HomeClass 11 PhysicsChapter 12
CLASS 11 PHYSICS NCERT · CHAPTER 12

Kinetic Theory

How molecular motion explains gas pressure, temperature and heat capacity.

CHAPTER 12

Kinetic Theory

Follow the molecular model from gas laws to pressure, temperature, heat capacities and collisions.

12.1

Molecular nature of matter

A gas law links measurable bulk quantities to a picture of discrete molecules.

  • Matter consists of molecules. Avogadro’s number NA≈6.02×10²³ mol⁻¹ counts molecules per mole; the molar mass divided by NA gives the mass of one molecule.
  • The ideal-gas equation is PV=μRT=NkBT, where μ is amount in moles, N=μNA, R=8.314 J mol⁻¹ K⁻¹ and kB=R/NA≈1.38×10⁻²³ J K⁻¹. Use absolute temperature.
  • At ordinary conditions, gas molecules are separated by distances much larger than their size. Their own volume and intermolecular forces are negligible compared with the container volume and kinetic energy, except during collisions.
  • The molecular volume fraction is roughly density of gas / density of liquid for the same material. For water vapour at 100 °C and 1 atm, 0.6/1000≈6×10⁻⁴: most of the volume is empty space.
  • Molecular masses are tiny and molecular number densities are huge. Water’s molecular mass is about 0.018/NA≈3×10⁻²⁶ kg; at liquid density its molecular volume is about 3×10⁻²⁹ m³, corresponding to a radius near 2 Å.
molecules: random motion between collisionscollisions change direction
At ordinary conditions, molecules travel many molecular diameters between brief collisions. The ideal-gas model neglects forces except during collisions.
NCERT Example 12.1

Molecular volume fraction in water vapour

Question. The density of water is 1000 kg m⁻³. The density of water vapour at 100 °C and 1 atm pressure is 0.6 kg m⁻³. Estimate the ratio of molecular volume to total vapour volume.

  1. Step 1. For the same mass, volume is inversely proportional to density. The vapour occupies 1000/0.6 times the liquid volume.
  2. Step 2. In liquid water, molecules are closely packed, so molecular volume is approximately the total volume.
  3. Step 3. The fraction in vapour is therefore the inverse expansion ratio: 0.6/1000.

Answer: 6×10⁻⁴.

NCERT Example 12.2

Size of a water molecule

Question. Estimate the volume and radius of a water molecule using the data in Example 12.1.

  1. Step 1. A water molecule has mass 0.018/6×10²³≈3×10⁻²⁶ kg.
  2. Step 2. At density 1000 kg m⁻³ its volume is m/ρ≈3×10⁻²⁹ m³.
  3. Step 3. Model it as a sphere: V=(4/3)πr³, so r=[3V/(4π)]^(1/3).

Answer: r≈2×10⁻¹⁰ m=2 Å.

NCERT Example 12.3

Mean separation of water molecules

Question. What is the average distance between atoms (interatomic distance) in water? Use the data in Examples 12.1 and 12.2.

  1. Step 1. The same mass occupies about 1.67×10³ times more volume as vapour than liquid.
  2. Step 2. Volume available per molecule rises by this factor, so a characteristic spacing rises by its cube root, approximately 12 (the text rounds the expansion to 10³, giving 10).
  3. Step 3. Using the textbook’s rounded estimate, separation is about 2×(10×2 Å).

Answer: About 40 Å using the NCERT order-of-magnitude estimate.

12.2–12.3

Gas laws and ideal-gas behaviour

The gas laws are limiting relations; real gases approach them when dilute and warm.

  • Boyle’s law: for fixed amount at fixed temperature, PV is constant. Charles’ law: at fixed pressure, V/T is constant. Avogadro’s law: at fixed P,T, V is proportional to molecule number N.
  • Combining these gives PV=μRT=NkBT. For a mixture of non-reacting ideal gases at common V,T, Ptotal=ΣPi (Dalton’s law); each Pi is the pressure that component would exert alone in the same volume and temperature.
  • Real gases deviate from PV=μRT because molecules have finite size and interact. The ideal limit is approached at low pressure (low number density) and high temperature, away from condensation.
  • For a fixed gas sample, PV/T is approximately constant in the ideal region. At low pressure, extrapolated PV/T tends to μR, independent of the gas species for equal mole amount.
  • A gas molecule’s speed depends on its mass at a given temperature: lighter molecules move faster. Equal partial-pressure ratios in the same vessel imply equal mole (and molecule-number) ratios, not equal mass densities.
gas 1: P₁gas 2: P₂mixtureP=P₁+P₂
Dalton’s law follows by adding the ideal-gas contributions NᵢkBT/V.
NCERT Example 12.4

Neon–oxygen mixture

Question. A vessel contains neon and oxygen. Their partial pressures are in the ratio 3:2. Estimate the ratios of molecule numbers and mass densities. Atomic mass Ne=20.2 u; molecular mass O₂=32.0 u.

  1. Step 1. At common V,T, Pi=Ni kBT/V, so NNe/NO₂=PNe/PO₂=3/2.
  2. Step 2. Mass density ratio is (NNe mNe)/(NO₂ mO₂)=(3/2)(20.2/32.0).

Answer: Number ratio 3:2; mass-density ratio ≈0.947:1.

12.4

Kinetic theory: pressure derivation

Pressure is the average normal momentum transfer from molecule–wall collisions.

  • Model N molecules of mass m in a cube of side ℓ and volume V=ℓ³. Molecules move randomly; collisions with walls and one another are elastic, and intermolecular forces are negligible except during brief collisions.
  • For one molecule with velocity component vx toward a wall of area A=ℓ², an elastic collision reverses vx. Its momentum change is −2mvx; the wall receives impulse 2mvx.
  • In time Δt, molecules within distance vxΔt can reach the wall. Half move toward it, so the number striking is (1/2)nAvxΔt. Momentum transferred is (2mvx)(1/2 nAvxΔt)=nmAvx²Δt.
  • Force is momentum transfer per time; dividing by A gives pressure contribution nm⟨vx²⟩. Isotropy means ⟨vx²⟩=⟨vy²⟩=⟨vz²⟩=⟨v²⟩/3. Therefore P=(1/3)nm⟨v²⟩.
  • Since n=N/V, PV=(1/3)Nm⟨v²⟩=(2/3)Etr, where Etr=N(1/2)m⟨v²⟩ is total translational kinetic energy. Vessel shape does not affect the result; pressure is scalar in equilibrium.
  • Real molecular collisions do not change the steady velocity distribution, so the ensemble average remains valid. The ideal-gas model assumes collisions are brief and elastic.
wall area Aside ℓmolecule
Elastic wall collisions reverse the normal velocity component and transfer momentum to the wall; the average transfer produces pressure.
Derivation: For one wall, Δp per hit=2mvx and hits per Δt=(1/2)nAvxΔt. Thus F/A=nm⟨vx²⟩. Random isotropic motion gives ⟨vx²⟩=⟨v²⟩/3, so P=(1/3)nm⟨v²⟩.
12.4

Temperature, molecular speed and mixtures

Kinetic theory gives temperature a direct microscopic meaning.

  • Equate PV=(2/3)Etr with PV=NkBT: Etr=(3/2)NkBT and average translational kinetic energy per molecule=(1/2)m⟨v²⟩=(3/2)kBT.
  • The rms speed is vrms=√⟨v²⟩=√(3kBT/m)=√(3RT/M), where M is molar mass in kg mol⁻¹. At the same temperature, every ideal-gas species has the same mean translational kinetic energy, but lighter molecules have greater rms speed.
  • For a mixture, pressures add: P=(1/3)Σni mi⟨vi²⟩. Thermal equilibrium makes each species’ mean translational energy (3/2)kBT, yielding Dalton’s law.
  • Compression by a moving piston transfers energy to molecules through collisions; the mean kinetic energy and temperature rise. During expansion, gas molecules transfer energy to the receding piston and cool if no compensating heat enters.
  • At STP, 1 mol of an ideal gas occupies about 22.4 L. Always distinguish gauge pressure from absolute pressure in gas-law calculations: Pabsolute=Pgauge+Patmosphere.
NCERT Example 12.5

Argon and chlorine at equal temperature

Question. A flask contains argon and chlorine in a 2:1 mass ratio at 27 °C. Find ratios of average kinetic energy per molecule and rms speeds. Atomic mass Ar=39.9 u; molecular mass Cl₂=70.9 u.

  1. Step 1. At equal temperature, each species has mean translational kinetic energy (3/2)kBT.
  2. Step 2. From (1/2)mvrms²=(3/2)kBT, vrms∝1/√m.

Answer: Mean kinetic energies Ar:Cl₂=1:1; vrms(Ar)/vrms(Cl₂)=√(70.9/39.9)≈1.33. The mass proportion of the mixture does not change these ratios.

NCERT Example 12.6

Uranium hexafluoride isotope speeds

Question. Uranium has isotopes of masses 235 u and 238 u. In uranium hexafluoride, which isotope moves faster and what is the percentage speed difference?

  1. Step 1. Molecular masses are 235+6×19=349 u and 238+6×19=352 u.
  2. Step 2. At a fixed temperature, rms speed is inversely proportional to square root of molecular mass.
  3. Step 3. Relative speed difference is [√(352/349)−1]×100%.

Answer: The ²³⁵UF₆ molecule is faster, by approximately 0.44%.

NCERT Example 12.7

Piston collisions and gas temperature

Question. (a) A ball rebounds from a bat moving toward it. Does it move faster or slower? (b) Explain why compression heats a gas. (c) What happens when compressed gas expands against a piston? (d) Does a heavy cricket bat help?

  1. Step 1. In the bat frame, the approaching ball’s speed is u+V. Elastic rebound reverses this relative velocity; transforming back gives rebound speed u+2V away from the wicket.
  2. Step 2. A piston moving inward is like the bat moving toward molecules: collisions increase their kinetic energy, raising temperature.
  3. Step 3. An expanding gas transfers energy to an outward-moving piston; its molecules lose mean kinetic energy and cool if heat does not enter.
  4. Step 4. A heavier bat is less slowed by impact and can transfer more momentum/energy to the ball.

Answer: (a) Faster (for an ideal massive bat); (b) compression heats; (c) expansion can cool; (d) a heavier bat can deliver a faster rebound.

12.5

Law of equipartition of energy

At thermal equilibrium, quadratic energy terms share energy equally on average.

  • For translation, ε=(1/2)mvx²+(1/2)mvy²+(1/2)mvz². Since the three directions are equivalent and ⟨ε⟩=(3/2)kBT, each term averages (1/2)kBT.
  • Equipartition law: each independent quadratic term in energy contributes (1/2)kBT per molecule at temperature T. A degree of freedom is an independent coordinate needed to specify motion/configuration.
  • A monatomic molecule has 3 translational degrees of freedom. A rigid linear diatomic molecule has 3 translational+2 rotational active degrees; rotation about its bond axis is negligible at ordinary temperatures in the classical treatment.
  • A rigid non-linear molecule has 3 translational+3 rotational degrees. Each vibrational mode has both quadratic kinetic and potential terms, contributing kBT when fully active.
  • Equipartition is a classical result. At low temperatures, quantum energy levels can freeze out rotational or vibrational modes; the effective number of active degrees depends on temperature.
MonatomicDiatomic, linearNon-lineartranslation: 3translation + rotation: 5translation + rotation: 6
Each independent quadratic degree of freedom contributes ½kBT per molecule in the classical equipartition model.
Proof for translation: Isotropy makes the average of each of the three equal quadratic terms the same. Their sum is the mean translational energy (3/2)kBT. Dividing equally among x, y and z gives (1/2)kBT per translational degree of freedom.
12.6

Specific heat capacities

Equipartition predicts heat capacities from the active molecular degrees of freedom.

  • At constant volume, no expansion work is done, so heat raises internal energy. For f active quadratic degrees per molecule, U=(f/2)NkBT and CV=(f/2)R per mole.
  • For an ideal gas, CP−CV=R (Mayer relation): at constant pressure, added heat supplies internal-energy increase plus expansion work. For a monatomic gas, CV=3R/2 and CP=5R/2; γ=CP/CV=5/3.
  • A rigid diatomic gas at ordinary temperature has f=5, so CV=5R/2, CP=7R/2 and γ=7/5. At high temperature active vibrations add energy and raise heat capacity; at low T rotations/vibrations may be frozen out.
  • For non-linear polyatomic molecules with six active translational/rotational degrees, CV=3R and CP=4R classically; vibrational modes add further contributions when excited.
  • In a solid, each atom vibrates in three dimensions. Each direction has quadratic kinetic and potential energy, so mean energy is 3kBT per atom and U=3RT per mole. Neglecting small solid expansion work gives molar C≈3R (Dulong–Petit law).
  • The classical solid prediction works near room temperature for many solids but fails at low temperature and for some materials (including carbon near room temperature). Quantum theory explains the fall in heat capacity as vibrational modes freeze out.
NCERT Example 12.8

Heating helium in a fixed-volume cylinder

Question. A fixed-capacity 44.8 L cylinder contains helium at standard temperature and pressure. How much heat raises its temperature by 15.0 °C? Take R=8.31 J mol⁻¹ K⁻¹.

  1. Step 1. At STP, 22.4 L per mole means the cylinder contains 2 mol.
  2. Step 2. Helium is monatomic, so CV=3R/2. The cylinder is fixed-volume, so Q=nCVΔT.
  3. Step 3. Q=2×(3/2×8.31)×15.0.

Answer: Q≈374 J.

12.7

Mean free path

Molecules move quickly, but frequent collisions interrupt their paths.

  • For hard-sphere molecules of diameter d, a molecule moving at average speed ⟨v⟩ sweeps a collision cylinder of cross-section πd². In time Δt the swept volume is πd²⟨v⟩Δt.
  • At number density n, collision count in this simplified stationary-target model is nπd²⟨v⟩Δt. Thus mean time between collisions τ≈1/(nπd²⟨v⟩) and mean free path ℓ=⟨v⟩τ≈1/(nπd²).
  • Other molecules also move. Using their mean relative speed gives the more accurate mean free path ℓ=1/(√2 πnd²), collision rate ≈√2 nπd²⟨v⟩ and mean collision time τ≈ℓ/⟨v⟩.
  • Since n=P/(kBT), ℓ=kBT/(√2πd²P): mean free path increases with temperature and decreases with pressure and molecular diameter squared. In an evacuated tube it can approach the tube length.
  • At STP, air with d≈2 Å has ℓ≈2.9×10⁻⁷ m, about 1500 molecular diameters. Mean free path underlies diffusion, viscosity and thermal conduction.
diameter dmean free path ℓswept collision area = πd²
A collision occurs when another molecular centre enters within distance d of the moving molecule’s centreline.
NCERT Example 12.9

Mean free path in water vapour

Question. Estimate the mean free path of a water molecule in water vapour at 373 K, using the STP air value ℓ≈2.9×10⁻⁷ m and the textbook assumptions.

  1. Step 1. Take molecular diameter approximately that used for air; thus ℓ∝1/n.
  2. Step 2. At fixed pressure, n∝1/T, so ℓ scales in proportion to T.
  3. Step 3. ℓ(373 K)≈2.9×10⁻⁷×373/273. The textbook rounds its STP value to 3×10⁻⁷ m.

Answer: About 4×10⁻⁷ m.

12

Points to ponder

  1. Pressure exists throughout a fluid, not only at its walls; an internal gas layer is in equilibrium because pressure acts on both sides.
  2. At ordinary conditions, intermolecular spacing is about ten times the interatomic spacing in liquids and solids. Mean free path is much larger: roughly 100 times that spacing and 1000 times molecular size.
  3. Count both quadratic terms of each vibration: kinetic and potential. One active vibrational mode contributes kBT.
  4. Air molecules do not settle because their kinetic energy greatly exceeds the gravitational potential-energy change over ordinary room heights. Density is only slightly greater at lower levels.
  5. In general ⟨v²⟩ is not equal to ⟨v⟩²; the mean of a square is not the square of a mean.
MODEL

Molecules in a container

Watch elastic wall collisions transfer momentum. Change the speed scale to see why higher temperature means greater mean kinetic energy; the particle paths are illustrative, not a molecular-scale numerical simulation.

1.0×

Mean kinetic energy ∝ T; wall impacts transfer normal momentum.

27 QUESTIONS · ALL SECTIONS

Quiz

Choose an answer to see why it is right.

0 / 27 AnsweredCorrect: 0 (0%)
1. The number of molecules in μ moles is:
One mole contains NA molecules.
2. Boltzmann constant is related to R by:
kB=R/NA.
3. The ideal-gas equation in molecule form is:
Use molecule count with kB.
4. At fixed temperature, Boyle’s law states:
For a fixed amount of gas, PV is constant.
5. Dalton’s law says the total pressure is:
Each ideal component contributes Pi.
6. Ideal-gas behaviour is approached at:
Dilute, warm gases reduce intermolecular corrections.
7. In an elastic collision with a wall, the normal velocity component:
Tangential components remain unchanged.
8. Molecules striking a wall per Δt in the simple cube derivation are:
Only half of the molecules in the swept layer move toward the wall.
9. For isotropic molecular motion, ⟨vx²⟩ is:
The three equal component averages sum to ⟨v²⟩.
10. The kinetic-theory pressure equation is:
Pressure is one-third density times mean molecular kinetic term.
11. Mean translational kinetic energy per ideal-gas molecule is:
Temperature sets the average translational energy.
12. At the same temperature, rms speed is greatest for:
vrms∝1/√m.
13. Gas-law temperature must be measured in:
The equation uses absolute temperature.
14. Each active quadratic degree contributes average energy:
Equipartition gives one-half kBT per quadratic term.
15. A rigid linear diatomic molecule has how many active translational and rotational degrees at ordinary temperature?
Three translation plus two rotations.
16. A vibration with quadratic kinetic and potential terms contributes:
Both terms each contribute ½kBT.
17. For an ideal gas, Mayer relation is:
Constant-pressure heating includes PV work.
18. Monatomic ideal-gas CV is:
Three translational degrees give fR/2.
19. For a classical solid, Dulong–Petit predicts molar heat capacity:
Three-dimensional atomic vibrations give U=3RT.
20. Why can classical heat-capacity predictions fail at low temperature?
Quantum energy levels prevent all modes from being thermally active.
21. Mean free path varies with molecular diameter as:
ℓ=1/(√2πnd²).
22. Number density of an ideal gas is:
From P=nkBT.
23. At fixed temperature, lowering pressure makes mean free path:
Lower pressure means lower number density.
24. At STP, air mean free path is roughly:
It is about 1500 molecular diameters.
25. In an equilibrium gas, pressure acts:
An internal layer is in equilibrium because pressure acts on both sides.
26. For ordinary gases, intermolecular spacing is about:
The separation is not enormous compared with condensed matter spacing.
27. Which statement is generally true?
The mean of a square is generally different from the square of a mean.
NCERT · EXERCISES 12.1–12.10

Exercises

Try each textbook question before opening its worked solution.

Ex 12.1NCERT

12.1Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.

Related concept: Review the matching notes

  1. Step 1. At STP, 1 mol occupies 22.4×10⁻³ m³. Molecule volume as a sphere: (π/6)d³=(π/6)(3×10⁻¹⁰)³ m³.
  2. Step 2. Molecular volume fraction is NA(πd³/6)/(22.4×10⁻³).

Answer: About 3.8×10⁻⁴ (roughly 0.04%).

Ex 12.2NCERT

12.2Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.

Related concept: Review the matching notes

  1. Step 1. Use PV=μRT with μ=1 mol, T=273 K and P=1.01×10⁵ Pa.
  2. Step 2. V=RT/P=(8.31×273)/(1.01×10⁵) m³. Convert m³ to litres by multiplying by 1000.

Answer: V≈2.25×10⁻² m³≈22.4 L (rounded textbook value).

Ex 12.3NCERT

12.3Figure 12.8 shows plot of PV/T versus P for 1.00×10⁻³ kg of oxygen gas at two different temperatures. (a) What does the dotted plot signify? (b) Which is true: T₁>T₂ or T₁

Related concept: Review the matching notes

  1. Step 1. (a) The dotted line is the ideal-gas limit, where PV/T=μR is independent of pressure. (b) The T₁ curve lies closer to this ideal line, so T₁>T₂.
  2. Step 2. (c) At P→0, PV/T=μR=(m/M)R for ideal gas. Oxygen amount is 1.00×10⁻³/0.0320 mol.
  3. Step 3. (d) Match μ for hydrogen: mH₂/MH₂=mO₂/MO₂, hence mH₂=(1.00×10⁻³)(2.02/32.0) kg.

Answer: (b) T₁>T₂. (c) ≈0.260 J K⁻¹. (d) No; the same intercept requires ≈6.31×10⁻⁵ kg=0.0631 g H₂.

Ex 12.4NCERT

12.4An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out (R=8.31 J mol⁻¹ K⁻¹, molecular mass O₂=32 u).

Related concept: Review the matching notes

  1. Step 1. Convert gauge to absolute: P₁=16 atm and P₂=12 atm. V=0.030 m³; T₁=300 K, T₂=290 K.
  2. Step 2. For each state n=PV/(RT). Amount removed is Δn=V(P₁/T₁−P₂/T₂)/R, using pressure in Pa.
  3. Step 3. Mass removed=Δn×0.032 kg mol⁻¹.

Answer: ≈0.14 kg O₂ (using 1 atm=1.013×10⁵ Pa).

Ex 12.5NCERT

12.5An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?

Related concept: Review the matching notes

  1. Step 1. At the bottom, absolute pressure P₁=P₀+ρgh≈1.013×10⁵+1000×9.8×40 Pa. At surface P₂=P₀.
  2. Step 2. For a fixed amount of gas, V₂=V₁(P₁/P₂)(T₂/T₁). Use kelvin: 285 K and 308 K.

Answer: About 5.3 cm³ (neglect surface tension and gas dissolution).

Ex 12.6NCERT

12.6Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27 °C and 1 atm pressure.

Related concept: Review the matching notes

  1. Step 1. Use PV=NkBT, with P=1.013×10⁵ Pa, V=25.0 m³, T=300 K.
  2. Step 2. N=PV/(kBT), where kB=1.38×10⁻²³ J K⁻¹.

Answer: N≈6.1×10²⁶ molecules.

Ex 12.7NCERT

12.7Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) 10 million kelvin (typical stellar-core temperature).

Related concept: Review the matching notes

  1. Step 1. Use mean translational thermal energy ⟨K⟩=(3/2)kBT.
  2. Step 2. Convert 27 °C to 300 K; substitute each temperature.
  3. Step 3. For electron-volts, divide joules by 1.602×10⁻¹⁹ J/eV.

Answer: (i) 6.2×10⁻²¹ J ≈0.039 eV; (ii) 1.24×10⁻¹⁹ J≈0.78 eV; (iii) 2.07×10⁻¹⁶ J≈1.29 keV.

Ex 12.8NCERT

12.8Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed the same in the three cases? If not, in which case is vrms the largest?

Related concept: Review the matching notes

  1. Step 1. PV=NkBT: equal P,V,T gives equal N.
  2. Step 2. At equal T, vrms∝1/√molecule mass. Compare molar masses: Ne≈20.2, Cl₂≈70.9, UF₆≈238+6×19≈352 u.

Answer: Equal molecule numbers; speeds are unequal. Neon has the largest rms speed.

Ex 12.9NCERT

12.9At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at −20 °C? (Atomic mass of Ar=39.9 u, of He=4.0 u.)

Related concept: Review the matching notes

  1. Step 1. Set 3kBTAr/mAr=3kBTHe/mHe, so TAr/mAr=THe/mHe.
  2. Step 2. THe=253 K; TAr=253×39.9/4.0.

Answer: TAr≈2525 K (about 2.5×10³ K).

Ex 12.10NCERT

12.10Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N₂=28.0 u).

Related concept: Review the matching notes

  1. Step 1. d=2r=2.0×10⁻¹⁰ m; n=P/(kBT), with P=2.026×10⁵ Pa and T=290 K.
  2. Step 2. Use ℓ=1/(√2πnd²), then vrms=√(3RT/M) with M=0.028 kg mol⁻¹.
  3. Step 3. Mean free time τ≈ℓ/vrms and frequency f≈1/τ. A rough duration for one collision is d/vrms; it is much shorter than τ.

Answer: n≈5.1×10²⁵ m⁻³; ℓ≈1.1×10⁻⁷ m; vrms≈508 m s⁻¹; τ≈2.2×10⁻¹⁰ s; f≈4.6×10⁹ s⁻¹. Collision duration d/vrms≈4×10⁻¹³ s, about 500 times shorter than the free-flight time.

POINT-WISE SUMMARY

Revision

Key definitions, assumptions and results, grouped by section.

12.1 · Molecular nature of matter

Review notes
  • PV=μRT=NkBT; kB=R/NA.
  • For dilute gases, molecular volume and intermolecular forces are small corrections.
  • Water molecule: mass ≈3×10⁻²⁶ kg; radius ≈2 Å.

12.2–12.3 · Gas laws and ideal-gas behaviour

Review notes
  • Boyle: PV const; Charles: V/T const; Avogadro: V∝N at fixed P,T.
  • Ideal gas: PV=NkBT; mixture pressure is sum of partial pressures.
  • Ideal behaviour improves as pressure falls and temperature rises.

12.4 · Kinetic theory: pressure derivation

Review notes
  • Pressure derivation: impulse per collision × collision rate, divided by wall area.
  • P=(1/3)nm⟨v²⟩; PV=(2/3)Etr.
  • Isotropy gives one-third of mean-square speed in each direction.

12.4 · Temperature, molecular speed and mixtures

Review notes
  • ⟨Ktrans⟩=(3/2)kBT, independent of gas identity.
  • vrms=√(3kBT/m)=√(3RT/M); at fixed T, vrms∝1/√M.
  • Gas-law calculations use absolute pressure and kelvin.

12.5 · Law of equipartition of energy

Review notes
  • Each active quadratic term contributes ½kBT per molecule.
  • Rigid molecule f: monatomic 3; linear 5; non-linear 6 (translation + rotation).
  • Each active vibration contributes kBT; quantum effects freeze modes at low T.

12.6 · Specific heat capacities

Review notes
  • Gas with f active quadratic terms: CV=fR/2; CP=CV+R; γ=CP/CV.
  • Monatomic: (CV,CP,γ)=(3R/2,5R/2,5/3); diatomic: (5R/2,7R/2,7/5).
  • Classical solid: U=3RT and C≈3R; quantum effects matter at low T.

12.7 · Mean free path

Review notes
  • Simple model: ℓ≈1/(nπd²); relative-motion correction: ℓ=1/(√2πnd²).
  • τ≈ℓ/⟨v⟩; n=P/(kBT).
  • Mean free path grows as density/pressure falls and is central to transport in gases.
THREE LEVELS · 27 QUESTIONS

Test yourself

Each level samples the chapter from molecular model to applications.

1. The number of molecules in μ moles is:
2. Boltzmann constant is related to R by:
3. The ideal-gas equation in molecule form is:
4. At fixed temperature, Boyle’s law states:
5. Dalton’s law says the total pressure is:
6. Ideal-gas behaviour is approached at:
7. In an elastic collision with a wall, the normal velocity component:
8. Molecules striking a wall per Δt in the simple cube derivation are:
9. For isotropic molecular motion, ⟨vx²⟩ is: