Chapter 03 · Kinetic Theory of Gases · Full book

Kinetic Theory of Gases: The Hidden World Behind Heat

Core intuition: Chapter Roadmap

Thermodynamics treats a gas as a smooth substance with pressure, temperature, and heat capacity. But a gas is really an enormous swarm of molecules in ceaseless, chaotic motion. This chapter rebuilds every macroscopic thermal quantity from that microscopic picture—and shows exactly where the rebuilding succeeds and where it breaks.

The molecular model — point molecules, no forces between collisions, elastic walls, and molecular chaos: the handful of idealisations that make a gas calculable.
Pressure from collisions — \(p = \tfrac{1}{3}\rho\langle v^2\rangle\), a steady macroscopic force assembled from a rain of random molecular impacts.
Temperature as energy — \(\tfrac{1}{2}m\langle v^2\rangle = \tfrac{3}{2}k_BT\): temperature is nothing but the mean translational kinetic energy of a molecule.
Equipartition and specific heats — each quadratic degree of freedom carries \(\tfrac{1}{2}k_BT\), fixing \(C_V\), \(C_P\), and \(\gamma\) for monatomic, diatomic, and polyatomic gases.
The speed distribution — the Maxwell–Boltzmann law \(f(v)\) and the three speeds it defines: most-probable \(v_p\), mean \(\bar v\), and root-mean-square \(v_\mathrm{rms}\).
Mean free path and transport — how far a molecule travels between collisions, and the diffusion, viscosity, and thermal conduction that follow from it.
Where classical theory fails — frozen degrees of freedom and the quantum activation rule \(k_BT \gtrsim h\nu\), the crack through which quantum mechanics first entered physics.

Master this chapter and the gas laws stop being empirical rules to memorise: they become theorems about molecules obeying Newton's laws.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

The Pressure Puzzle: A gas exerts a steady, constant pressure on the walls of its container—yet it is made of discrete molecules that hit the wall and bounce off at random times, at random locations, with random speeds. How does a rain of random, chaotic impacts produce a smooth, constant force?
The Temperature Paradox: Open a container of hydrogen beside a container of carbon dioxide. Both are at room temperature—\(300\;\mathrm{K}\)—yet hydrogen molecules move at nearly four times the speed of \(\mathrm{CO_2}\) molecules. If temperature is a measure of molecular speed, how can two gases at the same temperature have molecules moving so differently?
The Smell Mystery: Perfume molecules travel at roughly \(500\;\mathrm{m\,s^{-1}}\)—faster than a speeding bullet. Yet the scent from a bottle opened across the room takes several minutes to reach you. If the molecules are supersonic, why does the smell crawl?
The Frozen Rotation: A diatomic molecule like \(\mathrm{N_2}\) has two atoms bonded together; they can rotate about two independent axes and vibrate along the bond. Classical physics says every mode of motion should share the energy equally. Yet at room temperature, the vibrational mode contributes almost nothing to the heat capacity of nitrogen. Why does nature selectively “freeze out” perfectly legitimate degrees of freedom?

Each question exposes a crack in the macroscopic picture of heat.
By the end of this chapter, every one of them will be transparent.

Why a Microscopic Theory?

Thermodynamics, as developed in Chapter 14, is a remarkable achievement. It constrains every heat engine, every chemical reaction, every phase transition—without once asking what matter is made of. The First Law says energy is conserved; the Second Law says entropy never decreases; Carnot's theorem says no engine can beat a reversible one. All of this follows from two empirical postulates, with no mention of atoms.

Yet thermodynamics is deliberately silent on certain questions. It tells you that the internal energy \(U\) of an ideal gas depends only on temperature—but not why. It tells you that the heat capacities of monatomic and diatomic gases differ—but not how much, or why. It introduced the forward references \(C_V = \tfrac{f}{2}R\) and \(\gamma = (f+2)/f\) in Chapter 14 without derivation, because those results require a molecular picture to make sense.

Kinetic theory supplies that picture.

The central hypothesis is audacious in its simplicity: a gas is a vast collection of molecules in ceaseless, random motion, and all macroscopic properties—pressure, temperature, heat capacity—are statistical averages over that molecular chaos.

This is not a philosophical claim. It is a quantitative programme. We will derive the ideal gas law \(pV = nRT\) from Newton's second law applied to molecular collisions. We will show that temperature is precisely the mean translational kinetic energy per molecule, divided by \(\tfrac{3}{2}k_B\). We will explain why \(C_V\) has the values it does, and why \(\gamma\) takes the values \(\tfrac{5}{3}\), \(\tfrac{7}{5}\), and \(\tfrac{4}{3}\) for monatomic, diatomic, and polyatomic gases respectively. And we will honestly confront the places where classical kinetic theory fails— failures that eventually forced the birth of quantum mechanics.

Core intuition: The Programme of Kinetic Theory

Macroscopic quantities are averages over microscopic disorder.

Pressure \(\longleftrightarrow\) rate of momentum transfer to walls.
Temperature \(\longleftrightarrow\) mean translational kinetic energy per molecule.
Heat capacity \(\longleftrightarrow\) number of quadratic energy modes (equipartition).
Transport (viscosity, diffusion, conductivity) \(\longleftrightarrow\) mean free path between collisions.

Each link is derived, not assumed.

A Brief Historical Orientation

Boyle, Charles, and Gay-Lussac discovered the gas laws empirically in the seventeenth and eighteenth centuries. Clausius, Maxwell, and Boltzmann provided the molecular explanation in the nineteenth century—and were met with fierce resistance from physicists who doubted atoms existed. The argument was settled definitively in 1905 when Einstein's analysis of Brownian motion gave a quantitative prediction for the mean displacement of a pollen grain, confirmed experimentally by Perrin in 1908. By then, Boltzmann—whose entire career was spent defending the atomic hypothesis—had already died.

The lesson is not merely historical. It illustrates that a theory which explains empirical laws (rather than merely fitting them) earns a different kind of confidence. Kinetic theory does not just reproduce \(pV = nRT\); it derives it from mechanics, and predicts things thermodynamics alone never could.

✓ Checkpoint · Why a Microscopic Theory?

Thermodynamics derives the relation \(C_P - C_V = R\) for an ideal gas using only the First Law and the ideal gas law. What additional ingredient does kinetic theory provide that thermodynamics cannot supply on its own?
The individual values of \(C_V\) and \(C_P\); thermodynamics gives only their difference. Kinetic theory gives \(C_V = \tfrac{f}{2}R\) from equipartition, fixing both.
Name one macroscopic quantity that kinetic theory explains but classical thermodynamics cannot predict from its axioms alone.
Any one of: the numerical value of \(C_V\); the ratio \(\gamma\); the speed of sound \(v = \sqrt{\gamma RT/M}\); the mean free path.
(Trap) A student argues: “Thermodynamics already predicts \(pV = nRT\), so kinetic theory adds nothing for the ideal gas law.” Identify the flaw.
Thermodynamics takes \(pV = nRT\) as an empirical input (definition of an ideal gas). Kinetic theory derives it from Newton's laws applied to molecular collisions—a far stronger statement, because it also predicts when deviations occur (high density, low temperature) and why.

Molecular Model of an Ideal Gas

◈ The molecular gas box

live bench

Watch the particles rattle inside the cylinder: pressure is the drumbeat of their collisions on the walls, and \(PV=nRT\) is the tally of it all.

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↗ open in the Simulations page

Fig 15.1 Each molecular velocity resolves into \(v_x,v_y,v_z\). No direction is special, so the three mean-square components are equal — giving \(\langle v_x^2\rangle=\tfrac13\langle c^2\rangle\), the seed of the pressure derivation.

Before any calculation is possible, we must state precisely what we mean by an ideal gas. Vague talk of “molecules in random motion” is not enough; we need a model sharp enough to apply Newton's laws.

The Five Assumptions

Point molecules. Each molecule is a point mass \(m\) with no internal structure (for translational motion purposes). The total volume occupied by the molecules themselves is negligible compared with the volume \(V\) of the container.
No intermolecular forces except during collisions. Between collisions, molecules move in straight lines at constant velocity—Newton's first law applies. Forces act only during the brief, violent instant of a collision.
Elastic collisions. All collisions—molecule–molecule and molecule–wall—conserve kinetic energy. No energy is lost to deformation, heat radiation, or internal excitation during a collision.
Molecular chaos (Stosszahlansatz). The velocities of different molecules are statistically uncorrelated before a collision. There are no preferred directions: the velocity distribution is isotropic.
Large numbers. The number of molecules \(N\) is enormous (\(N \sim 10^{23}\) for a mole). Statistical averages are therefore equal to instantaneous values to extraordinary precision; fluctuations are negligible.

These five assumptions define the ideal gas. Real gases satisfy them well at low pressures and high temperatures— conditions under which molecules are far apart and move fast enough that their brief interaction time is small compared with the time between collisions.

Core intuition: What the Model Buys Us

Assumptions 1–3 allow us to track a single molecule with Newton's laws between collisions. Assumption 4 allows us to average over all molecules without worrying about correlations. Assumption 5 allows us to replace time-averages with ensemble-averages— the pressure on the wall is constant because enormous numbers of molecules are hitting it every second, not because any individual molecule is.

Notation Fixed for the Chapter

Throughout Chapter 15 we use:

\(N\) = total number of molecules in the sample.
\(n\) = number of moles; \(N = n N_A\).
\(N_A = 6.022 \times 10^{23}\;\mathrm{mol^{-1}}\) (Avogadro's number).
\(k_B = 1.381 \times 10^{-23}\;\mathrm{J\,K^{-1}}\) (Boltzmann's constant); \(k_B = R/N_A\).
\(m\) = mass of one molecule; \(M\) = molar mass (kg/mol); \(m = M/N_A\).
\(\vec{v}_i\) = velocity of molecule \(i\); \(v_i = |\vec{v}_i|\).
\(\langle v^2 \rangle = \dfrac{1}{N}\sum_{i=1}^{N} v_i^2\) = mean square speed.
\(v_\mathrm{rms} = \sqrt{\langle v^2 \rangle}\) = root-mean-square speed.

Why “Ideal” Is Not a Weakness

Students sometimes treat “ideal gas” as a euphemism for “wrong model.” This misunderstands the role of idealisation in physics.

The ideal gas model is exactly solvable and quantitatively accurate for noble gases at ordinary conditions—deviations are measured in fractions of a percent for helium at room temperature and atmospheric pressure. For diatomic gases like \(\mathrm{N_2}\) and \(\mathrm{O_2}\), it is accurate to better than \(1\%\) at standard conditions. Deviations become significant only at high pressures (molecules close enough to feel each other's attractions) or low temperatures (intermolecular potential energy becomes comparable to kinetic energy). The van der Waals equation handles these corrections; that is a topic for physical chemistry.

For JEE, NEET, and AP Physics, the ideal gas model is exact within the precision of every numerical problem you will encounter.

Birth of the formula: Orders of Magnitude: How Ideal Is Air?

At standard conditions (\(T = 300\;\mathrm{K}\), \(p = 10^5\;\mathrm{Pa}\)): Number density. From \(pV = Nk_BT\): \[ \frac{N}{V} = \frac{p}{k_BT} = \frac{10^5}{1.381\times10^{-23}\times 300} \approx 2.4\times10^{25}\;\mathrm{m^{-3}}. \] Average spacing between molecules. If each molecule occupies a cube of side \(d\), then \(d \sim (N/V)^{-1/3} \approx (2.4\times10^{25})^{-1/3} \approx 3.4\;\mathrm{nm}\). Molecular diameter. A nitrogen molecule has diameter \(\approx 0.37\;\mathrm{nm}\). Ratio: spacing/diameter \(\approx 3.4/0.37 \approx 9\). So molecules are roughly nine diameters apart on average. The volume they occupy is approximately \((1/9)^3 \approx 0.14\%\) of the container volume. Assumption 1 (point molecules) is justified to better than \(0.1\%\). The intermolecular potential (Lennard-Jones) has a range of \(\sim 0.4\;\mathrm{nm}\) and a depth of \(\sim 10^{-21}\;\mathrm{J}\). The mean kinetic energy per molecule is \(\tfrac{3}{2}k_BT \approx 6\times10^{-21}\;\mathrm{J}\) at \(300\;\mathrm{K}\)—six times larger than the potential well depth. Assumption 2 (no intermolecular forces between collisions) is well justified.

Misconception alert

“All molecules in a gas move at the same speed.” This is false, and the misconception has real consequences. Even in a gas at equilibrium, molecular speeds span a wide range—from nearly zero to several times \(v_\mathrm{rms}\). The Maxwell–Boltzmann distribution gives the precise probability of finding a molecule in any speed interval. The correct statement is: the distribution of speeds is fixed at fixed temperature—but individual molecules are constantly exchanging energy through collisions, so any given molecule's speed fluctuates wildly. This matters for chemistry: only molecules in the high-speed tail of the distribution have enough energy to surmount an activation barrier and react. Temperature raises the entire distribution, dramatically increasing the fraction in the tail—which is why reaction rates are so sensitive to temperature (Arrhenius equation).

✓ Checkpoint · The Ideal Gas Model

An ideal gas is compressed to half its volume at constant temperature. Which of the five model assumptions is most likely to be violated after compression, and why?
Assumption 1 (point molecules) and assumption 2 (no intermolecular forces between collisions): at half the volume, the number density doubles, molecules are closer together, their own volume is no longer negligible, and they spend a larger fraction of time within range of each other's forces.
Kinetic theory assumes molecular chaos (isotropic velocity distribution). A gas is placed in a gravitational field. Does this violate the assumption? Explain carefully.
Locally, no: at any given height the distribution remains isotropic (gravity is weak compared with molecular kinetic energies). The effect of gravity is to create a density gradient— more molecules at lower heights—but the velocity distribution at each height is still isotropic. The assumption holds locally; it fails globally only in the sense that density is not uniform.
(Trap) Two containers hold the same ideal gas at the same temperature. Container A has \(N\) molecules; container B has \(2N\) molecules in the same volume. A student claims: “Both have the same \(\langle v^2 \rangle\), so they have the same pressure.” Find the error.
\(\langle v^2 \rangle\) depends only on temperature, so it is indeed the same in both. But pressure is proportional to both \(\langle v^2 \rangle\) and \(N/V\) (the number density). Container B has twice the pressure. The student forgot that more molecules means more collisions per unit area per unit time.

Pressure from Molecular Collisions

◈ Pressure from collisions

live bench

Each rebound delivers a tiny impulse; averaged over countless molecules it becomes a steady push, \(P=\tfrac13\rho\overline{c^2}\). Change the speed and the pressure follows.

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Fig 15.2 A molecule of mass \(m\) rebounds elastically from the wall, reversing \(v_x\) and delivering impulse \(2mv_x\). Summing these tiny kicks over every molecule and every second yields the steady pressure on the wall.

Pressure feels continuous. Press your hand against a wall and the force is smooth, steady, uniform. Yet if the kinetic theory is correct, that smooth force is the net result of an enormous number of discrete, random impacts—molecules bouncing off the wall millions of times per second per square millimetre.

The first perplexing question asks how random impacts produce steady pressure. The answer is not philosophical; it is a calculation. We will derive the exact result from Newton's second law, one molecule at a time, and then average over the entire gas.

Step 1: One Molecule, One Wall

Consider a cubic container of side \(L\), volume \(V = L^3\), containing \(N\) molecules. Focus on a single molecule of mass \(m\) moving with velocity \(\vec{v} = (v_x,\, v_y,\, v_z)\).

The molecule travels toward the right-hand wall (perpendicular to \(x\)), hits it, and—by assumption 3 (elastic collision with a wall of infinite mass)—bounces back with \(x\)-component reversed: \[ v_x \;\longrightarrow\; -v_x. \] The \(y\)- and \(z\)-components are unchanged.

Change in momentum of the molecule from this collision: \[ \Delta p_x = m(-v_x) - m(v_x) = -2mv_x. \] By Newton's third law the wall receives impulse \(+2mv_x\).

Step 2: Time Between Successive Hits

After bouncing off the right wall, the molecule travels to the left wall (distance \(L\)), bounces, and returns—a total distance \(2L\) at speed \(|v_x|\).

Time between successive hits on the right wall: \[ \Delta t = \frac{2L}{|v_x|}. \]

Step 3: Force from One Molecule

Force is the rate of momentum transfer. This single molecule delivers impulse \(2mv_x\) every \(\Delta t\) seconds, so the average force it exerts on the right wall is: \[ F_1 = \frac{2mv_x}{\Delta t} = \frac{2mv_x \cdot |v_x|}{2L} = \frac{mv_x^2}{L}. \] (We write \(v_x^2\) because \(v_x \cdot |v_x| = v_x^2\) for \(v_x \gt 0\).)

Step 4: Force from All \(N\) Molecules

Summing over all \(N\) molecules: \[ F = \sum_{i=1}^{N} \frac{m v_{xi}^2}{L} = \frac{m}{L} \sum_{i=1}^{N} v_{xi}^2 = \frac{Nm}{L} \langle v_x^2 \rangle, \] where \(\langle v_x^2 \rangle = \frac{1}{N}\sum_i v_{xi}^2\) is the mean square \(x\)-component of velocity.

Step 5: Isotropy Connects Components to Speed

By assumption 4 (molecular chaos, isotropic distribution), no direction is preferred: \[ \langle v_x^2 \rangle = \langle v_y^2 \rangle = \langle v_z^2 \rangle. \] Since \(v^2 = v_x^2 + v_y^2 + v_z^2\), taking averages: \[ \langle v^2 \rangle = \langle v_x^2 \rangle + \langle v_y^2 \rangle + \langle v_z^2 \rangle = 3\langle v_x^2 \rangle. \] Therefore: \[ \langle v_x^2 \rangle = \frac{\langle v^2 \rangle}{3}. \]

Step 6: Pressure

Pressure is force per unit area; the wall has area \(A = L^2\): \[ p = \frac{F}{A} = \frac{F}{L^2} = \frac{Nm\langle v_x^2 \rangle}{L^3} = \frac{Nm\langle v^2 \rangle}{3V}. \]

Since \(\rho = Nm/V\) is the mass density of the gas: \[ \boxed{p = \frac{1}{3}\rho\langle v^2 \rangle = \frac{1}{3}\frac{Nm}{V}\langle v^2 \rangle.} \]

This is the fundamental result of kinetic theory. Every symbol in it is mechanical—mass, speed, volume—with no mention of temperature or heat. Temperature enters only when we interpret what \(\langle v^2 \rangle\) means, which is the business of a later section.

Step 7: The Ideal Gas Law Emerges

Write \(p = \tfrac{1}{3}(Nm/V)\langle v^2 \rangle\), so: \[ pV = \frac{1}{3}Nm\langle v^2 \rangle = \frac{2}{3}N \cdot \underbrace{\frac{1}{2}m\langle v^2 \rangle}_{\bar{K}_\text{trans}}, \] where \(\bar{K}_\text{trans}\) is the mean translational kinetic energy per molecule.

Below we will show that \(\bar{K}_\text{trans} = \tfrac{3}{2}k_BT\), which immediately gives: \[ pV = \frac{2}{3}N \cdot \frac{3}{2}k_BT = Nk_BT = nRT. \]

This is the ideal gas law, derived from first principles. \[ pV = nRT. \] This closes the forward reference planted in Chapter 14.

Birth of the formula: Why the Cube Assumption Does Not Matter

The derivation above used a cube of side \(L\). A student might worry: what if the container is a sphere, a cylinder, or an irregular shape? The result \(p = \tfrac{1}{3}\rho\langle v^2\rangle\) is independent of container geometry. Here is why. For any convex container, a molecule bouncing between two parallel walls separated by distance \(d_x\) in the \(x\)-direction has round-trip time \(\Delta t = 2d_x/|v_x|\), giving force \(mv_x^2/d_x\) on each wall. Pressure on that wall is \(mv_x^2/(d_x \cdot A_x)\), where \(A_x \cdot d_x = V\), so the contribution is \(mv_x^2/V\)—independent of shape. Summing over all molecules and all three directions gives the same \(p = Nm\langle v^2\rangle/(3V)\). For non-convex containers the argument requires averaging over all possible trajectories, but the result is unchanged. The cube is a convenience, not an assumption.

Misconception alert

“Pressure requires molecules to be moving toward the wall right now.” The derivation shows that only molecules with \(v_x \gt 0\) hit the right wall at any given moment—roughly half the total. A common error is to count only these molecules and forget to divide by the full volume \(V\) (which contains all \(N\) molecules). The factor of \(\tfrac{1}{3}\) in \(p = \tfrac{1}{3}\rho\langle v^2\rangle\) accounts for both the “only half hit each wall” factor (giving \(\tfrac{1}{2}\)) and the isotropy decomposition \(\langle v_x^2\rangle = \tfrac{1}{3}\langle v^2\rangle\). Do not rederive these factors independently and then multiply—the \(\tfrac{1}{3}\) already contains both.

Example 1

Estimate the root-mean-square speed of nitrogen molecules (\(M = 28\;\mathrm{g\,mol^{-1}}\)) at \(T = 300\;\mathrm{K}\), given that the pressure of nitrogen at STP is \(10^5\;\mathrm{Pa}\) and its density is \(\rho = 1.25\;\mathrm{kg\,m^{-3}}\). Do this without using \(\tfrac{3}{2}k_BT\) for kinetic energy.

Solution. From \(p = \tfrac{1}{3}\rho\langle v^2\rangle\): \[ \langle v^2 \rangle = \frac{3p}{\rho} = \frac{3 \times 10^5}{1.25} = 2.4\times10^5\;\mathrm{m^2\,s^{-2}}. \] \[ v_\mathrm{rms} = \sqrt{2.4\times10^5} \approx 490\;\mathrm{m\,s^{-1}}. \] This is the direct mechanical route—pressure and density alone, no temperature, no Boltzmann constant. It confirms that molecular speeds in a gas at room conditions are of order \(500\;\mathrm{m\,s^{-1}}\), setting the scale for the “smell mystery”.

✓ Checkpoint · Pressure from Collisions

A gas is heated at constant volume, doubling its absolute temperature. By what factor does \(v_\mathrm{rms}\) change?
\(v_\mathrm{rms} \propto \sqrt{T}\); doubling \(T\) increases \(v_\mathrm{rms}\) by \(\sqrt{2}\).
Two ideal gases, A (\(M = 4\;\mathrm{g\,mol^{-1}}\), helium) and B (\(M = 32\;\mathrm{g\,mol^{-1}}\), oxygen), are at the same temperature. Find the ratio \(v_\mathrm{rms,A}/v_\mathrm{rms,B}\).
At same \(T\): \(\tfrac{1}{2}Mv_\mathrm{rms}^2 = \tfrac{3}{2}RT\) (per mole), so \(v_\mathrm{rms} \propto M^{-1/2}\). Ratio \(= \sqrt{32/4} = \sqrt{8} = 2\sqrt{2} \approx 2.83\).
(Trap) A student writes: “Since only half the molecules move toward any given wall at once, the pressure formula should have \(\tfrac{1}{6}\), not \(\tfrac{1}{3}\).” Correct the error precisely.
The \(\tfrac{1}{2}\) from “only half move toward the wall” is already embedded in \(\langle v_x^2\rangle = \tfrac{1}{3}\langle v^2\rangle\). That isotropy relation does the counting correctly. Inserting an additional \(\tfrac{1}{2}\) double-counts and gives the wrong answer.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · RMS speed from temperature

Find the root-mean-square speed of oxygen molecules (\(M = 32\;\mathrm{g\,mol^{-1}}\)) at \(T = 300\;\mathrm{K}\).

Solution. Given: \(M = 32\;\mathrm{g\,mol^{-1}} = 0.032\;\mathrm{kg\,mol^{-1}}\), \(T = 300\;\mathrm{K}\).
Find: \(v_\mathrm{rms}\).
Setup: The temperature route uses \(v_\mathrm{rms} = \sqrt{3RT/M}\), which follows directly from \(\tfrac12 m\langle v^2\rangle = \tfrac32 k_BT\) once written per mole.
Solve: \[ v_\mathrm{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3(8.314)(300)}{0.032}} = \sqrt{2.34\times10^5} \approx 484\;\mathrm{m\,s^{-1}}. \] Answer: \(\boxed{v_\mathrm{rms} \approx 484\;\mathrm{m\,s^{-1}}}\)
Check: Dimensions: \(\sqrt{(\mathrm{J\,mol^{-1}K^{-1}})(\mathrm{K})/(\mathrm{kg\,mol^{-1}})} = \sqrt{\mathrm{J/kg}} = \sqrt{\mathrm{m^2\,s^{-2}}} = \mathrm{m\,s^{-1}}\) ✓. Comparison: nitrogen (\(M=28\)) gives \(\sim 490\;\mathrm{m\,s^{-1}}\) at the same temperature; oxygen is heavier, so it is slightly slower, as found. ✓

Example 2 · Mean kinetic energy per molecule and per mole

At \(T = 300\;\mathrm{K}\), find the average translational kinetic energy of a single molecule of any ideal gas, and the total translational kinetic energy of one mole.

Solution. Given: \(T = 300\;\mathrm{K}\).
Find: \(\bar K\) per molecule and \(U_\mathrm{tr}\) per mole.
Setup: Translational KE per molecule is \(\bar K = \tfrac32 k_BT\) (mass-independent); per mole multiply by \(N_A\), or equivalently use \(\tfrac32 RT\).
Solve: \[ \bar K = \tfrac32 k_BT = \tfrac32(1.381\times10^{-23})(300) \approx 6.21\times10^{-21}\;\mathrm{J}, \] \[ U_\mathrm{tr} = \tfrac32 RT = \tfrac32(8.314)(300) \approx 3.74\times10^{3}\;\mathrm{J}. \] Answer: \(\boxed{\bar K \approx 6.21\times10^{-21}\;\mathrm{J},\qquad U_\mathrm{tr} \approx 3.74\;\mathrm{kJ\,mol^{-1}}}\)
Check: Consistency: \(N_A\bar K = (6.022\times10^{23})(6.21\times10^{-21}) \approx 3.74\times10^{3}\;\mathrm{J}\), matching the per-mole value ✓. The result is independent of the gas's identity, as equipartition demands. ✓

Example 3 · Ratio of rms speeds at equal temperature

At the same temperature, find the ratio \(v_\mathrm{rms}(\mathrm{H_2})/v_\mathrm{rms}(\mathrm{O_2})\). Take \(M_{\mathrm{H_2}} = 2\;\mathrm{g\,mol^{-1}}\), \(M_{\mathrm{O_2}} = 32\;\mathrm{g\,mol^{-1}}\).

Solution. Given: same \(T\); \(M_{\mathrm{H_2}} = 2\), \(M_{\mathrm{O_2}} = 32\;\mathrm{g\,mol^{-1}}\).
Find: the speed ratio.
Setup: At fixed \(T\), \(v_\mathrm{rms} \propto 1/\sqrt{M}\), so the ratio depends only on the masses.
Solve: \[ \frac{v_\mathrm{rms}(\mathrm{H_2})}{v_\mathrm{rms}(\mathrm{O_2})} = \sqrt{\frac{M_{\mathrm{O_2}}}{M_{\mathrm{H_2}}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4. \] Answer: \(\boxed{v_\mathrm{rms}(\mathrm{H_2}) = 4\,v_\mathrm{rms}(\mathrm{O_2})}\)
Check: Hydrogen is \(16\times\) lighter, and the speed scales as the square root of the mass ratio, \(\sqrt{16}=4\) ✓. The lighter gas is faster, as expected. ✓

Example 4 · Pressure from number density

A vessel contains \(n = 2.5\times10^{25}\) molecules per cubic metre at \(T = 300\;\mathrm{K}\). Find the pressure.

Solution. Given: \(n = 2.5\times10^{25}\;\mathrm{m^{-3}}\), \(T = 300\;\mathrm{K}\).
Find: \(P\).
Setup: The microscopic form of the ideal gas law is \(P = n k_BT\), with \(n\) the number density.
Solve: \[ P = n k_BT = (2.5\times10^{25})(1.381\times10^{-23})(300) \approx 1.04\times10^{5}\;\mathrm{Pa}. \] Answer: \(\boxed{P \approx 1.04\times10^{5}\;\mathrm{Pa} \approx 1.02\;\mathrm{atm}}\)
Check: Sanity: an ideal gas at STP has number density \(\approx 2.69\times10^{25}\;\mathrm{m^{-3}}\) (Loschmidt) and \(P \approx 1\;\mathrm{atm}\); a slightly smaller density at \(300\;\mathrm{K}\) landing near \(1\;\mathrm{atm}\) is exactly right. ✓

Example 5 · Internal energy of a diatomic gas

Find the internal energy of \(3\;\mathrm{mol}\) of nitrogen (\(f = 5\): translation \(+\) rotation) at \(T = 350\;\mathrm{K}\).

Solution. Given: \(n = 3\;\mathrm{mol}\), \(f = 5\), \(T = 350\;\mathrm{K}\).
Find: \(U\).
Setup: By equipartition each active degree of freedom holds \(\tfrac12 k_BT\) per molecule, so \(U = \tfrac{f}{2}nRT\).
Solve: \[ U = \tfrac{5}{2}nRT = \tfrac{5}{2}(3)(8.314)(350) \approx 2.18\times10^{4}\;\mathrm{J}. \] Answer: \(\boxed{U \approx 2.18\times10^{4}\;\mathrm{J}}\)
Check: Limiting case: if the molecule were monatomic (\(f=3\)), \(U\) would be \(\tfrac35\) as large, \(\approx 1.31\times10^4\;\mathrm{J}\), smaller because two rotational modes are switched off ✓.

Problem bank

Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.

✎ Free practice questions

Three Speeds of Oxygen
At \(T = 27\,{}^\circ\mathrm{C}\), find \(v_\mathrm{rms}\), \(\bar{v}\), and \(v_p\) for oxygen molecules (\(M = 32\;\mathrm{g\,mol^{-1}}\)). Which of the three should be used if you want to compute the pressure of the gas?
\(T=300\;\mathrm{K}\): \(v_p=\sqrt{2RT/M}\approx395\;\mathrm{m\,s^{-1}}\); \(\bar{v}=\sqrt{8RT/\pi M}\approx446\;\mathrm{m\,s^{-1}}\); \(v_\mathrm{rms}=\sqrt{3RT/M}\approx484\;\mathrm{m\,s^{-1}}\). Use \(v_\mathrm{rms}\) for pressure: \(p=\tfrac{1}{3}\rho v_\mathrm{rms}^2\).
Heating a Rigid Container
A gas is enclosed in a rigid container. Its temperature is raised from \(300\;\mathrm{K}\) to \(1200\;\mathrm{K}\). By what factor do the following change: (a) \(v_\mathrm{rms}\), (b) mean KE per molecule, (c) total internal energy (monatomic gas), (d) pressure?
(a) \(v_\mathrm{rms}\propto\sqrt{T}\): factor \(\sqrt{4}=2\). (b) \(\bar{K}\propto T\): factor 4. (c) \(U\propto T\): factor 4. (d) \(p\propto T\) at constant \(V\): factor 4.
A Two-Gas Mixture
\(2\;\mathrm{mol}\) of helium and \(3\;\mathrm{mol}\) of argon are mixed in a \(10\;\mathrm{L}\) container at \(300\;\mathrm{K}\). Find: (a) total pressure, (b) total internal energy, (c) \(C_V\) of the mixture per mole of mixture.
(a) \(p=n_\mathrm{tot}RT/V=(5)(8.314)(300)/0.01=1.247\times10^6\;\mathrm{Pa}\). (b) \(U=\tfrac{3}{2}(2)RT+\tfrac{3}{2}(3)RT=\tfrac{15}{2}RT=\tfrac{15}{2}(8.314)(300)=18{,}707\;\mathrm{J}\). (c) \(C_{V,\mathrm{mix}}=(2\cdot\tfrac{3}{2}R+3\cdot\tfrac{3}{2}R)/5=\tfrac{3}{2}R\) per mole of mixture. (Both species are monatomic, so the mixture \(C_V\) is also \(\tfrac{3}{2}R\).)
Translational KE Equals \(\tfrac{3}{2}pV\)
A gas of molar mass \(M\) is in a container of volume \(V\) at pressure \(p\) and temperature \(T\). Show that the total translational KE of all the molecules equals \(\tfrac{3}{2}pV\). Does this result depend on \(f\)? Does it depend on what gas it is?
Total translational KE \(= N\cdot\tfrac{3}{2}k_BT = nN_A\cdot\tfrac{3}{2}(R/N_A)T = \tfrac{3}{2}nRT\). From \(pV = nRT\): total translational KE \(= \tfrac{3}{2}pV\). ✓ It does not depend on \(f\) (rotational and vibrational modes contribute to \(U\) but not to translational KE). It does not depend on the identity of the gas—only on \(p\) and \(V\).
The Molecular Beam
In a molecular beam experiment, molecules effuse from an oven through a small hole and travel to a detector. The speed distribution of molecules hitting the hole is not \(f(v)\) but is proportional to \(v\,f(v)\) (faster molecules hit the hole more often per unit time). (a) Explain physically why the factor of \(v\) appears. (b) Show that the mean speed of molecules in the beam is \(\langle v\rangle_\mathrm{beam} = \tfrac{3\pi}{8}\bar{v}\). (c) Is the beam hotter or cooler (in terms of mean KE) than the oven gas?
(a) The flux of molecules hitting a hole per unit time is proportional to \(v\) per molecule (faster molecules sweep out a larger volume per unit time and strike the hole more frequently). (b) Beam distribution: \(g(v) \propto v\,f(v) \propto v^3 e^{-mv^2/2k_BT}\). Normalising and evaluating \(\langle v\rangle_\mathrm{beam} = \int_0^\infty v\,g(v)\,dv / \int_0^\infty g(v)\,dv\) with \(\int_0^\infty v^4 e^{-\alpha v^2}\,dv = \tfrac{3\sqrt{\pi}}{8\alpha^{5/2}}\) and \(\bar{v} = \sqrt{8k_BT/\pi m}\) gives \(\langle v\rangle_\mathrm{beam} = \tfrac{3\pi}{8}\bar{v}\). ✓ (c) \(\langle v^2\rangle_\mathrm{beam} = 4k_BT/m\), so \(\langle KE\rangle_\mathrm{beam} = 2k_BT\) versus \(\tfrac{3}{2}k_BT\) for the oven. The beam is effectively hotter: faster molecules are over-sampled.

Chapter test

A paper drawn at random from this chapter's bank. Choose the exam you are training for — the marking scheme, pace and difficulty mix follow the real pattern. Work on paper; when you finish (or the clock runs out), the answers are revealed and you mark yourself honestly.

Pattern

Questions

Clock

Free practice questions

Mean Free Path Meets the Quantum Limit
At what pressure does the mean free path of nitrogen equal its de Broglie wavelength \(\lambda_\mathrm{dB} = h/mv_\mathrm{rms}\)? Take \(d = 3.7\times10^{-10}\;\mathrm{m}\), \(T = 300\;\mathrm{K}\). Comment on whether ideal gas theory is valid at this pressure.
\(\lambda = k_BT/(\sqrt{2}\,\pi d^2 p)\); \(\lambda_\mathrm{dB} = h/\sqrt{3mk_BT}\). Setting equal and solving for \(p\): \(p = k_BT\sqrt{3mk_BT}/(\sqrt{2}\,\pi d^2 h)\). With \(m = 28\times1.66\times10^{-27} = 4.65\times10^{-26}\;\mathrm{kg}\), \(k_BT = 4.14\times10^{-21}\;\mathrm{J}\), \(h = 6.63\times10^{-34}\;\mathrm{J\,s}\): \(\lambda_\mathrm{dB} \approx 2.8\times10^{-11}\;\mathrm{m}\), giving \(p\) of order \(10^8\;\mathrm{Pa}\sim10^3\;\mathrm{atm}\). At such pressures ideal gas theory breaks down completely; intermolecular forces dominate and the van der Waals equation is needed.
The Isothermal Atmosphere
Assume Earth's atmosphere is an ideal gas at uniform temperature \(T\) with mean molar mass \(M = 29\;\mathrm{g\,mol^{-1}}\). (a) Show that pressure varies with height \(h\) as \(p(h) = p_0\,e^{-Mgh/RT}\) (the barometric formula). (Hint: consider hydrostatic equilibrium \(dp = -\rho g\,dh\) and use the ideal gas law to eliminate \(\rho\).) (b) At what height does the pressure drop to \(1/e\) of its sea-level value? Evaluate numerically at \(T = 250\;\mathrm{K}\). (c) The actual atmosphere is not isothermal. State one consequence of this for the real pressure profile.
(a) Hydrostatic: \(dp = -\rho g\,dh\). Ideal gas: \(\rho = pM/RT\). So \(dp/p = -(Mg/RT)\,dh\). Integrating from \(0\) to \(h\): \(p = p_0\exp(-Mgh/RT)\). ✓ (b) Pressure drops to \(p_0/e\) when \(Mgh/RT = 1\): \(h = RT/(Mg) = (8.314)(250)/(0.029\times9.8) \approx 7.3\;\mathrm{km}\). (c) The real atmosphere has a temperature lapse rate (\(\approx\)6–10 K km\(^{-1}\)); pressure drops more steeply at low altitude (cooler, denser air) than the isothermal model predicts.