PhysicsISC Class 11 · Chapter 12

Kinetic Theory of Gases ISC Class 11: Notes and Numericals

Updated 15 October 2026 · Fact-checked

Kinetic Theory of Gases is Chapter 12 of ISC Class 11 Physics. It covers the equation of state of a perfect gas, the assumptions of kinetic theory, the derivation of pressure p = ⅓ρc², the kinetic interpretation of temperature, rms speed, degrees of freedom, the law of equipartition and specific heats of gases, mean free path and Avogadro's number.

What Kinetic Theory of Gases is about

A gas is a huge number of tiny molecules in rapid, random motion. Kinetic theory explains the measurable properties of a gas (pressure, temperature, specific heat) from this picture.

Perfect gas. Experiment gives the equation of state $p V = n R T = N k T$ , where n is the number of moles, N the number of molecules, R = 8.31 J mol⁻¹ K⁻¹ and $k = \frac{R}{N _{A}} = 1.38 \times 1 0^{- 23}$ J K⁻¹ (Boltzmann constant). Avogadro's number $N_{A} = 6.02 \times 1 0^{23}$ mol⁻¹: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.

Pressure. From the assumptions (point-like molecules, elastic collisions, no forces between collisions, random motion), the momentum given to the walls leads to $p = \frac{1}{3} ρ \overline{c^{2}}$ , so the root mean square speed is $c_{r m s} = \frac{3 p}{ρ} = \frac{3 R T}{M}$ . It grows as √T and, at fixed temperature, does not depend on pressure.

Temperature. Combining with pV = nRT (an experimental result that kinetic theory cannot derive) gives the average kinetic energy: $\frac{3}{2} k T$ per molecule and $\frac{3}{2} R T$ per mole. Temperature is a measure of the average translational KE of the molecules.

Equipartition. A molecule has a number of degrees of freedom f (3 for monatomic, 5 for a rigid diatomic, 6 for a non-linear polyatomic). The law of equipartition (statement only) gives each degree $\frac{1}{2} k T$ , so $C_{v} = \frac{f}{2} R$ , $C_{p} = (\frac{f}{2} + 1) R$ and $γ = 1 + \frac{2}{f}$ . The mean free path, the average distance between collisions, is $λ = \frac{1}{2 π d ^{2} n}$ .

Weightage: Kinetic Theory of Gases belongs to unit 9 of the ISC Class XI Physics syllabus. CISCE weights units 7-9 together (Properties of Bulk Matter; Heat and Thermodynamics; Behaviour of Perfect Gases and Kinetic Theory of Gases) at 20 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Physics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Equation of state of a perfect gas: A perfect (ideal) gas obeys Boyle's law (pV constant at fixed T), Charles's law (V ∝ T at fixed p) and the pressure law (p ∝ T at fixed V) exactly. Together they give the equation of state: pV = nRT. Writing n = N/N_A and $k = \frac{R}{N _{A}}$ : pV = NkT, or p = n'kT, where n' = N/V is the number of molecules per unit volume. Real gases behave nearly ideally at low pressure and high temperature, when molecules are far apart. Work done on a gas in compression is $- \int p d V$ (isothermal: $n R T ln \frac{V _{1}}{V _{2}}$ done on the gas).; Worked example
Find the number of gas molecules in 1 m³ at STP (1.013 × 10⁵ Pa, 273 K). (k = 1.38 × 10⁻²³ J K⁻¹)
Try it first, then show the solutionSolution
$n^{'} = \frac{p}{k T} = \frac{1.013 \times 1 0 ^{5}}{1.38 \times 1 0 ^{- 23} \times 273}$
$n^{'} = 2.69 \times 1 0^{25} m^{- 3}$
Answer: About $2.7 \times 1 0^{25}$ molecules per m³ (Loschmidt's number), for any gas
Assumptions of the kinetic theory: (1) A gas consists of a very large number of identical molecules, which behave as perfectly elastic rigid spheres. (2) The size of a molecule is negligible compared with the distance between molecules (point masses); the volume of the molecules is negligible compared with the volume of the gas. (3) There are no forces between molecules except during collisions, so between collisions they move in straight lines with uniform velocity. (4) Collisions with each other and with the walls are elastic, and the time of a collision is negligible compared with the time between collisions. (5) The molecules move randomly in all directions with all possible speeds; in the steady state the number density is uniform. (6) Newton's laws of motion apply to the molecules.; Worked example
Which assumptions of kinetic theory fail for a real gas at very high pressure, and why does the gas then deviate from pV = nRT?
Try it first, then show the solutionSolution
At high pressure the molecules are crowded: their own volume is no longer negligible compared with the container's volume.
They are also close enough for intermolecular attractions to act between collisions.
Answer: The 'negligible volume' and 'no intermolecular forces' assumptions fail, so the gas departs from ideal behaviour
Pressure of a gas: p = ⅓ρc²: Consider N molecules, each of mass m, in a cube of side l. A molecule with x-component of velocity $c_{x}$ hits wall A and rebounds elastically: change of momentum $= 2 m c_{x}$ . It returns to A after travelling 2l, so it hits A $\frac{c _{x}}{2 l}$ times per second. Force on A from this molecule $= 2 m c_{x} \times \frac{c _{x}}{2 l} = \frac{m c _{x}^{2}}{l}$ . Total force $= \frac{m}{l} \sum c_{x}^{2}$ ; pressure $p = \frac{m}{l ^{3}} \sum c_{x}^{2}$ . Since motion is random, $\sum c_{x}^{2} = \sum c_{y}^{2} = \sum c_{z}^{2} = \frac{1}{3} \sum c^{2} = \frac{1}{3} N \overline{c^{2}}$ . So $p = \frac{1}{3} \frac{m N}{V} \overline{c^{2}} = \frac{1}{3} ρ \overline{c^{2}}$ , where ρ = mN/V is the density and $\overline{c^{2}}$ the mean square speed. Also $p V = \frac{1}{3} m N \overline{c^{2}}$ , and $p = \frac{2}{3} E$ , where E is the translational KE per unit volume.; Worked example
The density of nitrogen at STP is 1.25 kg m⁻³. Find the rms speed of its molecules. (p = 1.013 × 10⁵ Pa)
Try it first, then show the solutionSolution
$c_{r m s} = \frac{3 p}{ρ} = \frac{3 \times 1.013 \times 1 0 ^{5}}{1.25}$
$c_{r m s} = 493 m s^{- 1}$
Answer: About 493 m s⁻¹, faster than the speed of sound in air
Kinetic interpretation of temperature: For one mole: $p V = \frac{1}{3} M \overline{c^{2}}$ (M = molar mass). The experimental equation of state gives pV = RT. (pV = nRT is an experimental result and cannot be derived from kinetic theory; combining the two is what gives temperature its meaning.) So $\frac{1}{3} M \overline{c^{2}} = R T$ , and the translational kinetic energy per mole is $\frac{1}{2} M \overline{c^{2}} = \frac{3}{2} R T$ . Per molecule, dividing by $N_{A}$ : $\frac{1}{2} m \overline{c^{2}} = \frac{3}{2} k T$ . Temperature is proportional to the average translational kinetic energy of the molecules; at absolute zero this energy would be zero. All gases at the same temperature have the same average KE per molecule, whatever their mass.; Worked example
Find the average translational kinetic energy of a gas molecule at 300 K, and the total translational KE of one mole. (k = 1.38 × 10⁻²³ J K⁻¹, R = 8.31 J mol⁻¹ K⁻¹)
Try it first, then show the solutionSolution
Per molecule: $\frac{3}{2} k T = 1.5 \times 1.38 \times 1 0^{- 23} \times 300 = 6.21 \times 1 0^{- 21} J$
Per mole: $\frac{3}{2} R T = 1.5 \times 8.31 \times 300 = 3740 J$
Answer: $6.2 \times 1 0^{- 21}$ J per molecule; about 3.7 kJ per mole
RMS speed: effect of temperature and pressure: From the results above, $c_{r m s} = \overline{c^{2}} = \frac{3 p}{ρ} = \frac{3 R T}{M} = \frac{3 k T}{m}$ . Effect of temperature: $c_{r m s} \propto T$ ; quadrupling the absolute temperature doubles it. Effect of pressure: at constant temperature, increasing p increases ρ in the same proportion (Boyle's law), so p/ρ and hence $c_{r m s}$ are unchanged. Effect of molar mass: at the same temperature $c_{r m s} \propto \frac{1}{M}$ , so light gases move faster (this is why hydrogen gradually escapes from the Earth's atmosphere).; Worked example
Find the rms speed of oxygen molecules at 27 °C, and the temperature at which it would be twice as great. (M = 0.032 kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹)
Try it first, then show the solutionSolution
$c_{r m s} = \frac{3 \times 8.31 \times 300}{0.032} = 483 m s^{- 1}$
Doubling needs four times T: 4 × 300 = 1200 K (927 °C)
Answer: 483 m s⁻¹; 1200 K
Degrees of freedom: The degrees of freedom of a molecule are the number of independent ways in which it can store energy (independent coordinates needed to describe its motion). A monatomic molecule (He, Ar) has only translational motion along x, y and z: f = 3. A diatomic molecule (O₂, N₂) treated as rigid has 3 translational and 2 rotational degrees (rotation about the molecular axis contributes nothing, as the moment of inertia about it is negligible): f = 5. At high temperatures its atoms also vibrate, adding 2 more (kinetic and potential energy of vibration): f = 7. A non-linear polyatomic molecule (H₂O, CH₄) has 3 translational and 3 rotational degrees: f = 6 (ignoring vibration). A linear polyatomic molecule such as CO₂ has 5 (rigid).; Worked example
State the degrees of freedom of (a) argon, (b) nitrogen at room temperature, (c) nitrogen at very high temperature, (d) methane (ignoring vibration).
Try it first, then show the solutionSolution
(a) Monatomic: 3 translational.
(b) Rigid diatomic: 3 + 2 = 5. (c) With vibration: 5 + 2 = 7.
(d) Non-linear polyatomic: 3 + 3 = 6.
Answer: 3; 5; 7; 6
Law of equipartition of energy and specific heats of gases: Law of equipartition of energy (statement only): in thermal equilibrium at absolute temperature T, the total energy of a molecule is shared equally among its degrees of freedom, each having an average energy of $\frac{1}{2} k T$ . So a molecule with f degrees of freedom has $\frac{f}{2} k T$ , and one mole has internal energy $U = \frac{f}{2} R T$ . Then $C_{v} = \frac{d U}{d T} = \frac{f}{2} R$ , $C_{p} = C_{v} + R = (\frac{f}{2} + 1) R$ and $γ = \frac{C _{p}}{C _{v}} = 1 + \frac{2}{f}$ . Values: monatomic (f = 3): $C_{v} = \frac{3}{2} R$ , γ = 5/3 ≈ 1.67; rigid diatomic (f = 5): $C_{v} = \frac{5}{2} R$ , γ = 7/5 = 1.4; diatomic with vibration (f = 7): γ = 9/7 ≈ 1.29; non-linear polyatomic (f = 6): $C_{v} = 3 R$ , γ = 4/3 ≈ 1.33. These agree well with measurements for monatomic and simple diatomic gases.; Worked example
Find the ratio γ for a mixture of 1 mole of helium and 1 mole of oxygen (treat oxygen as rigid).
Try it first, then show the solutionSolution
Total $C_{v}$ = (3/2)R + (5/2)R = 4R for 2 mol, i.e. 2R per mole
Total $C_{p}$ = (5/2)R + (7/2)R = 6R for 2 mol, i.e. 3R per mole
γ = 3R/2R = 1.5
Answer: γ = 1.5
Mean free path and Avogadro's number: Molecules of finite diameter d collide with one another. The mean free path λ is the average distance a molecule travels between successive collisions: $λ = \frac{1}{2 π d ^{2} n}$ , where n is the number of molecules per unit volume; using p = nkT, $λ = \frac{k T}{2 π d ^{2} p}$ . It increases with temperature and decreases with pressure and molecular size. At STP it is about $1 0^{- 7}$ m, hundreds of molecular diameters, which is why a smell spreads slowly despite molecular speeds of hundreds of m s⁻¹. Avogadro's hypothesis: equal volumes of all gases at the same temperature and pressure contain the same number of molecules. Avogadro's number $N_{A} = 6.022 \times 1 0^{23}$ mol⁻¹, the number of particles in one mole; one mole of any ideal gas at STP occupies 22.4 L.; Worked example
Estimate the mean free path of nitrogen molecules at STP, taking the molecular diameter as 3 × 10⁻¹⁰ m and n = 2.69 × 10²⁵ m⁻³.
Try it first, then show the solutionSolution
$λ = \frac{1}{2 π ( 3 \times 1 0 ^{- 10} ) ^{2} \times 2.69 \times 1 0 ^{25}}$
$λ = \frac{1}{1.08 \times 1 0 ^{7}} = 9.3 \times 1 0^{- 8} m$
Answer: About $9 \times 1 0^{- 8}$ m, roughly 300 molecular diameters

Physics in real life: Kinetic Theory of Gases

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Why the atmosphere keeps nitrogen but loses hydrogen

Suppose the upper air is at about 300 K. Compare the rms speeds of hydrogen (M = 2 g mol⁻¹) and nitrogen (M = 28 g mol⁻¹) with the Earth's escape velocity of 11.2 km s⁻¹.

Find both rms speeds.

Try it first, then show the solutionSolution

H₂: $c_{r m s} = \frac{3 \times 8.31 \times 300}{0.002} = 1934 m s^{- 1}$
N₂: $c_{r m s} = \frac{3 \times 8.31 \times 300}{0.028} = 517 m s^{- 1}$

Answer: Hydrogen's rms speed is nearly four times nitrogen's, so far more of its fastest molecules exceed 11.2 km s⁻¹ and escape over millions of years.

Memory hook: Light gas, fast molecules, leaky atmosphere.

Energy of the air in a room

Imagine a classroom 5 m × 4 m × 3 m filled with air at 27 °C and 1.013 × 10⁵ Pa.

Estimate the number of air molecules and their total translational kinetic energy.

Try it first, then show the solutionSolution

$N = \frac{p V}{k T} = \frac{1.013 \times 1 0 ^{5} \times 60}{1.38 \times 1 0 ^{- 23} \times 300} = 1.47 \times 1 0^{27}$
Total KE = $\frac{3}{2} p V = 1.5 \times 1.013 \times 1 0^{5} \times 60 = 9.1 \times 1 0^{6} J$

Answer: About 1.5 × 10²⁷ molecules carrying about 9 MJ of random kinetic energy.

Memory hook: Total translational KE = 1.5 pV.

Why a smell spreads slowly

Suppose perfume molecules move at about 500 m s⁻¹ and have a mean free path of about 9.3 × 10⁻⁸ m in air.

Roughly how many collisions does each molecule make per second?

Try it first, then show the solutionSolution

Collisions per second ≈ speed/mean free path = 500/9.3 × 10⁻⁸
≈ 5.4 × 10⁹ per second

Answer: About five billion collisions a second, so each molecule zigzags randomly and the smell takes minutes to cross a room.

Memory hook: Fast molecules, short paths, slow spread.

Topics in this chapter

Kinetic Theory of Gases formula sheet

Equation of state: $p V = n R T = N k T, k = \frac{R}{N _{A}}$; k = $1.38 \times 1 0^{- 23}$ J K⁻¹.
Pressure of a gas: $p = \frac{1}{3} ρ \overline{c^{2}} = \frac{1}{3} \frac{m N}{V} \overline{c^{2}}$; Also p = (2/3) × KE per unit volume.
RMS speed: $c_{r m s} = \frac{3 p}{ρ} = \frac{3 R T}{M} = \frac{3 k T}{m}$; ∝ √T; independent of p at fixed T.
Kinetic energy: $\frac{1}{2} m \overline{c^{2}} = \frac{3}{2} k T; E_{m o l e} = \frac{3}{2} R T$; Translational KE.
Equipartition: $\frac{1}{2} k T per degree of freedom; U = \frac{f}{2} R T per mole$; Statement only.
Specific heats from f: $C_{v} = \frac{f}{2} R, C_{p} = (\frac{f}{2} + 1) R, γ = 1 + \frac{2}{f}$; f = 3, 5, 6 (monatomic, diatomic, polyatomic).
Mean free path: $λ = \frac{1}{2 π d ^{2} n} = \frac{k T}{2 π d ^{2} p}$; ≈ $1 0^{- 7}$ m at STP.
Avogadro's number: $N_{A} = 6.022 \times 1 0^{23} m o l^{- 1}$; 1 mol at STP = 22.4 L.

Kinetic Theory of Gases: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Molecule in a cubical container (derivation of pressure)

Deriving p = ⅓ρc².

Draw

A cube of side l with axes x, y, z along its edges; shade the wall A perpendicular to the x-axis.
A molecule moving towards A with velocity c, resolved into $c_{x}$ , $c_{y}$ , $c_{z}$ .
Show it rebounding from A with x-component − $c_{x}$ and travelling back across the cube a distance l.

Label

side l
wall A
$c_{x}$
$c_{y}$
$c_{z}$
momentum change 2m $c_{x}$
time between hits 2l/ $c_{x}$

Watch out: Write the three key steps under the figure: Δp = 2mcₓ; hits per second cₓ/2l; force = mcₓ²/l.

Degrees of freedom of a diatomic molecule

Explaining why f = 5 for a rigid diatomic molecule.

Draw

Two spheres joined by a short rod along the z-axis, with x, y, z axes through the centre of mass.
Three arrows for translation along x, y and z.
Two curved arrows for rotation about the x- and y-axes; mark rotation about the z-axis (the bond) as negligible.

Label

atom
bond (z-axis)
translation x, y, z
rotation about x
rotation about y
rotation about z: negligible I

Watch out: Add that vibration along the bond adds two more degrees at high temperature (f = 7).

How the school exam asks Kinetic Theory of Gases

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Assumptions and definitions	List the assumptions; define rms speed, degrees of freedom, mean free path; state equipartition.	Number the assumptions; give precise one-line definitions with formulae.
Derivation of pressure	Derive p = ⅓ρc² and hence c_rms = √(3p/ρ).	Follow the four steps: momentum change, collision rate, force, averaging over three directions.
Kinetic interpretation of temperature	Show KE per molecule = (3/2)kT; explain the meaning of temperature.	Combine pV = ⅓Mc² with pV = RT; state that pV = nRT is experimental.
RMS speed numericals	c_rms of a gas at a temperature; ratio for two gases; temperature for a given speed.	Use M in kg mol⁻¹ and T in kelvin; use ratios c ∝ √(T/M).
Equipartition and specific heats	Find Cv, Cp, γ for gases of different atomicity and for mixtures.	Count f, then Cv = fR/2; for mixtures add n·Cv and n·Cp separately.
Mean free path	Calculate λ; effect of pressure and temperature; why diffusion is slow.	Use λ = kT/(√2πd²p); λ ∝ T/p.

Tricks to remember Kinetic Theory of Gases

Short memory aids. Say each one aloud once, then use it on a question the same day.

One-third rho c squared
p = ⅓ρc²: one-third because only one of three directions pushes on a given wall.
Use it when: Remembering the pressure formula.
Three-halves k T
Three directions, half kT each: (3/2)kT per molecule.
Use it when: Kinetic energy and temperature.
f = 3, 5, 6
Mono 3, Di 5, Poly 6 (non-linear); γ = 1 + 2/f gives 1.67, 1.4, 1.33.
Use it when: Specific heat ratios.
Speed squared goes with T over M
c_rms² ∝ T/M: four times hotter, twice as fast; sixteen times heavier, four times slower.
Use it when: Ratio questions.
Pressure leaves speed alone
Squeeze a gas at constant T: denser, but molecules just as fast.
Use it when: Effect of pressure on c_rms.

Kinetic Theory of Gases: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: How rms speed depends on temperature · 1 mark · MCQExam level

The rms speed of gas molecules is proportional to:
(a) T
(b) √T
(c) T²
(d) 1/T

Try it first, then show the solutionSolution

$c_{r m s} = 3 R T / M$ , so it varies as √T (T in kelvin).

Answer: (b)

Example 2: Same temperature, same KE · 1 mark · Assertion-reasonExam level

Assertion: At the same temperature, hydrogen and oxygen molecules have the same average translational kinetic energy.
Reason: The average translational kinetic energy per molecule, (3/2)kT, depends only on the absolute temperature.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

(3/2)kT contains no mass term, so all gases at the same T have the same average KE; the lighter H₂ molecules simply move faster.

Answer: (a)

Example 3: Hydrogen versus oxygen · 2 marks · Short answerHarder exam level

Find the ratio of the rms speeds of hydrogen and oxygen molecules at the same temperature. [2]

Try it first, then show the solutionSolution

$c_{r m s} \propto \frac{1}{M}$ at the same T
$\frac{c _{H_{2}}}{c _{O_{2}}} = \frac{32}{2} = 4$

Answer: 4 : 1

Example 4: rms speed from pressure and density · 3 marks · Short answerHarder exam level

Show that $c_{r m s} = \frac{3 p}{ρ}$ and find it for oxygen at 0 °C, whose density at 1.013 × 10⁵ Pa is 1.43 kg m⁻³. Would it change if the pressure were doubled at the same temperature? [3]

Try it first, then show the solutionSolution

From $p = \frac{1}{3} ρ \overline{c^{2}}$ : $\overline{c^{2}} = \frac{3 p}{ρ}$ , so $c_{r m s} = \frac{3 p}{ρ}$ .
$c_{r m s} = \frac{3 \times 1.013 \times 1 0 ^{5}}{1.43} = 461 m s^{- 1}$
Doubling p at constant T doubles ρ (Boyle's law), so p/ρ and $c_{r m s}$ are unchanged.

Answer: 461 m s⁻¹; unchanged

Example 5: γ of a gas mixture · 3 marks · Short answerHarder exam level

A container holds 2 mol of argon and 3 mol of nitrogen (rigid diatomic). Find the molar heat capacities $C_{v}$ and $C_{p}$ and the ratio γ for the mixture. (R = 8.31 J mol⁻¹ K⁻¹) [3]

Try it first, then show the solutionSolution

$C_{v}$ (mixture) = $\frac{2 \times \frac{3}{2} R + 3 \times \frac{5}{2} R}{2 + 3} = \frac{10.5 R}{5} = 2.1 R$
$C_{p}$ (mixture) = $C_{v} + R = 3.1 R$
γ = 3.1R/2.1R = 1.48

Answer: $C_{v}$ = 2.1R = 17.5 J mol⁻¹ K⁻¹; $C_{p}$ = 3.1R = 25.8 J mol⁻¹ K⁻¹; γ ≈ 1.48

Example 6: Pressure and the meaning of temperature · 5 marks · Long answerHarder exam level

(a) Using the kinetic theory of gases, derive $p = \frac{1}{3} ρ \overline{c^{2}}$ . [3]
(b) Hence show that the average kinetic energy of a molecule is $\frac{3}{2} k T$ . [1]
(c) Find the total translational KE of 0.25 mol of a gas at 400 K. (R = 8.31 J mol⁻¹ K⁻¹) [1]

Try it first, then show the solutionSolution

(a) A molecule hitting wall A with velocity component $c_{x}$ changes its momentum by $2 m c_{x}$ and strikes A $\frac{c _{x}}{2 l}$ times a second: force $= \frac{m c _{x}^{2}}{l}$ .
Total pressure on A: $p = \frac{m}{l ^{3}} \sum c_{x}^{2}$ . Random motion: $\sum c_{x}^{2} = \frac{1}{3} N \overline{c^{2}}$ .
$p = \frac{1}{3} \frac{m N}{V} \overline{c^{2}} = \frac{1}{3} ρ \overline{c^{2}}$
(b) Per mole $p V = \frac{1}{3} M \overline{c^{2}} = R T$ (experiment), so $\frac{1}{2} M \overline{c^{2}} = \frac{3}{2} R T$ ; dividing by $N_{A}$ : $\frac{1}{2} m \overline{c^{2}} = \frac{3}{2} k T$ .
(c) E = (3/2)nRT = 1.5 × 0.25 × 8.31 × 400 = 1246.5 J

Answer: (c) About 1.25 kJ

Example 7: Matching speeds · 3 marks · Short answerHOTS / competency

At what temperature will the rms speed of oxygen molecules equal that of hydrogen molecules at 27 °C? [3]

Try it first, then show the solutionSolution

Equal $c_{r m s}$ needs equal T/M: $\frac{T _{O_{2}}}{32} = \frac{300}{2}$
$T_{O_{2}} = 300 \times 16 = 4800 K$
= 4527 °C

Answer: 4800 K

Example 8: Mean free path in a vacuum chamber · 3 marks · Short answerHOTS / competency

A vacuum chamber 0.5 m wide contains nitrogen at 1.0 Pa and 300 K (molecular diameter 3 × 10⁻¹⁰ m, k = 1.38 × 10⁻²³ J K⁻¹). Find the mean free path. Below what pressure would a molecule typically cross the chamber without hitting another molecule? [3]

Try it first, then show the solutionSolution

$λ = \frac{k T}{2 π d ^{2} p} = \frac{1.38 \times 1 0 ^{- 23} \times 300}{1.414 \times π \times ( 3 \times 1 0 ^{- 10} ) ^{2} \times 1.0}$
λ = 4.14 × 10⁻²¹/4.0 × 10⁻¹⁹ = 1.0 × 10⁻² m, only about 1 cm, so molecules still collide with each other far more often than with the walls.
λ ∝ 1/p: for λ ≥ 0.5 m the pressure must fall by a factor of about 48, to below about 0.02 Pa.

Answer: About 1 cm; below about 0.02 Pa

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

A cylinder of helium

Suppose 2 g of helium (M = 4 g mol⁻¹) is stored in a 10-litre cylinder at 27 °C. Treat it as an ideal monatomic gas. (R = 8.31 J mol⁻¹ K⁻¹)

(i) · 1 mark
Find the pressure in the cylinder.
Answer
Answer: About 1.25 × 10⁵ Pa
n = 0.5 mol; p = nRT/V = 0.5 × 8.31 × 300/0.01 = 1.25 × 10⁵ Pa.
(ii) · 1 mark
Find the rms speed of the atoms.
Answer
Answer: About 1370 m s⁻¹
$c_{r m s} = 3 \times 8.31 \times 300/0.004 = 1367$ m s⁻¹.
(iii) · 2 marks
Find the total internal energy of the gas.
Answer
Answer: About 1870 J
Monatomic: U = (3/2)nRT = 1.5 × 0.5 × 8.31 × 300 = 1870 J (all translational).
(iv) · 1 mark
If the absolute temperature were doubled, what would happen to the rms speed?
Answer
Answer: It would increase by √2, to about 1930 m s⁻¹
c_rms ∝ √T.

Identifying a gas from its γ

Suppose a student measures γ = 1.40 for an unknown gas at room temperature. (R = 8.31 J mol⁻¹ K⁻¹)

(i) · 1 mark
How many degrees of freedom does a molecule of the gas have?
Answer
Answer: 5
γ = 1 + 2/f = 1.4 gives f = 5.
(ii) · 1 mark
What kind of molecule is it likely to be?
Answer
Answer: Diatomic (rigid), such as N₂ or O₂
3 translational + 2 rotational degrees.
(iii) · 1 mark
Find its $C_{v}$ and $C_{p}$ .
Answer
Answer: 20.8 and 29.1 J mol⁻¹ K⁻¹
Cv = 5R/2; Cp = 7R/2.
(iv) · 2 marks
Find the heat needed to warm 1 mol by 10 K at constant volume, and say how γ would change at very high temperatures.
Answer
Answer: About 208 J; γ falls towards 9/7 ≈ 1.29
Q = 1 × 20.8 × 10 = 208 J. At high T vibrations add 2 degrees (f = 7), so Cv rises to 7R/2 and γ falls.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the average translational kinetic energy of a gas molecule at 127 °C and the temperature at which it would be three times as large. (k = 1.38 × 10⁻²³ J K⁻¹) (2 marks)

Step	Marks
T = 400 K: KE = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 400 = 8.28 × 10⁻²¹ J	1
KE ∝ T, so it triples at 3 × 400 = 1200 K (927 °C)	1

Find the rms speed of nitrogen molecules at 27 °C. By what factor would it change if the gas were heated to 927 °C? (M = 0.028 kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹) (3 marks)

Step	Marks
$c_{r m s} = 3 R T / M$ with T = 300 K	1
= √(3 × 8.31 × 300/0.028) = 517 m s⁻¹	1
927 °C = 1200 K = 4 × 300 K, so c_rms doubles	1

Derive an expression for the pressure exerted by an ideal gas on the walls of its container, stating the assumptions used. (5 marks)

Step	Marks
Assumptions: point molecules, elastic collisions, no intermolecular forces, random motion	1
Change of momentum of one molecule at wall A = 2mcₓ; collisions per second cₓ/2l	1
Force by one molecule = mcₓ²/l; total pressure p = (m/l³)Σcₓ²	1
Random motion: Σcₓ² = (1/3)Nc̄²	1
p = (1/3)(mN/V)c̄² = (1/3)ρc̄²	1

Where students lose marks in Kinetic Theory of Gases

The slips examiners see most often, and what to write instead.

Mistake: Using molar mass in grams
Students put M = 32 instead of 0.032 kg mol⁻¹ in √(3RT/M), giving speeds about 30 times too small.
Wrong working
$c_{r m s} = 3 \times 8.31 \times 300/32 = 15.3$ m s⁻¹.
Right working
$c_{r m s} = 3 \times 8.31 \times 300/0.032 = 483$ m s⁻¹.
Do this: Always convert M to kg mol⁻¹.
Usually costs: 1-2 marks.
Mistake: Claiming that pV = nRT follows from kinetic theory
Students derive p = ⅓ρc² and then claim this proves the gas laws.
Wrong working
From kinetic theory we derive pV = nRT.
Right working
pV = nRT is experimental; combining it with pV = ⅓Mc² gives KE = (3/2)RT, which is the kinetic interpretation of temperature.
Do this: Say 'combining with the experimental gas equation'.
Usually costs: 1 mark.
Mistake: Thinking rms speed rises with pressure
Students reason 'more pressure, faster molecules' even at constant temperature.
Wrong working
Doubling the pressure doubles c_rms.
Right working
At constant T, ρ doubles with p, so √(3p/ρ) is unchanged.
Do this: Remember c_rms depends only on T and M.
Usually costs: 1 mark.
Mistake: Counting degrees of freedom wrongly
Students give a diatomic molecule 6 degrees (three rotations) or forget the 'rigid' condition.
Wrong working
O₂: 3 translational + 3 rotational = 6, so γ = 4/3.
Right working
Rotation about the bond axis contributes nothing: f = 5, γ = 1.4.
Do this: Linear molecules have only 2 rotational degrees.
Usually costs: 1 mark.
Mistake: Averaging speeds instead of squares
Students compute the rms speed of a set of molecules as the average speed.
Wrong working
Speeds 2, 4, 6 m s⁻¹: c_rms = 4 m s⁻¹.
Right working
c_rms = √((4 + 16 + 36)/3) = √18.7 = 4.32 m s⁻¹.
Do this: Square, average, then take the square root.
Usually costs: 1 mark.

Practice questions on Kinetic Theory of Gases

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The degrees of freedom of a rigid diatomic gas molecule are:
(a) 3
(b) 5
(c) 6
(d) 7
Hint
3 translations + 2 rotations.
Answer
(b)
Q2 · 1 mark · MCQExam level
Boltzmann's constant k is equal to:
(a) R × N_A
(b) R/N_A
(c) N_A/R
(d) R/T
Hint
Gas constant per molecule.
Answer
(b)
Q3 · 1 mark · Very short answerExam level
State the law of equipartition of energy.
Hint
Per degree of freedom.
Answer
In thermal equilibrium, the energy of a molecule is shared equally among its degrees of freedom, each having an average of ½kT.
Q4 · 1 mark · Very short answerExam level
Define mean free path.
Hint
Between collisions.
Answer
The average distance travelled by a molecule between two successive collisions.
Q5 · 2 marks · Short answerExam level
State any two assumptions of the kinetic theory of gases.
Hint
Size, forces, collisions.
Answer
Any two: molecules are point masses of negligible volume; there are no forces between molecules except during collisions; collisions are perfectly elastic and of negligible duration; molecules move randomly in all directions.
Q6 · 2 marks · Short answerHarder exam level
Find the rms speed of carbon dioxide molecules at 57 °C. (M = 0.044 kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹)
Hint
√(3RT/M), with T in kelvin.
Answer
T = 330 K: c_rms = √(3 × 8.31 × 330/0.044) = 432 m s⁻¹.
Q7 · 2 marks · Short answerHarder exam level
How does the rms speed of a gas change if (a) its pressure is doubled at constant temperature, (b) its absolute temperature is doubled at constant pressure?
Hint
c ∝ √T.
Answer
(a) No change. (b) It increases by a factor √2.
Q8 · 2 marks · Short answerHarder exam level
Find $C_{v}$ , $C_{p}$ and γ for a non-linear polyatomic gas, ignoring vibrations. (R = 8.31 J mol⁻¹ K⁻¹)
Hint
f = 6.
Answer
Cv = 3R = 24.9 J mol⁻¹ K⁻¹; Cp = 4R = 33.2 J mol⁻¹ K⁻¹; γ = 4/3.
Q9 · 2 marks · Short answerHarder exam level
How many molecules are there per cubic metre of a gas at 1 atm (1.013 × 10⁵ Pa) and 300 K? (k = 1.38 × 10⁻²³ J K⁻¹)
Hint
n = p/kT.
Answer
n = 1.013 × 10⁵/(1.38 × 10⁻²³ × 300) = 2.45 × 10²⁵ m⁻³.
Q10 · 3 marks · Short answerHarder exam level
At what temperature is the average translational KE of a gas molecule equal to 1.6 × 10⁻¹⁹ J (the energy of an electron accelerated through 1 V)? (k = 1.38 × 10⁻²³ J K⁻¹)
Hint
(3/2)kT = E.
Answer
T = 2E/3k = 2 × 1.6 × 10⁻¹⁹/(3 × 1.38 × 10⁻²³) = 7.7 × 10³ K.
Q11 · 3 marks · Short answerHarder exam level
The mean free path of a gas molecule is 9.3 × 10⁻⁸ m at 1 atm. Find it at 10⁻³ atm and at 2 atm, at the same temperature, and explain the dependence.
Hint
λ ∝ 1/p.
Answer
λ = kT/(√2πd²p) ∝ 1/p: 9.3 × 10⁻⁵ m at 10⁻³ atm; 4.65 × 10⁻⁸ m at 2 atm. Lower pressure means fewer molecules per unit volume and fewer collisions.
Q12 · 5 marks · Long answerHarder exam level
(a) List four assumptions of the kinetic theory of gases. [2]
(b) Starting from $p = \frac{1}{3} ρ \overline{c^{2}}$ and pV = nRT, show that the average KE per molecule is $\frac{3}{2} k T$ . [2]
(c) Why can pV = nRT not be derived from kinetic theory alone? [1]
Hint
Temperature is not defined in the mechanical model.
Answer
(a) Point molecules; no forces except in collisions; elastic collisions of negligible duration; random motion with uniform density. (b) For one mole pV = ⅓Mc̄² = RT, so ½Mc̄² = (3/2)RT; dividing by N_A gives ½mc̄² = (3/2)kT. (c) Kinetic theory gives pV in terms of molecular speeds; temperature enters only through the experimental gas equation, which links pV to T.
Q13 · 3 marks · Short answerHOTS / competency
A mixture contains hydrogen and oxygen at the same temperature. Find the ratio of (a) the average KE per molecule and (b) the total translational KE per gram, of hydrogen to oxygen.
Hint
Per gram means per number of molecules in a gram.
Answer
(a) 1 : 1 (both (3/2)kT). (b) One gram contains 1/2 mol of H₂ and 1/32 mol of O₂; KE per gram ∝ moles, so H₂ : O₂ = (1/2) : (1/32) = 16 : 1.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Pressure of an ideal gas from kinetic theory?Show answer
p = ⅓ρc̄².
2.Average KE of a molecule?Show answer
(3/2)kT.
3.rms speed formula?Show answer
√(3RT/M) = √(3p/ρ).
4.Value of Boltzmann's constant?Show answer
1.38 × 10⁻²³ J K⁻¹.
5.γ for a monatomic gas?Show answer
5/3.
6.Energy per degree of freedom?Show answer
½kT per molecule.
7.Mean free path formula?Show answer
λ = 1/(√2πd²n).
8.Avogadro's number?Show answer
6.022 × 10²³ mol⁻¹.

Best order to study it

Learn the equation of state pV = nRT = NkT and Avogadro's hypothesis.
Learn the assumptions of kinetic theory.
Derive p = ⅓ρc² step by step.
Combine with pV = nRT to get KE = (3/2)kT and c_rms; learn the effects of T and p.
Learn degrees of freedom and the statement of equipartition; derive Cv, Cp and γ for each type of gas.
Learn the mean free path and its dependence on pressure and temperature.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you list the assumptions of kinetic theory?
Can you derive p = ⅓ρc²?
Can you show that KE per molecule = (3/2)kT?
Can you find c_rms for a gas at a given temperature?
Can you explain why c_rms does not depend on pressure at constant T?
Can you state equipartition and count degrees of freedom?
Can you find Cv, Cp and γ for gases and mixtures?
Can you calculate the mean free path?

Kinetic Theory of Gases: frequently asked questions

Do I need to derive the law of equipartition of energy?

No. The syllabus asks for the statement only; the derivation is not required. You must apply it to find the specific heats of monatomic, diatomic and polyatomic gases.

Is the derivation of p = ⅓ρc² required?

Yes. The syllabus says derive p = ⅓ρc̄² from the assumptions, and then obtain c_rms = √(3p/ρ) and the effect of temperature and pressure on it.

Can kinetic theory derive the ideal gas equation?

No. The ISC syllabus stresses that pV = nRT is an experimental result. Combining it with the kinetic-theory result for pV gives the kinetic interpretation of temperature: KE = (3/2)RT per mole.

Why does a diatomic molecule have only two rotational degrees of freedom?

Its moment of inertia about the line joining the two atoms is negligible, so rotation about that axis stores practically no energy. Only rotations about the two perpendicular axes count.

Why are measured specific heats of some gases higher than equipartition predicts at high temperature?

At high temperatures molecules also vibrate, adding kinetic and potential energy of vibration (two more degrees of freedom for a diatomic molecule), so Cv increases and γ decreases.

What is the difference between mean speed and rms speed?

Mean speed is the average of the molecular speeds; rms speed is the square root of the average of their squares. The rms speed is always a little larger and is the one linked to kinetic energy and pressure.

Based on: CISCE ISC 2028 Physics syllabus (Classes XI and XII); CISCE Regulations for the ISC Examination Year 2028 (Class XI internal examination)