so the total internal energy is
\[U = 3k_B T \times N_A = 3RT\]
where \(U\) is the internal energy per mole (J), \(R\) is the universal gas constant, and \(T\) is the absolute temperature (K).
Now we find the specific heat. When a solid is heated at constant pressure, the heat supplied is \(\Delta Q = \Delta U + P\Delta V\). But a solid barely expands, so \(\Delta V\) is negligible and the \(P\Delta V\) work term drops out, leaving \(\Delta Q \approx \Delta U\). Hence its molar specific heat is:
\[C = \frac{\Delta Q}{\Delta T} = \frac{\Delta U}{\Delta T} = 3R \qquad (12.37)\]
Table 12.3 lists the measured specific heats of some common solids at room temperature.
Table 12.3 — Specific Heat Capacity of some solids at room temperature and atmospheric pressure:
| Substance | Specific heat (J kg⁻¹ K⁻¹) | Molar specific heat (J mol⁻¹ K⁻¹) |
|---|---|---|
| Aluminium | 900.0 | 24.4 |
| Carbon | 506.5 | 6.1 |
| Copper | 386.4 | 24.5 |
| Lead | 127.7 | 26.5 |
| Silver | 236.1 | 25.5 |
| Tungsten | 134.4 | 24.9 |
As the last column shows, the molar specific heats of these solids cluster near the predicted \(3R \approx 24.9\ \text{J mol}^{-1}\text{K}^{-1}\) — good agreement at ordinary temperatures. Carbon, with only \(6.1\), is the notable exception, again a hint that classical physics is incomplete and that quantum effects matter at the atomic scale.
12.7 Mean Free Path
Here is a puzzle. We found that gas molecules move very fast — around the speed of sound, hundreds of metres per second. Yet if someone opens a gas cylinder or a bottle of perfume in one corner of a room, the smell takes several minutes to reach the far corner. A cloud of smoke can hang together in still air for hours. If the molecules are so fast, why is the spreading so slow?
The answer is that a molecule almost never travels in a straight line for long. Molecules have a finite (though tiny) size, so they keep bumping into one another. Each collision knocks a molecule off in a new direction. Instead of racing straight across the room, a molecule staggers along a jagged, zig-zag path, as suggested in Fig. 12.7a — taking a very long time to make real progress in any one direction.
Figure to come
Fig. 12.7a – The zig-zag path of a single gas molecule undergoing many collisions, showing that although each segment is fast, the net displacement across the room is small.
This zig-zag motion is why we introduced, back in Section 12.2, the idea of the mean free path — the average distance a molecule travels between two successive collisions. We now derive a formula for it.
Suppose the molecules of a gas are tiny spheres, each of diameter \(d\). Focus on one particular molecule moving with the average speed \(\langle v \rangle\). This molecule will collide with any other molecule whose centre comes within a distance \(d\) of its own centre.
As the chosen molecule moves, in a small time \(\Delta t\) it sweeps out a cylinder. The cylinder’s cross-section is a circle of radius \(d\) (area \(\pi d^2\)) and its length is \(\langle v \rangle \Delta t\). So the molecule sweeps a volume
\[\pi d^2\, \langle v \rangle\, \Delta t\]
and any other molecule whose centre lies inside this swept volume gets hit, as shown in Fig. 12.7.
Figure to come
Fig. 12.7 – A molecule of diameter d sweeping out a cylinder of radius d and length \(\langle v \rangle \Delta t\) in time \(\Delta t\); molecules whose centres fall inside the cylinder collide with it.
If \(n\) is the number of molecules per unit volume (the number density), then the number of molecules in this swept volume — that is, the number of collisions in time \(\Delta t\) — is
\[n\, \pi d^2\, \langle v \rangle\, \Delta t.\]
Dividing by \(\Delta t\), the rate of collisions (collisions per second) is \(n\pi d^2 \langle v \rangle\). The average time between two successive collisions, called the collision time \(\tau\), is just the reciprocal of this rate:
\[\tau = \frac{1}{n\pi \langle v \rangle d^2} \qquad (12.38)\]
The mean free path \(l\) is then simply the average speed multiplied by this time between collisions:
\[l = \langle v \rangle\, \tau = \frac{1}{n\pi d^2} \qquad (12.39)\]
Here \(l\) is the mean free path (m), \(n\) is the number density (m⁻³), \(d\) is the molecular diameter (m), and \(\langle v \rangle\) is the average molecular speed (m s⁻¹). Notice something neat: the average speed \(\langle v \rangle\) has cancelled out of \(l\) — the mean free path depends only on how crowded the gas is and how big the molecules are.
In deriving this, we quietly assumed that all the other molecules stand still while our chosen molecule moves. In reality every molecule is moving, and what really matters for collisions is the average relative velocity between molecules, \(\langle v_r \rangle\). Replacing \(\langle v \rangle\) by \(\langle v_r \rangle\) in Eq. (12.38), a more careful treatment introduces a factor of \(\sqrt{2}\) and gives the standard result:
\[l = \frac{1}{\sqrt{2}\, n\pi d^2} \qquad (12.40)\]
Let us put in numbers for air. Taking the average speed \(\langle v \rangle = 485\ \text{m s}^{-1}\), and using STP conditions, the number density is
\[n = \frac{6.02 \times 10^{23}}{22.4 \times 10^{-3}} = 2.7 \times 10^{25}\ \text{m}^{-3}.\]
Taking the molecular diameter \(d = 2 \times 10^{-10}\ \text{m}\), we get
\[\tau = 6.1 \times 10^{-10}\ \text{s}, \qquad l = 2.9 \times 10^{-7}\ \text{m} \approx 1500\, d \qquad (12.41)\]
So a molecule travels, on average, about 1500 times its own diameter before it hits something — an enormous distance on the molecular scale, but still tiny compared with a room.
As Eq. (12.40) shows, the mean free path depends inversely on both the number density \(n\) and the molecular size (through \(d^2\)). This has a striking consequence: in a highly evacuated tube, \(n\) becomes very small, so the mean free path can grow as large as the length of the tube itself — the molecules simply fly from wall to wall without meeting one another.
Finally, the mean free path is more than a curiosity. Through kinetic theory, the bulk measurable properties of a gas — its viscosity (internal friction), its heat conductivity, and its rate of diffusion — can all be related back to microscopic parameters like the size of a molecule. Historically, it was through exactly these relations that scientists first estimated how big molecules actually are.
12.8 Summary
- The ideal gas equation connecting pressure (\(P\)), volume (\(V\)), and absolute temperature (\(T\)) is
\[PV = \mu RT = k_B NT\]
where \(\mu\) is the number of moles and \(N\) is the number of molecules. \(R\) and \(k_B\) are universal constants, with
\[R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}, \qquad k_B = \frac{R}{N_A} = 1.38 \times 10^{-23}\ \text{J K}^{-1}.\]
Real gases satisfy the ideal gas equation only approximately — the agreement is better at low pressures and high temperatures.
- Kinetic theory of an ideal gas gives the relation
\[P = \tfrac{1}{3}\, n\, m\, \overline{v^2}\]
where \(n\) is the number density of molecules, \(m\) the mass of one molecule, and \(\overline{v^2}\) the mean of the squared speed. Combined with the ideal gas equation, it yields a kinetic interpretation of temperature:
\[\tfrac{1}{2} m\, \overline{v^2} = \tfrac{3}{2} k_B T, \qquad v_{rms} = \left(\overline{v^2}\right)^{1/2} = \sqrt{\frac{3k_B T}{m}}.\]
This tells us that the temperature of a gas is a measure of the average kinetic energy of a molecule, independent of the nature of the gas or molecule. In a mixture of gases at a fixed temperature, the heavier molecule has the lower average speed.
- The translational kinetic energy is
\[E = \tfrac{3}{2}\, k_B NT.\]
This leads to the relation
\[PV = \tfrac{2}{3}\, E.\]
- The law of equipartition of energy states that if a system is in equilibrium at absolute temperature \(T\), the total energy is distributed equally among its different energy modes of absorption, the energy in each mode being \(\tfrac{1}{2}k_B T\). Each translational and each rotational degree of freedom corresponds to one energy mode and has energy \(\tfrac{1}{2}k_B T\). Each vibrational frequency has two modes of energy (kinetic and potential), with a corresponding energy of
\[2 \times \tfrac{1}{2}k_B T = k_B T.\]
Using the law of equipartition of energy, the molar specific heats of gases can be determined, and the values agree with the experimental specific heats of several gases. The agreement can be improved by including vibrational modes of motion.
The mean free path \(l\) is the average distance covered by a molecule between two successive collisions:
\[l = \frac{1}{\sqrt{2}\, n\, \pi\, d^2}\]
where \(n\) is the number density and \(d\) the diameter of the molecule.
Physical Quantities in this Chapter
| Physical Quantity | Symbol | SI Unit | Remarks |
|---|---|---|---|
| Pressure | \(P\) | Pa (N m⁻²) | force per unit area on the walls |
| Volume | \(V\) | m³ | — |
| Absolute temperature | \(T\) | K | measured on the kelvin scale |
| Number of moles | \(\mu\) | mol | amount of substance |
| Number of molecules | \(N\) | dimensionless | \(N = nV\) |
| Number density | \(n\) | m⁻³ | \(n = N/V\) |
| Universal gas constant | \(R\) | J mol⁻¹K⁻¹ | \(8.314\); \(R = N_A k_B\) |
| Boltzmann constant | \(k_B\) | J K⁻¹ | \(1.38\times10^{-23}\); \(k_B = R/N_A\) |
| Avogadro number | \(N_A\) | mol⁻¹ | \(6.02\times10^{23}\) |
| Molar mass | \(M_0\) | kg mol⁻¹ | mass of one mole |
| Mass of one molecule | \(m\) | kg | \(m = M_0/N_A\) |
| Mean square speed | \(\overline{v^2}\) (or \(\langle v^2\rangle\)) | m²s⁻² | average of the squared speeds |
| Root mean square speed | \(v_{rms}\) | m s⁻¹ | \(\sqrt{3k_B T/m}\) |
| Average (mean) speed | \(\langle v \rangle\) | m s⁻¹ | plain average of speeds |
| Internal (translational) energy | \(E\), \(U\) | J | \(E = \tfrac{3}{2}k_B NT\) |
| Molar specific heat (constant volume) | \(C_v\) | J mol⁻¹K⁻¹ | \(C_v = dU/dT\) |
| Molar specific heat (constant pressure) | \(C_p\) | J mol⁻¹K⁻¹ | \(C_p = C_v + R\) |
| Ratio of specific heats | \(\gamma\) | dimensionless | \(\gamma = C_p/C_v\) |
| Degrees of freedom | \(f\) | dimensionless | independent ways to store energy |
| Mean free path | \(l\) | m | \(1/(\sqrt{2}\,n\pi d^2)\) |
| Collision time | \(\tau\) | s | average time between collisions |
| Molecular diameter | \(d\) | m | assumed spherical molecule |
12.9 NCERT Questions
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be \(3\ \text{Å}\).
Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, \(0\ ^\circ\text{C}\)). Show that it is \(22.4\) litres.
Figure 12.8 shows a plot of \(PV/T\) versus \(P\) for \(1.00\times10^{-3}\ \text{kg}\) of oxygen gas at two different temperatures.
Figure to come
Fig. 12.8 – A graph of \(PV/T\) (J K⁻¹) on the y-axis versus \(P\) on the x-axis, showing two curves labelled \(T_1\) and \(T_2\) that both approach the same value on the y-axis at low pressure, with a horizontal dotted line marking that common value.
a. What does the dotted plot signify?
b. Which is true: $T_1 > T_2$ or $T_1 < T_2$?
c. What is the value of $PV/T$ where the curves meet on the $y$-axis?
d. If we obtained similar plots for $1.00\times10^{-3}\ \text{kg}$ of hydrogen, would we get the same value of $PV/T$ at the point where the curves meet on the $y$-axis? If not, what mass of hydrogen yields the same value of $PV/T$ (for the low-pressure, high-temperature region of the plot)? (Molecular mass of $H_2 = 2.02\ \text{u}$, of $O_2 = 32.0\ \text{u}$, $R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$.)An oxygen cylinder of volume \(30\) litre has an initial gauge pressure of \(15\) atm and a temperature of \(27\ ^\circ\text{C}\). After some oxygen is withdrawn from the cylinder, the gauge pressure drops to \(11\) atm and its temperature drops to \(17\ ^\circ\text{C}\). Estimate the mass of oxygen taken out of the cylinder. (\(R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}\), molecular mass of \(O_2 = 32\ \text{u}\).)
An air bubble of volume \(1.0\ \text{cm}^3\) rises from the bottom of a lake \(40\ \text{m}\) deep at a temperature of \(12\ ^\circ\text{C}\). To what volume does it grow when it reaches the surface, which is at a temperature of \(35\ ^\circ\text{C}\)?
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity \(25.0\ \text{m}^3\) at a temperature of \(27\ ^\circ\text{C}\) and \(1\) atm pressure.
Estimate the average thermal energy of a helium atom at (i) room temperature (\(27\ ^\circ\text{C}\)), (ii) the temperature on the surface of the Sun (\(6000\ \text{K}\)), and (iii) the temperature of \(10\) million kelvin (the typical core temperature of a star).
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal numbers of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is \(v_{rms}\) the largest?
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at \(-20\ ^\circ\text{C}\)? (Atomic mass of Ar \(= 39.9\ \text{u}\), of He \(= 4.0\ \text{u}\).)
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at \(2.0\) atm and temperature \(17\ ^\circ\text{C}\). Take the radius of a nitrogen molecule to be roughly \(1.0\ \text{Å}\). Compare the collision time with the time the molecule moves freely between two successive collisions. (Molecular mass of \(N_2 = 28.0\ \text{u}\).)
12.10 Check Your Concepts
Explain why the forces between molecules can be neglected for most of the time in a gas, but not in a solid or a liquid. Relate your answer to the average spacing between molecules in each state.
What is meant by an ideal gas? State the conditions of pressure and temperature under which a real gas behaves most nearly like an ideal gas, and explain physically why these conditions help.
In the kinetic-theory derivation of gas pressure, the mean square speed splits equally among the three directions, introducing a factor of \(\tfrac{1}{3}\). Explain physically where this factor of \(\tfrac{1}{3}\) comes from.
Two different ideal gases are kept at the same temperature. Show, using the kinetic interpretation of temperature, that the average translational kinetic energy per molecule is the same for both, even though their molecular masses differ.
At the same temperature, do lighter or heavier gas molecules have the greater rms speed? Justify your answer using the relation between kinetic energy and temperature.
State the law of equipartition of energy. Explain why a single vibrational mode contributes \(k_B T\) to the average energy of a molecule, whereas a rotational degree of freedom contributes only \(\tfrac{1}{2}k_B T\).
Explain why the molar specific heat at constant pressure \(C_p\) is always greater than the molar specific heat at constant volume \(C_v\) for a gas, yet the two are practically equal for a solid.
The measured molar specific heat of chlorine gas is noticeably larger than the value predicted for a rigid diatomic molecule. Give a reason for this difference.
Using the law of equipartition of energy, explain why the molar specific heat of most solids at ordinary temperature is close to \(3R\). Name the empirical law that first expressed this result, and mention one element that is an exception.
Gas molecules move at speeds comparable to the speed of sound, yet a gas released at one end of a room takes minutes to be smelt at the other end. Using the idea of mean free path, explain this slow spreading.
12.11 Practice with Numericals
Calculate the root mean square speed of hydrogen molecules at \(300\ \text{K}\). (Take molar mass of \(H_2 = 2.0\times10^{-3}\ \text{kg mol}^{-1}\), \(R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}\).)
A fixed mass of an ideal gas at \(2.0\) atm and \(300\ \text{K}\) is allowed to expand to twice its original volume while its temperature rises to \(450\ \text{K}\). Find the new pressure of the gas.
Find (i) the average translational kinetic energy of a single molecule of an ideal gas at \(300\ \text{K}\), and (ii) the total translational kinetic energy of one mole of the gas at the same temperature. (Take \(k_B = 1.38\times10^{-23}\ \text{J K}^{-1}\), \(R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}\).)
For a rigid diatomic gas, write down \(C_v\), \(C_p\), and \(\gamma\) in terms of \(R\). Then find the heat required to raise the temperature of \(3\) moles of the gas by \(20\ \text{K}\) at constant volume. (Take \(R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}\).)
A container holds \(4\ \text{g}\) of helium (monatomic) and \(28\ \text{g}\) of nitrogen (rigid diatomic) at the same temperature. Find the ratio of the total internal energies of the two gases. (Molar mass of He \(= 4\ \text{g mol}^{-1}\), of \(N_2 = 28\ \text{g mol}^{-1}\).)
The rms speed of oxygen molecules at a certain temperature is \(480\ \text{m s}^{-1}\). Find the rms speed of hydrogen molecules at the same temperature. (Molar mass of \(O_2 = 32\ \text{g mol}^{-1}\), of \(H_2 = 2\ \text{g mol}^{-1}\).)
A gas has a number density \(n = 2.7\times10^{25}\ \text{m}^{-3}\) and its molecules have diameter \(d = 3.0\times10^{-10}\ \text{m}\). Estimate the mean free path of a molecule using \(l = \dfrac{1}{\sqrt{2}\,n\pi d^2}\).
Calculate the number of molecules present in \(2.0\ \text{g}\) of hydrogen gas. (Take molar mass of \(H_2 = 2.0\ \text{g mol}^{-1}\), \(N_A = 6.02\times10^{23}\ \text{mol}^{-1}\).)
Equal numbers of moles of a monatomic gas and a rigid diatomic gas are each supplied the same amount of heat at constant volume. Determine which gas undergoes the greater rise in temperature, and by what factor. (Use \(C_v = \tfrac{3}{2}R\) for the monatomic gas and \(C_v = \tfrac{5}{2}R\) for the diatomic gas.)
Using the law of equipartition (\(C = 3R\)), estimate the molar specific heat of a solid, and hence find the heat required to raise the temperature of \(0.5\) mole of the solid by \(30\ \text{K}\). (Take \(R = 8.31\ \text{J mol}^{-1}\text{K}^{-1}\).)
12.12 Points to Ponder
The pressure of a fluid is not exerted only on the walls of its container. Pressure exists everywhere inside a fluid. Any layer of gas within the volume of a container stays in equilibrium precisely because the pressure is the same on both sides of that layer.
Do not form an exaggerated picture of how far apart molecules are in a gas. At ordinary pressures and temperatures, the intermolecular distance in a gas is only about 10 times the interatomic distance in solids and liquids. What is really different is the mean free path: in a gas it is about 100 times the interatomic distance, and about 1000 times the size of a molecule.
The law of equipartition of energy is stated thus: in thermal equilibrium, the energy for each degree of freedom is \(\tfrac{1}{2}k_B T\). Each quadratic (squared) term in the total energy expression of a molecule counts as one degree of freedom. So each vibrational mode gives 2 (not 1) degrees of freedom — a kinetic energy term and a potential energy term — corresponding to an energy of
\[2 \times \tfrac{1}{2}k_B T = k_B T.\]
The molecules of air in a room do not all fall and settle on the floor under gravity, because of their high speeds and constant collisions. In equilibrium there is only a very slight increase in density at lower heights (as happens in the atmosphere). This effect is small because, for ordinary heights, the gravitational potential energy \(mgh\) of a molecule is far less than its average kinetic energy \(\tfrac{1}{2}mv^2\).
\(\langle v^2 \rangle\) is not always equal to \((\langle v \rangle)^2\). The average of a squared quantity is not necessarily the square of the average. Can you find your own examples that illustrate this statement?