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Chapter 12 - Thermodynamics

Socho winter morning hai, tum apne dono haathon ko ragad rahe ho. Thodi der me garam lagta hai. Kaam (rubbing) se garmi bani. Ulta, pressure cooker me steam se piston hilta hai - garmi se kaam. Isi khel ko samajhne ka naam hai thermodynamics. Yeh bulk systems ki kahani hai, molecules ko individually track nahi karte, bas pressure, volume, temperature, internal energy jaise kuch macroscopic variables se poora system samajhte hain.

Thermo = heat + work interconversion. Macroscopic science, few variables P,V,T,U,m. Not concerned with centre-of-mass KE. Internal disordered motion only.

12.1 Introduction - Why Thermodynamics?

Maan lo tumne ek bullet chalai. Mechanics me hum uski kinetic energy dekhte hain. Jab bullet lakdi me ghus ke ruk jaati hai, uski KE kahan gayi? Lakdi aur bullet garam ho jaate hain. Yani ordered motion disorder me badal gaya. Yeh thermodynamics ka domain hai.

Old picture me log sochte the heat ek invisible fluid hai - caloric - jo garam body se thandi body me behta hai jaise paani upar se neeche tank me. Par Rumford ne 1798 me dekha ki top ko drill karne se paani ubal jaata hai, aur drill sharp ho ya blunt, heat utni hi banti hai jitna kaam horse kar raha hai. So heat fluid nahi, energy ka ek form hai. Kaam se heat ban sakti hai, heat se kaam.

Isliye thermodynamics bana - jo batata hai kaise energy ek form se dusre me badlegi, par kya limitation hai. First law energy bachao ka hai, second law direction batata hai.

Caloric fluid theory discarded. Rumford boring cannon experiment proved work to heat conversion. Thermo = laws governing thermal energy conversion.

Story: Socho mom kitchen me chakki chala rahi hai. Chakki ghumane me mehnat (work), par aata aur patthar dono garam. Same as rubbing palms. Energy gayab nahi hui, form badla. That is thermodynamics in your kitchen.

12.2 Thermal Equilibrium - Adiabatic vs Diathermic Walls

Socho do dabbe hain, A me gas aur B me gas. Dono ko ek insulating wall - adiabatic wall - se alag rakho. Adiabatic matlab garmi cross nahi kar sakti, chahe wall hile bhi. Ab A ka pressure, volume kuch bhi ho, B se equilibrium me hi hai kyunki interaction hi nahi. Koi farak nahi padta.

Ab us wall ko hata ke conducting wall laga do - diathermic wall. Yeh heat ko jaane deti hai. Ab dekho, dheere dheere dono gases ke pressure volume change honge jab tak dono ka kuch final state na aa jaye jahan heat flow ruk jaye. Us state ko bolte hain thermal equilibrium. Simple matlab - ab dono ka temperature same.

Is example se ek baat clear: agar do alag systems ek hi state tak pahunch ke ruke, to unme koi property common hai jisko hum temperature bolte hain. Par yeh formal kaise define hota hai? Uske liye zeroth law chahiye.

Adiabatic wall = no heat flow. Diathermic = allows heat flow. Thermal equilibrium = temps equal. Equilibrium state = macroscopic variables P,V,T,m constant in time.

12.3 Zeroth Law of Thermodynamics - Temperature Ki Definition

Maan lo A aur B alag hain adiabatic wall se, par dono C ke contact me hain via conducting walls. Dono A aur B eventually C ke saath equilibrium me aa jayenge. Ab A aur B ke beech ki adiabatic wall hata ke conducting wall laga do aur C ko alag kar do. Dekhoge A aur B ab bhi equilibrium me hain, change nahi hota.

Yahi observation hai zeroth law: If two systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.

Iska matlab ek physical quantity hai jo equilibrium pe same hoti hai - usi ko temperature T kehte hain. Isliye agar TA = TC aur TB = TC, to TA = TB. Simple, par bahut powerful. Isi se thermometer kaam karta hai. 1931 me Fowler ne isko zeroth naam diya kyunki first aur second pehle se the.

Zeroth law: A eq C, B eq C => A eq B. Defines temperature. Basis of thermometry.

1. Zeroth Law & Thermal Equilibrium

Revision Points — After Diagram 1:
• Adiabatic wall = insulating, no heat, Diathermic = conducting allows heat
• Thermal equilibrium = no net heat flow, temperatures equal
• Zeroth law gives definition of T, TA=TC and TB=TC => TA=TB
• Equilibrium state: macroscopic variables constant in time
• Intensive vs extensive concept introduced here

12.4 Heat, Internal Energy and Work - Teen Alag Cheezen

Socho ek dabbe me bahut saare molecules hain. Har molecule bhag raha hai, rotate kar raha hai, vibrate kar raha hai. Un sabki kinetic + potential energy ka sum jo centre-of-mass ke frame me dekha jaye, usko bolte hain internal energy U. Agar pura dabba truck pe 100 km/h ja raha hai, uski kinetic energy U me count nahi hogi. Sirf disordered andar ki energy.

Ab U ko change kaise karte hain? Do tarike. Pehla - dabbe ko garam body se touch karo. Temperature difference se energy flow hogi - isko bolte hain heat. Dusra - piston ko dabao, weight rakho. Bina temperature difference ke energy transfer - isko bolte hain work.

Important: Heat aur work energy in transit hain, state variable nahi hain. Yeh bolna galat hai ki gas me itni heat hai. Sahi hai: gas ki itni internal energy hai. Par yeh bolna sahi hai: itni heat supply ki, itna work kiya. Jaise bank account me balance hota hai (U), par deposit kaise aaya - cash (Q) ya transfer (W) - woh path pe depend karta hai.

U = sum of molecular KE+PE in CM frame, state variable, path independent. Q and W = modes of energy transfer, not state variables, path dependent. Q>0 heat supplied to system, W>0 work done BY system.

Story: Socho tum chai garam kar rahe ho gas pe - woh heat se U badh rahi hai. Aur socho cycle pump se tyre me hawa bhar rahe ho - piston dabane se work se U badh raha hai, pump garam. Dono se same kaam, internal energy badhi, par raaste alag.

Common Mistake: Students bolte hain 'body me heat hai'. Galat. Body me internal energy hai. Heat tabhi naam aata hai jab flow ho raha hai temperature difference ki wajah se. Exam me statement check karo, wordings trap karte hain.

2. First Law - Heat, Work & Internal Energy

Revision Points — After Diagram 2:
• U is state function, depends only on P,V,T, not history
• Q is energy due to temperature difference, W is energy by mechanical means
• Sign: Q>0 added to system, W>0 done by system
• U includes translational, rotational, vibrational disordered motion
• Centre-of-mass KE not included in U

12.5 First Law of Thermodynamics - Energy Conservation

Ab general principle lagao - energy conserve hoti hai. Agar system ko heat di aur usne work bahar kiya, to bachi hui energy andar store hogi U ke roop me. Toh heat = U change + work. Yeh hai first law. Simple.

Maan lo 1g paani liquid se vapour banta hai. Latent heat 2256 J lagti hai. Vapour ka volume 1 cm3 se 1671 cm3 ban jata hai. Atmospheric pressure pe work P dV = 1.013e5 * 1670e-6 = 169.2 J. Toh U change = 2256 - 169.2 = 2086.8 J. Matlab zyada tar heat internal energy badhane me gayi, thodi work me.

Alternative form: Q - W = U change. Left side path pe depend karta hai, par combination hamesha same kyunki right side state function hai. Isothermal expansion of ideal gas me U change 0 hota hai, to Q=W. Heat pura work me badal gaya, par iska matlab ye nahi ki 100 percent efficient engine bana sakte, kyunki cycle me wapas aana padta hai.

First Law: Q = U change + W. Conservation of energy. For gas, W = P dV at constant P. U change path independent, Q and W path dependent. U=0 for isothermal ideal gas, for cyclic process U=0 => Qnet=Wnet.

Numerical Hack: Sign confusion? Remember - Q added to system positive, W done BY system positive. If work done ON system, W negative. Many textbooks opposite sign use Q = U - W, but here we follow Q = U + W with W by system. Stick to one convention in paper, mention clearly.

12.6 Specific Heat Capacity - Solid Ka 3R

Socho ek substance ko heat di, temperature badha. Heat capacity S = Q / T change. Par yeh mass pe depend karega. Isliye specific heat capacity s = (1/m)(Q/T change) J/kgK, aur molar specific heat C = (1/mu)(Q/T change) J/molK.

Ab interesting baat - gas ke liye do specific heats: Cv constant volume pe, Cp constant pressure pe. Kyunki pressure constant pe heat ka kuch part work me chala jata hai, toh zyada heat chahiye same T change ke liye. Isliye Cp > Cv.

Solid ke liye simple model: Har atom 3 directions me oscillator, har direction me average energy kB T. Toh 1 atom ki average energy 3kB T, 1 mole ki U = 3RT. Constant pressure pe volume change negligible, toh Q approx U change, toh C = U/T = 3R = 24.94 J/molK. Table dekho to Al, Cu, Fe sabke around 24-26 hota hai room temperature pe, carbon exception. Low temperature pe fail hota hai, quantum effect.

Paani ki specific heat 4186 J/kgK bahut high hai, isliye harbour town ka climate mild hota hai, desert me nahi. Coolant me high s chahiye taki zyada heat absorb kare bina zyada garam hue.

s = (Q/mT), C = (Q/mu T). For gases: Cv at V const, Cp at P const. For solid: C=3R Dulong-Petit law. Water s=4186 J/kgK = 1 cal/gK. 1 cal = 4.186 J, defined from 14.5 to 15.5C water.

12.6.1 Cp - Cv = R Proof

Socho 1 mole ideal gas. First law: Q = U + P dV. Constant volume pe dV=0, toh Q = U = Cv dT. Matlab Cv = (U/T).

Constant pressure pe: Q = Cv dT + P dV. Par ideal gas PV=RT => P dV = R dT at constant P. Toh Q = Cv dT + R dT = (Cv+R)dT. Par definition se Q = Cp dT. Compare karo: Cp = Cv + R. So Cp - Cv = R. Independent of gas type for ideal gas.

Isliye gamma = Cp/Cv >1. Monoatomic ke liye Cv = (3/2)R, Cp = (5/2)R, gamma=5/3=1.67. Diatomic room temp pe Cv=(5/2)R, Cp=(7/2)R, gamma=1.4. Exam me yaad rakhna.

Derivation: At V const Q=Cv dT=U. At P const Q=Cp dT=U+P dV=Cv dT+R dT => Cp-Cv=R. gamma=Cp/Cv.

3. Specific Heat & Cp-Cv Relation

Revision Points — After Diagram 3:
• s independent of mass, depends on nature, T, process
• C solid predicted 3R, agrees at ordinary T except carbon, diamond
• Calorie old unit, now J used, mechanical equivalent J=4.186 J/cal
• Cp>Cv because at constant P heat does work P dV
• gamma = Cp/Cv ratio important for adiabatic

12.7 Thermodynamic State Variables and Equation of State

Socho system ka equilibrium state describe karna hai. Kaunse variables chahiye? Pressure, volume, temperature, mass, composition. Yeh sab state variables - sirf state pe depend, path pe nahi.

Equation of state unko connect karta hai. Ideal gas ke liye PV = mu RT. Agar mu fixed hai, to sirf do independent hain - jaise P aur V. P-V curve fixed T pe isotherm kehlata hai.

Ab ek naya classification: extensive aur intensive. Socho system ko do equal parts me baanto. Jo variables aadhe ho jaye - jaise U, V, M mass - woh extensive (size pe depend). Jo same rahe - jaise P, T, density rho - woh intensive. Check karo: Q = U + P dV - dono side extensive hona chahiye. P (intensive)*dV(extensive) = extensive, sahi hai. Dimensional check jaisa kaam karta hai numerical me mistake pakadne ke liye.

Non-equilibrium example: Agar partition suddenly hatao to gas free expansion karegi, pressure uniform nahi. Tab P,V,T define hi nahi kar sakte beech me. Final equilibrium pe hi state variables wapas define honge.

State variables: P,V,T,m. Equation of state: PV=mu RT for ideal gas. Extensive: U,V,M proportional size. Intensive: P,T,density independent of size. Check extensivity for equation consistency.

Story: Socho chai ka cup. Uski volume double karo (do cup) to internal energy bhi double - extensive. Par temperature same rahegi - intensive. Jaise class me students count extensive, average age intensive.

12.8 Thermodynamic Processes - Quasistatic Is Key

Maan lo gas cylinder me piston pe weight rakha hai. Bahar ka pressure gas ke pressure ke barabar hai, temperature bhi surroundings ke barabar. Equilibrium hai.

Ab bahar ka pressure thoda kam karo weight hata ke. Piston bahar accelerate karega, gas non-equilibrium se jayegi, pressure uniform nahi. Final me wapas equilibrium. Yeh fast process quasistatic nahi hai.

Quasistatic process ek idealisation hai - infinitely slow process jahan har stage pe system equilibrium me rahe. Har step me pressure difference infinitesimally small. Tab hi hum P-V diagram pe path draw kar sakte hain. Real processes agar slow hain, friction nahi, temperature gradient nahi, to quasistatic ke close hote hain. Aage se hum quasistatic hi deal karenge unless stated.

Special processes: Isothermal T constant, Isobaric P constant, Isochoric V constant, Adiabatic Q=0 insulated.

Quasistatic = infinitely slow, equilibrium at each stage, P_ext = P_sys +- dP, T_sur = T_sys +- dT. Non-quasistatic = free expansion, explosive reaction, intermediate states not on P-V-T surface.

12.8.1 Isothermal Process

Isothermal me T fixed. Ideal gas ke liye PV = constant (Boyle's law). P-V hyperbola.

Work: V1 se V2 tak, small step W = P dV, integrate karo P=mu RT / V. Toh W = integral P dV = mu RT ln(V2/V1) from V1 to V2. Internal energy T pe only depend karti ideal gas me, toh U change 0. First law se Q = W. Agar V2>V1 expansion, W>0 gas work karta, heat absorb karta. Compression me W<0, heat release.

Example: Metallic cylinder ko bade reservoir me rakho jiska temperature fixed hai, dheere expand karo - isothermal. Surroundings ka temperature change nahi hota kyunki reservoir bada hai, par infinitesimal difference hamesha rahega heat flow ke liye, warna quasistatic nahi.

Isothermal: T const, PV=const. W=Q=mu RT ln(V2/V1). For V2>V1 W>0 expansion, V2

12.8.2 Adiabatic Process

Adiabatic me system insulated, Q=0. Toh first law se U change = -W. Agar gas work karta expand hoke, internal energy ghatti, temperature girti.

Ideal gas ke liye PV^gamma = const, where gamma=Cp/Cv. Steeper than isothermal curve. Work: W = integral P dV from V1 to V2 using P=K/V^gamma. Integrate karne pe W = (P1V1 - P2V2)/(gamma-1) = mu R(T1-T2)/(gamma-1). Agar expansion W>0 to T2T1 heating - isiliye cycle pump garam ho jata hai.

Example: Rapid expansion jahan heat exchange ka time nahi, jaise tyre se hawa nikalna thanda lagta hai, ya insulated cylinder me slow par insulated expansion.

Adiabatic: Q=0, PV^gamma=const, T V^{gamma-1}=const, T^gamma P^{1-gamma}=const. W=(P1V1-P2V2)/(gamma-1)=R(T1-T2)/(gamma-1). W>0 => T drops, W<0 => T rises.

12.8.3 Isochoric and Isobaric and Cyclic

Isochoric V constant. Piston locked, rigid vessel. No work, W=0. Toh Q = U change = mu Cv dT. Heat sirf internal energy badhane me gayi. Temperature change specific heat at constant V se decide.

Isobaric P constant. Weight on piston constant. W = P(V2-V1) = mu R(T2-T1). Q = U + W = mu Cv dT + mu R dT = mu Cp dT. Toh Cp = Cv+R consistent.

Cyclic process: System wapas initial state pe. U state function hai, toh U change 0 full cycle ke liye. Toh Qnet = Wnet. P-V diagram pe loop ka area = net work. Clockwise cycle work positive (engine), anticlockwise work negative (refrigerator). Bahut numericals isi pe aate hain - area nikal lo.

Isochoric: V const, W=0, Q=U=mu Cv dT. Isobaric: P const, W=P dV=mu R dT, Q=mu Cp dT. Cyclic: U=0, Wnet=integral P dV= area enclosed, Qnet=Wnet.

4. Thermodynamic Processes & P-V Work

Revision Points — After Diagram 4:
• Isothermal: PV=const, W=mu RT ln(V2/V1), U=0
• Adiabatic: PV^gamma=const, Q=0, W=(P1V1-P2V2)/(gamma-1)
• Isochoric: W=0, Q=U
• Isobaric: W=P dV
• Cyclic: U=0, W=area, Qnet=Wnet

12.9 Heat Engines - Kaam Nikalna Garmi Se

Heat engine ek device hai jo cyclic process se heat ko work me badalta hai. Socho steam engine, petrol engine. Working substance (steam, fuel-air mixture) cycle se guzarti hai.

Basic features: (1) Working substance system hai. (2) Cycle ke kuch steps me hot reservoir T1 se Q1 heat absorb karta hai. (3) Kuch steps me cold reservoir T2 ko Q2 heat reject karta hai. (4) Difference W = Q1 - Q2 surroundings ko work ke roop me milta hai piston wheels etc via shaft. Schematic dekho diagram me.

Efficiency eta = W/Q1 = (Q1-Q2)/Q1 = 1 - Q2/Q1. Q2=0 hota to eta=1 100 percent hota, par experience kehta hai aisa kabhi nahi hota. Kuch heat waste hamesha cold ko dena padta hai. First law mana nahi karta Q2=0 ko, par second law mana karta hai. Conversion mechanism alag engines me alag - external furnace vs internal combustion - par principle same.

Heat engine: cyclic device, absorbs Q1 from hot T1, rejects Q2 to cold T2, delivers W=Q1-Q2. eta=W/Q1=1-Q2/Q1. eta=1 impossible. Requires two reservoirs.

Trick: Efficiency kabhi 1 nahi ho sakti second law ki wajah se. Numerical me agar Q2=0 diya ho aur 100 percent efficiency maang rahe, check karo - idealised nahi, impossible case hai. Real engine me eta less than 1 always.

5. Heat Engine vs Refrigerator

Revision Points — After Diagram 5:
• Engine: Hot to Q1 to Engine to W + Q2 to Cold, W=Q1-Q2
• eta=W/Q1=1-Q2/Q1 less than 1 always
• Refrigerator reverse: Cold to Q2 to +W to Q1 to Hot
• COP alpha=Q2/W for fridge, =Q1/W for heat pump
• alpha can be greater than 1, but eta cannot exceed 1
• Q1=W+Q2 energy conservation always holds

12.10 Refrigerators and Heat Pumps - Ulta Engine

Fridge engine ka ulta hai. Working substance (gas usually) thandi jagah se Q2 heat nikalta hai T2 pe, uspe bahar se W work karte hain electricity se, aur garam reservoir T1 ko Q1 = W+Q2 heat de dete hain. Cycle me steps: (a) gas high se low pressure expand - thandi vapour-liquid mixture, (b) thandi space se heat absorb vapour ban, (c) external work se vapour garam, (d) surroundings ko heat release karke initial state.

Coefficient of performance alpha = Q2/W = Q2/(Q1-Q2). Yeh efficiency nahi hai. Efficiency 1 se kam hoti, par alpha greater than 1 ho sakta hai. Koi upper bound first law se nahi, par second law kehta hai alpha infinite nahi ho sakta - matlab W=0 se fridge nahi chalega. Bina electricity ke thandi se garam me heat nahi jayega.

Heat pump same device, purpose alag. Fridge ka purpose thandi jagah ko thanda rakhna, heat pump ka purpose garam jagah ko garam karna (room heating when outside cold). Heat pump ka COP = Q1/W. AC cooling mode me refrigerator hai, heating mode me heat pump.

Refrigerator: extracts Q2 from cold T2, needs W, rejects Q1=W+Q2 to hot T1. alpha=Q2/W. Heat pump COP = Q1/W. W not zero required. Second law forbids alpha infinite.

Story: Socho fridge band karo to kitchen garam ho jata hai kyunki Q1 = W+Q2 - andar se Q2 nikala, par W ka heat bhi plus Q1 ke roop me kitchen me hi aaya. So net heating. Isiliye fridge ko room cooler ki tarah use nahi kar sakte darwaza khol ke.

12.11 Second Law of Thermodynamics - Direction Ka Law

First law kehta hai energy conserve hogi. Par bahut processes jo energy conserve karte hain, phir bhi nature me nahi hote. Jaise table pe padi book khud jump karke height pe chali jaye table thanda ho ke. Energy conserve hogi - table ki internal energy kam, book ki potential badhi - par aisa kabhi nahi dekha.

Isliye second law chahiye. Yeh kehta hai heat engine ki efficiency kabhi 1 nahi ho sakti, fridge ka COP infinite nahi ho sakta. Matlab kuch direction forbidden hai.

Do statements famous hain. Kelvin-Planck: No process is possible whose sole result is absorption of heat from a reservoir and complete conversion into work. Matlab single reservoir se heat leke pura work nahi bana sakte. Q2 zero nahi ho sakta.

Clausius: No process is possible whose sole result is transfer of heat from colder object to hotter object without work input. Garmi naturally hot se cold jaati hai, cold se hot nahi bina kaam ke.

Dono statements equivalent hain, prove kiya ja sakta hai. Simple meaning: 100 percent engine impossible, bina kaam ke fridge impossible.

First law allows many processes but nature doesn't. Second law gives limitation to eta and alpha. Kelvin-Planck: No 100 percent conversion of heat to work from single reservoir. Clausius: No spontaneous cold to hot heat transfer. Both equivalent.

12.12 Reversible and Irreversible Processes

Maan lo system i se f tak jata hai, heat Q absorb karta, work W karta. Kya hum reverse kar sakte hain jisse system aur surroundings dono initial state pe wapas aa jaye, koi aur change na ho universe me? Agar haan, to reversible, warna irreversible.

Nature ke most processes irreversible hain. Examples: Vessel ka base oven pe garam, hatane pe heat base se other parts me jaake uniform hota, phir thanda hota surroundings me. Reverse nahi hoga - vessel ka ek part khud thanda ho ke base garam nahi karega spontaneously. Free expansion irreversible. Petrol-air combustion irreversible. Gas cylinder se leakage diffusing pura room me - wapas cylinder me nahi jayega khud.

Irreversibility do reasons: (1) Many processes take system to non-equilibrium states, (2) Dissipative effects friction, viscosity. Jaise moving body ruk ke floor aur body ko garam karta, work heat me badla. Reverse me heat se body move nahi hogi - woh second law violate karega. Ideal reversible process quasistatic hona chahiye aur dissipative zero. Jaise frictionless movable piston se ideal gas ka quasistatic isothermal expansion reversible hai.

Reversible: can be reversed to restore both system and surroundings, no other change. Requires quasistatic + no dissipation. Irreversible: spontaneous processes, free expansion, combustion, diffusion, stirring liquid, friction. Most natural processes irreversible.

Exam Tip: Reversible process hypothetical construct. Real processes close if slow, frictionless, no finite T difference. For a process to be reversible quasistatic, T difference between system and reservoir must be infinitesimal at each stage.

12.13 Carnot Engine - Ideal Sabse Efficient

Socho hot reservoir T1 aur cold reservoir T2. Maximum efficiency kitni ho sakti hai aur kaunsa cycle use karna chahiye? Carnot ne 1824 me socha, basic concepts firm hone se pehle hi sahi jawab diya. Ideal engine reversible hona chahiye, kyunki irreversibility efficiency ghatati hai.

Reversible engine jo do temperatures ke beech kaam kare usko Carnot engine kehte hain. Usko kaise banaye? Heat absorption/release isothermal hona chahiye (reversible heat transfer ke liye infinitesimal T difference). Aur temperature T1 se T2 change adiabatic hona chahiye (bina heat flow ke, warna series of reservoirs chahiye hote do se zyada).

Isliye Carnot cycle me 4 steps: 1 to 2 isothermal expansion T1 pe Q1 absorb, 2 to 3 adiabatic expansion T1 se T2, 3 to 4 isothermal compression T2 pe Q2 release, 4 to 1 adiabatic compression T2 se T1. Working substance ideal gas lete hain simplicity ke liye, P-V diagram Fig me loop.

Work calculation: W1 to 2 = Q1 = mu R T1 ln(V2/V1), W2 to 3 = R(T1-T2)/(gamma-1), W3 to 4 = Q2 = mu R T2 ln(V3/V4) negative (work done on gas), W4 to 1 = R(T2-T1)/(gamma-1). Total W = mu R T1 ln(V2/V1) - mu R T2 ln(V3/V4). Adiabatic relations se V3/V4 = V2/V1. Toh eta = W/Q1 = 1 - T2/T1. Simple! Aur result working substance pe independent hai - koi bhi system lo, eta same.

Carnot cycle: two isotherms + two adiabats. Steps: 1-2 isothermal exp T1 Q1=W1 to 2, 2-3 adiabatic exp, 3-4 isothermal comp T2 Q2=W3 to 4, 4-1 adiabatic comp. eta Carnot =1-T2/T1 =1-Q2/Q1 => Q1/T1=Q2/T2. Reversible engine max efficient.

Carnot Theorem and Absolute Temperature Scale

Carnot theorem ke do parts: (a) T1,T2 ke beech kaam karne wala koi engine Carnot se zyada efficient nahi ho sakta, (b) Carnot efficiency working substance ki nature pe independent hai.

Proof by contradiction: Socho irreversible engine I ki efficiency Carnot R se zyada hai. Dono ko couple karo I engine ki tarah kaam kare, R refrigerator ki tarah, same Q1 source ko return kare. Calculation se dekho net result cold reservoir se heat W' -W nikal ke pura work me badal gaya bina source change ke - Kelvin-Planck violate. Isliye eta I greater than eta R galat.

Second part se important result: Q1/T1 = Q2/T2 Carnot cycle me. Yeh universal relation hai, nature of system se independent. Isi ko use karke absolute thermodynamic temperature scale define hoti hai jo ideal gas temperature ke equal aati hai ideal gas working substance ke liye. Isliye Kelvin scale truly universal hai.

Carnot theorem: No engine between T1,T2 more efficient than Carnot, efficiency independent of substance. Proof uses Kelvin-Planck violation if eta I greater than eta R. Universal relation Q1/T1=Q2/T2 leads to absolute temperature scale.

6. Second Law & Carnot Cycle

Revision Points — After Diagram 6:
• Kelvin-Planck: Cannot convert all heat from single reservoir to work
• Clausius: Cannot transfer heat cold to hot without work
• Carnot cycle = reversible, max eta = 1 - T2/T1, depends only on reservoirs
• Reversible = quasistatic + non-dissipative, frictionless piston
• Irreversible examples: free expansion, combustion, diffusion, friction

Gold Reference Table - All Formulas Correct

Concept	Formula (Correct)	Daily Example
Zeroth Law	A eq C and B eq C => A eq B, T defines equilibrium	Two chai cups both in equilibrium with room thermometer, so equal temp
Internal Energy	U = sum (KE+PE) of molecules in CM frame, state function	Pressure cooker steam - disordered motion, not truck KE
First Law	Q = U change + W, W = P dV (const P)	Rubbing palms: Work to U to heat, geyser heating water
Specific Heat	s = (1/m) Q/dT, C = (1/mu) Q/dT	Water 4186 J/kgK high, so coolant and harbour climate mild
Cp - Cv	Cp - Cv = R, gamma = Cp/Cv	Gas heated at constant P needs extra heat for work P dV
Solid Heat Capacity	C = 3R = 24.94 J/molK (Dulong-Petit)	Aluminium, copper atoms vibrating 3D
Equation of State	PV = mu RT, Isotherm: P-V curve T const	Cycle tyre pressure-volume relation
Extensive vs Intensive	Extensive: U,V,M; Intensive: P,T,density	Two cups tea: U double, T same
Isothermal Work	W = mu R T ln(V2/V1), U=0, Q=W	Slow expansion in large water bath, temp constant
Adiabatic	PV^gamma=const, TV^{gamma-1}=const, Q=0, W=(P1V1-P2V2)/(gamma-1)=mu R(T1-T2)/(gamma-1)	Air escaping from tyre cools, pump heats on compression
Isochoric	V const, W=0, Q=U=mu Cv dT	Rigid pressure cooker locked volume, heat goes to U
Isobaric	P const, W=P(V2-V1)=mu R dT, Q=mu Cp dT	Weight on piston constant, heating expands
Cyclic	U=0, Wnet=integral P dV=Area of loop, Qnet=Wnet	Steam engine repeated cycle, area = work output
Heat Engine Efficiency	eta=W/Q1=1-Q2/Q1, W=Q1-Q2	Petrol engine: Q1 from fuel, Q2 waste to air, W to wheels
Refrigerator COP	alpha=Q2/W=Q2/(Q1-Q2), Q1=W+Q2	Fridge extracts heat from inside, needs electricity W
Heat Pump COP	alpha p=Q1/W	AC heating room in winter, can be greater than 1
Second Law K-P	No engine can convert all heat from one reservoir to work	100 percent efficient engine impossible
Second Law Clausius	No spontaneous cold to hot heat transfer without W	Fridge needs power to move heat cold to hot
Carnot Efficiency	eta Carnot=1-T2/T1, Q1/T1=Q2/T2, No engine greater than Carnot	Ideal limit of all engines between two temps
Sign Convention	Q greater than 0 added to system, W greater than 0 done by system, U=Q-W in some books but here Q=U+W	Expansion W positive, compression W negative

Final Revision - Exam Ready

1. Basics - Zeroth, Equilibrium, Variables:
• Zeroth law: A eq C, B eq C => A eq B => defines T
• Thermal equilibrium: no net heat flow, T equal, diathermic allows heat, adiabatic blocks
• State variables: P,V,T,m, state function path independent, U, P, V, T are state, Q,W not
• Extensive: U,V,M halves on splitting, Intensive: P,T,density remains same, check equation extensivity
• Ideal gas equation PV=mu RT, isotherm PV=const, equation of state connects variables

2. First Law and Processes - The Workhorse:
• First law: Q = U change + W, U path independent, Q-W path independent combination
• W=P dV const P, general W=integral P dV area under P-V
• Isochoric: W=0, Q=U=mu Cv dT
• Isobaric: W=P dV=mu R dT, Q=mu Cp dT
• Isothermal: U=0 ideal gas, W=Q=mu R T ln(V2/V1), W greater than 0 expansion, W less than 0 compression
• Adiabatic: Q=0, PV^gamma=const, TV^{gamma-1}=const, W=(P1V1-P2V2)/(gamma-1)=mu R(T1-T2)/(gamma-1), T drops on expansion

3. Specific Heats - Numbers to Remember:
• s=Q/m dT J/kgK, C=Q/mu dT J/molK
• Water: 4186 J/kgK =1 cal/gK, 1 cal=4.186 J from 14.5 to 15.5 C
• Solid Dulong-Petit: C=3R=24.94 J/molK at ordinary T, fails low T and for carbon
• Cp-Cv=R derivation: Cv=U/dT at V const, Cp dT=Cv dT+R dT
• gamma=Cp/Cv, monoatomic gamma=5/3=1.67 Cv=3R/2, diatomic gamma=1.4 Cv=5R/2 room T
• High s means more heat needed for same dT, used for coolant and climate moderation

4. Engines and Refrigerators - Efficiency and COP:
• Heat engine: Q1 from hot T1, Q2 to cold T2, W=Q1-Q2, eta=W/Q1=1-Q2/Q1 less than 1
• No 100 percent engine, Q2 never zero, needs two reservoirs
• Refrigerator reverse: Q2 from cold, W input, Q1=W+Q2 to hot, alpha=Q2/W, can be greater than 1
• Heat pump same as fridge, purpose heating, COPp=Q1/W=1+alpha
• Energy conservation Q1=W+Q2 always, sign: Q greater than 0 added, W greater than 0 by system
• AC = refrigerator in cooling, heat pump in heating mode

5. Second Law, Carnot and Numerical Checklist:
• Kelvin-Planck: No process whose sole result is absorption of heat from reservoir and complete conversion to work
• Clausius: No process whose sole result is transfer of heat from colder to hotter without work
• Both statements equivalent, second law forbids eta=1 and alpha infinite
• Reversible: quasistatic + no dissipation, equilibrium at every step, can be reversed restoring system+surroundings
• Irreversible: free expansion, explosive chemical, diffusion, friction/viscosity, stirring
• Carnot cycle: 1-2 isothermal exp T1 Q1, 2-3 adiabatic exp T1 to T2, 3-4 isothermal comp T2 Q2, 4-1 adiabatic comp T2 to T1, clockwise work positive
• eta Carnot=1-T2/T1=1-Q2/Q1, Q1/T1=Q2/T2 universal, independent of working substance, max efficiency, no engine greater than Carnot
• Numerical checklist: Identify process type, write given P,V,T, use PV=mu RT to find mu, apply first law with sign, for cyclic area=work, for adiabatic use gamma, for efficiency use Kelvin temps always, convert cal to J 4.19, geyser problem rate = power = m s dT / time, fuel consumption = power / combustion heat