Kinetic Theory

This chapter builds kinetic theory from the ground up in six connected steps. It opens with the molecular nature of matter — the atomic hypothesis, molecular size, and how spacing differs between solids, liquids and gases. It revisits the behaviour of gases from Chapter 10, recasting the ideal gas equation in molecular language via the Boltzmann constant, and recovers Boyle's, Charles' and Avogadro's laws and Dalton's law of partial pressures as consequences of one equation. The heart of the chapter is the kinetic theory of an ideal gas: pressure derived from molecules bouncing off a wall, then the kinetic interpretation of temperature connecting a macroscopic quantity to a molecular one. From there it introduces degrees of freedom and the law of equipartition of energy, predicts the molar specific heats of monatomic, diatomic and polyatomic gases (and solids), and closes with the mean free path, explaining why gases diffuse slowly despite molecules moving at hundreds of metres per second.

The atomic hypothesis and the size of a molecule

Matter is made of atoms and molecules in perpetual, restless motion — attracting weakly at moderate separation, repelling strongly if squeezed. This Atomic Hypothesis is ancient (Kanada in India, sixth century BCE; Democritus in Greece, fourth century BCE), but it took Dalton's laws of definite and multiple proportions to turn it into a testable theory, now confirmed by electron microscope images. An atom is about an angstrom across, $1\ \text{Å} = 10^{-10}\ \text{m}$. In solids and liquids, tightly packed atoms sit only a few angstroms apart ($\approx 2\ \text{Å}$), so the interatomic force — long-range attraction, short-range repulsion — is significant. In a gas, molecules are typically tens of angstroms apart, so intermolecular forces are negligible except during a collision. Gas molecules therefore spend almost all their time moving freely in straight lines, occasionally colliding elastically — the entire physical foundation of kinetic theory. A gas's stillness is deceptive: it is a dynamic equilibrium, molecules colliding and exchanging speeds incessantly, with only average properties — pressure, temperature — staying constant.

The gas laws and the ideal gas equation

For a fixed mass of gas at low pressure and temperature well above liquefaction, pressure, volume and absolute temperature obey a simple relation, introduced in Chapter 10 and reinterpreted here at the molecular level.

where: $P$ is pressure, $V$ is volume, $T$ is absolute temperature, $\mu$ is the number of moles, $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ is the universal gas constant, $N$ is the number of molecules, and $k_B = R/N_A = 1.38\times10^{-23}\ \text{J K}^{-1}$ is the Boltzmann constant, $N_A = 6.02\times10^{23}\ \text{mol}^{-1}$ being Avogadro's number. A gas obeying this exactly is, by definition, an ideal gas; real gases approach it at low pressure and high temperature. Equivalent forms: $P = k_B n T$ ($n = N/V$, number density) and $P = \rho RT/M_0$ ($\rho$ mass density, $M_0$ molar mass). Three familiar gas laws are Equation (1) with one variable fixed. Hold $\mu, T$: $PV = \text{constant}$ — Boyle's law. Hold $\mu, P$: $V \propto T$ — Charles' law. Compare samples at the same $P, V, T$: $N$ must match — Avogadro's hypothesis, equal volumes at the same $T, P$ hold equal numbers of molecules. One mole of any ideal gas at STP ($273\ \text{K}$, $1\ \text{atm}$) occupies $22.4$ litres, the molar volume. For a mixture of non-reacting ideal gases sharing $V, T$, the total pressure is the sum of what each would exert alone:

where: each $P_i = \mu_i RT/V$ is the partial pressure of gas $i$. This is Dalton's law of partial pressures: each species contributes independently to the momentum delivered to the walls.

Pressure of a gas: the molecular picture

Kinetic theory replaces the empirical gas laws with a mechanical model: a gas is a very large number of molecules moving freely between elastic collisions, treated as point particles since their size is tiny compared with their spacing. Each molecule striking a wall reverses its perpendicular velocity component while the parallel components are unchanged; summed over the huge number of collisions per second, this delivered momentum is the force, and force per area is pressure:

where: $n$ is number density, $m$ is molecular mass, and $\overline{v^2}$ is the mean of the squared speed (not the square of the mean speed) — connecting the easily measured quantity pressure to purely microscopic quantities.

Kinetic interpretation of temperature

Multiplying Equation (3) by $V$ and using $nV = N$ gives $PV = \frac{2}{3}N\left(\frac{1}{2}m\overline{v^2}\right) = \frac{2}{3}E$, since the internal energy of an ideal gas is purely translational kinetic energy, $E = N \times \frac{1}{2}m\overline{v^2}$. Comparing with the ideal gas equation $PV = k_BNT$ gives $E = \frac{3}{2}k_BNT$, and dividing by $N$,

where: the left side is the average kinetic energy of one molecule and the right side involves only absolute temperature $T$. This is one of the most important results in thermal physics: the average kinetic energy of a gas molecule is proportional to absolute temperature and depends on nothing else — not pressure, volume, or the identity of the gas. Two gases at the same temperature share the same average molecular kinetic energy; the lighter molecule simply moves faster to compensate for its smaller mass. Taking the square root of $\overline{v^2}$ gives the root mean square speed,

For nitrogen at $T = 300\ \text{K}$, with $m = 4.65\times10^{-26}\ \text{kg}$, this gives $v_{rms} = 516\ \text{m/s}$ — comparable to the speed of sound in air, no coincidence since sound propagates through molecular collisions.

Degrees of freedom and the equipartition of energy

Kinetic theory so far accounts only for translational motion — three directions of flight, contributing terms $\frac{1}{2}mv_x^2$, $\frac{1}{2}mv_y^2$, $\frac{1}{2}mv_z^2$, each averaging $\frac{1}{2}k_BT$. A molecule with more than one atom can also rotate, and if not perfectly rigid, its atoms can vibrate. Each independent way a molecule can move — each degree of freedom — corresponds to one squared (quadratic) term in its energy. A monatomic molecule (argon, neon) has only the three translational degrees of freedom. A diatomic molecule such as $\text{O}_2$ or $\text{N}_2$, treated as a rigid dumbbell, adds two independent rotation axes (rotation about the bond itself carries negligible moment of inertia) — five in total. If the atoms also oscillate along the bond like a spring, that vibrational mode adds a kinetic term $\frac{1}{2}m(dy/dt)^2$ and a potential term $\frac{1}{2}ky^2$ — two quadratic terms, not one. The law of equipartition of energy — first proved by Maxwell — states that in thermal equilibrium at temperature $T$, energy is shared equally among every quadratic term, each averaging $\frac{1}{2}k_BT$:

where: each translational or rotational degree of freedom gives one quadratic term, $\frac{1}{2}k_BT$; each vibrational mode gives two (kinetic and potential), a full $k_BT$. Count the quadratic terms, multiply by $\frac{1}{2}k_BT$ each — this is the master tool for predicting specific heats below.

Specific heat capacities of gases and solids

Applying equipartition, molecule by molecule, predicts molar specific heats directly from molecular structure. A monatomic gas has three translational degrees of freedom, so $U = \frac{3}{2}k_BT \times N_A = \frac{3}{2}RT$, giving $C_v = \frac{3}{2}R$. Using $C_p - C_v = R$ (true for any ideal gas), $C_p = \frac{5}{2}R$ and $\gamma = 5/3 \approx 1.67$. A rigid diatomic gas has five degrees of freedom, so $C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$, $\gamma = 7/5 = 1.4$ — derived in full below. A vibrating diatomic gains one more mode contributing a full $k_BT$ per mole, giving $C_v = \frac{7}{2}R$, $C_p = \frac{9}{2}R$, $\gamma = 9/7$. A general polyatomic molecule (non-linear, so all three rotation axes carry appreciable moment of inertia) has three translational, three rotational and $f$ vibrational modes: $C_v = (3+f)R$, $C_p = (4+f)R$, $\gamma = (4+f)/(3+f)$. Ignoring vibration, predicted $C_v$ is $12.5\ \text{J mol}^{-1}\text{K}^{-1}$ for monatomic gases, $20.8$ for diatomic, $24.93$ for triatomic, with $C_p - C_v = 8.31\ \text{J mol}^{-1}\text{K}^{-1}$ throughout — matching $R$ and agreeing well with measurement. Gases like $\text{Cl}_2$ show larger measured heats than this because their vibrational modes are already active and add their own $k_BT$. The same reasoning extends to solids: each atom vibrates in three dimensions about a fixed mean position, and a one-dimensional oscillation carries average energy $2 \times \frac{1}{2}k_BT = k_BT$, so the average energy per atom is $3k_BT$, and $U = 3RT$ per mole. Since a solid's volume barely changes with temperature, essentially all heat supplied goes into raising $U$, so $C = 3R \approx 24.9\ \text{J mol}^{-1}\text{K}^{-1}$ — the Dulong and Petit law, which agrees well with most solids at ordinary temperature (carbon is a notable exception).

Mean free path

Gas molecules travel at speeds comparable to sound, yet a whiff of gas released in one corner of a kitchen takes minutes to reach the other. Molecules have finite size $d$, so they collide constantly, their paths deflected at every encounter. The mean free path $l$ is the average distance travelled between successive collisions. Model molecules as hard spheres of diameter $d$. A molecule moving at average speed $\langle v \rangle$ collides with any molecule whose centre comes within $d$ of its own; in time $\Delta t$ it sweeps a collision cylinder of volume $\pi d^2\langle v\rangle\Delta t$, giving $n\pi d^2\langle v\rangle\Delta t$ collisions. Accounting for the fact that all molecules move (not just the one under study) replaces $\langle v\rangle$ by the mean relative speed, introducing a factor of $\sqrt{2}$:

where: $n$ is number density and $d$ is molecular diameter. Mean free path grows if the gas is more rarefied or the molecules smaller — in a highly evacuated tube, $l$ can become comparable to the tube's own length. For air at STP, $n = 2.7\times10^{25}\ \text{m}^{-3}$ and $d = 2\times10^{-10}\ \text{m}$ give $l \approx 2.9\times10^{-7}\ \text{m} \approx 1500\,d$, about a hundred times the interatomic spacing. It is this large mean free path, not molecular slowness, that makes gas diffusion far slower than raw molecular speed suggests, and why gases disperse to fill any unconfined container.

Board level - molecular volume and radius of a water molecule (extends NCERT Examples 12.1 and 12.2)

The density of liquid water is $1000\ \text{kg m}^{-3}$. The density of water vapour at $100\ ^\circ\text{C}$ and $1\ \text{atm}$ is $0.6\ \text{kg m}^{-3}$. Estimate (a) the fraction of the total volume occupied by the molecules themselves in the vapour, and (b) the radius of a single water molecule.

Step 1. Compare volumes for the same mass of water. Volume is inversely proportional to density, so the vapour occupies $1000/0.6 = 1667$ times the volume of the same mass of liquid water. Since molecules in the closely packed liquid essentially fill the space (fractional molecular volume $\approx 1$), the fractional molecular volume in the vapour is smaller by the same factor: $1/1667 = 6\times10^{-4}$.

Step 2. Find the mass of a single water molecule. One mole of water ($\text{H}_2\text{O}$), molar mass $18\ \text{g} = 0.018\ \text{kg}$, contains $N_A = 6\times10^{23}$ molecules, so $m = 0.018/(6\times10^{23}) = 3\times10^{-26}\ \text{kg}$.

Step 3. Find the molecular volume using the liquid-phase density (molecules are essentially touching there). $V_{molecule} = m/\rho = (3\times10^{-26})/1000 = 3\times10^{-29}\ \text{m}^3$.

Step 4. Convert volume to radius using $V = \frac{4}{3}\pi r^3$. $r^3 = \frac{3V}{4\pi} \approx 7.2\times10^{-30}\ \text{m}^3$, so $r \approx 1.9\times10^{-10}\ \text{m} \approx 2\ \text{Å}$. Answer: molecular volume fraction $\approx 6\times10^{-4}$; radius of a water molecule $\approx 2\times10^{-10}\ \text{m}$ ($2\ \text{Å}$).

Competitive level - percentage speed difference between uranium isotopes in $\text{UF}_6$ gas (extends NCERT Example 12.6, JEE/NEET pattern)

Natural uranium hexafluoride gas contains molecules built from the two uranium isotopes of mass $235\ \text{u}$ and $238\ \text{u}$, both combined with six fluorine atoms of mass $19\ \text{u}$ each. At a fixed temperature, which isotope's molecule has the larger average speed, and what is the percentage difference in $v_{rms}$ between them?

Step 1. Compute the molecular mass of each isotopologue. $^{235}\text{UF}_6$: $235 + 6(19) = 349\ \text{u}$. $^{238}\text{UF}_6$: $238 + 114 = 352\ \text{u}$.

Step 2. Relate speed and mass at fixed $T$. Since $\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_BT$ is the same for both isotopologues (temperature depends only on average kinetic energy, not on which molecule carries it), $m\overline{v^2}$ is constant, so $v_{rms} \propto 1/\sqrt{m}$ — the lighter $^{235}\text{UF}_6$ has the larger $v_{rms}$.

Step 3. Compute the ratio of speeds. $\dfrac{v_{349}}{v_{352}} = \sqrt{\dfrac{352}{349}} = \sqrt{1.0086} = 1.0044$.

Step 4. Express as a percentage difference. $(1.0044-1)\times100\% = 0.44\%$. Answer: the lighter $^{235}\text{UF}_6$ molecule moves about $0.44\%$ faster than $^{238}\text{UF}_6$, at any common temperature. Timing note: under two minutes. The problem reduces to one ratio, $\sqrt{m_2/m_1}$ — never compute $v_{rms}$ in absolute units, since temperature cancels.

Competitive level - mean free path and collision frequency of nitrogen at 2.0 atm (extends NCERT Exercise 12.10, JEE pattern)

Estimate the mean free path and the collision frequency of a nitrogen molecule in a cylinder of nitrogen gas at $2.0\ \text{atm}$ and $17\ ^\circ\text{C}$. Take the radius of a nitrogen molecule to be $1.0\ \text{Å}$, and compare the mean time between collisions to the time the molecule spends actually colliding.

Step 1. Find the number density from the ideal gas equation. $T = 290\ \text{K}$, $P = 2.0\ \text{atm} = 2.03\times10^5\ \text{Pa}$. From $n = P/(k_BT)$: $n = (2.03\times10^5)/[(1.38\times10^{-23})(290)] = 5.06\times10^{25}\ \text{m}^{-3}$.

Step 2. Find the mean free path using $d = 2r = 2.0\times10^{-10}\ \text{m}$. From $l = 1/(\sqrt{2}n\pi d^2)$: $\sqrt{2}\,n\pi d^2 = 1.414\times(5.06\times10^{25})\times(1.26\times10^{-19}) = 9.0\times10^6\ \text{m}^{-1}$, giving $l \approx 1.1\times10^{-7}\ \text{m}$.

Step 3. Find $v_{rms}$ using $m_{N_2} = 4.65\times10^{-26}\ \text{kg}$. $v_{rms} = \sqrt{3k_BT/m} = \sqrt{3(1.38\times10^{-23})(290)/(4.65\times10^{-26})} \approx 508\ \text{m/s}$.

Step 4. Get the collision frequency and compare with the collision duration. Frequency $= v_{rms}/l \approx 4.6\times10^9\ \text{s}^{-1}$, so $\tau = l/v_{rms} \approx 2.2\times10^{-10}\ \text{s}$, while collision duration $\sim d/v_{rms} \approx 3.9\times10^{-13}\ \text{s}$. Answer: $l \approx 1.1\times10^{-7}\ \text{m}$, collision frequency $\approx 4.6\times10^9\ \text{s}^{-1}$, and the free-flight time is about $560$ times longer than the collision itself. Timing note: about three minutes. Chain: $P, T \to n \to l \to v_{rms} \to$ frequency. Convert to kelvin and pascals first, and never drop the $\sqrt{2}$ factor.

The single most common numerical slip is confusing molecular diameter $d$ with radius $r$ in the mean free path formula — it needs $d^2$, so using $r$ underestimates the collision cross-section by a factor of four. A close second is dropping the $\sqrt{2}$ factor in $l = 1/(\sqrt{2}n\pi d^2)$, which comes from averaging over the relative velocity of colliding molecules rather than treating targets as stationary. Students frequently forget that $C_p - C_v = R$ holds for every ideal gas — monatomic, diatomic or polyatomic — and try to re-derive it each time instead of using it directly. Another frequent error is writing $\overline{v^2}$ as $(\overline{v})^2$; the mean of the squares is not the square of the mean. Finally, many state the equipartition theorem as "each degree of freedom gets $k_BT$" instead of $\frac{1}{2}k_BT$, and forget that a vibrational mode counts as two quadratic terms and hence a full $k_BT$.

• State the ideal gas equation in both forms, $PV = \mu RT$ and $PV = k_BNT$, converting between moles, molecules and the Boltzmann and universal gas constants.
• Derive Boyle's, Charles' and Avogadro's laws and Dalton's law of partial pressures as consequences of the ideal gas equation.
• Reproduce the derivation of $P = \frac{1}{3}nm\overline{v^2}$ from a molecule elastically colliding with a container wall.
• Combine kinetic pressure with the ideal gas equation to derive $\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_BT$ and explain why temperature is molecular.
• Calculate $v_{rms}$ for any gas at any temperature, and compare speeds of different gases or isotopes.
• Count translational, rotational and vibrational degrees of freedom of a molecule and apply the equipartition of energy.
• Derive $C_v$, $C_p$ and $\gamma$ for monatomic and diatomic gases and for solids (Dulong and Petit law) from equipartition.
• Compute mean free path and collision frequency, and explain why gases diffuse slowly despite high molecular speeds.
• Distinguish $\overline{v^2}$ from $(\overline{v})^2$, and universal translational energy from gas-dependent total internal energy.

Ask the tutor: if you are unsure how to combine the kinetic-theory pressure formula with the ideal gas equation, or want to walk through a mean free path calculation step by step, ask the tutor to work one through with your own numbers.