Class 11 Physics · Chapter 11
Short answer:
Class 11 Physics Chapter 11 — Thermodynamics ka core idea ek line me: heat (Q), work (W) aur internal energy (ΔU) teeno aapas me first law se judte hain — ΔQ = ΔU + ΔW. Isothermal process me ΔT = 0 (isliye ΔU = 0), adiabatic me Q = 0, isochoric me W = 0, aur Carnot engine ki efficiency η = 1 − T₂/T₁ hoti hai (T Kelvin me). Neeche is chapter ka poora NCERT solutions set — sabhi 9 exercise questions, formulas table, common mistakes aur board-pattern PYQs — Hinglish me diya gaya hai.
Class 11 Physics Chapter 11 "Thermodynamics" wahi chapter hai jo aage chal ke Class 11 ke hi agle chapter "Kinetic Theory" aur Class 12 ke heat-engine/entropy based numericals ki neev banata hai — agar ye chapter concept-wise clear ho jaaye, to kinetic theory of gases class 11 derivation waala portion (jaha molecules ke microscopic motion se pressure-temperature derive hota hai) kaafi easy lagta hai.
Ye class 11 physics ncert solutions series ka hissa hai, jaha har chapter step-by-step CBSE marking-scheme style me solve kiya ja raha hai. Isi series me Units and Measurements, Motion in a Straight Line, Motion in a Plane (projectile motion), Laws of Motion, Work Energy Power, System of Particles and Rotational Motion, Gravitation, Mechanical Properties of Solids aur Fluids, Thermal Properties of Matter, Oscillations aur Waves ki solutions bhi available hain — agar tumhe class 11 physics chapter 2 motion in a straight line ncert solutions, units and measurements class 11 important questions, gravitation class 11 numericals with solutions, laws of motion class 11 ncert exemplar, system of particles and rotational motion important formulas, oscillations class 11 physics important questions, waves class 11 physics doppler effect formula, mechanical properties of fluids class 11 numericals ya motion in a plane projectile motion class 11 solutions chahiye to un chapters ki solutions bhi cross-check kar sakte ho.
Rationalised syllabus note (2026-27 session): current NCERT structure me "Physical World" wala standalone chapter hata diya gaya hai, isliye Class 11 Physics ab 14 chapters ka hai aur numbering seedha "Units and Measurements" se Chapter 1 ke roop me shuru hoti hai — is hisaab se Thermodynamics naye numbering me Chapter 11 baithta hai. Ye confirmation multiple registry-grade (third-party aggregator) sources se cross-verify kiya gaya hai; primary ncert.nic.in PDF is run me directly fetch nahi ho paayi, isliye final numbering apne school ki current-session copy ya ncert.nic.in ki latest PDF se ek baar zaroor match kar lena, khaaskar agar tum class 11 physics all chapters pdf 2026-27 download kar rahe ho. Is chapter ke andar heat-engine aur refrigerator wale kuch derivation-heavy portions simplify kiye gaye hain (poora chapter delete nahi hua), isliye ye class 11 physics deleted syllabus 2026-27 list me chapter-level pe nahi, topic-trim level pe aata hai. Textbook ab bhi do physical parts (Part I mechanics, Part II thermal/oscillations/waves) me chapta hai, par ye sirf printing/binding split hai, syllabus unit nahi.
Chapter ka core focus: thermal equilibrium, zeroth law, first law of thermodynamics, specific/molar heat capacities, thermodynamic processes (isothermal, adiabatic, isochoric, isobaric, cyclic), heat engines, refrigerators aur second law of thermodynamics. Neeche thermodynamics class 11 physics notes ki tarah formulas table, uske baad saare 9 official NCERT exercise questions fully solved milenge.
Chapter 11 Summary — 5 Minute Revision
Is chapter me humne dekha ki thermodynamics heat, work aur internal energy ke beech macroscopic relationship study karta hai — thermal equilibrium aur zeroth law se shuru karke, first law of thermodynamics (ΔQ = ΔU + ΔW) tak. Alag-alag thermodynamic processes — isothermal, adiabatic, isochoric, isobaric, cyclic — apni-apni special conditions ke saath define hue. Specific heat capacity aur molar specific heat capacities (C_p, C_v) ka relation C_p − C_v = R nikla. Heat engines aur refrigerators ke through second law of thermodynamics (Kelvin-Planck aur Clausius statements) samjha, aur Carnot engine ki ideal efficiency η = 1 − T₂/T₁ derive hui. Numericals me PV work calculation (isothermal aur adiabatic), heat-capacity based Q calculation, aur engine efficiency/coefficient of performance ke sawaal practice kiye gaye.
In-Text Questions — Solutions
Zeroth law of thermodynamics ka statement do, aur bataiye ye thermometer ke concept se kaise related hai.
Zeroth law: agar system A aur system B, dono alag-alag ek third system C ke thermal equilibrium me hain, to A aur B bhi aapas me thermal equilibrium me honge.
Isi law se temperature ko ek measurable, well-defined quantity maana jaata hai — thermometer khud ek reference system (C) ki tarah kaam karta hai jo kisi body ke thermal equilibrium me aakar uska temperature reading deta hai.
Ek gas ko isochoric process se garam kiya jaata hai. Is process me gas dwara kiya gaya work kitna hoga, aur kyun?
Isochoric (constant volume) process me volume change (ΔV) zero hota hai, isliye:
W = PΔV = P × 0 = 0
Poori supplied heat sirf internal energy badhane me use hoti hai: ΔQ = ΔU.
Adiabatic process me Q = 0 kyun hota hai — do practical examples do jaha process ko approximately adiabatic maana jaata hai.
Adiabatic process ya to (i) itna fast hota hai ki heat exchange ke liye time hi nahi milta, ya (ii) perfectly insulated walls me hota hai jisse heat bahar/andar nahi ja sakti.
Examples: (1) bicycle pump me hawa ka sudden compression (piston tezi se push karne par gas garam ho jaata hai, Q ≈ 0), (2) sound waves ka air me propagation — compressions/rarefactions itni fast hoti hain ki heat exchange negligible rehta hai.
First law of thermodynamics energy conservation ka hi ek roop kyun kaha jaata hai?
First law kehta hai ΔQ = ΔU + ΔW — yaani system ko di gayi heat, na to create hoti hai na destroy, wo sirf do jagah jaati hai: system ki internal energy badhane me, ya system dwara external work karne me. Ye energy conservation ke general principle ka hi thermodynamic version hai.
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–Q9)
12.1 Ek geyser paani ko 3.0 litre/min ki rate se 27°C se 77°C tak garam karta hai. Agar geyser gas burner par chalta hai aur combustion ki heat of combustion 4×10⁴ J/g hai, to fuel consumption ki rate kya hogi?
Mass flow rate, m = 3.0 L/min = 3000 g/min (paani ki density ≈ 1 g/mL).
Temperature rise, ΔT = 77°C − 27°C = 50°C
Specific heat of water, c = 4.2 J/g°C
Heat required per minute, Q = mcΔT = 3000 × 4.2 × 50 = 6.3 × 10⁵ J/min
Fuel consumption rate = Q ÷ (heat of combustion) = (6.3 × 10⁵) ÷ (4 × 10⁴) = 15.75 g/min
Answer: fuel consumption rate ≈ 15.75 g/min.
12.2 Kamre ke temperature par 2.0×10⁻² kg nitrogen (N₂) ka temperature 45°C se raise karne ke liye, constant pressure par kitni heat supply karni hogi? (Molecular mass of N₂ = 28; R = 8.3 J mol⁻¹K⁻¹)
Mass = 2.0 × 10⁻² kg = 20 g. Molar mass M = 28 g/mol.
Moles, n = 20/28 = 0.714 mol
N₂ diatomic gas hai, isliye molar heat capacity at constant pressure:
C_p = (7/2)R = 3.5 × 8.3 = 29.05 J mol⁻¹K⁻¹
Q = n C_p ΔT = 0.714 × 29.05 × 45 ≈ 933.9 J
Answer: Q ≈ 933.9 J (≈ 934 J) heat supply karni hogi.
12.3 Samjhaiye: (a) Do bodies jinke temperature T₁ aur T₂ alag hain, thermal contact me laane par zaroori nahi ki mean temperature (T₁+T₂)/2 par settle ho. (b) Chemical/nuclear plant ke coolant ki specific heat high honi chahiye. (c) Driving ke dauraan car tyre ka air pressure badh jaata hai. (d) Harbour town ka climate same latitude ke desert town se zyada temperate hota hai.
(a) Final equilibrium temperature dono bodies ke heat capacity (mass × specific heat) par depend karta hai, sirf initial temperature par nahi. Agar dono bodies ki thermal capacities barabar na ho, to equilibrium temperature simple average se hat jaata hai.
(b) High specific heat matlab thodi si temperature rise me coolant zyada heat absorb kar sakta hai. Isse coolant plant se heat efficiently le jaata hai bina khud bahut zyada garam hue.
(c) Driving ke dauraan road-friction aur engine heat se tyre ke andar ki air garam ho jaati hai. Tyre ka volume lagbhag constant rehta hai, isliye Gay-Lussac's law (P ∝ T, constant V par) ke mutabik pressure badh jaata hai.
(d) Harbour town ke paas bada water body (samundar) hota hai jiski specific heat bahut high hoti hai — ye slowly heat absorb/release karta hai aur temperature ke extreme swings ko moderate karta hai. Desert town me aisa moderating water body nahi hota, isliye wahan din-raat aur seasons ke beech temperature variation zyada extreme hota hai.
12.4 Ek insulated cylinder me movable piston ke saath 3 moles hydrogen (H₂) STP par hai. Piston bhi insulated hai. Agar gas ko compress karke original volume ka half kar diya jaaye, to pressure kis factor se badhega?
Cylinder poori tarah insulated hai, isliye process adiabatic hai. H₂ diatomic gas hai, γ = 7/5 = 1.4.
Adiabatic relation:
P₁V₁^γ = P₂V₂^γ
Diya hai V₂ = V₁/2, isliye:
P₂/P₁ = (V₁/V₂)^γ = 2^1.4 ≈ 2.639
Answer: pressure lagbhag 2.64 guna (≈ 2.639×) badh jaata hai.
12.5 Gas ko adiabatically state A se state B tak le jaane me system par 22.3 J work kiya jaata hai. Agar gas ko A se B tak ek doosre path se le jaaya jaaye jisme net heat absorbed 9.35 cal hai, to us case me system dwara kiya gaya net work kitna hoga? (1 cal = 4.19 J)
Path 1 (adiabatic): Q = 0. Work done ON the system = 22.3 J, isliye work done BY the gas, W = −22.3 J.
First law: ΔU = Q − W = 0 − (−22.3) = +22.3 J
Chunki ΔU ek state function hai (path-independent), state A se B tak ΔU dono paths ke liye same rahega, yaani ΔU = 22.3 J.
Path 2: Q = 9.35 cal = 9.35 × 4.19 = 39.18 J
W = Q − ΔU = 39.18 − 22.3 = 16.88 J
Answer: is doosre path me system dwara kiya gaya net work ≈ 16.9 J hai.
12.6 Do cylinders A aur B (equal capacity) stopcock se connected hain. A me STP par gas hai, B poori tarah evacuated hai, poora system thermally insulated hai. Stopcock achanak khol diya jaata hai. (a) A aur B me final pressure kya hoga? (b) Internal energy me kya change hoga? (c) Temperature me kya change hoga? (d) Kya system ke intermediate states P-V-T surface par lie karte hain?
Ye free expansion (Joule expansion) ka case hai — gas vacuum me expand hota hai, koi external opposing pressure nahi hai.
(a) Gas ab dono cylinders (double volume) me fail jaata hai. Free expansion me koi work nahi hota aur system insulated hai (Q = 0), isliye ΔU = 0 ⇒ ideal gas ke liye temperature constant rehta hai. PV = nRT se, V double hone par aur T constant rehne par, final pressure initial pressure ka half ho jaata hai dono cylinders me.
(b) ΔU = Q − W = 0 − 0 = 0
(c) Chunki ideal gas ki internal energy sirf temperature par depend karti hai aur ΔU = 0, isliye ΔT = 0 — temperature same rehta hai.
(d) Nahi. Free expansion ke dauraan gas turbulent aur non-uniform hota hai — intermediate states equilibrium me nahi hote, isliye P-V-T surface (jo sirf equilibrium states represent karta hai) par nahi lie karte. Sirf initial aur final equilibrium states hi surface par honge.
12.7 Ek steam engine per minute 5.4×10⁸ J work deliver karta hai aur apne boiler se 3.6×10⁹ J heat per minute leta hai. Engine ki efficiency kya hai? Per minute kitni heat waste hoti hai?
Efficiency, η = W/Q₁ = (5.4×10⁸)/(3.6×10⁹) = 0.15 = 15%
Heat wasted = Q₁ − W = 3.6×10⁹ − 5.4×10⁸ = 3.06×10⁹ J/min
Answer: efficiency = 15%; wasted heat = 3.06 × 10⁹ J per minute.
12.8 Ek electric heater kisi system ko 100 W ki rate se heat supply karta hai. Agar system 75 J/s ki rate se work perform karta hai, to internal energy kis rate se badh rahi hai?
Rate of heat supply, dQ/dt = 100 W
Rate of work done by system, dW/dt = 75 J/s
dU/dt = dQ/dt − dW/dt = 100 − 75 = 25 W
Answer: internal energy 25 J/s (25 W) ki rate se badh rahi hai.
12.9 Ek thermodynamic system ko original state D se intermediate state E tak Fig. 11.13 (P-V diagram) me dikhaye gaye linear process se le jaaya jaata hai. Uske baad volume ko E se F tak isobaric process se wapas original value tak reduce kiya jaata hai. D se E se F tak gas dwara kiya gaya total work calculate kijiye.
Method (CBSE marking-scheme style):
P-V diagram par kisi bhi process me gas dwara kiya gaya work us curve ke neeche ka area hota hai.
Step 1 — D se E (linear/sloped process): Ye ek trapezium ka area hai:
W(D→E) = ½ (P_D + P_E) × (V_E − V_D)
Step 2 — E se F (isobaric, volume wapas V_D tak compress): Ye ek rectangle ka area hai, aur compression hai isliye negative:
W(E→F) = P_E × (V_D − V_E)
Total work:
W(D→E→F) = W(D→E) + W(E→F)
Note: Final numeric answer P_D, V_D, P_E, V_E ki exact figure-coordinates (jo Fig. 11.13/12.13 me graph par diye hote hain) par depend karta hai — apni print copy ki diagram se exact values padhkar upar diye formula me daal do; method aur setup yahi rahega.
Important Equations — Ek Nazar Me
| Concept | Formula / Statement |
|---|---|
| Zeroth Law | Agar A, C ke thermal equilibrium me hai aur B bhi C ke thermal equilibrium me hai, to A aur B bhi aapas me thermal equilibrium me honge (temperature isi se define hoti hai). |
| First Law of Thermodynamics | ΔQ = ΔU + ΔW |
| Specific heat capacity | C = ΔQ / (mΔT) |
| Molar C at constant volume | C_v = (f/2)R (f = degrees of freedom) |
| Molar C at constant pressure | C_p = C_v + R |
| Mayer's relation | C_p − C_v = R |
| Ratio of specific heats | γ = C_p / C_v |
| Work — isothermal process | W = nRT ln(V₂/V₁) = 2.303 nRT log₁₀(V₂/V₁) |
| Work — adiabatic process | W = (P₁V₁ − P₂V₂) / (γ − 1) |
| Adiabatic relations | PV^γ = const; TV^(γ−1) = const; T^γ P^(1−γ) = const |
| Efficiency of heat engine | η = W/Q₁ = 1 − Q₂/Q₁ |
| Carnot engine efficiency | η = 1 − T₂/T₁ (T Kelvin me) |
| Coefficient of performance (refrigerator) | α = Q₂/W = Q₂/(Q₁ − Q₂) |
| Carnot refrigerator COP | α = T₂/(T₁ − T₂) |
| Kelvin-Planck statement | Koi bhi engine 100% efficient nahi ho sakta — poora heat, work me convert nahi ho sakta bina kuch heat reject kiye. |
| Clausius statement | Bina external work ke, heat khud-ba-khud cold body se hot body me transfer nahi ho sakta. |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Sign convention ka confusion: NCERT convention me W gas dwara kiya gaya work hota hai, isliye ΔQ = ΔU + ΔW likhte waqt agar gas compress ho raha hai to W negative hoga — students isse ulta le lete hain.
- Isothermal aur adiabatic ko mix karna: Isothermal me ΔT = 0 isliye ΔU = 0; adiabatic me Q = 0 par ΔU zero nahi hota — dono conditions bilkul alag hain, exam me swap mat karo.
- C_p aur C_v ka galat use: Constant volume process me Q = nC_vΔT use hoga, constant pressure me Q = nC_pΔT — dono formula interchange karna common numerical mistake hai.
- Kelvin scale bhoolna: Carnot efficiency ya COP ke formula me temperature hamesha Kelvin (absolute) me daalo — Celsius daalne par η ya α galat aa jaata hai.
- Quasi-static assumption ignore karna: W = ∫PdV sirf quasi-static/reversible process ke liye directly valid hai — sudden/free expansion (jaise free-expansion type numericals) me ye formula apply nahi hota kyunki intermediate states equilibrium me nahi hote.
- log ka sign bhool jaana: Isothermal work W = nRT ln(V₂/V₁) me agar gas compress ho raha hai (V₂ < V₁), to ln negative aayega, matlab W negative — yaani work gas par kiya ja raha hai, gas dwara nahi. Ye sign numericals me yaad rakhna zaroori hai.
Board-Style Important Questions
- 1 mark: Zeroth law of thermodynamics ka statement likhiye. Ye kis physical quantity ko define karne me help karta hai?
Answer: Agar system A, system C ke thermal equilibrium me hai, aur system B bhi C ke thermal equilibrium me hai, to A aur B bhi aapas me thermal equilibrium me honge. Ye law temperature ko ek well-defined physical quantity ke roop me define karta hai. - 2 marks: Ek ideal gas ke liye Mayer's relation C_p − C_v = R derive kijiye.
Answer: Constant volume par, dQ = dU (kyunki dV = 0) ⇒ C_v = (dU/dT)_V. Constant pressure par, dQ = dU + PdV ⇒ C_p = (dU/dT)_P + P(dV/dT)_P. Ek mole ideal gas ke liye PV = RT ⇒ P(dV/dT) = R. Chunki ideal gas ki U sirf T par depend karti hai, (dU/dT)_P = (dU/dT)_V = C_v. IsliyeC_p − C_v = R
- 3 marks: Isothermal aur adiabatic process me koi teen antar likhiye.
Answer: (i) Isothermal me temperature constant rehta hai (ΔT = 0); adiabatic me heat exchange zero hota hai (Q = 0). (ii) Isothermal process slow aur perfectly conducting walls me hota hai; adiabatic process fast ya perfectly insulated walls me hota hai. (iii) P-V diagram par isothermal curve (PV = const), adiabatic curve (PV^γ = const) se kam steep hoti hai. - 3 marks: Ek Carnot engine source se 600 K par heat leta hai aur sink ko 300 K par heat reject karta hai. Iski efficiency nikaliye.
Answer:η = 1 − T₂/T₁ = 1 − 300/600 = 0.5 = 50%
- 5 marks: Second law of thermodynamics ke Kelvin-Planck aur Clausius statements likhiye, aur Carnot engine ke liye efficiency ka expression derive kijiye.
Answer: Kelvin-Planck statement: Koi bhi cyclic process aisa possible nahi jiska sole result ho ki heat ek reservoir se liya jaaye aur poora work me convert ho jaaye, bina kisi heat ko reject kiye. Clausius statement: Koi bhi cyclic process aisa possible nahi jiska sole result heat ka cold body se hot body me transfer ho, bina external work input ke. Carnot engine do isothermal aur do adiabatic processes ka cycle hai; heat Q₁ source (T₁) se liya jaata hai, Q₂ sink (T₂) ko diya jaata hai. Poore cycle me ΔU = 0, isliye W = Q₁ − Q₂. Efficiency η = W/Q₁ = 1 − Q₂/Q₁. Reversible cycle ke liye Q₂/Q₁ = T₂/T₁, isliyeη = 1 − T₂/T₁
Aksar Poochhe Jaane Wale Sawaal
Class 11 Physics Thermodynamics chapter me kitne exercise questions hain?
NCERT ke is chapter me total 9 exercise questions hain (original numbering me 12.1 se 12.9 tak). Rationalised 2026-27 syllabus me chapter number badalkar 11 ho gaya hai, par exercise content same hai.
Class 11 Physics ke saare chapters ka PDF 2026-27 session ke liye kahan se lein?
Sabse reliable source official ncert.nic.in hai jaha current session ki latest PDF free milti hai. Chapter numbering final verify karne ke liye wahi source treat karo, kyunki third-party sites me numbering thoda outdated reh sakta hai.
2026-27 syllabus me Class 11 Physics ka koi chapter delete hua hai kya?
Rationalised syllabus me 'Physical World' wala standalone chapter (purana Chapter 1) fully hata diya gaya hai — ab book seedha 'Units and Measurements' se shuru hoti hai aur total 14 chapters bache hain. Kuch aur chapters me poora chapter nahi, balki select topics/derivations trim hue hain (jaise is Thermodynamics chapter me heat-engine/refrigerator ke kuch sections simplify hue hain).
First law of thermodynamics kya kehta hai?
First law energy conservation ka thermodynamic version hai — jo heat (ΔQ) system ko diya jaata hai, wo ya to system ki internal energy (ΔU) badhata hai ya system dwara external work (ΔW) me convert hota hai: ΔQ = ΔU + ΔW.
Thermodynamics aur Kinetic Theory of Gases (agla chapter) me kya farak hai?
Thermodynamics macroscopic level pe kaam karta hai — pressure, volume, temperature jaise bulk quantities se. Kinetic theory of gases microscopic level pe jaata hai aur molecules ki motion se hi pressure/temperature jaise properties derive karta hai — isliye kinetic theory of gases class 11 derivation is chapter ka natural next step hai.
CBSE board exam me Thermodynamics chapter ki weightage kitni hoti hai?
Exact chapter-wise weightage har saal CBSE ke official curriculum document/sample paper ke saath thoda change ho sakti hai, isliye yahan koi fixed number claim nahi kar rahe — class 11 physics chapter wise weightage cbse ke liye current session ka official CBSE curriculum PDF hi check karo. Jo consistent rehta hai wo ye hai ki Thermodynamics se numericals (heat engine efficiency, PV work) almost har saal poochhe jaate hain.
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.
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