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Theory · Physics Circle Master Book

Kinetic Theory of Gases

01
CONCEPT

Pressure & Kinetic Energy of a Gas

In one line
Gas pressure is the result of molecules bouncing off the walls. The average kinetic energy of the molecules depends only on the temperature.
ASSUMPTIONS

14.1.1 Assumptions of kinetic theory

  1. A gas is made of a very large number of identical molecules.
  2. Molecules are point masses: their volume is negligible compared with the volume of the gas.
  3. They move randomly in all directions with all possible speeds.
  4. Collisions with each other and with the walls are perfectly elastic.
  5. No force acts between molecules except during a collision; collision time is negligible.
  6. Gravity on the molecules is neglected.
PRESSURE

14.1.2 Pressure of an ideal gas

vxAL
A molecule hitting wall A of a cube of side L
  1. Change in momentum at the wall: Δp = 2mvx.
  2. Time between hits on the same wall: 2Lvx.
  3. Force by one molecule = mvx²L ⇒ wall force ∝ v² ∝ T.
  4. Random motion: vx² = vy² = vz² = v²3.
P = 13mNVv2rms = 13ρv2rms
PRESSURE AND ENERGY

14.1.3 Pressure and kinetic energy

P = 23E  ·  E = translational KEvolume
  1. Total translational KE of a gas = 32PV.
  2. Pressure rises if: more molecules (P ∝ N), smaller volume (P ∝ 1V), higher temperature (P ∝ v2rms ∝ T).
  3. Masses halved and speeds doubled ⇒ P doubles (P ∝ mv²).
KINETIC ENERGY AND TEMPERATURE

14.1.4 Kinetic energy and temperature

Emolecule = 32kBT  ·  Emole = 32RT
Amount of gasTranslational KE
1 molecule32kBT
N molecules32NkBT
n moles32nRT
1 gram32RTM
  1. Average translational KE per molecule depends only on T, not on the gas: H₂ and O₂ at the same T have equal average KE.
  2. KE ∝ T (in kelvin): to double KE, double T.
  3. At T = 0 K the translational motion stops.
  4. KE per gram depends on M: smaller M ⇒ larger KE per gram.
  5. Temperature is a measure of the average translational KE of the molecules.
N = 3PV2Emolecule  ·  32kBT = eV (electron through V volt)
Practise PYQs on this topicPressure & Kinetic Energy of a Gas · 39 questions with solutions
02
CONCEPT

Molecular Speeds

In one line
Gas molecules have a spread of speeds. Three averages are used: the most probable, the average and the root mean square speed.
RMS SPEED

14.2.1 Root mean square speed

vrms = √v₁² + v₂² + …N = √3Pρ
vrms = √3RTM = √3kBTm
  1. vrms ∝ √T (in kelvin) and ∝ 1 ÷ √M.
  2. At the same T: lighter molecules move faster. H₂ is 4 times faster than O₂.
  3. vrms does not depend on pressure alone at constant T (P ∝ ρ).
  4. Same speed for two gases ⇒ T₁M₁ = T₂M₂.
  5. For H₂ at NTP: vrms ≈ 1840 m/s.
THREE SPEEDS

14.2.2 Most probable, average, rms

vmp = √2RTM  ·  vav = √8RTπM  ·  vrms = √3RTM
vrms > vav > vmp
vrms : vav : vmp ≈ 1.73 : 1.60 : 1.41
  1. For a set of numbers: vav = sum ÷ N; vrms = √(mean of squares).
  2. vmp = the speed held by the largest number of molecules.
MAXWELL DISTRIBUTION

14.2.3 How the speeds are spread

vdN/dvvmpvavvrmsT₁T₂ > T₁
At higher T the peak moves right and the curve flattens
  1. Area under the curve = total number of molecules (stays the same).
  2. Peak is at vmp. The curve is not symmetric.
  3. Higher T: peak shifts to higher speed and becomes lower and broader.
SPEED OF SOUND

14.2.4 Compared with vrms

vsound = √γRTM  ·  vrmsvsound = √3γ
  1. A planet keeps an atmosphere only if vrms < escape speed. The Moon has none.
Practise PYQs on this topicMolecular Speeds & Maxwell Distribution · 90 questions with solutions
03
CONCEPT

Law of Equipartition of Energy

In one line
In thermal equilibrium, energy is shared equally among all the degrees of freedom: each gets ½ kBT per molecule.
THE LAW

14.3.1 Law of equipartition

per molecule: 12kBT per degree
per mole: 12RT per degree
  1. Degrees of freedom (f) are counted as in the Thermodynamics chapter: 3 translational, plus rotational, plus 2 for each vibrational mode.
  2. Translational KE alone is always (32)kBT, whatever the molecule.
monoatomic3/2kTdiatomic (rigid)5/2kTdiatomic (+ vibration)7/2kTeach box = ½ kT
Energy per molecule = (f2)kBT
ENERGY FOR f DEGREES

14.3.2 Energy for f degrees of freedom

AmountEnergy
1 moleculef2kBT
N moleculesf2NkBT
1 molef2RT
n molesf2nRT
  1. Monoatomic: (32)kBT. Rigid diatomic: (52)kBT. Non-rigid diatomic: (72)kBT.
  2. Rotational energy of a diatomic molecule = 2 × ½kBT = kBT, which is less than its translational (32)kBT.
  3. The energies of individual molecules are spread out (Maxwell–Boltzmann); the law gives the average.
WHY Cv CHANGES WITH T

14.3.3 Why Cv changes with T

  1. At very low T only translation is active: Cv = (32)R for H₂.
  2. At room T rotation joins: Cv = (52)R.
  3. At high T vibration joins: Cv = (72)R.
Practise PYQs on this topicLaw of Equipartition of Energy · 5 questions with solutions
04
CONCEPT

Mean Free Path & Real Gases

In one line
The mean free path is the average distance a molecule travels between two collisions. Real gases behave ideally only at low pressure and high temperature.
MEAN FREE PATH

14.4.1 Mean free path

λ = average free path between collisions
λ = 1√2 πd²n = kBT√2 πd²P = m√2 πd²ρ
  1. d = diameter of a molecule, n = number of molecules per unit volume.
  2. λ ∝ 1d² and ∝ 1n.
  3. Constant P: λ ∝ T. Constant T: λ ∝ 1 ÷ P.
  4. Closed vessel (fixed n): λ does not change with T.
COLLISION FREQUENCY

14.4.2 Collision frequency

ν = vavλ  ·  mean free time τ = λvav
  1. Fixed n: ν ∝ v ∝ √T.
  2. Same T: ν ∝ n (more molecules ⇒ more collisions).
  3. Heating at constant volume ⇒ more collisions per second.
REAL GASES

14.4.3 Real gases

PPV/RTidealCO₂H₂1
Compressibility factor PVRT for one mole
  1. For an ideal gas PV ÷ RT = 1 at all pressures.
  2. Real gases approach ideal behaviour at low pressure and high temperature; they deviate most at high pressure and low temperature.
  3. Easily liquefied gases (CO₂) deviate the most.
VAN DER WAALS EQUATION

14.4.4 Van der Waals equation

(P + aV²)(V − b) = RT
  1. b: volume correction for the size of the molecules. a: pressure correction for the attraction between molecules.
  2. Unit of a: N m⁴ mol⁻²; unit of b: m³ mol⁻¹.
Tc = 8a27Rb  ·  Pc = a27b²  ·  Vc = 3b
  1. Critical temperature: above it a gas cannot be liquefied by pressure alone (CO₂: 31.1°C).
GRAHAM’S LAW

14.4.5 Graham’s law of diffusion

r ∝ 1√ρ ∝ 1√M
Practise PYQs on this topicMean Free Path & Real Gases · 26 questions with solutions
EXAM-DAY CARD

Just see this before the exam

Kinetic Theory of Gases
P = (13)ρv2rms
pressure
P = (23)E
energy
E = (32)kBT
KE per molecule
vrms = √(3RTM)
rms speed
½kBT per f
equipartition
λ = 1√2 πd²n
mean free path
FORMULA SHEET

Whole chapter on one page

Pressure

P = (13)ρv2rms

P = (23)E

wall force ∝ T

KEtotal = (32)PV

Kinetic energy

(32)kBT per molecule

(32)RT per mole

KE ∝ T (kelvin)

N = 3PV2E

Speeds

vrms = √(3RTM)

vav = √(8RTπM)

vmp = √(2RTM)

vrms > vav > vmp

Equipartition

½kBT per degree

E = (f2)kBT

mono 3, dia 5, +2 per vib

Cv = (f2)R

Mean free path

λ = 1√2 πd²n

λ = kBT√2 πd²P

ν = vavλ

fixed n: ν ∝ √T

Real gases

ideal at low P, high T

(P + aV²)(V − b) = RT

Tc = 8a27Rb

r ∝ 1√M

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Using °C in vrms ∝ √T or KE ∝ T

Thinking KE per molecule depends on the gas

vrms depends on P at fixed T

λ changes with T in a closed vessel

CORRECT ✓

Always use kelvin

Average KE = (32)kBT for every gas

At fixed T, P ∝ ρ: vrms unchanged

Fixed n ⇒ λ fixed; ν ∝ √T