Kinetic Theory: Class 11 Physics · Chapter 12

Physics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 22 Oct 2026 · Updated 22 Oct 2026

85 min
1 simulation
18 practice questions
24 MCQs

What you will learn in Kinetic Theory

Describe the molecular picture of matter and estimate molecular sizes and spacings from densities
Use the ideal gas equation in the forms PV = μRT = Nk_BT and apply Boyle's, Charles' and Dalton's laws
Derive P = ⅓nm⟨v²⟩ from molecular impacts on a wall and state the assumptions of the kinetic model
Interpret temperature as the average translational kinetic energy of a molecule and calculate rms speeds
State the law of equipartition of energy and count the degrees of freedom of different molecules
Predict the molar heat capacities of gases and solids and explain where the predictions fail
Derive and use the mean free path l = 1/(√2 nπd²)

Original companion text aligned to NCERT Physics Class 11, Chapter 12, pp. 244–258. Not affiliated with or endorsed by NCERT.

12.1 Introduction

Pump up a bicycle tyre on a hot afternoon and it turns hard. Inside, roughly $10^{23}$ air molecules are racing about at hundreds of metres per second and drumming on the rubber. Kinetic theory (अणुगति सिद्धांत) explains how a gas behaves by starting from exactly this picture: a swarm of tiny molecules in fast, random motion.

Why should so simple a picture work? Forces between atoms reach only a few atomic sizes. Gas molecules are usually far apart, so they feel these forces only in brief collisions and simply coast in between.

The model pays off handsomely. From it we shall derive the gas laws, discover that temperature measures molecular energy, predict how much heat a gas absorbs per kelvin, and estimate how far a molecule flies between collisions. Most of this was worked out in the nineteenth century by Clausius, Maxwell and Boltzmann.

12.2 Molecular nature of matter

In his famous lectures (1963), Richard Feynman asked a thought question: if all scientific knowledge were lost and only one sentence could be handed on, which sentence would carry the most? His choice was the atomic hypothesis. In our words: everything is built from atoms, tiny particles that never stop moving, that pull on one another when slightly separated and push apart when crowded together. Most of this chapter grows out of that sentence.

From historyNCERT Box: Atomic Hypothesis in Ancient India and Greece · p. 245

Atoms before experiments

Indian philosophers of the Vaisheshika school, traditionally traced to the sage Kanada, reasoned their way to atoms more than two thousand years ago. Their smallest unit, the paramanu (परमाणु), could not be split, created or destroyed. Combinations explained variety: a pair of paramanus made a dvyanuka, dyads grouped further into a tryanuka, and what a substance was like depended on which paramanus it held and in what numbers. Earth, water, fire and air each had their own kind of paramanu; space (akasha), by contrast, was continuous, with no atoms at all.

In Greece, Democritus (about 460–370 BCE), following his teacher Leucippus, called the smallest pieces atomos, meaning uncuttable. He imagined atoms of many shapes and sizes and even guessed that jagged atoms make a taste sour while smooth, round ones make it sweet. Without experiments to decide between rival guesses, the idea stayed a philosophy for two thousand years. What finally turned it into science was a humble instrument: the chemist's balance.

Two chemical regularities needed explaining. First, a compound always contains its elements in a fixed proportion by mass: pure water, from a Himalayan spring or from a laboratory reaction, has $8 g$ of oxygen for every $1 g$ of hydrogen. This is the law of definite proportions. Second, if the same two elements can make several compounds, the masses of one that combine with a fixed mass of the other are in a ratio of small whole numbers. With $12 g$ of carbon, carbon monoxide has $16 g$ of oxygen and carbon dioxide has $32 g$ , a ratio of $1 : 2$ . This is the law of multiple proportions.

Dalton's explanation (1808): an element is a collection of one kind of atom, every hydrogen atom alike and unlike any oxygen atom, and a compound forms when atoms link in small whole-number groups. Fixed, whole-number mass ratios then follow at once.

Gases supplied a clue about volumes. Gay-Lussac found that reacting gases combine in simple ratios by volume: two litres of hydrogen and one litre of oxygen give two litres of steam, all measured at the same temperature and pressure. In 1811 Amedeo Avogadro explained this with a bold idea, Avogadro's law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. The volume ratio $2 : 1 : 2$ is then a ratio of molecules, as in $2 H_{2} + O_{2} \overset{}{\to} 2 H_{2} O$ . This works only if hydrogen and oxygen travel as two-atom molecules, so chemists came to speak of a molecular theory of matter.

Timeline

From a guess to a measured fact

c. 400 BCE
Democritus names the atom
Matter is pictured as uncuttable particles moving through empty space.
1662
Boyle's law
Robert Boyle shows that the pressure of trapped air varies inversely as its volume.
1738
Pressure from impacts
Daniel Bernoulli explains gas pressure as the result of particles striking the walls.
1808–1811
Dalton, Gay-Lussac, Avogadro
Atomic theory, combining volumes and the equal-volumes hypothesis.
1857–1860
Kinetic theory takes shape
Clausius introduces the mean free path; Maxwell finds the spread of molecular speeds.
1908
Atoms confirmed
Jean Perrin's measurements of Brownian motion match Einstein's 1905 theory and settle the debate.

Even around 1900 a few leading scientists doubted that atoms were real. The doubts faded after Perrin's experiments. Today electron microscopes and the scanning tunnelling microscope (invented in 1981) produce images in which single atoms can be picked out.

Modern measurements put numbers on this picture (Table 12.1). The natural unit is the ångström: 1 Å is $10^{- 10} m$ . Notice where the big jump comes. Molecules in a gas are only about ten times farther apart than in a liquid, yet each flies roughly twenty times that spacing before meeting another molecule. That flight distance, the mean free path, is the topic of Section 12.7.

Table 12.1Typical lengths in matterNCERT p. 245

Length	Typical value	What it explains
Size of an atom	about 1 Å	the basic scale
Spacing of atoms in a solid	2–3 Å, centre to centre	atoms locked in place: a fixed shape
Spacing in a liquid	about the same as in a solid	atoms slide past one another: liquids flow
Spacing in a gas at 1 atm	tens of Å	molecules feel forces only in collisions
Mean free path in a gas at 1 atm	several hundred Å (about 700 Å in room air)	a gas spreads out unless a container stops it

The force between two atoms changes character with distance (Fig. 12.1). Pushed closer than their natural spacing, they repel strongly. A little farther apart they attract. Farther still, the force dies away. This short reach explains why solids and liquids hold together while a gas behaves as if its molecules hardly interact.

2/3Slightly beyond $r_{0}$ they attract: this holds solids and liquids together.

Fig. 12.1The force between two atoms: strong repulsion when squeezed, weak attraction a little apart, almost nothing far away.NCERT p. 245

A graph of force against separation: positive values mean repulsion, negative values attraction. The curve plunges from a large repulsion at small separation, crosses zero at the natural spacing r0, dips to a shallow attractive minimum and then flattens to zero. In the first step two atoms are squeezed together with arrows pushing them apart, and a dot sits on the steep repulsive part of the curve. In the second step the atoms are a few ångström apart with arrows pulling them together, and the dot sits at the bottom of the attractive dip. In the last step the atoms are far apart with no arrows, and the dot is on the flat tail where the force is almost zero.

SimulationSame molecules, three states

A closed container of water molecules with a thermometer and a temperature slider from minus 20 to 140 degrees Celsius. Below 0 degrees the molecules sit in a regular pattern and jiggle in place; between 0 and 100 degrees they slide past one another as a liquid; above 100 degrees they fly apart and fill the container as a gas. A label names the state.

Fig. 12.2Drag the temperature slider and watch how the spacing and motion of the water molecules change (sizes and spacings are not to scale).

Try this

Cool the water below 0 °C. Do the molecules stop moving, or do they jiggle in place?
Heat it past 100 °C. How does the spacing in the gas compare with the spacing in the liquid?
In which state does a molecule travel farthest before it meets another?

Watch a busy railway platform from a footbridge: the crowd's overall density hardly changes, yet every person in it is moving. A gas in a stoppered bottle is like that, only far busier. Its molecules collide billions of times a second and change speed at every collision; only averages such as the pressure and the mean energy stay steady. This is a dynamic equilibrium.

Zoom further in and the uncuttable atom turns out to have parts: electrons around a nucleus of protons and neutrons, which are themselves built from quarks, and nobody yet knows whether quarks are the final layer. None of that matters here. Treating each molecule as a tiny ball in endless motion is enough to explain gases and, with one extra idea, solids.

12.3 Behaviour of gases

Gases follow the simplest rules of the three states. Press the plunger of a sealed syringe of air and the pressure climbs; warm the syringe and the air pushes back harder. As long as the gas is thin and well above its liquefying temperature, all such changes fit one equation (Chapter 10):

P V = K T

(12.1)

where

$T$: the absolute (kelvin) temperature
$K$: a constant for the given sample

The constant $K$ depends on how much gas there is: double the amount at the same $P$ and $T$ and the volume doubles, so $K$ doubles. In molecular terms $K$ grows in step with the number of molecules $N$ , and we write $K = N k_{B}$ . Experiments show that $k_{B}$ is universal: hydrogen, helium and carbon dioxide all give the same value. It is the Boltzmann constant.

\frac{P_{1} V_{1}}{N_{1} T_{1}} = \frac{P_{2} V_{2}}{N_{2} T_{2}} = k_{B}

(12.2)

Read Eq. (12.2) the other way round: two samples of different gases with the same $P$ , $V$ and $T$ must contain the same $N$ . Avogadro's hypothesis has become a consequence of the gas law. The number of molecules in one mole is the Avogadro number $N_{A} \approx 6.02 \times 10^{23}$ . At $273 K$ and $1 a t m$ one mole of any gas fills about $22.4 L$ , and its mass in grams equals its molecular mass in u.Chemists now usually take standard pressure as $1 b a r = 10^{5} P a$ , a little below 1 atm; at 273 K and 1 bar a mole fills 22.7 L. With the moles counted, the gas law takes its familiar form:

ideal gas equation

P V = μ R T

(12.3)

where

$μ$: the number of moles
$R = N_{A} k_{B}$: the universal gas constant, $8.314 J m o l^{- 1} K^{- 1}$

μ = \frac{M}{M_{0}} = \frac{N}{N_{A}}

(12.4)

where

$M$: the mass of the sample
$M_{0}$: the molar mass
$N$: the number of molecules

Putting $μ = N / N_{A}$ into Eq. (12.3) gives $P V = N k_{B} T$ . Dividing by $V$ brings in the number density $n = N / V$ , the number of molecules per unit volume:

P = n k_{B} T

where

$n$: molecules per unit volume
$k_{B}$: $1.38 \times 10^{- 23} J / K$

Writing $μ = M / M_{0}$ instead, and the mass density as $ρ = M / V$ , gives a form that needs no molecule counting:

P = \frac{ρ R T}{M_{0}}

(12.5)

where

$ρ$: the mass density of the gas

Definition

Ideal gas

A model gas whose molecules have no size and exert no forces except in collisions, so that it would obey $P V = μ R T$ under all conditions. Real gases approach it when they are thin and hot.

How close does a real gas come? A handy test is the ratio $Z = P V / μ R T$ , which is exactly $1$ for an ideal gas. Fig. 12.3 sketches $Z$ against pressure for one real gas at three temperatures. When the gas is cold and compressed, its molecules are crowded: their attractions and their own sizes matter, and $Z$ strays from $1$ . As the pressure falls, every curve heads back to $Z = 1$ , because widely spaced molecules hardly interact.

3/3Hotter gases stay closer to $Z = 1$ , and every curve meets it as $P \to 0$ .

Fig. 12.3Real gases approach ideal behaviour at low pressure and high temperature (a sketch, not data for a particular gas).NCERT Fig. 12.1 · p. 247

Axes show the ratio Z = PV/μRT upwards and pressure across. A dashed horizontal line at Z = 1 is labelled ideal gas. In the first step only this line is drawn. In the second step a curve for low temperature starts at Z = 1, dips well below the line and rises again at high pressure. In the third step curves for a medium and a high temperature are added: the medium one dips only a little, the high one rises gently above 1, and a ring at the left marks where all three curves meet the ideal line as the pressure goes to zero.

Hold both the amount of gas and its temperature steady, and Eq. (12.3) leaves the product $P V$ unchanged:

Boyle's law

P V = constant

(12.6)

Squeeze a sealed syringe to half its volume at room temperature and the pressure inside doubles: that is Boyle's law, $P V$ constant at fixed $T$ . Leave a balloon on a sunny windowsill and it swells, its volume rising in step with the kelvin temperature while the pressure stays near atmospheric: that is Charles' law, $V \propto T$ at fixed $P$ . Real gases obey both closely when thin and warm, and stray near liquefaction (Fig. 12.4).

Air is a mixture, and each of its gases pushes on the walls as if the others were not there. Put $μ_{1}$ moles of one gas, $μ_{2}$ of another and so on, none of them reacting, into a closed vessel (volume $V$ , temperature $T$ ). As far as the gas law is concerned a molecule is a molecule, so only the total amount counts:

P V = (\sum_{i} μ_{i}) R T

(12.7)

where

$\sum_{i} μ_{i}$: $μ_{1} + μ_{2} + \dots$ , the total number of moles

\begin{aligned} P & = μ_{1} \frac{R T}{V} \\ + μ_{2} \frac{R T}{V} + \dots \end{aligned}

(12.8)

Dalton's law of partial pressures

P = P_{1} + P_{2} + \dots

(12.9)

Read the first term on its own: $μ_{1} R T / V$ is what a gauge would show if gas 1 were alone in the vessel. We call it the partial pressure $P_{1}$ of that gas. The reading for the mixture is simply the sum of what each gas would give by itself: Dalton's law of partial pressures.

1/3At fixed $T$ , $P \propto 1 / V$ : each curve is one temperature, and a hotter gas lies farther out.

Fig. 12.4The gas laws that follow from the ideal gas equation.NCERT Figs. 12.2, 12.3 · p. 247

The first step shows pressure against volume for a fixed amount of gas: two curves that fall as the volume grows, the outer one labelled hotter and the inner one cooler. The second step shows volume against kelvin temperature at fixed pressure: two straight lines that, extended as dashed lines, pass through 0 K, the steeper one labelled lower pressure. The third step shows three bars of pressure: gas 1 alone, gas 2 alone, and the mixture, whose bar is the first two stacked on top of each other.

The next three examples use densities to estimate how much room molecules actually take up.

Example 12.1NCERT Example 12.1 · p. 247

How empty is a vapour?

Liquid ethanol has a density of $789 k g / m^{3}$ . Its vapour at the normal boiling point ( $78^{\circ} C$ and $1 a t m$ ) has a density of $1.60 k g / m^{3}$ . Take the volume of all the molecules (the molecular volume) to be the volume of the liquid they would form. What fraction of the vapour's volume do the molecules themselves fill?

Try it yourself first. Then reveal the solution one step at a time.

Example 12.2NCERT Example 12.2 · p. 248

The size of a molecule

Use the data of Example 12.1 to estimate the radius of an ethanol molecule ( $C_{2} H_{5} O H$ , molar mass $46 g / m o l$ ).

Try it yourself first. Then reveal the solution one step at a time.

Example 12.3NCERT Example 12.3 · p. 248

How far apart are vapour molecules?

Estimate the average distance between neighbouring molecules in ethanol vapour at $78^{\circ} C$ and $1 a t m$ , and compare it with the size of a molecule.

Try it yourself first. Then reveal the solution one step at a time.

Example 12.4NCERT Example 12.4 · p. 248

A welding gas mixture

A welding cylinder holds a non-reacting mixture of argon (atomic mass $39.9 u$ ) and carbon dioxide (molecular mass $44.0 u$ ). Their partial pressures are in the ratio $4 : 1$ . Find the ratio, argon to carbon dioxide, of (a) the numbers of molecules and (b) the mass densities.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

How many moles of air are in a

2.0 L

bottle at

27^{\circ} C

and

1.0 a t m

? Take

1 a t m = 1.01 \times 10^{5} P a

mol

Show answer

Answer 0.081 mol

μ = (1.01 \times 10^{5}) (2.0 \times 10^{- 3}) / (8.314 \times 300) \approx 0.081 m o l

. Forgetting to convert litres to cubic metres gives an answer 1000 times too big.

12.4 Kinetic theory of an ideal gas

Now we build the model behind the gas laws. Its rules are few:

Huge numbers. Even a cupful of air holds about $10^{22}$ molecules, so averages over them are extremely reliable.
Points in empty space. Each molecule is tiny compared with the gaps around it, so for most of the time it feels nothing and flies straight at a steady speed.
Brief, elastic encounters. A molecule changes velocity only when it meets another molecule or a wall. These collisions conserve kinetic energy as well as momentum, and each is over in a moment compared with the flight time between collisions.
No favourite direction. On average the molecules move equally in every direction.

12.4.1 Pressure of an ideal gas

Pressure is what the wall feels when molecules bounce off it. To find it we count the momentum delivered to the wall each second (Fig. 12.5).

2/3Only molecules within $v_{x} Δ t$ of the wall, and moving towards it, arrive in time $Δ t$ .

Fig. 12.5Pressure comes from the momentum molecules deliver to the wall.NCERT Fig. 12.4 · p. 249

A flat wall of area A stands at the right. In the first step one molecule approaches the wall at a slant and leaves at the mirror-image slant: only the part of its velocity towards the wall reverses, and an arrow beside the wall shows the impulse 2mvx it receives. In the second step a shaded slab of thickness vx times delta t lies against the wall; some molecules in it move towards the wall (highlighted, arrows to the right) and others away from it (grey, arrows to the left), and a label says only those moving towards the wall hit it. In the third step many molecules move in random directions throughout the container, and the result P = one third n m times the mean square speed is shown.

DerivationPressure from molecular impacts

Take a cubical box of side $l$ and set the $x$ , $y$ and $z$ axes along its edges. Focus on one face, the one at right angles to the $x$ -axis; its area is $A = l^{2}$ .

A molecule of mass $m$ and velocity $(v_{x}, v_{y}, v_{z})$ bounces elastically off this wall. Only the component normal to the wall reverses; $v_{y}$ and $v_{z}$ are unchanged. Its $x$ -momentum goes from $m v_{x}$ to $- m v_{x}$ , a change of

Δ p_{x} = - 2 m v_{x}

so, by conservation of momentum, the wall receives $2 m v_{x}$ from each such hit.

Which molecules manage to reach the wall during a short interval $Δ t$ ? One travelling at $v_{x}$ covers $v_{x} Δ t$ in that time, so only those starting in the thin slab of thickness $v_{x} Δ t$ beside the wall (volume $A v_{x} Δ t$ ) can arrive, and of these, on average, only the half heading towards the wall do. If $n$ molecules per unit volume have this velocity, the number of hits is $\frac{1}{2} n A v_{x} Δ t$ and the momentum delivered is

\begin{aligned} Q & = (2 m v_{x}) \\ \times (\frac{1}{2} n A v_{x} Δ t) \end{aligned}

(12.10)

Force is momentum delivered per unit time, and pressure is force per unit area:

P = \frac{Q}{A Δ t} = n m v_{x}^{2}

(12.11)

Real molecules do not share one velocity: Eq. (12.11) is the contribution of one velocity group. Adding the groups replaces $v_{x}^{2}$ by its average over all the molecules, with $n$ now the total number density:

P = n m ⟨ v_{x}^{2} ⟩

(12.12)

The gas has no preferred direction (it is isotropic), so the three components share the same mean square. Since $v^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ ,

\begin{aligned} ⟨ v_{x}^{2} ⟩ & = ⟨ v_{y}^{2} ⟩ = ⟨ v_{z}^{2} ⟩ \\ = \frac{1}{3} ⟨ v^{2} ⟩ \end{aligned}

(12.13)

Putting this into Eq. (12.12):

pressure of an ideal gas

P = \frac{1}{3} n m ⟨ v^{2} ⟩

(12.14)

where

$⟨ v^{2} ⟩$: the mean square speed of the molecules

Two worries about this derivation. First, we used a cube; would a round cylinder or a bottle give a different pressure? No. Only a small patch of wall entered the argument, and $A$ and $Δ t$ dropped out of the answer. This agrees with Pascal's law (Chapter 9): a fluid at rest has the same pressure everywhere.

Second, we ignored molecules hitting one another. A collision does change a molecule's velocity, but in a steady gas the collisions shuffle molecules between velocity groups without changing how many sit in each group, like passengers swapping seats on a full train. The average $⟨ v_{x}^{2} ⟩$ , which is all the formula uses, stays the same, provided each collision is brief compared with the free flight between collisions.

InteractiveGas in a box

Molecules of helium, nitrogen or carbon dioxide move in a one-litre box closed on the right by a piston. Sliders set the temperature from 100 to 600 kelvin, the volume from 0.4 to 1 litre and the number of molecules drawn. Molecules flash when they strike the piston, and a dial shows the pressure worked out from those impacts, with a tick for the value nkBT predicts. A histogram shows the spread of molecular speeds with the predicted curve and a line at the rms speed. A table lists the measured and predicted pressure, the number of hits, the rms speed and the mean kinetic energy per molecule.

Fig. 12.6Press Play. Each drawn molecule stands for

10^{20}

real ones; the dial adds up the momentum they deliver to the piston.

Try this

Experiments with the gas box

Let the count run for half a minute. How close does the measured pressure get to the predicted $n k_{B} T$ ?
Halve the volume at a fixed temperature. What happens to the pressure, and to the rate of hits on the piston?
Double the absolute temperature at fixed volume. By what factor does $v_{rms}$ grow? By what factor does the pressure grow?
Switch from nitrogen to helium at the same temperature. Which readings change, and which stay the same?

Quick check

1 of 3

In the derivation, why does each bounce give the wall an impulse of

2 m v_{x}

rather than

m v_{x}

The

x

-momentum changes from

+ m v_{x}

- m v_{x}

The molecule strikes the wall twiceThe wall recoils with speed

v_{x}

Only half the molecules move towards the wall

Show answer

Answer (A) The $x$ -momentum changes from $+ m v_{x}$ to $- m v_{x}$

The change is

(- m v_{x}) - (+ m v_{x}) = - 2 m v_{x}

, so the wall gets

2 m v_{x}

. The factor

\frac{1}{2}

for molecules moving towards the wall enters separately, when we count hits.

12.4.2 Kinetic interpretation of temperature

DerivationTemperature and molecular energy

Multiply Eq. (12.14) by $V$ and write $N = n V$ for the number of molecules:

P V = \frac{1}{3} N m ⟨ v^{2} ⟩

(12.15a)

P V = \frac{2}{3} N (\frac{1}{2} m ⟨ v^{2} ⟩)

(12.15b)

Inside the bracket sits the mean translational kinetic energy per molecule. In an ideal gas the molecules exert no forces on one another between collisions, so there is no potential energy to add, and the total translational kinetic energy is

E = N \times \frac{1}{2} m ⟨ v^{2} ⟩

(12.16)

so that Eq. (12.15b) becomes

P V = \frac{2}{3} E

(12.17)

Compare this with the gas law written as $P V = N k_{B} T$ :

E = \frac{3}{2} N k_{B} T

(12.18)

average kinetic energy of a molecule

\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T

(12.19)

Pause on Eq. (12.19), the key result of the chapter. A thermometer in a gas is, in effect, measuring how vigorously its molecules move. At $300 K$ every gas, light helium or heavy xenon, has the same average translational energy per molecule, about $6.2 \times 10^{- 21} J$ , and doubling the kelvin temperature doubles it; pressure and volume do not enter. The constant $k_{B}$ is the exchange rate between the thermometer's scale and molecular energy.

Two bonuses follow. Eq. (12.18) shows that the internal energy of an ideal gas changes only when its temperature changes, a fact used constantly in thermodynamics (Chapter 11). And since the model was built to agree with $P V = N k_{B} T$ , every gas law that follows from that equation now has a molecular explanation.Here $E$ is only the energy of moving from place to place. Molecules that rotate or vibrate store extra energy, which the next section counts.

In a mixture of non-reacting ideal gases each gas adds its own share to Eq. (12.14):

\begin{aligned} P & = \frac{1}{3} n_{1} m_{1} ⟨ v_{1}^{2} ⟩ \\ + \frac{1}{3} n_{2} m_{2} ⟨ v_{2}^{2} ⟩ \\ + \dots \end{aligned}

(12.20)

In equilibrium the molecules of every gas have the same average kinetic energy, $\frac{3}{2} k_{B} T$ , so each term is $n_{i} k_{B} T$ and

P = (\sum_{i} n_{i}) k_{B} T

(12.21)

So the partial pressures add, just as Dalton found by experiment, but now the rule follows from molecular motion.

Eq. (12.19) also tells us how fast molecules move. The square root of the mean square speed is the root mean square (rms) speed:

\begin{aligned} v_{rms} & = \sqrt{⟨ v^{2} ⟩} \\ = \sqrt{\frac{3 k_{B} T}{m}} \\ = \sqrt{\frac{3 R T}{M_{0}}} \end{aligned}

On a $40^{\circ} C$ summer afternoon ( $T = 313 K$ ), nitrogen molecules ( $M_{0} = 0.028 k g / m o l$ ) have an rms speed of about $528 m / s$ , roughly one and a half times the speed of sound in that air (about $355 m / s$ ). At a given temperature lighter molecules are faster, since $v_{rms} \propto 1 / \sqrt{m}$ .

Example 12.5NCERT Example 12.5 · p. 250

Molecules in biogas

Biogas from a village digester is mainly methane ( $C H_{4}$ , $16.0 u$ ) and carbon dioxide ( $C O_{2}$ , $44.0 u$ ). Suppose a sample at $35^{\circ} C$ is 60 % methane by volume. Find the ratio, methane to carbon dioxide, of (a) the average kinetic energy per molecule and (b) the rms speeds.

Try it yourself first. Then reveal the solution one step at a time.

Example 12.6NCERT Example 12.6 · p. 251

Sorting neon atoms

Neon contains atoms of mass $20 u$ and $22 u$ . In 1913 F. W. Aston tried to separate them by making neon diffuse again and again through a porous material; the change he found was tiny. Which atoms are faster on average, and by what percentage do their rms speeds differ?

Try it yourself first. Then reveal the solution one step at a time.

Fig. 12.7 shows why a porous wall sorts molecules. The pores are so narrow that molecules pass through one at a time, and each kind gets through at a rate proportional to its average speed. Lighter molecules therefore leak faster: the rate of diffusion of a gas varies as one over the square root of its molecular mass (Graham's law). One pass changes the mixture only a little, so the process is repeated through many stages.

2/3Faster, lighter molecules reach the pores more often, so the gas that passes is richer in them.

Fig. 12.7Lighter molecules get through a porous wall more often (the difference is exaggerated here).NCERT Fig. 12.5 · p. 251

A box is split by a porous wall drawn as a column of short blocks with gaps. In the first step the left side holds ten lighter molecules (filled circles) and ten heavier molecules (hollow rings) and the right side is empty. In the second step five lighter and two heavier molecules have passed through to the right, leaving five lighter and eight heavier on the left, because the lighter ones move faster and reach the pores more often. In the third step an arrow below says the stage is repeated again and again to build up the enrichment.

Example 12.7NCERT Example 12.7 · p. 251

Bouncing off a moving wall

(a) A ball moving at $10 m / s$ hits a very heavy wall head-on and rebounds elastically. Find its rebound speed if the wall is (i) at rest, (ii) moving towards the ball at $2 m / s$ , (iii) moving away from it at $2 m / s$ .

(b) Use (a) to explain why the barrel of a bicycle pump warms up as you pump. (c) What happens to a gas that pushes a piston outwards? (d) Swung at the same speed, why does a heavier hockey stick send the ball off faster?

Try it yourself first. Then reveal the solution one step at a time.

2/3A wall moving in at $V$ returns the ball at $u + 2 V$ : a gas being compressed heats up.

Fig. 12.8A moving wall changes the rebound speed: this is how a piston heats or cools a gas.

A ball arrives from the left at 10 m/s along the top row and leaves to the left along the bottom row after bouncing off a heavy wall on the right. In the first step the wall is at rest and the ball leaves at 10 m/s. In the second step the wall moves left towards the ball at 2 m/s and the ball leaves at 14 m/s, with a longer arrow and the note that the ball speeds up. In the third step the wall moves right, away from the ball, at 2 m/s and the ball leaves at 6 m/s, with a shorter arrow and the note that it slows down.

12.5 Law of equipartition of energy

A molecule moving with velocity $(v_{x}, v_{y}, v_{z})$ has translational kinetic energy

\begin{aligned} ε_{t} & = \frac{1}{2} m v_{x}^{2} + \frac{1}{2} m v_{y}^{2} \\ + \frac{1}{2} m v_{z}^{2} \end{aligned}

(12.22)

In thermal equilibrium at temperature $T$ , Eq. (12.19) says the average of this sum is

⟨ ε_{t} ⟩ = \frac{3}{2} k_{B} T

(12.23)

No direction is special, so each of the three terms takes an equal share:

\begin{aligned} ⟨ \frac{1}{2} m v_{x}^{2} ⟩ & = \frac{1}{2} k_{B} T \\ ⟨ \frac{1}{2} m v_{y}^{2} ⟩ & = \frac{1}{2} k_{B} T \\ ⟨ \frac{1}{2} m v_{z}^{2} ⟩ & = \frac{1}{2} k_{B} T \end{aligned}

(12.24)

How many numbers does it take to say where something is? A bead sliding on a straight wire needs one coordinate, a carrom coin on the board needs two, and a fly in a room needs three. These counts are the degrees of freedom (स्वातंत्र्य कोटि) of the motion. A gas molecule is like the fly: its three numbers match the three squared terms in Eq. (12.22). Each describes movement of the whole molecule from place to place, which we call translation, and Eq. (12.24) has already told us that each one averages $\frac{1}{2} k_{B} T$ .

Definition

Degree of freedom

An independent way a molecule can move: for translation, one coordinate needed to fix its position. When counting energy, each squared term in the molecule's energy is one mode that can hold energy.

ExampleA carrom coin sliding on the board has two translational degrees of freedom.

For our purposes argon and helium atoms are featureless points: translation is all they can do. A dumbbell-shaped molecule such as $N_{2}$ or $O_{2}$ has more options. It can tumble end over end about its centre of mass, and there are two independent ways to do so, about either of two axes at right angles to the bond (Fig. 12.9). What about twirling around the bond itself, like a spindle? Its moment of inertia is so tiny that, by quantum mechanics, this spin cannot take up energy at ordinary temperatures, so it is not counted. For a rigid dumbbell the energy is therefore

\begin{aligned} ε & = ε_{t} + ε_{r} \\ = ε_{t} + \frac{1}{2} I_{1} ω_{1}^{2} \\ + \frac{1}{2} I_{2} ω_{2}^{2} \end{aligned}

(12.25)

where

$ω_{1}, ω_{2}$: angular speeds about the two axes
$I_{1}, I_{2}$: the moments of inertia about them

Each rotation adds one more squared term, now in an angular speed.

2/4Two rotations, about axes perpendicular to the bond: two squared terms in angular speed.

Fig. 12.9The ways a diatomic molecule can move and store energy.NCERT Fig. 12.6 · p. 252

Two atoms joined by a bond sit in the middle. In the first step three arrows from the centre point along x, y and z, labelled 3 translational. In the second step two dashed axes through the centre, one vertical and one slanting into the page, each with a curved arrow, are labelled 2 rotational. In the third step a dashed axis runs along the bond with a small curved arrow, labelled spin about the bond, too small to count. In the last step the bond becomes a spring and the atoms move apart, with outward arrows, labelled vibration: kinetic and potential energy, 2 terms.

Real bonds are springy, though: the two atoms can oscillate along the bond like masses joined by a spring. In light, stiff molecules such as $C O$ this vibration is barely active at room temperature, but in heavier, floppier ones such as $C l_{2}$ it already takes up a good share of energy (Table 12.3). A vibration adds two squared terms, one for kinetic and one for potential energy:

ε_{v} = \frac{1}{2} m {(\frac{d y}{d t})}^{2} + \frac{1}{2} k y^{2}

(12.26)

where

$y$: the stretch of the bond from its natural length
$k$: the force constant of the bond
$m$: the effective mass of the vibrating pair

Then $ε = ε_{t} + ε_{r} + ε_{v}$ . Notice the difference: translation and rotation give one squared term per degree of freedom, but each vibrational mode gives two.

LawLaw of equipartition of energy

In thermal equilibrium at absolute temperature $T$ , the energy of a molecule is shared equally among all the squared terms in its energy, each having an average value of $\frac{1}{2} k_{B} T$ . So each translational and each rotational degree of freedom holds $\frac{1}{2} k_{B} T$ on average, and each vibrational mode holds $2 \times \frac{1}{2} k_{B} T = k_{B} T$ .

Quick check

1 of 3

C O

molecule translates, rotates and vibrates. How many squared terms in its energy hold energy?

5678

Show answer

Answer (C) 7

3 translational + 2 rotational + 2 vibrational (kinetic and potential) = 7. Answering 6 counts the vibration only once.

12.6 Specific heat capacity

Equipartition turns a count of squared terms into a prediction of heat capacity. Recall from Chapter 11 that the molar specific heat capacity at constant volume, $C_{v}$ , is the rise in internal energy of one mole per kelvin, and that for an ideal gas $C_{p} - C_{v} = R$ .

12.6.1 Monatomic gases

A monatomic molecule has only the three translational degrees of freedom, so its average energy is $\frac{3}{2} k_{B} T$ . One mole contains $N_{A}$ molecules, so

\begin{aligned} U & = \frac{3}{2} k_{B} T \times N_{A} \\ = \frac{3}{2} R T \end{aligned}

(12.27)

C_{v} = \frac{d U}{d T} = \frac{3}{2} R

(12.28)

For any ideal gas,

C_{p} - C_{v} = R

(12.29)

C_{p} = \frac{5}{2} R

(12.30)

γ = \frac{C_{p}}{C_{v}} = \frac{5}{3}

(12.31)

12.6.2 Diatomic gases

A rigid diatomic molecule has five degrees of freedom, three translational and two rotational, so

\begin{aligned} U & = \frac{5}{2} k_{B} T \times N_{A} \\ = \frac{5}{2} R T \end{aligned}

(12.32)

C_{v} = \frac{5}{2} R, C_{p} = \frac{7}{2} R

(12.33)

γ = \frac{7}{5}

(12.34)

If the molecule also vibrates, the vibrational mode adds $k_{B} T$ per molecule, making $\frac{5}{2} k_{B} T + k_{B} T = \frac{7}{2} k_{B} T$ in all:

\begin{aligned} U & = \frac{7}{2} k_{B} T \times N_{A} \\ = \frac{7}{2} R T \\ C_{v} & = \frac{7}{2} R \\ C_{p} & = \frac{9}{2} R \\ γ & = \frac{9}{7} \end{aligned}

(12.35)

Which picture is right? Both, at different temperatures. Fig. 12.10 sketches $C_{v}$ for hydrogen. Very cold, it behaves like a monatomic gas; around room temperature $C_{v} \approx \frac{5}{2} R$ ; only when it is very hot does vibration add the last $R$ . Classical equipartition cannot explain these steps. Quantum mechanics can: rotations and vibrations take up energy only in packets, and at low temperatures $k_{B} T$ is too small to supply them.

2/3Rotation has switched on: $C_{v} \approx \frac{5}{2} R$ .

Fig. 12.10Molar heat capacity of hydrogen (schematic): modes switch on as the temperature rises.

A graph of Cv divided by R against temperature on a logarithmic scale, with dashed levels at 3/2, 5/2 and 7/2. In the first step a flat line at 3/2 is labelled translation. In the second step the curve rises smoothly to 5/2, labelled plus rotation, with a dot marking room temperature on this plateau. In the third step it rises again to 7/2 at very high temperatures, labelled plus vibration.

12.6.3 Polyatomic gases

A non-linear polyatomic molecule, such as $H_{2} O$ or $C H_{4}$ , can rotate about three independent axes. It has 3 translational and 3 rotational degrees of freedom, plus some number $f$ of vibrational modes. Each molecule then holds $\frac{3}{2} k_{B} T + \frac{3}{2} k_{B} T + f k_{B} T$ on average, so per mole

\begin{aligned} U & = (3 + f) k_{B} T N_{A} \\ = (3 + f) R T \end{aligned}

\begin{aligned} C_{v} & = (3 + f) R \\ C_{p} & = (4 + f) R \\ γ & = \frac{4 + f}{3 + f} \end{aligned}

(12.36)

The relation $C_{p} - C_{v} = R$ holds for every ideal gas, whatever its molecules. A linear molecule such as $C O_{2}$ is a special case: like a diatomic molecule it has only two rotational degrees of freedom, so a rigid $C O_{2}$ molecule would have $C_{v} = \frac{5}{2} R$ . Table 12.2 collects the predictions when vibrations are ignored or counted, and Table 12.3 compares them with measurements.

Table 12.2Molar heat capacities predicted by equipartitionNCERT Table 12.1 · p. 254

Type of molecule	Squared terms	$C_{v}$ ( $J m o l^{- 1} K^{- 1}$ )	$C_{p}$ ( $J m o l^{- 1} K^{- 1}$ )	$γ$
Monatomic	3	$\frac{3}{2} R = 12.5$	$\frac{5}{2} R = 20.8$	1.67
Rigid diatomic	5	$\frac{5}{2} R = 20.8$	$\frac{7}{2} R = 29.1$	1.40
Diatomic with vibration	7	$\frac{7}{2} R = 29.1$	$\frac{9}{2} R = 37.4$	1.29
Rigid non-linear polyatomic	6	$3 R = 24.9$	$4 R = 33.3$	1.33

Table 12.3Measured molar heat capacities near

298 K

(

C_{p}

from NIST data;

C_{v} = C_{p} - R

)NCERT Table 12.2 · p. 254

Gas	Type	$C_{p}$ ( $J m o l^{- 1} K^{- 1}$ )	$C_{v}$ ( $J m o l^{- 1} K^{- 1}$ )	$γ$
Krypton, $K r$	monatomic	20.8	12.5	1.67
Carbon monoxide, $C O$	diatomic	29.1	20.8	1.40
Chlorine, $C l_{2}$	diatomic	34.0	25.6	1.32
Carbon dioxide, $C O_{2}$	linear triatomic	37.1	28.8	1.29
Ethane, $C_{2} H_{6}$	polyatomic	52.5	44.2	1.19

Krypton and carbon monoxide match the rigid predictions almost exactly. Chlorine, carbon dioxide and ethane overshoot them: each has low-frequency vibrations, from heavy atoms or bending motions, that are already partly excited at room temperature. Yet counting every vibration in full would overshoot the measurements the other way. The truth lies in between, because the vibrations are only partly active, just as the quantum picture of Fig. 12.10 suggests. Equipartition works well at ordinary temperatures once we know which modes are fully active.

Example 12.8NCERT Example 12.8 · p. 254

Warming a flask of nitrogen

A rigid $5.0 L$ steel flask holds nitrogen at $27^{\circ} C$ and $2.0 a t m$ . How much heat is needed to warm the gas to $87^{\circ} C$ ? Treat nitrogen as an ideal rigid diatomic gas and take $1 a t m = 1.013 \times 10^{5} P a$ . How much heat would be needed if the gas were kept at constant pressure instead?

Try it yourself first. Then reveal the solution one step at a time.

12.6.4 Specific heat capacity of solids

Equipartition also works for many solids. Picture each atom held in place by spring-like bonds to its neighbours (Fig. 12.11). It vibrates about its mean position in three independent directions. Each direction is an oscillator with a kinetic and a potential term, worth $2 \times \frac{1}{2} k_{B} T = k_{B} T$ , so each atom holds $3 k_{B} T$ on average. For one mole,

\begin{aligned} U & = 3 k_{B} T \times N_{A} \\ = 3 R T \end{aligned}

When a solid is heated at atmospheric pressure, almost all the heat goes into $U$ : the solid hardly expands, so the work $P Δ V$ in $Δ Q = Δ U + P Δ V$ is tiny. Hence

molar heat capacity of a solid

\begin{aligned} C & = \frac{Δ Q}{Δ T} = \frac{Δ U}{Δ T} \\ = 3 R \end{aligned}

(12.37)

3/3Three directions give $3 k_{B} T$ per atom, so $U = 3 R T$ per mole.

Fig. 12.11Each atom of a solid oscillates in three directions.

Fifteen atoms sit on a square grid, joined to their neighbours by straight lines that stand for spring-like bonds; the middle atom is highlighted. In the first step the lattice is at rest. In the second step the middle atom moves sideways and a double arrow shows it oscillating along one direction, which carries kinetic and potential energy worth kBT on average. In the third step two more double arrows show oscillation in the other two directions, giving 3kBT per atom.

Table 12.4Heat capacities of some solids near

25^{\circ} C

NCERT Table 12.3 · p. 254

Solid	Molar mass ( $g / m o l$ )	$c$ ( $J k g^{- 1} K^{- 1}$ )	$C$ ( $J m o l^{- 1} K^{- 1}$ )	$C / R$
Iron	55.85	449	25.1	3.02
Gold	196.97	129	25.4	3.06
Zinc	65.38	388	25.4	3.06
Nickel	58.69	445	26.1	3.14
Silicon	28.09	712	20.0	2.41
Diamond	12.01	508	6.1	0.73

Iron, gold, zinc and nickel all sit close to $3 R \approx 25 J m o l^{- 1} K^{- 1}$ , as Dulong and Petit found by experiment in 1819. Per kilogram their specific heats differ more than threefold, yet per mole they nearly agree: what matters is the number of atoms. Silicon, and especially diamond, fall well short. Their light atoms and very stiff bonds vibrate so fast that, by quantum mechanics, the vibrations are only partly active at room temperature. Cooled enough, every solid's heat capacity drops towards zero.

Quick check

1 of 4

Which gas would you expect to have

γ \approx 1.67

nitrogencarbon dioxideargonsteam

Show answer

Answer (C) argon

Argon is monatomic, so

C_{v} = \frac{3}{2} R

and

γ = \frac{5}{3}

. Nitrogen is diatomic (

\frac{7}{5}

); carbon dioxide and steam have even more modes and smaller

γ

12.7 Mean free path

Open a bottle of attar at the front of a still classroom. The scent molecules leave at a few hundred metres per second, yet a student at the back may not notice the smell for many seconds. The reason is collisions: a scent molecule never travels far before striking an air molecule, so its path is a zigzag that wanders only slowly away from the bottle. (In a real room gentle air currents usually carry the scent faster than diffusion alone.)

To estimate the distance between collisions, model the molecules as hard spheres of diameter $d$ . Two molecules collide whenever their centres come within $d$ of each other (Fig. 12.12).

2/3In time $Δ t$ a molecule sweeps a tube of radius $d$ and volume $π d^{2} ⟨ v ⟩ Δ t$ .

Fig. 12.12Counting collisions with the tube a molecule sweeps out.NCERT Fig. 12.7 · p. 255

In the first step two equal spheres touch, with their centres marked and a double arrow showing that the centres are a distance d apart. In the second step one molecule moves to the right and sweeps out a long tube of radius d and length the average speed times delta t; three molecules whose centres lie inside the tube are highlighted as the ones it will hit, while others outside are drawn in grey. In the third step a zigzag line shows one molecule's path, with a dot at every collision, labelled free paths between collisions.

DerivationHow far between collisions?

Follow one molecule moving with average speed $⟨ v ⟩$ , and for now imagine all the others at rest. In a time $Δ t$ it sweeps out a tube of radius $d$ and length $⟨ v ⟩ Δ t$ , and it hits every molecule whose centre lies inside that tube, of volume $π d^{2} ⟨ v ⟩ Δ t$ .

With $n$ molecules per unit volume, it makes $n π d^{2} ⟨ v ⟩ Δ t$ collisions in $Δ t$ . The collision rate is therefore $n π d^{2} ⟨ v ⟩$ , and the average time between collisions is

τ = \frac{1}{n π d^{2} ⟨ v ⟩}

(12.38)

The average distance travelled between collisions is the mean free path:

l = ⟨ v ⟩ τ = \frac{1}{n π d^{2}}

(12.39)

In fact the other molecules move too, and what matters is the relative speed. A careful average over the speeds raises the collision rate by a factor $\sqrt{2}$ , so

mean free path

l = \frac{1}{\sqrt{2} n π d^{2}}

(12.40)

How long is it? For air at $27^{\circ} C$ and $1 a t m$ , $n = P / k_{B} T \approx 2.4 \times 10^{25} m^{- 3}$ . Take an effective molecular diameter $d \approx 3.7 \times 10^{- 10} m$ , the value that fits the measured viscosity of air. Eq. (12.40) then gives $l \approx 6.7 \times 10^{- 8} m$ , about $67 n m$ or roughly 180 molecular diameters. With a mean speed near $470 m / s$ , the time between collisions is $τ = l / ⟨ v ⟩ \approx 1.4 \times 10^{- 10} s$ : each molecule collides about seven billion times a second.

Pump most of the air out of a vessel and the free flights stretch enormously: at $10^{- 4} P a$ and room temperature the mean free path in air is about $70 m$ , longer than any laboratory chamber. In such a vacuum molecules fly from wall to wall without meeting one another. That is why an electron microscope needs a high vacuum: its electrons must cross the column without hitting gas molecules. In general, halving the number density doubles $l$ , and halving the molecular diameter quadruples it.

Example 12.9NCERT Example 12.9 · p. 255

Thin air at 30 km

A research balloon floats $30 k m$ up, where the air pressure is about $1.2 k P a$ and the temperature about $227 K$ . Starting from the sea-level estimate above ( $l \approx 67 n m$ at $300 K$ and $101 k P a$ ), find the mean free path at the balloon.

Try it yourself first. Then reveal the solution one step at a time.

These numbers correct a common picture of a gas. The molecules are not enormously far apart: in room air the typical spacing, $(k_{B} T / P)^{1 / 3} \approx 3.4 n m$ , is only about ten times the spacing in a solid or liquid. What is really long is the mean free path, about twenty times that spacing and nearly two hundred molecular diameters. This long free flight gives a gas its typical behaviour: it spreads out to fill any container and cannot be held without one.

Kinetic theory also ties the mean free path to properties measured in bulk. Viscosity, thermal conductivity and diffusion all depend on how far molecules carry momentum, energy or themselves between collisions. In 1865 Josef Loschmidt combined the measured viscosity of air with Maxwell's theory to make one of the first estimates of the size of molecules.

InteractiveMean free path lab

Hard-sphere molecules move through a box that repeats on every side, seen from the front, with nearer molecules drawn darker. One orange molecule leaves a zigzag trail with a kink at every collision. Sliders set the number of molecules, their diameter and the temperature. A table compares the mean free path from the formula 1 over root 2 n pi d squared with the value measured from all the collisions, and shows the number of collisions, the collisions per molecule per second and the rms speed.

Fig. 12.13Press Play. Lengths are in units where the box is 100 units wide; the orange molecule leaves a trail.

Try this

Experiments with the mean free path lab

Note the measured mean free path, then halve the number of molecules (250 to 125). Does it double, as Eq. (12.40) predicts?
Double the diameter, from 3 to 6 units. By what factor should the mean free path change? Check it.
Raise the temperature. The molecules speed up: does the mean free path change, or only the collision rate?

Since $l \propto 1 / n$ , halving the molecules doubles it; since $l \propto 1 / d^{2}$ , doubling $d$ makes it four times shorter. Heating at fixed $n$ leaves $l$ unchanged: the molecules cover the same distance between collisions, only faster, so the collision rate rises in proportion to the speed. In the most crowded settings the measured value comes out a little shorter than the formula, because the molecules' own size starts to matter.

Quick check

1 of 3

The pressure of a gas is doubled at constant temperature. Its mean free path

doublesfalls to a quarterhalvesis unchanged

Show answer

Answer (C) halves

n = P / k_{B} T

doubles, and

l \propto 1 / n

, so

l

halves. A quarter would need

d

to double.

Recap

Kinetic Theory: key points and summary

Matter is made of molecules in ceaseless motion. Atoms attract at a few ångström and repel when squeezed; in a gas these forces act only during collisions (Section 12.2).
Ideal gas equation: $P V = μ R T = N k_{B} T$ , with $R = 8.314 J m o l^{- 1} K^{- 1}$ and $k_{B} = R / N_{A} = 1.38 \times 10^{- 23} J / K$ . Real gases obey it best at low pressure and high temperature (Section 12.3).
Boyle's law, Charles' law, Avogadro's hypothesis and Dalton's law of partial pressures all follow from it (Fig. 12.4).
Momentum delivered to the walls gives $P = \frac{1}{3} n m ⟨ v^{2} ⟩$ ((12.14)).
With the gas law, $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ : temperature measures the average translational kinetic energy of a molecule, whatever the gas (Section 12.4.2).
$P V = \frac{2}{3} E$ with $E = \frac{3}{2} N k_{B} T$ , so the internal energy of an ideal gas depends only on $T$ ((12.18)).
$v_{rms} = \sqrt{3 k_{B} T / m}$ : at one temperature lighter molecules are faster, which is how porous barriers separate isotopes (Example 12.6).
A gas compressed by a moving piston warms up; a gas that pushes a piston out cools (Example 12.7).
Equipartition: each squared term in a molecule's energy averages $\frac{1}{2} k_{B} T$ , and a vibrational mode holds $k_{B} T$ (Law).
Molar heat capacities: monatomic $C_{v} = \frac{3}{2} R$ , $γ = \frac{5}{3}$ ; rigid diatomic $\frac{5}{2} R$ , $\frac{7}{5}$ ; non-linear polyatomic $(3 + f) R$ ; always $C_{p} - C_{v} = R$ (Section 12.6).
Solids: $C \approx 3 R$ per mole (Dulong–Petit), with light, stiff solids such as diamond falling short (Table 12.4).
Mean free path $l = 1 / (\sqrt{2} n π d^{2})$ : about $70 n m$ in room air, much longer than the spacing between molecules (Section 12.7).

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In your NCERT textbook	Practice question here
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12.2 Molecular nature of matter3 questions
12.3 Behaviour of gases4 questions
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12.4.2 Kinetic interpretation of temperature4 questions
12.5 Law of equipartition of energy3 questions
12.6.4 Specific heat capacity of solids4 questions
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Glossary · 10

Key terms (English–Hindi)

Avogadro numberआवोगाद्रो संख्याavogadro sankhya: The number of particles in one mole, $N_{A} \approx 6.02 \times 10^{23}$ .First seen in the chapter →

Boltzmann constantबोल्ट्ज़मान नियतांकboltzmann niyatank: The gas constant per molecule, $k_{B} = R / N_{A} = 1.38 \times 10^{- 23} J / K$ .First seen in the chapter →

degree of freedomस्वातंत्र्य कोटिswatantrya koti: An independent way a molecule can move; in energy counting, each squared term in its energy.First seen in the chapter →

ideal gasआदर्श गैसaadarsh gas: A model gas that obeys $P V = μ R T$ exactly; real gases approach it at low pressure and high temperature.

kinetic theoryअणुगति सिद्धांतanugati siddhant: The model that explains the properties of a gas from the motion of its molecules, using Newton's laws and averages.First seen in the chapter →

law of equipartition of energyऊर्जा के समविभाजन का नियमurja ke samvibhajan ka niyam: In thermal equilibrium each squared term in a molecule's energy has an average of $\frac{1}{2} k_{B} T$ .

mean free pathमाध्य मुक्त पथmadhya mukt path: The average distance a molecule travels between collisions, $l = 1 / (\sqrt{2} n π d^{2})$ .First seen in the chapter →
molar specific heat capacityमोलर विशिष्ट ऊष्मा धारिताmolar vishisht ushma dharita: Heat needed to warm one mole by one kelvin: $C_{v}$ at constant volume, $C_{p}$ at constant pressure.First seen in the chapter →

partial pressureआंशिक दाबaanshik daab: The pressure one gas in a mixture would exert if it alone filled the container at the same temperature.First seen in the chapter →

rms speedवर्ग-माध्य-मूल चालvarg madhya mool chaal: The square root of the mean of the squared molecular speeds, $v_{rms} = \sqrt{3 k_{B} T / m}$ .First seen in the chapter →

Kinetic Theory: Class 11 Physics · Chapter 12

What you will learn in Kinetic Theory

12.1 Introduction

12.2 Molecular nature of matter

12.3 Behaviour of gases

12.4 Kinetic theory of an ideal gas

12.4.1 Pressure of an ideal gas

12.4.2 Kinetic interpretation of temperature

12.5 Law of equipartition of energy

12.6 Specific heat capacity

12.6.1 Monatomic gases

12.6.2 Diatomic gases

12.6.3 Polyatomic gases

12.6.4 Specific heat capacity of solids

12.7 Mean free path

Kinetic Theory: key points and summary

Practice questions with answers (NCERT “Exercises”)

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Key terms (English–Hindi)

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