Kinetic Theory

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 12.

If molecules move at 500 m/s, why does perfume take minutes?

Open a bottle of perfume in one corner of a room and the scent reaches the far corner within a minute or two. Yet the molecules in the air are moving faster than the speed of sound, around 500 m s⁻¹. If they travelled in straight lines, the scent would arrive in a hundredth of a second. This chapter explains the gap between those two numbers. A gas is a crowd of tiny particles in constant random motion, and all of its everyday properties, pressure, temperature, specific heat and the slowness of diffusion, come from averaging over that crowd.

From atoms to the ideal gas

The idea that matter is made of atoms in perpetual motion, attracting at short range and repelling when squeezed too close, grew from Dalton's atomic theory in the early 1800s. Gay-Lussac found that gases combine in simple whole-number ratios of volumes. Avogadro explained why: equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

An atom is about 1 Å ($10^{-10}$ m) across. In solids and liquids the atoms are about 2 Å apart, nearly touching. In a gas at ordinary conditions they are tens of ångströms apart, so the forces between them matter only during brief collisions. That is why a gas compresses easily, a liquid hardly, and a solid almost not at all.

One mole contains the Avogadro number $N_{A} = 6.02 \times 10^{23}$ molecules. The Boltzmann constant $k_{B} = R/N_{A} = 1.38 \times 10^{-23}$ J K⁻¹ is the gas constant per molecule, and R = 8.314 J mol⁻¹ K⁻¹.

Real gases at low pressure and well above their liquefaction temperature obey one combined law, and an ideal gas is one that obeys it exactly at all pressures and temperatures. It can be written in four equivalent forms:

Boyle's law ($PV = \text{constant}$ at fixed T), Charles' law ($V \propto T$ at fixed $P$) and Dalton's law of partial pressures (the pressure of a mixture of non-reacting gases is the sum of the pressures each would exert alone) all follow from these. One mole of any ideal gas occupies 22.4 L at STP (0 °C and 1 atm). Real gases come closest to ideal behaviour at low pressure and high temperature, when the molecules are far apart and fast.

Worked example: oxygen in a cylinder. A 10 L cylinder holds oxygen at 5 atm and 27 °C. What is the mass of oxygen? Take $M_{0} = 32$ g mol⁻¹, R = 8.314 J mol⁻¹ K⁻¹ and 1 atm = $1.013 \times 10^{5}$ Pa.

The common wrong step is to put in 27 for T, or litres for V together with pascals for P. The gas law needs kelvin and cubic metres whenever R is in joules.

Kinetic theory of pressure: where the 1/3 comes from

The model rests on a handful of assumptions about an ideal gas.

Now take a cube of gas with n molecules per unit volume, each of mass m. A molecule moving with velocity component $v_{x}$ strikes a wall at right angles to the x direction and bounces back, so its momentum changes by $2mv_{x}$. In a time $\Delta t$ only molecules within a distance $v_{x}\Delta t$ of the wall, and moving towards it, can hit it. Only half of the molecules move toward the wall rather than away from it. So the number hitting a unit area is $(1/2) n v_{x} \Delta t$, and the pressure, which is force per unit area, is

$$P = \dfrac{(1/2)\, n v_{x}\Delta t\,(2mv_{x})}{\Delta t} = n m v_{x}^{2}$$

The molecules have different speeds, so $v_{x}^{2}$ is replaced by its average. Because the motion is random, the three directions are equally likely, and $\bar{v}{x}^{2} = \bar{v}{y}^{2} = \bar{v}_{z}^{2} = \bar{v}^{2}/3$. Therefore

$$P = (1/3)\, n m \bar{v}^{2}$$

Intermolecular collisions do not change this result, since the elastic collisions only shuffle momentum between molecules. Multiplying by V gives $PV = (2/3)\,N\,((1/2)\,m\,\bar{v}^{2})$, which says that the pressure is two thirds of the translational kinetic energy per unit volume.

Temperature is molecular kinetic energy

Compare $PV = (2/3)\,N\,((1/2)\,m\,\bar{v}^{2})$ with the ideal gas law $PV = Nk_{B}T$:

$$\dfrac{1}{2} m \bar{v}^{2} = \dfrac{3}{2} k_{B} T$$

The average translational kinetic energy of a molecule depends only on the absolute temperature. It does not depend on the gas, the pressure or the volume. That is the deepest meaning of temperature: a measure of the random kinetic energy of the molecules. Because the energy per molecule is the same, a lighter molecule must move faster. The root mean square speed is

$$v_{\text{rms}} = \sqrt{\dfrac{3k_{B}T}{m}} = \sqrt{\dfrac{3RT}{M_{0}}} = \sqrt{\dfrac{3P}{\rho}}$$

Here $M_{0}$ is the molar mass and must be in kg mol⁻¹.

Worked example: speeds and temperature. Find the rms speed of nitrogen at 27 °C. At what temperature would it be twice as large?

There are two common wrong steps: using the molar mass in grams (an error of $\sqrt{1000}$ in the answer), and doubling the temperature instead of quadrupling it.

Worked example: energy of a molecule. Find the average translational kinetic energy of one molecule at 27 °C, and of one mole of molecules.

The energy per molecule is the same for hydrogen and for oxygen at this temperature. Only the speeds differ.

Pause and predict: hydrogen and oxygen are mixed in one container at 300 K. Which has the greater average kinetic energy per molecule, and which moves faster?

Check your answer: the average kinetic energy is identical, $(3/2) k_{B} T$. Hydrogen molecules are lighter, so they move four times faster ($\sqrt{32/2} = 4$).

Why light gases leave the atmosphere. At 300 K a hydrogen molecule's rms speed is about 1.9 km s⁻¹. Earth's escape speed is 11.2 km s⁻¹. These molecules are not fast enough on average, yet Earth keeps little hydrogen, while heavier gases stay.

Beyond the textbook. For the exam: $v_{\text{rms}} = \sqrt{3RT/M_{0}}$, and lighter gases move faster at the same T. In reality: molecules in a gas do not all have the rms speed. They follow a spread called the Maxwell distribution, in which the most probable speed ($\sqrt{2RT/M_{0}}$) is smaller than the mean speed ($\sqrt{8RT/\pi M_{0}}$), which is in turn smaller than $v_{\text{rms}}$. Hydrogen escapes from the top of Earth's atmosphere because the fastest molecules in the tail of this distribution can reach escape speed, and the upper atmosphere is much hotter than 300 K. The Moon, with an escape speed of only 2.4 km s⁻¹, holds on to almost no gas at all. Why it matters: an average is not a limit. A small fraction of fast molecules matters over billions of years.

Degrees of freedom and equipartition

A degree of freedom is an independent way a molecule can store energy, one quadratic term in its energy expression. A free particle can move in three directions, so it has 3 translational degrees of freedom. A diatomic molecule such as $O_{2}$ or $N_{2}$ is like a dumb-bell. It can also spin about two axes perpendicular to the line joining the atoms, giving 2 rotational degrees. Spin about the line joining the atoms is ignored, because the moment of inertia about it is tiny. A bond that can stretch adds a vibrational mode, which has two quadratic terms, kinetic and potential.

The law of equipartition of energy: in thermal equilibrium at temperature T, every quadratic term in the molecule's energy carries an average energy of $(1/2) k_{B} T$. A translational or rotational degree therefore adds $(1/2) k_{B} T$, and a vibrational mode adds $k_{B} T$. Counting the degrees gives the internal energy per mole and so the specific heats, with $C_{P} = C_{V} + R$ for any ideal gas:

(The values of $C_{V}$ and $C_{P}$ are in J mol⁻¹ K⁻¹.) These predictions match experiments well for monatomic and diatomic gases at room temperature.

Worked example: heating oxygen. Two moles of oxygen (a rigid diatomic gas) are at 300 K. Find the internal energy, and the heat needed to raise the temperature by 10 K at constant volume and at constant pressure.

The common wrong step is to use the monatomic value (3/2) R because "it is just a gas". The molecule's structure sets $C_{V}$.

Solids. In a solid every atom vibrates about a fixed position in three directions. Each direction has a kinetic and a potential term, so there are 6 quadratic terms, giving $3k_{B}T$ per atom and $U = 3RT$ per mole. The molar specific heat is therefore $C = 3R \approx 24.9$ J mol⁻¹ K⁻¹, the Dulong–Petit law, and it holds for most solids near room temperature.

Beyond the textbook. For the exam: a rigid diatomic gas has 5 degrees of freedom and $\gamma$ = 1.40, while diamond is an exception to Dulong–Petit. In reality: the degrees of freedom you can count depend on temperature. Quantum mechanics allows a motion to absorb energy only if the energy supplied is large enough to excite it. At very low temperatures hydrogen therefore behaves like a monatomic gas, since its rotation is frozen out. Vibration switches on at temperatures well above room temperature. Diamond's atoms are bound so stiffly that their vibrations are also not fully excited at 300 K. Why it matters: the standard numbers (5 degrees of freedom for a diatomic gas, 3R for a solid) are the right ones at room temperature, and the exceptions have the same reason.

Mean free path: the zigzag walk of a molecule

A molecule collides with others after a short flight. The average distance between two successive collisions is the mean free path $\lambda$. Treat molecules as hard spheres of diameter d. A molecule sweeps out a cylinder of cross-section $\pi d^{2}$ and collides with any molecule whose centre is in it. If all the other molecules were at rest, the path would be $1/(n\pi d^{2})$. They are moving too, which raises the collision rate by a factor $\sqrt{2}$, so

$$\lambda = \dfrac{1}{\sqrt{2}\, n \pi d^{2}}$$

Since $n = P/(k_{B}T)$, the mean free path varies as $\lambda \propto T/P$. It becomes longer when the gas is thinner or hotter.

Worked example: air at STP. Take n = $2.7 \times 10^{25}$ m⁻³ and d = $2 \times 10^{-10}$ m. Find $\lambda$ and the number of collisions per second for a molecule moving at about 500 m s⁻¹.

Dropping the $\sqrt{2}$ gives $2.9 \times 10^{-7}$ m, about 40% too large. This is the number the question will tempt you with. If the pressure is reduced to a thousandth at the same temperature, n falls by 1000 and $\lambda$ becomes about 0.2 mm.

This answers the opening puzzle. Each molecule suffers billions of collisions per second, so its path is a zigzag of steps about $10^{-7}$ m long. The net distance covered is far smaller than the length of the path, and the scent needs minutes to cross a room. The same logic is behind the diffusion of gases through a porous wall. Lighter molecules get through faster, in proportion to $1/\sqrt{M_{0}}$.

Beyond the textbook. For the exam: gas molecules at the same temperature have different rms speeds in proportion to $1/\sqrt{M_{0}}$, so lighter molecules diffuse faster. In reality: this is used to enrich uranium. Uranium hexafluoride gas made with $^{235}$U has a molar mass of 349 g mol⁻¹, and with $^{238}$U it is 352 g mol⁻¹, so the lighter molecules move only $\sqrt{352/349} \approx 1.004$ times faster, a gain of about 0.4% per pass. Plants therefore needed hundreds to thousands of stages in series. Why it matters: it shows how a tiny speed difference between molecules is turned into a practical separation.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Thermodynamics uses $C_{V}$, $C_{P}$ and $\gamma$, and this chapter explains their values from molecular motion.
• Same ideas elsewhere: Thermal Properties of Matter introduced the gas laws and the Kelvin scale that this chapter explains.
• Next: Oscillations and Waves follow, and the speed of sound in a gas depends on the same quantities, $\gamma$, T and $M_{0}$.