Class 11 Physics · Chapter 12
Short answer:
Kinetic Theory (Class 11 Physics, Chapter 12) gas ke macroscopic properties — pressure, temperature, volume — ko uske molecules ke random motion se explain karta hai. Core result: P = (1/3) n m⟨v²⟩ aur average kinetic energy per molecule Ē = (3/2) kBT — matlab temperature sirf molecules ki average KE ka measure hai, kisi gas ki "nature" ka nahi. Yehi kinetic theory of gases class 11 derivation is chapter ka sabse zyada exam-asked part hai, saath me RMS speed, law of equipartition of energy (Cv, Cp, γ) aur mean free path.
Kinetic Theory Class 11 Physics ka woh chapter hai jo Thermodynamics (Chapter 11) ke "macroscopic" P, V, T ko microscopic level tak le jaata hai — yaani gas ke andar itne saare molecules kaise random directions me udd rahe hain, aur unka collective motion hi pressure aur temperature ban jaata hai. Jo student pichhle chapter me Thermodynamics ki zeroth/first law samajh chuka hai, uske liye ye chapter naturally agla step hai. Class 11 physics ncert solutions dhoondhne waale students ke liye ye chapter high-yield hai kyunki isme numericals (Boyle's law, RMS speed, mean free path) aur theory (equipartition, degrees of freedom) dono equally important hain, aur Thermodynamics ke saath milkar poora Thermal Physics unit banate hain. Neeche har concept step-by-step derivation ke saath, real NCERT-style exercises poori working ke saath, aur common mistakes cover kiye gaye hain.
Chapter 12 Summary — 5 Minute Revision
Kinetic theory postulates karta hai ki gas ke molecules continuous random motion me hote hain aur wall se elastic collisions karte hain — isi se pressure derive hota hai: P = (1/3) n m⟨v²⟩. Ideal gas equation PV = NkBT se combine karne par milta hai Ē = (3/2) kBT — average KE per molecule sirf temperature par depend karta hai, gas ki nature/mass par nahi. RMS speed vrms = √(3RT/M) hai. Law of equipartition of energy kehta hai har degree of freedom (1/2)kBT energy leta hai — isi se monatomic (Cv=3/2R), diatomic (Cv=5/2R ya 7/2R with vibration) aur polyatomic gases ke specific heats nikalte hain. Chapter mean free path (λ = 1/(√2πd²n)) aur real gas behaviour (Van der Waals equation) ke saath close hota hai.
In-Text Questions — Solutions
STP par ek gas ka volume 2 L hai (P = 1 atm). Isi temperature par pressure ko 4 atm kar diya jaaye to naya volume kya hoga?
Boyle's Law (T constant): P₁V₁ = P₂V₂
1 × 2 = 4 × V₂
V₂ = 2/4 = 0.5 L
Naya volume 0.5 L hoga.
Rigid diatomic molecule (jaise N₂, koi vibration nahi) ke liye degrees of freedom kitne hote hain, aur C_v kya hoga?
Rigid diatomic molecule me 3 translational + 2 rotational degrees of freedom hote hain (rotation about the bond axis negligible hai kyunki moment of inertia ~0).
f = 5
Law of equipartition se: Cv = (f/2)R = (5/2)R ≈ 20.8 J mol⁻¹K⁻¹
Helium (M = 4u) aur Oxygen (M = 32u) same temperature T par hain. Kis gas ke molecules ki average kinetic energy zyada hogi?
Average KE per molecule: Ē = (3/2)kBT — is formula me mass ka koi role nahi hai, sirf T pe depend karta hai.
Dono gases ki average KE per molecule barabar hogi (same T par). Lekin vrms alag hoga — lighter He molecules zyada tezi se move karenge (vrms ∝ 1/√M) taaki unki KE bhi utni hi ho.
Mean free path λ kin-kin factors par depend karta hai?
λ = 1 / (√2 π d² n)
Yahan d = molecule ka diameter, n = number density (molecules/volume). Matlab λ molecule size (d²) aur density (n) dono ke inversely proportional hai — jyada pressure (jyada n) ya bada molecule size → chhota mean free path.
Poore Class 11 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q12)
Oxygen gas ke ek molecule ka diameter roughly 3 Å liya jaaye. STP par oxygen ke actual volume ka kitna fraction molecules khud occupy karte hain, estimate karo.
Ek molecule ka radius r = 1.5 Å = 1.5 × 10⁻¹⁰ m
Vmolecule = (4/3)πr³ = (4/3)(3.14)(1.5×10⁻¹⁰)³ ≈ 1.41 × 10⁻²⁹ m³
1 mole (NA = 6.023×10²³ molecules) ka total molecular volume:
Vtotal molecules = 6.023×10²³ × 1.41×10⁻²⁹ ≈ 8.5 × 10⁻⁶ m³
STP par molar volume = 22.4 L = 22.4 × 10⁻³ m³
Fraction = (8.5×10⁻⁶)/(22.4×10⁻³) ≈ 3.8 × 10⁻⁴
Matlab molecules ka apna volume total gas volume ka sirf ≈ 3.8×10⁻⁴ (0.038%) hai — baaki khaali space hai. Isi wajah se gas easily compress ho jaata hai.
Molar volume dikhao ki STP par 22.4 litre hota hai.
Ideal gas equation (1 mole ke liye): PV = RT ⟹ V = RT/P
STP: T = 273 K, P = 1 atm = 1.013 × 10⁵ Pa, R = 8.31 J mol⁻¹K⁻¹
V = (8.31 × 273) / (1.013 × 10⁵)
V = 2268.6 / 101300 = 0.0224 m³
V = 22.4 × 10⁻³ m³ = 22.4 litre
1.00×10⁻³ kg oxygen gas ke liye PV/T vs P ka graph do temperatures T₁ aur T₂ par plot hai — dono curves y-axis (P→0) par ek hi point pe milte hain. (a) Y-intercept ki value kya hai? (b) Isi value ko dene wala hydrogen ka mass kitna hoga?
P→0 par PV/T ki value = μR, jahan μ = number of moles (temperature-independent — ye ideal gas ki universal property hai, isliye dono curves same point pe milte hain).
μ (oxygen) = mass/M = (1.00×10⁻³ kg)/(32×10⁻³ kg/mol) = 1/32 mol
PV/T (at P→0) = μR = (1/32) × 8.31 ≈ 0.26 J/K
Hydrogen ke liye same PV/T value paane ke liye same number of moles chahiye (μ = 1/32 mol), kyunki R constant hai:
mass(H₂) = μ × M(H₂) = (1/32) × 2.02×10⁻³ kg
mass(H₂) ≈ 6.3 × 10⁻⁵ kg
Oxygen cylinder (volume 30 L) ka initial pressure 15 atm, temperature 27°C hai. Kuch oxygen nikaalne ke baad pressure 11 atm aur temperature 17°C ho jaata hai. Kitna oxygen (mass) nikaala gaya?
Ideal gas equation: PV = (mass/M)RT ⟹ mass = PVM/(RT)
Initial state: P₁ = 15 atm = 15×1.013×10⁵ Pa = 1.5195×10⁶ Pa, T₁ = 300 K
m₁ = (1.5195×10⁶ × 0.030 × 0.032) / (8.31 × 300) ≈ 0.585 kg
Final state: P₂ = 11 atm = 1.1143×10⁶ Pa, T₂ = 290 K
m₂ = (1.1143×10⁶ × 0.030 × 0.032) / (8.31 × 290) ≈ 0.444 kg
Mass nikaala gaya:
Δm = m₁ − m₂ = 0.585 − 0.444 ≈ 0.141 kg
Ek air bubble (volume 1.0 cm³) lake ke 40 m gehre bottom se surface tak uthta hai. Bottom par temperature 12°C aur surface par 35°C hai. Bubble ka volume surface par kya hoga?
Bottom pressure = Atmospheric + water column pressure:
P₁ = P₀ + ρgh = 1.013×10⁵ + (1000 × 9.8 × 40) = 1.013×10⁵ + 3.92×10⁵ = 4.933×10⁵ Pa
Surface pressure P₂ = P₀ = 1.013×10⁵ Pa
T₁ = 285 K, T₂ = 308 K
Combined gas law: P₁V₁/T₁ = P₂V₂/T₂
V₂ = (P₁V₁T₂)/(T₁P₂) = (4.933×10⁵ × 1.0 × 308)/(285 × 1.013×10⁵)
V₂ ≈ 5.26 cm³
25.0 m³ capacity wale room me 27°C temperature aur 1 atm pressure par total air molecules (N₂, O₂, water vapour etc. milaakar) estimate karo.
N = PV/(kBT)
P = 1.013×10⁵ Pa, V = 25 m³, T = 300 K, kB = 1.38×10⁻²³ J/K
N = (1.013×10⁵ × 25) / (1.38×10⁻²³ × 300)
N = (2.5325×10⁶) / (4.14×10⁻²¹) ≈ 6.11 × 10²⁶ molecules
Helium atom ki average thermal energy nikaalo (i) room temperature 27°C par (ii) Sun ki surface temperature 6000 K par (iii) star ke core temperature 10⁷ K par.
Average thermal energy per atom: Ē = (3/2)kBT
(i) T = 300 K:
Ē = (3/2)(1.38×10⁻²³)(300) ≈ 6.2 × 10⁻²¹ J
(ii) T = 6000 K:
Ē = (3/2)(1.38×10⁻²³)(6000) ≈ 1.24 × 10⁻¹⁹ J
(iii) T = 10⁷ K:
Ē = (3/2)(1.38×10⁻²³)(10⁷) ≈ 2.07 × 10⁻¹⁶ J
Note: energy sirf temperature par depend karti hai, isliye teenon jagah formula same hai, sirf T badalta hai.
Teen equal-capacity vessels me same temperature-pressure par Neon (monatomic), Chlorine (diatomic) aur Uranium hexafluoride (polyatomic) gas bhare hain. Kya teenon me molecules ki number equal hai? Kya v_rms bhi same hai?
Number of molecules: Avogadro's law se, same V, T, P par har ideal gas me equal number of molecules hote hain (N = PV/kBT — is formula me gas ki nature involve nahi hoti). Isliye teenon vessels me molecules equal hain.
v_rms: vrms = √(3RT/M) — ye molar mass M par depend karta hai, jo teenon gases ka alag hai (Ne ≈ 20u, Cl₂ ≈ 71u, UF₆ ≈ 352u).
Chunki vrms ∝ 1/√M, sabse halka gas Neon ka v_rms sabse zyada hoga.
Argon gas cylinder me kis temperature par argon atom ka rms speed, −20°C par helium atom ke rms speed ke barabar hoga? (M_Ar = 39.9 u, M_He = 4.0 u)
vrms = √(3RT/M) — dono equal karne par:
3RTAr/MAr = 3RTHe/MHe
TAr = THe × (MAr/MHe)
THe = −20°C = 253 K
TAr = 253 × (39.9/4.0) = 253 × 9.975 ≈ 2524 K
Nitrogen gas (2.0 atm, 17°C) me ek molecule ka mean free path aur collision frequency nikaalo. Molecule ka radius ≈ 1.0 Å, M(N₂) = 28.0 u lo. Collision time ko free-motion time se compare bhi karo.
Number density:
n = P/(kBT) = (2×1.013×10⁵)/(1.38×10⁻²³ × 290) ≈ 5.06 × 10²⁵ /m³
Mean free path (d = 2 × radius = 2×10⁻¹⁰ m):
λ = 1/(√2 π d² n) = 1/(1.414 × 3.14 × 4×10⁻²⁰ × 5.06×10²⁵)
λ ≈ 1.11 × 10⁻⁷ m
v_rms:
vrms = √(3RT/M) = √(3×8.31×290/0.028) ≈ 508 m/s
Collision frequency: ν = vrms/λ = 508/(1.11×10⁻⁷) ≈ 4.6 × 10⁹ collisions/sec
Collision time τ = d/vrms = (2×10⁻¹⁰)/508 ≈ 3.9 × 10⁻¹³ s, jabki time between collisions (1/ν) ≈ 2.2 × 10⁻¹⁰ s — matlab collision time, free-motion time se ≈ 500 gunaa chhota hai, isliye molecule zyadatar time freely move karta hai.
1 metre lambi horizontal tube (ek end closed) me 76 cm lambi mercury thread hai jo 15 cm air column trap karti hai. Tube ko vertically ghumaakar open end neeche kar diya jaaye to kya hoga?
Horizontal me: Trapped air pressure = atmospheric (mercury horizontal hone se koi extra weight-pressure nahi jodta): P₁ = 76 cmHg, L₁ = 15 cm
Vertical (open end neeche): Ab mercury apna weight se neeche gir ne ki koshish karega, kuch mercury tube se bahar nikal sakta hai. Maan lo mercury ki length x cm bahar nikal jaati hai:
Naya air column: L₂ = 15 + x, naya mercury: m = 76 − x
Pressure balance (mercury column top se bottom tak pressure x-length jitna kam hota hai, jo bahar spill hua hai): P₂ = x (cmHg)
Boyle's Law: P₁L₁ = P₂L₂
76 × 15 = x(15 + x)
x² + 15x − 1140 = 0
x = [−15 + √(225 + 4560)]/2 = [−15 + 69.2]/2 ≈ 27.1
Naya air column length ≈ 15 + 27.1 = 42.1 cm, aur mercury ≈ 76 − 27.1 ≈ 48.9 cm reh jaata hai (baaki mercury tube se bahar spill ho jaata hai).
Ek apparatus se hydrogen ki diffusion rate 28.7 cm³/s hai. Same conditions me ek doosri gas ki diffusion rate 7.2 cm³/s measure hoti hai. Gas identify karo.
Graham's Law of diffusion: R₁/R₂ = √(M₂/M₁)
28.7/7.2 = √(M/2.02)
(3.986)² = M/2.02
15.89 = M/2.02
M ≈ 32.1 u
Molecular mass ≈ 32 u — ye gas Oxygen (O₂) hai.
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Boyle's Law (T const) | PV = constant |
| Charles' Law (P const) | V/T = constant |
| Gay-Lussac's Law (V const) | P/T = constant |
| Ideal gas equation | PV = nRT = NkBT |
| Dalton's Law of partial pressures | P = P1 + P2 + P3 + … |
| Pressure from kinetic theory | P = (1/3) n m ⟨v²⟩ = (1/3) ρ ⟨v²⟩ |
| PV in terms of average KE | PV = (1/3) N m ⟨v²⟩ = (2/3) N Ē |
| Average KE per molecule | Ē = (3/2) kBT |
| RMS speed | vrms = √(3kBT/m) = √(3RT/M) |
| Boltzmann constant | kB = R/NA ≈ 1.38 × 10⁻²³ J/K |
| Law of equipartition of energy | Energy per degree of freedom = (1/2) kBT |
| Monatomic gas (f = 3) | Cv = (3/2)R, Cp = (5/2)R, γ = 5/3 |
| Diatomic gas — rigid (f = 5) | Cv = (5/2)R, Cp = (7/2)R, γ = 7/5 |
| Diatomic gas — with vibration (f = 7) | Cv = (7/2)R, Cp = (9/2)R, γ = 9/7 |
| Polyatomic gas (f = 6 + 2f′, f′ = vibrational modes) | Cv = (3 + f′)R, Cp = (4 + f′)R |
| Mayer's relation | Cp − Cv = R |
| Mean free path | λ = 1 / (√2 π d² n) |
| Real gas equation (Van der Waals) | (P + a/V²)(V − b) = RT |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- P = (1/3) n m⟨v²⟩ me ⟨v²⟩ (mean of squares) hai, na ki (average speed)² — students often ⟨v⟩² ko ⟨v²⟩ samajh kar galat answer nikaalte hain.
- Gas law numericals me Celsius ko Kelvin me convert karna bhool jaate hain (T = °C + 273) — ye single mistake poora numerical galat kar deta hai.
- Diatomic gas ke degrees of freedom hamesha 5 nahi hote — high temperature par vibrational modes activate hone se f = 7 ho jaata hai; isse C_v/C_p/γ galat calculate hote hain.
- k_B (per-molecule constant, 1.38×10⁻²³ J/K) aur R (per-mole constant, 8.31 J/mol K) ko confuse kar dete hain — N (number of molecules) wale formula me k_B aur n (moles) wale formula me R use karna hai, mix karne se answer Avogadro number ke factor se galat ho jaata hai.
- Ye maan lete hain ki heavier gas molecules ki average kinetic energy zyada hoti hai — jabki Ē = (3/2)k_BT sirf temperature par depend karta hai, mass par nahi (sirf v_rms mass par depend karta hai).
- Mean free path ya molecular size ke numericals me diameter aur radius ko interchange kar dete hain, ya atm ko Pascal me convert karna bhool jaate hain — dono se answer poore order of magnitude se galat ho jaata hai.
Board-Style Important Questions
- 1 mark: Ideal gas ke ek molecule ki average kinetic energy kis factor par depend karti hai?
- 2 marks: Boyle's Law aur Charles' Law se ideal gas equation kaise derive hoti hai, likho.
- 3 marks: Kinetic theory ka use karke gas ke pressure ka expression P = (1/3)nm⟨v²⟩ derive karo.
- 3 marks: Law of equipartition of energy state karo aur ispar use karke diatomic gas (rigid rotor) ka C_v/C_p nikaalo.
- 5 marks: Kinetic theory ke postulates likho aur inse ideal gas equation PV = Nk_BT derive karo. Isi se v_rms ka expression bhi nikaalo.
Aksar Poochhe Jaane Wale Sawaal
Kinetic theory of gases class 11 derivation me pressure ka formula kaise banta hai?
Ek cube-shaped container me gas molecules wall se elastic collision karte hain. Ek molecule ka momentum change aur uske collisions ki frequency calculate karke total force nikalte hain, jisse P = (1/3) n m⟨v²⟩ milta hai — poori derivation is chapter ke exercise solutions me step-by-step di gayi hai.
Class 11 Physics ke saare chapters ka PDF kahan milega (2026-27 session ke liye)?
Class 11 physics all chapters pdf 2026-27 ke liye NCERT ki official website (ncert.nic.in) hi authoritative source hai — rationalised syllabus ke hisaab se ab 14 chapters hain (Physical World standalone chapter drop ho chuka hai, syllabus Units and Measurements se directly start hota hai).
2026-27 session me class 11 physics deleted syllabus me kya-kya hata hai?
Sabse bada change: purana Chapter 1 'Physical World' fully drop ho gaya hai — ab syllabus 14 chapters ka hai, 15/16 nahi. Baaki chapters me poore topics nahi, balki kuch derivations/sections trim hue hain (jaise Gravitation me geostationary satellite detail, Thermodynamics me heat engine ka kuch part) — exact list official NCERT PDF se cross-check karna best hai.
Kinetic Theory chapter, Thermodynamics (previous chapter) se kaise related hai?
Thermodynamics macroscopic quantities (P, V, T, heat, work) ke laws deta hai; Kinetic Theory unhi quantities ko microscopic level (molecules ka random motion) se explain karta hai. Dono chapters together 'Thermal Physics' unit banate hain, isliye thermodynamics class 11 physics notes ke saath ye chapter revise karna helpful hota hai.
Kinetic Theory ke baad next chapter (Oscillations) me kya naya hai?
Oscillations chapter periodic motion (SHM) introduce karta hai — energy, phase, aur time period ke concepts. Kinetic theory me energy/motion ka jo intuition banta hai (average KE, random motion) wahi oscillations class 11 physics important questions solve karte waqt kaam aata hai.
Kinetic Theory me kaunse concepts se related dusre chapters ke numericals bhi practice karne chahiye?
Gas laws aur units ka strong base rakhne ke liye units and measurements class 11 important questions revise karo; motion se related formulas (velocity, KE) ke liye class 11 physics chapter 2 motion in a straight line ncert solutions helpful hain; aur pressure/force concepts ke liye laws of motion class 11 ncert exemplar practice karna accha rahega.
Class 11 Physics — Saare Chapters
Board exam tak sirf revision karna hai?
Class 11 Physics ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
NCERT Kaksha ek swatantra shaikshik platform hai aur NCERT ya CBSE se aadhikarik roop se sambaddh (officially affiliated) nahi hai. Kisi galti ki jaankari dene ke liye contact kijiye.