Thermodynamics: Class 11 Physics Chapter 11 Hinglish mein

Physics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 22 Oct 2026 · Updated 22 Oct 2026

100 min
1 simulation
16 practice questions
34 MCQs

Is chapter mein aap kya seekhenge

Thermal equilibrium ko samjhao aur zeroth law se temperature ke idea ko justify karo
Internal energy (ek state variable) ko heat aur work se alag pehchano, jo transit mein energy hain
First law of thermodynamics ko sahi signs ke saath apply karo, constant pressure ke liye ΔW = PΔV bhi
Specific aur molar heat capacities define karo, solids ke liye C = 3R samjhao aur ideal gas ke liye Cp − Cv = R derive karo
Quasi-static isothermal, adiabatic, isobaric, isochoric aur cyclic processes describe karo aur har ek ke liye W, Q aur ΔU calculate karo
Second law ke Kelvin–Planck aur Clausius forms state karo aur reversible ko irreversible processes se alag pehchano
Carnot cycle ko analyse karo, η = 1 − T2/T1 derive karo aur samjhao ki do reservoirs ke beech koi bhi engine isse behtar kyun nahi ho sakta

NCERT Physics Class 11, Chapter 11, pp. 226–243 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

11.1 Introduction

Gali ka ek knife-grinder apne stone wheel ko pedal karta hai, uspar ek bhothri kitchen knife dabata hai, aur sparks ki bauchhaar udti hai. Ek minute baad blade itna garam ho jaata hai ki pakda nahi jaata. Uske pairon ka kiya hua work steel ko garam karne mein lag gaya. Pressure cooker ulti taraf chalta hai: steam vent par rakhe bhaari weight ko upar dhakelti hai aur seeti bajwati hai, yaani flame se aayi energy aakhir mein work karti hai. Thermodynamics (ऊष्मागतिकी) heat, work aur matter ke andar stored energy ke beech aise hi exchanges ki physics hai.

2/2Flame ki heat se steam banti hai, aur steam vent par rakhe weight ko utha kar work karti hai.

Fig. 11.1Energy work aur matter ki internal energy ke beech dono taraf jaati hai.

Left mein ek grinding wheel ghoom raha hai aur uspar ek knife blade dabaya gaya hai, contact point se sparks ud rahe hain; label par likha hai work se garam blade. Right mein ek pressure cooker flame par rakha hai. Step 1 grinding wheel ko highlight karta hai. Step 2 cooker ko highlight karta hai: uska weight vent se uthta hai aur steam bahar nikalti hai; label par likha hai heat se utha hua weight.

Bahut samay tak heat ko ek substance maana jaata tha. Athaarahvi sadi ke scientists caloric ki baat karte the, ek weightless fluid jo bodies mein samaa jaata tha aur garam bodies se thandi bodies mein tab tak risata tha jab tak dono mein barabar na ho jaaye. Yeh idea garam aur thande paani ko milane ko kaafi achhe se samjha deta tha. Lekin friction ko yeh nahi samjha paata tha, jo bina kisi limit ke heat paida karta dikhta tha.

Faisla karne wala observation Munich ke ek arsenal se aaya. 1798 mein Benjamin Thompson, jinhe Count Rumford ke naam se zyada jaana jaata hai, ne ek brass cannon barrel ko jaan-boojh kar bhothri drill se bore karwaya, jise ghode ghuma rahe the. Jab tak ghode kaam karte rahe, barrel garam bana raha, itna garam ki uske charon taraf rakhe paani ke tank ko ubaal de. Bhothri drill bahut kam metal nikalti hai, isliye fluid wali picture ke hisaab se usse bahut kam caloric nikalna chahiye tha, aur koi bhi stored supply aakhir mein khatam ho jaani chahiye thi. Dono predictions galat nikli: heat bas ghodon ke kiye work ke saath chalti rahi. Rumford ne nateeja nikala ki heat koi stored fluid nahi balki motion ka ek roop hai, aur yahan work usmein badal raha tha.

Timeline

Caloric se laws of thermodynamics tak

1798
Rumford ki cannon boring
Friction se bani heat kiye gaye work ke saath badhti hai, kisi stored fluid ke saath nahi.
1824
Sadi Carnot ke engines
Carnot poochte hain ki ek heat engine zyada se zyada kitna work de sakta hai (Section 11.11).
1840s
Mayer aur Joule
Dikhaya jaata hai ki work aur heat energy ke aapas mein badle ja sakne wale roop hain; Joule girte weights aur paddle wheels se is conversion ko maapte hain.
1850–1851
Clausius aur Kelvin
Second law do equivalent forms mein state kiya jaata hai (Section 11.9).
1930s
Zero number wala law
Ralph Fowler sabse basic law ko zeroth law naam dete hain (Section 11.3).

Is chapter ke laws unnees-vi sadi ke scientists ne dhoondhe the, jinmein se kai ko atoms ke asli hone par shak tha, aur in laws ko atoms ki kabhi zaroorat nahi padi. Thermodynamics kuch bulk quantities ke saath kaam karta hai jinhe aam instruments padh sakte hain: pressure, volume, temperature, mass aur composition. Iski tulna ek litre hawa ke molecule-by-molecule description se karo, jise lagbhag $2.5 \times 10^{22}$ particles ko follow karna padega. Agle chapter ki kinetic theory beech mein aati hai: woh molecules use karti hai, lekin sirf unka average behaviour.

Dono subjects alag sawaal bhi poochte hain. Mechanics poori body ko follow karta hai: uska centre of mass kahan jaata hai aur forces aur torques ke under woh kaise ghoomti hai. Thermodynamics andar dekhta hai: body kitni garam hai, kitni compressed hai, uske molecules ki jiggling mein kitni energy bandhi hai. Cricket ball bat se kitni tez nikalti hai, yeh mechanics ka sawaal hai. Catch hone par ball aur wicketkeeper ke gloves kitne garam hote hain, yeh thermodynamics ka sawaal hai.

Quick check

2 mein se 1

Rumford ke cannon-boring observations ne kya dikhaya?

bani hui heat kiye gaye work ke saath badhti rahibhothri drill ne sabse zyada stored caloric chhodabrass iron se zyada heat store karta haiheat thandi bodies se garam bodies ki taraf flow karti hai

Answer dekho

Answer (A) bani hui heat kiye gaye work ke saath badhti rahi

Jab tak ghode kaam karte rahe, heat banti rahi, aur kisi stored supply ke khatam hone ka koi nishan nahi tha. Isse pata chala ki heat work se paida hone wali energy ka ek roop hai, metal se nichoda gaya koi fluid nahi.

11.2 Thermal equilibrium

Paani ki ek sealed bottle jo kal se room mein rakhi hai, bahut boring haalat mein hai: kisi bhi pal uska pressure, volume ya temperature check karo, same values milengi. Thermodynamics ise thermodynamic equilibrium ki state kehta hai, jismein bulk variables (pressure, volume, temperature, mass aur composition) time ke saath badalna band kar dete hain. Is shabd ka matlab mechanics ke forces aur torques ke balance se alag hai; table par padi kitaab mechanical equilibrium mein hai, chahe woh garam ho ya thandi.

Dhyaan do ki yeh baat sirf averages ke baare mein hai. Flask ke andar molecules ab bhi idhar-udhar bhaagte hain aur har second arbon baar takraate hain; mechanical sense mein unmein se koi bhi equilibrium mein nahi hai. Jo badalna band hua hai woh bulk description hai.

System waisa hi rehta hai ya nahi, yeh uski walls par depend karta hai. Foam ke cool box mein pack ki gayi thandi bottle ghanton thandi rehti hai; wahi bottle garam paani ke steel tumbler mein rakho to jaldi gunguni ho jaati hai. Foam lagbhag ek adiabatic wall hai, jo heat ko rokti hai, jabki patla steel ya copper ek diathermic wall hai, jo heat ko jaane deti hai. Fig. 11.2 do gas samples, A aur B, ke liye yahi farq dikhata hai. Adiabatic partition ke peeche har gas apna pressure aur volume rakh sakti hai, chahe woh padosi se kitna bhi alag ho. Diathermic partition ke through energy heat ke roop mein flow karti hai, dono gases badalti hain, aur aakhir mein nayi states mein pahunchti hain jo badalna band kar deti hain. Tab A aur B ek doosre ke saath thermal equilibrium mein hote hain.

2/3Copper (diathermic) wall heat ko A se B tak jaane deti hai: A thandi hoti hai, B garam hoti hai.

Fig. 11.2Heat wall paar kar sakti hai ya nahi, isse tay hota hai ki A aur B ek common state tak pahunchenge ya nahi. (A aur B ek hi gas ke barabar samples hain.)NCERT Fig. 11.1 · p. 227

Gas ke do boxes paas-paas rakhe hain, left mein A aur right mein B, dono mein ek-ek thermometer. Step 1: ek moti foam wall unhe alag karti hai; A 70 °C padhta hai, B 20 °C, aur readings wahi rehti hain. Step 2: foam ki jagah patli copper wall lagti hai; ek heat arrow A se B ki taraf point karta hai aur readings 58 °C aur 32 °C ho jaati hain. Step 3: dono thermometers 45 °C padhte hain aur label kehta hai ki ab koi change nahi.

Flow rukne ke baad A aur B mein kya common hai? Experience turant jawab deta hai: dono barabar garam hain. Is intuition ko ek maapne laayak quantity mein badalna agle law ka kaam hai.

Quick check

2 mein se 1

Thermodynamic equilibrium wali gas mein molecules rest par hote hain.

SahiGalat

Answer dekho

Answer Galat

Thermodynamic equilibrium ka matlab hai ki bulk variables ab nahi badalte. Molecules chalte aur takraate rehte hain; sirf unke averages steady hote hain.

11.3 Zeroth law of thermodynamics

Kaise pata karein ki do bodies jo kabhi ek doosre ko chhui nahi, chhoone par heat exchange karengi ya nahi? Ek beech wale ki madad lo. Ek teesri body C, jaise ek thermometer, ko pehle A ke saath aur phir B ke saath equilibrium mein aane do (Fig. 11.3). Agar dono baar C same state mein pahunchti hai, to experiment dikhata hai ki A aur B ko aakhir mein contact mein rakhne par woh bilkul bhi heat exchange nahi karte: dono ki states nahi badalti.

LawZeroth law of thermodynamics

Agar do systems mein se har ek kisi teesre system ke saath thermal equilibrium mein hai, to woh dono bhi ek doosre ke saath thermal equilibrium mein hain.

Yeh itna obvious lagta hai ki kehne ki zaroorat hi nahi, aur bahut samay tak kisi ne kaha bhi nahi. First aur second laws ko number mil chuke the jab physicists ne notice kiya ki dono chupchaap isi par depend karte hain. 1930s mein Ralph Fowler ne ise zeroth law kehne ka sujhaav diya, kyunki logically yeh baaki sab se pehle aata hai.

Zeroth law hi temperature ko ek meaningful number banata hai. Yeh 'ke saath equilibrium mein hai' ko ek equals sign ki tarah behave karwata hai, jo transitive hota hai. Isliye bodies apne aap classes mein bant jaati hain, har class mein woh saari bodies hoti hain jo aapas mein equilibrium mein hongi, aur har class ko ek number se label kiya ja sakta hai. Yahi label temperature $T$ hai: thermal equilibrium wali bodies ka $T$ same hota hai, aur jo body kisi doosri ke saath heat exchange karegi uska alag hota hai. Thermometer bas ek suvidhajanak body C hai jo jaldi settle hoti hai aur apna label ek scale par dikhati hai. Ek thermometer se doosre tak in scales ko milana thermometry hai, jisse tum pichhle chapter mein mil chuke ho.

3/3B ko A mein utaaro: koi net heat flow nahi hota, bilkul jaisa zeroth law predict karta hai.

Fig. 11.3Thermometer C zeroth law ke teesre system ka role nibhata hai.NCERT Fig. 11.2 · p. 228

Left mein paani ki baalti A, right mein copper block B aur ek thermometer C. Step 1: C paani mein khada hai aur 31 °C padhta hai. Step 2: C ko copper block par le jaate hain aur woh phir se 31 °C padhta hai. Step 3: block ko paani mein utaara jaata hai, thermometer ko side mein rakh diya jaata hai, aur label kehta hai ki koi net heat flow nahi hai; T A barabar T B.

Try karke dekho

Ek thermometer, do liquids

Tumhe chahiye

ek thermometer
nal ke paani se bhara steel tumbler
cooking oil ki band bottle
ek steel ka chammach aur ek lakdi ka chammach

Pehle guess karo

Paani aur oil raat bhar same room mein rakhe rahe hain. Thermometer kya dikhayega?

oil zyada garam padhegapaani zyada garam padhegadono same padhengeyeh amounts par depend karta hai

Nal ke paani se bhara steel tumbler aur cooking oil ki band bottle ek hi table par raat bhar chhod do.
Thermometer ko ek minute paani mein rakho aur reading note karo. (Digital cooking thermometer sabse safe choice hai.)
Use ponchho, oil kholo aur oil mein reading lo.
Bottle band karo, use das minute paani mein khada rakho, phir paani ki reading dobara lo.
Aakhir mein ek hi drawer se nikale steel ke chammach aur lakdi ke chammach ko chhoo kar dekho. Kaun sa zyada thanda lagta hai?

Table 11.1Tumhari readings

Measurement	Prediction	Reading (°C)
Paani
Oil
Bottle rakhne ke baad paani

Tumhari entries apne aap save hoti hain.

Paani aur oil same padhte hain, kyunki dono room ki hawa, yaani common 'third body', ke saath equilibrium mein aa chuke the. Bottle ko paani mein rakhne se kuch nahi badalta, jaisa zeroth law predict karta hai. Steel ka chammach lakdi wale se zyada thanda lagta hai jabki dono room temperature par hain: steel tumhari ungliyon se heat tezi se le jaata hai. Skin heat flow ki rate mehsoos karti hai, temperature nahi.

11.4 Heat, internal energy aur work

Temperature batata hai ki do bodies ke chhoone par heat kis taraf flow karegi: garam se thandi ki taraf, jab tak temperatures barabar na ho jaayein. Aage badhne ke liye hamein teen aur ideas chahiye: internal energy, heat aur work.

Kisi bhi cheez ko zoom karke dekho, ek eent, paani ka glass, balloon ki hawa, aur tumhe molecules milenge jo random tarike se idhar-udhar bhaag rahe hain aur ek doosre ko kheench rahe hain. Is saari chhupi activity ki energy, yaani random motion ki kinetic energy plus molecular forces ki potential energy, internal energy $U$ (आंतरिक ऊर्जा) kehlati hai. Ise us frame mein gina jaata hai jismein poori cheez rest par ho: poori cheez ke chalne ki kinetic energy mechanics ka hissa hai, $U$ ka nahi (Fig. 11.4).

Definition

Internal energy

Kisi system ke molecules ki random motions ki kinetic energies aur unke interactions ki potential energies ka sum, us frame mein maapa gaya jismein system ka centre of mass rest par ho.

ExampleChalti train par gas ke sealed can ka $U$ utna hi hota hai jitna platform par, agar uska temperature same ho.

2/3Har molecule ko same forward velocity milti hai. Yeh ordered motion mechanics hai: $U$ nahi badalta.

Fig. 11.4

U

mein sirf box ke relative random motion ginta hai; box ki apni motion nahi ginti, jab tak koi collision use bikhra na de.NCERT Fig. 11.3 · p. 229

Gas molecules ka ek box, har molecule par random motion dikhane wala chhota arrow, ek trolley par rakha hai. Step 1: trolley rest par hai; label kehta hai ki molecules ki random motion internal energy U hai. Step 2: trolley right ki taraf chali gayi hai aur ek bada arrow poore box ki motion dikhata hai; molecular arrows waise hi hain aur label kehta hai ki U nahi badla. Step 3: trolley wall se takra kar ruk gayi hai; molecular arrows lambe hain, gas ka rang garam dikhta hai aur label kehta hai ki U badhta hai.

Humne $U$ ko molecules ke through imagine kiya, lekin thermodynamics ise bas ek aur bulk property maanta hai. $U$ itna useful isliye hai kyunki system ki present condition ise poori tarah fix kar deti hai; system wahan kis raaste se pahuncha, isse farq nahi padta. Kisi diye gaye pressure aur temperature par ek litre nitrogen ka $U$ same hoga, chahe use thande se garam kiya gaya ho, bade volume se dabaya gaya ho ya kisi aur tarike se woh state mili ho. Aisi quantity ko state variable kehte hain; pressure, volume aur temperature bhi state variables hain (Section 11.7). Gas mein molecular forces kamzor hote hain, isliye $U$ lagbhag poora kinetic energy hota hai. Jaisa agla chapter dikhata hai, ismein molecules ka container mein travel, unka rotation aur, kaafi garam hone par, unke atoms ka vibration shaamil hai.

SimulationMolecular motion ke roop mein internal energy

Paani ke molecules ka ek band container, jismein molecules chalte hue dots ki tarah bane hain, saath mein ek thermometer aur 5 se 120 degree Celsius tak ka temperature slider. Temperature badhane se dots tez chalte hain; 100 degree ke upar woh ek doosre se alag ho kar gas ki tarah container ko bhar dete hain.

Fig. 11.5Temperature slider ko drag karo aur molecules ko dekho. Tez random motion ka matlab zyada internal energy hai.

Try karo

Temperature ko 20 °C se 90 °C tak badhao. Molecules ki speed ka, aur isliye U ka, kya hota hai?
100 °C se aage jao. Molecules ek doosre se alag ho jaate hain; unhe alag kheenchne mein bhi energy lagti hai, isiliye boiling ke liye itni heat chahiye (Example 11.1).

Maan lo tum cylinder ki gas ki internal energy badhana chahte ho. Tumhare paas theek do options hain (Fig. 11.6): uske saath koi zyada garam cheez lagao, jaise hot plate, ya piston ko andar dhakelo. Dono kaam karte hain, aur dono ulte bhi chal sakte hain: thande surroundings energy kheench lete hain, aur jo gas apne piston ko bahar dhakelti hai woh aisa karne mein apna kuch $U$ kharch karti hai.

3/4Piston ko andar dhakelne se gas par work hota hai, koi temperature difference nahi chahiye: $U$ phir badhta hai.

Fig. 11.6Heat aur work do alag raaste hain jinse energy system mein aati ya usse jaati hai.NCERT Fig. 11.4 · p. 229

Piston wala gas ka ek vertical cylinder, aur right mein ek gauge jo internal energy U ka level dikhata hai. Step 1: gas rest par hai aur gauge beech ke level par hai. Step 2: cylinder hot plate par khada hai, heat arrows upar gas mein point karte hain aur gauge badhta hai. Step 3: ek arrow piston ko neeche dhakelta hai, gas compress hoti hai aur gauge aur badhta hai. Step 4: piston upar jaata hai aur uspar work out label wala arrow hai, heat arrows walls se bahar point karte hain, aur gauge girta hai.

Isse do roz ke shabdon ko ek precise matlab milta hai. Heat (ऊष्मा) woh energy hai jo system aur uske surroundings ke beech temperature difference ki wajah se transfer hoti hai. Work kisi bhi aur tarike se transfer hui energy hai: force ke under chalta piston, liquid ko hilata paddle, rassi se uthaya gaya weight.

Compare

Boundary paar karne ke energy ke do tarike

Heat

Temperature difference se chalti hai
Apne aap sirf garam se thande ki taraf flow karti hai
Example: flame par paani ka pan

Work

Koi temperature difference nahi chahiye
Force ek displacement ke through act karta hai
Example: cycle-pump ka handle dabana

Dono

Joules mein maape jaate hain
Process describe karte hain, state nahi
Internal energy $U$ badalte hain

Roz ki baat-cheet mein hum kehte hain ki garam iron mein 'bahut heat hai'. Thermodynamics zyada sakht hai: iron mein bahut internal energy hai, aur heat woh hai jo tumhari thandi shirt ko chhoone par usse bahar flow karti hai. In dono ko alag rakhne se is chapter ka zyaadatar confusion door ho jaata hai.

11.5 First law of thermodynamics

Heat aur work hi do tarike hain jinse energy system ki boundary paar karti hai, isliye conservation of energy ek simple hisaab-kitaab ka rule deta hai. Likho

$Δ Q$ surroundings dwara system ko di gayi heat ke liye,
$Δ W$ system dwara surroundings par kiye gaye work ke liye,
$Δ U$ system ki internal energy mein change ke liye.

LawFirst law of thermodynamics

System ko di gayi heat uski internal energy mein badhotri plus system dwara surroundings par kiye gaye work ke barabar hoti hai.

First law of thermodynamics

Δ Q = Δ U + Δ W

(11.1)

Ise ghar ke budget ki tarah padho: heat ke roop mein aayi income ya to bachayi jaati hai, jisse $U$ badhta hai, ya phir work ke roop mein kharch ho jaati hai. First law aur kuch nahi, bas aise system ke liye conservation of energy hai jo work ke saath heat bhi exchange kar sakta hai. Signs hisaab rakhte hain, isliye unhe ek baar fix kar lena faydemand hai (Table 11.2).

Table 11.2First law ke liye sign convention (physics)

Quantity	Positive jab	Negative jab
$Δ Q$	heat system mein flow karti hai	heat bahar flow karti hai
$Δ W$	system work karta hai (woh expand hota hai)	uspar work hota hai (woh compress hota hai)
$Δ U$	internal energy badhti hai	internal energy ghatti hai

Tumhari chemistry book system par kiye gaye work ko positive ginti hai aur yahi law $Δ U = q + w$ likhti hai. Physically kuch alag nahi hai; sirf work ke saath laga sign alag hai. Rearrange karne par hamara form yeh banta hai

Δ Q - Δ W = Δ U

(11.2)

Is version mein ek surprise chhupa hai. $P$ – $V$ diagram par same do states ke beech koi bhi do routes banao (Fig. 11.7). Work, yaani har route ke neeche ka area, saaf alag hai, aur heat bhi alag hai. Phir bhi $Δ Q - Δ W$ bilkul same aata hai, kyunki yeh $Δ U$ ke barabar hai aur $U$ ek state variable hai. Heat aur work safar ke hain; unka difference do end points ka hai.

3/4 $i$ se seedhe $f$ tak jao, gas ke expand hone ke saath pressure lagaataar girta hai: $W_{2}$ chhota hai.

Fig. 11.7Kiya gaya work

P

–

V

diagram par path ke neeche ka area hai, isliye yeh route par depend karta hai.

Pressure aur volume ka ek graph jismein initial state i high pressure aur chhote volume par hai aur final state f low pressure aur bade volume par. Step 1: sirf do points. Step 2: route 1 high pressure par right jaata hai aur phir seedha neeche; uske neeche ka bada area shaded hai aur W1 label hai. Step 3: route 2 i se f tak ek seedhi line hai; uske neeche ka chhota area shaded hai aur W2 label hai. Step 4: route 1 se jaana aur route 2 se wapas aana ek triangle gherta hai, jiska label W1 minus W2 hai, yaani loop ka net work.

Agar $Δ U = 0$ ho, jaise ideal gas ke slow isothermal expansion mein (Section 11.8.2), to $Δ Q = Δ W$ : andar li gayi heat ka har joule seedhe work ke roop mein aage chala jaata hai.

DerivationConstant pressure ke against kiya gaya work

Maan lo pressure $P$ wali gas area $A$ wale piston ko ek chhoti doori $Δ x$ tak bahar dhakelti hai. Piston par force $P A$ hai, isliye gas itna work karti hai

\begin{aligned} Δ W & = P A Δ x \\ = P Δ V \end{aligned}

kyunki $A Δ x$ gas ke volume mein badhotri $Δ V$ hai. Agar volume ke ek finite change ke dauraan pressure fixed rahe, to yahi expression poore change par laagu hota hai, aur first law ban jaata hai

Δ Q = Δ U + P Δ V

(11.3)

Compression mein $Δ V < 0$ , isliye $Δ W < 0$ : gas par work hota hai.

Example 11.1NCERT 11.5 · p. 231

Boiling ki heat kahan jaati hai?

Ek laboratory mein $1.00 m o l$ ( $46.1 g$ ) ethanol ko uske normal boiling point, ${78.4}^{\circ} C$ , par atmospheric pressure, $1.013 \times 10^{5} P a$ , ke under ubaala jaata hai. Ismein $38.6 k J$ heat lagti hai. Vapour ko ideal gas maano aur liquid ke volume (lagbhag $0.06 L$ ) ko negligible maano. Atmosphere ko peeche dhakelne mein ethanol dwara kiya gaya work, aur uski internal energy mein badhotri nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek gas

600 J

heat absorb karti hai aur surroundings par

250 J

work karti hai. Uski internal energy kitni badalti hai?

Answer dekho

Answer 350 J

Δ U = Δ Q - Δ W = 600 - 250 = 350 J,

yaani badhotri.

11.6 Specific heat capacity

Kisi body ko thodi si heat $Δ Q$ do aur uska temperature $Δ T$ badhta hai. Is ratio ko body ki heat capacity kehte hain:

S = \frac{Δ Q}{Δ T}

(11.4)

Paani ki baalti ki heat capacity ek cup paani se kahin zyada hoti hai, isliye $S$ ek object ko describe karta hai, material ko nahi. Mass $m$ se divide karne par sirf substance ki property milti hai, uski specific heat capacity (विशिष्ट ऊष्मा धारिता):

s = \frac{S}{m} = \frac{1}{m} \frac{Δ Q}{Δ T}

(11.5)

Definition

Specific heat capacity

Kisi substance ke unit mass ka temperature ek kelvin badhane ke liye zaroori heat; SI unit $J k g^{- 1} K^{- 1}$ .

ExampleLagbhag $4.2 k J$ se $1 k g$ paani $1 K$ garam hota hai.

Value substance par aur thodi si uske temperature par depend karti hai. Agar matter ko moles mein ginein, yaani moles ki sankhya $μ$ se divide karein, to molar specific heat capacity milti hai:

C = \frac{S}{μ} = \frac{1}{μ} \frac{Δ Q}{Δ T}

(11.6)

Iska unit $J m o l^{- 1} K^{- 1}$ hai, aur $s$ ki tarah yeh is par depend nahi karti ki tumhare paas kitna substance hai. $s$ aur $C$ dono is par bhi depend kar sakte hain ki heat kin conditions mein di gayi, jaisa gases mein dikhega. Moles mein ginne ka fayda yeh hai ki isse ek kamaal ki regularity saamne aati hai (Table 11.3).

Table 11.3

25^{\circ} C

ke aas-paas chhe solids ki heat capacities (standard data)NCERT Table 11.1 · p. 231

Solid	$M$ ( $g m o l^{- 1}$ )	$s$ ( $J k g^{- 1} K^{- 1}$ )	$C = s M$ ( $J m o l^{- 1} K^{- 1}$ )
Iron	55.85	449	25.1
Zinc	65.38	388	25.4
Nickel	58.69	444	26.1
Gold	196.97	129	25.4
Silicon	28.09	705	19.8
Beryllium	9.01	1825	16.4

Table ki specific heats mein chaudah guna tak ka farq hai, phir bhi chaaron metals ki molar values ek hi value ke kareeb hain: $25 J m o l^{- 1} K^{- 1} .$ Gold ke atoms bhaari hote hain, isliye ek kilogram gold mein unki sankhya kam hoti hai; atom ke hisaab se gold ko lagbhag utni hi heat chahiye jitni iron ko. Neeche ke halke, mazboot bonds wale elements peeche reh jaate hain, aur diamond sabse zyada peeche hai, common value ke lagbhag ek chauthai par. Ek simple model common value ko samjhata hai.

DerivationSolid ki molar heat capacity

Solid ko $N$ atoms ka model maano, har atom ek fixed position ke aas-paas vibrate kar raha hai. Ek one-dimensional oscillator energy do roop mein store karta hai, kinetic aur potential, aur law of equipartition of energy (Chapter 12) har roop ko average $\frac{1}{2} k_{B} T$ deta hai. Isliye vibration ki har direction average $k_{B} T$ rakhti hai, aur teen directions mein vibrate karne wala atom $3 k_{B} T$ rakhta hai.

Ek mole ke liye $N = N_{A}$ , aur $k_{B} N_{A} = R$ , isliye

\begin{aligned} U & = 3 k_{B} T \times N_{A} \\ = 3 R T \end{aligned}

Atmospheric pressure par solid ko garam karne se uska volume mushkil se badalta hai, isliye Eq. (11.3) mein work term $P Δ V$ negligible hai aur $Δ Q \approx Δ U$ . Isliye

\begin{aligned} C & = \frac{Δ Q}{Δ T} = \frac{Δ U}{Δ T} \\ = 3 R \end{aligned}

(11.7)

Numerically $3 R = 24.9 J m o l^{- 1} K^{- 1},$ jo Table 11.3 ke metals se match karta hai. Yeh rule low temperatures par fail ho jaata hai, jahan har solid ka $C$ zero ki taraf girta hai, aur diamond jaise halke, mazboot solids ke liye room temperature par bhi. In failures ko samjhane ke liye quantum physics ki zaroorat padi.

Example 11.2

Heat capacity se metal ki pehchaan

Ek jeweller ke paas kisi unknown pure metal ki $0.250 k g$ ki bar hai. Use ${20.0}^{\circ} C$ se ${60.0}^{\circ} C$ tak garam karne mein $2.35 k J$ lagte hain. (a) Uski specific heat capacity nikalo. (b) $C = 3 R$ maan kar uska molar mass estimate karo aur batao ki yeh kaun sa metal ho sakta hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Paani ki specific heat capacity

Heat ki purani unit, calorie, itne baarik temperatures ke jode se kyun define ki gayi hai? Kyunki paani ko har degree ke liye bilkul same heat nahi chahiye. Uski specific heat $0^{\circ} C$ ke paas sabse zyada hoti hai, tees ke beech wale degrees mein ek halke minimum tak girti hai aur $100^{\circ} C$ ki taraf phir badhti hai, kul change 1% se kam (Fig. 11.8). Isliye 'ek gram aur ek degree ke liye heat' wali unit ko batana padta tha ki kaun si degree. Calorie ko woh heat tay kiya gaya jo ek gram paani ko ${14.5}^{\circ} C$ aur ${15.5}^{\circ} C$ ke beech ki ek degree garam karne ke liye chahiye, aur yeh maatra $4.186 J$ hai.

Fig. 11.8

0^{\circ} C

aur

100^{\circ} C

ke beech paani ki specific heat 1% se kam badalti hai (values steam tables se).NCERT Fig. 11.5 · p. 232

Specific heat capacity, joules per gram per kelvin mein, ka 0 se 100 degree Celsius tak temperature ke against curve. Yeh 0 °C par lagbhag 4.22 se shuru hota hai, 30 aur 40 °C ke beech lagbhag 4.18 ke chaude minimum tak girta hai, aur 100 °C par phir lagbhag 4.22 tak badhta hai. Ek patli peeli band 14.5 se 15.5 °C ka interval dikhati hai jo calorie define karne mein use hota hai.

SI units mein paani ki specific heat hai lagbhag $4186 J k g^{- 1} K^{- 1},$ jo asaamanya roop se badi hai: bahut kam common substances per kilogram per kelvin itni energy store karte hain. Joule ke zamaane mein $4.186 J$ per calorie wale number ka ek bhaari naam tha, 'mechanical equivalent of heat', aur ise maapna ek bada milestone tha. Aaj yeh bas energy ki do units ke beech exchange rate hai, jaise rupaye aur paise, aur kyunki SI heat aur work dono ke liye joule use karta hai, purana naam ab use nahi hota.

Ideal gas ki specific heats

Solid ya liquid ke liye isse khaas farq nahi padta ki tum use kaise garam karte ho. Gas ke liye bahut farq padta hai (Fig. 11.9). Piston ko pin kar do to heat ka har joule internal energy ke roop mein gas mein hi rehta hai. Piston ko ek steady load ke against upar uthne do to heat ka kuch hissa work ke roop mein phir bahar chala jaata hai, isliye temperature kam badhta hai. Isliye gas ki do principal molar heat capacities hoti hain, constant volume par $C_{v}$ aur constant pressure par $C_{p}$ , jismein $C_{p} > C_{v}$ . Ideal gas ke liye yeh gap theek $R$ hai:

2/2Pinned: poora $Q$ $U$ badhata hai, isliye $Δ T$ zyada hai. Free: $Q$ ka kuch hissa load uthata hai, isliye $Δ T$ kam hai.

Fig. 11.9Constant pressure par heat ka kuch hissa work karta hai, isliye har kelvin ke liye zyada heat chahiye:

C_{p} > C_{v}

Identical burners par gas ke do cylinders, dono mein ek-ek thermometer. Left piston pinned hai isliye volume fixed hai; right piston par load hai aur woh chal sakta hai, isliye pressure fixed hai. Step 1: Q label wale barabar heat arrows dono mein jaate hain. Step 2: left thermometer zyada upar chadha hai aur label kehta hai ki saari heat U mein jaati hai; right piston ek arrow ke saath upar utha hai, uska thermometer kam chadha hai, aur label kehta hai ki heat U plus work mein jaati hai.

C_{p} - C_{v} = R

(11.8)

Derivation

C_{p} - C_{v} = R

ka proof

Ideal gas ka ek mole lo aur Eq. (11.3), $Δ Q = Δ U + P Δ V$ , apply karo. Constant volume par $Δ V = 0$ , isliye

\begin{aligned} C_{v} & = {(\frac{Δ Q}{Δ T})}_{v} \\ = \frac{Δ U}{Δ T} \end{aligned}

(11.9)

Subscript batata hai ki kya fixed rakha gaya hai. Right side par iski zaroorat nahi: ideal gas ka $U$ sirf $T$ ka function hai, isliye $Δ U / Δ T$ ko farq nahi padta ki $V$ constant rakha gaya ya $P$ . Constant pressure par,

\begin{aligned} C_{p} & = {(\frac{Δ Q}{Δ T})}_{p} \\ = \frac{Δ U}{Δ T} \\ + P {(\frac{Δ V}{Δ T})}_{p} \end{aligned}

(11.10)

Pehla term wahi hai jo Eq. (11.9) mein hai, phir se isliye kyunki $U$ sirf $T$ par depend karta hai. Ek mole ke liye $P V = R T$ , isliye fixed $P$ par temperature change $Δ T$ volume ko $Δ V = R Δ T / P$ badalta hai, aur

P {(\frac{Δ V}{Δ T})}_{p} = R

(11.11)

Eqs. (11.9) aur (11.11) ko Eq. (11.10) mein rakhne par $C_{p} = C_{v} + R$ milta hai, jo Eq. (11.8) hai.

Yeh ek general rule ka sabse saaf case hai: heat capacity substance ke saath-saath process par bhi depend karti hai. Gas ko isothermally garam karo to $Δ T = 0$ , isliye $Δ Q / Δ T$ unbounded hai; use adiabatically compress karo to uska temperature $Δ Q = 0$ ke saath badhta hai, isliye ratio zero hai. Conditions batane ke baad hi heat capacity ki ek definite value hoti hai.

Example 11.3

Helium ko do tarah se garam karna

$0.50 m o l$ helium, jo $C_{v} = \frac{3}{2} R$ wali monatomic ideal gas hai, ko $40 K$ garam kiya jaata hai, (a) ek rigid steel cylinder mein aur (b) aise cylinder mein jiska piston free chalta hai taaki pressure constant rahe. Har case mein zaroori heat aur (b) mein kiya gaya work nikalo. Value lo: $R = 8.314 J m o l^{- 1} K^{- 1} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Equipartition use karke, room temperature par ek typical solid ki molar heat capacity kya hai? Value lo:

R = 8.31 J m o l^{- 1} K^{- 1} .

J/(mol K)

Answer dekho

Answer 24.9 J/(mol K)

C = 3 R = 3 \times 8.31 = 24.9 J m o l^{- 1} K^{- 1},

yaani Dulong–Petit value.

11.7 Thermodynamic state variables aur equation of state

Kisi equilibrium state ko pakka karne ke liye bas numbers ki ek chhoti list chahiye. Cylinder ki gas ke liye yeh list hai uska pressure, volume, temperature aur mass, aur hawa jaise mixture ke liye uska composition bhi. Yeh uske state variables hain. Inke pata hote hi woh sab bhi pata chal jaata hai jo thermodynamics gas ke baare mein bata sakta hai, uski internal energy bhi.

Har state ko is tarah describe nahi kiya ja sakta. Ek partition mein chhed karo jiske ek taraf gas aur doosri taraf vacuum ho, to ek pal ke liye gas jet ki tarah tezi se bahar nikalti hai: chhed ke paas ghani, baaki jagah patli, aur poore sample ke liye koi ek pressure ya temperature nahi (Fig. 11.10). Spark ke turant baad motorcycle engine ke fuel–air mixture ke saath bhi yahi hota hai. Jab gas shaant ho kar uniform pressure aur temperature par, surroundings ke saath equilibrium mein aa jaati hai, tabhi state variables use phir se describe karte hain.

2/3Partition toot jaata hai: gas vacuum mein tezi se ghusti hai, uneven aur equilibrium se bahut door.

Fig. 11.10State variables sirf equilibrium states ko describe karte hain; beech wale step ki bhaagti gas ke koi state variables nahi hain.NCERT Fig. 11.6 · p. 233

Partition se bata hua ek insulated box, left mein gas molecules aur right mein vacuum. Step 1: molecules left half mein barabar faile hain; label kehta hai ki P, V aur T defined hain. Step 2: partition hat gaya hai aur molecules beech mein jama hain, arrows right ki taraf point karte hain; label kehta hai ki koi ek P ya T nahi hai. Step 3: molecules poore box mein barabar faile hain aur label kehta hai ki phir se equilibrium.

State variables ek doosre se independent nahi hain. Unke beech ka relation system ka equation of state hai. Ideal gas ke liye yeh ideal gas equation hai:

Ideal gas ka equation of state

P V = μ R T

Gas ki fixed maatra ke liye sirf do variables free hote hain, jaise $P$ aur $V$ ; teesra unse nikal aata hai. Ek fixed temperature par $P$ ke against $V$ ka graph isotherm hota hai. Real gases zyada complicated equations of state follow karti hain, khaas kar dense hone par.

State variables do tarah ke hote hain. $30^{\circ} C$ wala ek litre doodh do glasses mein barabar daalo. Har glass mein aadha volume, aadha mass aur aadhi internal energy hai, lekin har glass ka doodh ab bhi $30^{\circ} C$ par hai, same pressure aur same density ke saath. Jo quantities is tarah bant jaati hain woh extensive hain: woh system ke size ke saath badhti hain. Jo quantities har part mein same rehti hain, jaise temperature, pressure aur density, woh intensive hain.

Compare

System ko aadha kaato

Intensive (same rehte hain)

pressure $P$
temperature $T$
density $ρ$

Extensive (aadhe ho jaate hain)

volume $V$
mass $M$
internal energy $U$

Yeh idea ek error check ka kaam bhi karta hai, units check karne jaisa: sahi equation mein har term ek hi kind ka hota hai. Ise $Δ Q = Δ U + P Δ V$ par test karo. $Δ U$ extensive hai; $P Δ V$ intensive guna extensive hai, jo extensive hai; aur $Δ Q$ , bhale state variable nahi hai, system double hone par double ho jaata hai. Teeno match karte hain.

11.8 Thermodynamic processes

Process system ko ek state se doosri state tak le jaata hai, aur woh kaun sa path follow karta hai isse tay hota hai ki kitni heat aur work shaamil hain. Kuch standard processes zyaadatar situations cover kar lete hain. Lekin pehle, is par ek condition ki cheezein kitni tezi se hone di jaayein.

11.8.1 Quasi-static process

Socho ek room mein chupchaap rakhi gas jiske piston par ret ka load hai. Saari ret ek saath gira do (Fig. 11.11) aur piston uchhal kar upar jaata hai. Piston ke paas wali gas patli ho jaati hai isse pehle ki neeche wali gas tak 'khabar pahunche', isliye ek pal ke liye koi ek pressure ya temperature hota hi nahi. Thandi gas ko achanak garam plate se lagane par bhi woh isi tarah bikhar jaati hai. Aisi states ko analyse karna mushkil hai.

Iska ilaaj hai sabr. Ret ka ek-ek daana hatao aur har daane ke baad ruko: piston dheere-dheere upar sarakta hai, aur har pal gas uniform rehti hai, uska pressure upar ke load se bas ek daane ke weight jitna alag. Temperature bhi isi tarah dheere se badla ja sakta hai, gas ko reservoirs ki ek lambi line se garam karke, jismein har reservoir pichhle se zara sa garam ho. Infinitely chhote steps ki limit mein gas sirf equilibrium states se guzarti hai. Aise idealised process ko quasi-static (स्थैतिककल्प) kehte hain, matlab 'lagbhag static'.

3/3Ek-ek daana hatao: gas equilibrium states ki ek seedhi (staircase) se guzarti hai jo ek smooth curve ke saath chalti hai.

Fig. 11.11Sirf woh process jo itna slow ho ki equilibrium mein rahe,

P

–

V

diagram par curve ki tarah banaya ja sakta hai.NCERT Fig. 11.7 · p. 234

Left mein gas ka ek cylinder jiske piston par ret ka dher hai; right mein ek chhota pressure aur volume graph jismein states A aur B hain. Step 1: dher piston ko neeche dabata hai aur sirf state A marked hai. Step 2: poora dher hat gaya hai, piston upar uchhal gaya hai, ghoomte nishan gas ki halchal dikhate hain, aur graph par question mark wali ek dotted line A ko B se jodti hai. Step 3: ek-ek daana hatane ke baad thodi si ret bachi hai, aur graph A se B tak ek baarik staircase dikhata hai jo smooth curve ke saath chalti hai.

Koi bhi real process infinitely slow nahi hota, isliye yeh ek idealisation hai, bilkul mechanics ki frictionless surfaces ki tarah. Phir bhi dheere chalta piston aur sirf chhote temperature differences paar karti heat iske bahut kareeb aate hain, aur aage sab kuch yahi ideal maan kar chalta hai jab tak hum kuch aur na kahein.

Chaar special processes physics aur engineering mein baar-baar aate hain (Table 11.4). Isothermal process (समतापीय प्रक्रम) mein temperature fixed rakha jaata hai, jaise gas ki patli steel bottle ko paani ke bade tank mein dubo kar: paani itna zyada hai ki gas ke saath badli gayi heat uska temperature lagbhag nahi badalti. Isobaric process mein pressure fixed hota hai, aur isochoric process mein volume fixed hota hai. Adiabatic process (रुद्धोष्म प्रक्रम) mein system insulated hota hai, isliye koi heat na andar aati hai na bahar jaati hai. Table is section mein aage derive kiye gaye results ko ek jagah rakhti hai.

Table 11.4Ideal gas ke liye chaar special processesNCERT Table 11.2 · p. 234

Process	Kya fixed hai	$Δ U$	$W$ (gas dwara)	$Q$
Isothermal	$T$	$0$	$μ R T \ln (V_{2} / V_{1})$	$Q = W$
Adiabatic	koi heat flow nahi	$- W$	$μ R (T_{1} - T_{2}) / (γ - 1)$	$0$
Isobaric	$P$	$μ C_{v} Δ T$	$P Δ V = μ R Δ T$	$μ C_{p} Δ T$
Isochoric	$V$	$μ C_{v} Δ T$	$0$	$μ C_{v} Δ T$

11.8.2 Isothermal process

Constant temperature par ideal gas equation $P V = constant$ ban jaati hai: volume aadha karo to pressure double ho jaata hai. Yeh Boyle's law hai, jise ab ek khaas thermodynamic process ki tarah dekha ja raha hai.

DerivationIsothermal expansion mein kiya gaya work

Maan lo temperature $T$ par ideal gas ke $μ$ moles quasi-statically $(P_{1}, V_{1})$ se $(P_{2}, V_{2})$ tak jaate hain. Ek bahut chhote volume change $d V$ par pressure lagbhag constant hota hai, isliye gas work $d W = P d V$ karti hai. In sab tukdon ko jodne ka matlab integrate karna hai:

\begin{aligned} W & = \int_{V_{1}}^{V_{2}} P d V \\ = μ R T \int_{V_{1}}^{V_{2}} \frac{d V}{V} \end{aligned}

Yahan humne $P = μ R T / V$ use kiya aur constants ko integral ke bahar le liya. $1 / V$ ka integral $\ln V$ hai, isliye

Isothermal work

W = μ R T \ln \frac{V_{2}}{V_{1}}

(11.12)

Ideal gas ke liye $U$ sirf $T$ par depend karta hai, isliye $Δ U = 0$ aur first law se $Q = W$ milta hai. Expansion ( $V_{2} > V_{1}$ ) mein $W > 0$ : gas bath se heat leti hai aur poori ki poori work ke roop mein aage de deti hai. Compression mein $W < 0$ : gas par kiya gaya work heat ban kar bath mein chala jaata hai.

Example 11.4

Hawa ko dheere se dabana

Ek badi glass syringe mein $27^{\circ} C$ par $0.020 m o l$ hawa hai. Use ek bade water bath mein rakh kar uske plunger ko $500 m L$ se $200 m L$ tak bahut dheere andar dhakela jaata hai. Hawa ko ideal gas maan kar nikalo (a) hawa dwara kiya gaya work, (b) bath ke saath exchange hui heat aur (c) final pressure.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

11.8.3 Adiabatic process

Adiabatic process mein koi heat boundary paar nahi karti, isliye first law $Δ U = - Δ W$ ban jaata hai. Jab gas work karti hai to poora kharch apni internal energy se karti hai, aur ideal gas thandi ho jaati hai; jab uspar work hota hai, to woh garam hoti hai. Ek chhota calculus argument, jo is chapter ke bahar hai, dikhata hai ki quasi-static adiabatic change mein ideal gas yeh follow karti hai

Adiabatic process

P V^{γ} = constant

(11.13)

jahan

$γ = C_{p} / C_{v}$: specific heats ka ratio (molar ya per unit mass)

Kyunki $C_{p} > C_{v}$ , $γ$ 1 se bada hai: helium jaisi monatomic gas ke liye lagbhag $1.67$ aur hawa ke liye $1.40$ . Same adiabat par kinhi bhi do states ke beech,

P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}

(11.14)

$P = μ R T / V$ rakhne par ek equivalent form milta hai, $T V^{γ - 1} = constant$ , jo hamein Carnot engine ke liye chahiye hoga. Kyunki $γ > 1$ , adiabat par volume ke saath pressure isotherm se zyada tezi se girta hai, isliye kisi bhi point se guzarne wala adiabat zyada steep curve hota hai (Fig. 11.12). Adiabatically expand hone wali gas thandi hoti hai aur ek neeche wale isotherm par pahunchti hai.

3/3C ek thande isotherm (dotted) par hai: gas ne apne work ka kharch $U$ se uthaya.

Fig. 11.12Same starting point se adiabat isotherm se zyada steep girta hai (yahan

γ = 5 / 3

ke liye banaya gaya).NCERT Fig. 11.8 · p. 235

Pressure aur volume ka ek graph jismein starting state A high pressure par hai. Step 1: ek solid blue isotherm, label T1, A se neeche chaar guna volume wali state B tak jaata hai. Step 2: A se ek dashed vermillion adiabat zyada steep gir kar same final volume par state C tak pahunchta hai, jo B se neeche hai. Step 3: ek dotted thanda isotherm, label T2 less than T1, C se guzarta hai.

DerivationAdiabatic change mein kiya gaya work

Ek ideal gas ko adiabatically $(P_{1}, V_{1}, T_{1})$ se $(P_{2}, V_{2}, T_{2})$ tak le jao. Raaste mein $P = K / V^{γ}$ , jahan $K$ Eq. (11.13) ka constant hai, isliye

\begin{aligned} W & = \int_{V_{1}}^{V_{2}} P d V \\ = K \int_{V_{1}}^{V_{2}} \frac{d V}{V^{γ}} \\ = \frac{K}{1 - γ} [V^{1 - γ}]_{V_{1}}^{V_{2}} \end{aligned}

(11.15)

Dono limits par evaluate karo, $V_{2}$ par $K = P_{2} V_{2}^{γ}$ aur $V_{1}$ par $K = P_{1} V_{1}^{γ}$ likh kar, taaki $K V_{2}^{1 - γ} = P_{2} V_{2}$ aur $K V_{1}^{1 - γ} = P_{1} V_{1}$ :

Adiabatic work

\begin{aligned} W & = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} \\ = \frac{μ R (T_{1} - T_{2})}{γ - 1} \end{aligned}

(11.16)

Aakhri step mein $P V = μ R T$ use hota hai. Agar gas work karti hai ( $W > 0$ ), to $T_{2} < T_{1}$ aur woh thandi hoti hai; agar uspar work hota hai ( $W < 0$ ), to $T_{2} > T_{1}$ aur woh garam hoti hai. Yeh is result se match karta hai: $W = - Δ U = μ C_{v} (T_{1} - T_{2}),$ kyunki $C_{v} = R / (γ - 1)$ , jo $C_{p} - C_{v} = R$ aur $γ = C_{p} / C_{v}$ se nikalta hai.

Example 11.5

Diesel engine ko spark plug ki zaroorat kyun nahi

Diesel engine mein $27^{\circ} C$ aur $1.0 \times 10^{5} P a$ wali hawa ko itni tezi se uske volume ke solahve hisse tak compress kiya jaata hai ki mushkil se koi heat bahar nikalti hai. Compression ko $γ = 1.40$ ke saath quasi-static aur adiabatic maano. Nikalo (a) final pressure, (b) final temperature aur (c) ek mole hawa par kiya gaya work.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Try karke dekho

Adiabatic heating mehsoos karo

Tumhe chahiye

ek cycle pump

Pehle guess karo

Nozzle band karke tum handle ko kai baar tezi se andar dhakelte ho. Nozzle ke paas barrel ka kya hota hai?

woh thanda ho jaata haikuch mehsoos nahi hotawoh garam ho jaata haiwoh geela ho jaata hai

Cycle pump ke nozzle ko apne angoothe se kas kar band karo.
Handle ko tezi se andar dhakelo aur wapas uchhalne do, das baar.
Nozzle ke paas barrel ko chhoo kar dekho, phir handle ke paas.
Nozzle khula rakh kar dohrao aur compare karo.

Table 11.5Tumne kya mehsoos kiya

Trial	Prediction	Barrel zyada garam?
Nozzle band
Nozzle khula

Tumhari entries apne aap save hoti hain.

Nozzle band hone par nozzle ke paas barrel saaf taur par garam hota hai. Har tez stroke band hawa ko heat ke leak hone se pehle compress kar deta hai, isliye tumhara work hawa ki internal energy mein jaata hai, lagbhag ek adiabatic compression, aur phir garam hawa metal ko garam karti hai. Nozzle khula hone par zyaadatar hawa dabne ke bajaye bahar nikal jaati hai, aur barrel bahut kam garam hota hai. Dono trials mein thodi garmi plunger seal aur barrel ke beech friction se bhi aati hai.

Quick check

3 mein se 1

Do identical gas samples same state se shuru hote hain. Ek isothermally apne double volume tak expand hota hai, doosra adiabatically double volume tak. Kaun kam pressure par khatam hota hai?

adiabatic walaisothermal waladono same pressure par khatam hote hainyeh gas par depend karta hai

Answer dekho

Answer (A) adiabatic wala

Isothermally pressure aadha ho jaata hai. Adiabatically yeh

2^{γ}

factor se girta hai, jo 2 se zyada hai kyunki gas thandi bhi hoti hai. 'Yeh gas par depend karta hai' chunne ka mann karta hai kyunki

γ

badalta hai, lekin har gas ke liye

2^{γ} > 2

hai.

11.8.4 Isochoric process

Volume lock kar do to gas kuch bhi dhakel nahi sakti, isliye $Δ W = 0$ . Tab heat ka har joule $U$ mein jaata hai: $Δ Q = Δ U = μ C_{v} Δ T$ . $C_{v}$ ko theek isi tarah define kiya gaya tha, isliye $C_{v}$ tay karta hai ki gas kitni garam hogi. Dhoop mein chhode gaye band steel tiffin box ki hawa lagbhag isi tarah garam hoti hai, aur uska pressure uske kelvin temperature ke proportion mein badhta hai.

11.8.5 Isobaric process

Ab pressure fixed rakho aur gas ko garam hote hue expand hone do. Woh work karti hai

\begin{aligned} W & = P (V_{2} - V_{1}) \\ = μ R (T_{2} - T_{1}) \end{aligned}

(11.17)

jabki uski internal energy $μ C_{v} Δ T$ badhti hai. Heat ko dono bill bharne hain, kul $μ (C_{v} + R) Δ T = μ C_{p} Δ T$ , isliye constant pressure par $C_{p}$ temperature ki badhotri tay karta hai (Example 11.3). Khuli khidki wale room mein garam hoti hawa atmospheric pressure par rehti hai, aur khule pan se nikalti steam bhi.

11.8.6 Cyclic process

Engines khud ko dohraate hain: har cycle ke baad working gas theek wahin wapas aa jaati hai jahan se shuru hui thi. Kyunki $U$ sirf state par depend karta hai, poori cycle mein $U$ mein koi change nahi hota, aur tab first law kehta hai ki andar li gayi net heat diye gaye net work ke barabar honi chahiye. $P$ – $V$ diagram par cycle ek closed loop hoti hai. Expand hone wale leg par work uske neeche ka area hai; compress hone wale leg par, us leg ke neeche ka area negative ginta hai. Net work loop se ghira hua area hai, jo loop clockwise chalne par positive hota hai, jaise Fig. 11.7 ke jaane-aane wale route mein. Har engine jise tum jaante ho, motorcycle se le kar thermal power station tak, ek cycle mein chalta hai.

Example 11.6

Ek rectangular cycle

Engine cylinder ki gas ko $P$ – $V$ diagram par ek rectangle ke chakkar lagwaye jaate hain. Woh $2.0 \times 10^{5} P a$ par $1.0 L$ se $3.0 L$ tak expand hoti hai; constant volume par uska pressure $1.0 \times 10^{5} P a$ tak kam kiya jaata hai; usi pressure par use wapas $1.0 L$ tak compress kiya jaata hai; aur constant volume par uska pressure badha kar shuruaat tak laaya jaata hai. Nikalo (a) har cycle mein gas dwara kiya gaya net work, (b) uske dwara absorb ki gayi net heat aur (c) 25 cycles per second par uska power output.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Neeche diye lab mein chaaron processes try karo. Process koi bhi ho aur gas koi bhi chuno, teeno bars hamesha balance karte hain: $Q = Δ U + W$ .

InteractiveProcess lab

Ideal gas ka ek mole 20 litres aur 300 kelvin se shuru hota hai. Buttons se isothermal, adiabatic, isobaric ya isochoric process aur monatomic ya diatomic gas chuni jaati hai; ek slider final volume set karta hai, ya isochoric case mein final temperature. Play state ko pressure aur volume graph par uske path par chalata hai aur work ko shade karta hai, jabki ek cylinder piston aur gas ka temperature dikhata hai. Bars Q, U mein change aur W dikhate hain, jo hamesha Q barabar delta U plus W ko satisfy karte hain.

Fig. 11.13Ek process aur ek gas chuno, final volume (ya temperature) set karo aur Play dabao. Shaded area work hai; bars first law check karte hain.

Quick check

3 mein se 1

Har process ko us cheez se match karo jo fixed rehti hai.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

isothermal : temperature
isobaric : pressure
isochoric : volume
adiabatic : koi heat exchange nahi

Greek roots madad karte hain: iso = same, therm = heat/temperature, baros = weight/pressure, choros = space. Adiabatic ka matlab hai heat ke liye 'paar na kiya ja sakne wala'.

11.9 Second law of thermodynamics

First law energy conservation hai, aur yeh kisi bhi aisi machine ko rule out karta hai jo kuch nahi se energy bana de. Phir bhi kai aise events jo energy ko bilkul conserve karte, kabhi dikhte hi nahi. Ek cyclist dhalaan par aadhe raaste brake laga kar rukta hai, aur brake pads aur rims garam ho jaate hain. First law ko ulte case se koi aitraaz nahi: rims apne aap thande ho sakte the aur woh energy cycle ko wapas dhalaan par chadhane mein laga sakte the. Aisa kabhi nahi hota. Energy conservation ke alawa koi aur principle ise rokta hoga, aur woh principle hai second law of thermodynamics.

Second law sabse saaf un devices mein dikhta hai jo heat ko work mein badalte hain. Ek heat engine cycle mein chalta hai: woh hot reservoir se heat $Q_{1}$ leta hai, work $W$ deta hai aur cold reservoir ko heat $Q_{2}$ reject karta hai. Uski efficiency hai

Heat engine ki efficiency

η = \frac{W}{Q_{1}} = 1 - \frac{Q_{2}}{Q_{1}}

Refrigerator ulta chalta hai. Uspar work $W$ kiya jaata hai taaki woh thandi jagah se heat $Q_{2}$ kheenche, aur woh $Q_{1} = Q_{2} + W$ garam surroundings mein daal deta hai. Uska coefficient of performance hai

Coefficient of performance

α = \frac{Q_{2}}{W}

Second law kehta hai ki $η$ kabhi 1 tak nahi pahunch sakta aur $α$ kabhi infinite nahi ho sakta. Do classic statements ise precisely kehte hain (Fig. 11.14).

LawSecond law: Kelvin–Planck statement

Cycle mein kaam karne wale kisi device ke liye yeh impossible hai ki woh ek hi reservoir se heat le aur use poora work mein badal de, bina kisi aur effect ke.

LawSecond law: Clausius statement

Heat ka thandi body se garam body tak bina kisi aur effect ke, jaise bahar se diya gaya work, jaana impossible hai.

1/4Ek real engine: $Q_{1}$ andar, work $W$ bahar, aur kuch heat $Q_{2}$ hamesha cold reservoir ko reject hoti hai.

Fig. 11.14Kelvin–Planck perfect engine ko mana karta hai; Clausius perfect refrigerator ko mana karta hai.

Upar hot reservoir, neeche cold reservoir aur beech mein ek device. Step 1, real engine, allowed: heat Q1 neeche engine mein aati hai, work W right ki taraf nikalta hai aur heat Q2 neeche cold reservoir mein jaati hai. Step 2, perfect engine: heat Q andar aati hai aur poori work ban kar nikalti hai, cold reservoir ko kuch nahi; ek bada cross ise forbidden dikhata hai. Step 3, refrigerator, allowed: heat Q2 cold reservoir se upar uthti hai, work W right se andar aata hai aur heat Q1 hot reservoir tak jaati hai. Step 4, perfect refrigerator: heat Q bina work ke seedhe thande se garam tak jaati hai; ek bada cross ise forbidden dikhata hai.

Dono statements alag dikhte hain, ek engines ke baare mein aur ek refrigerators ke baare mein, phir bhi har ek doosre ko imply karta hai: ek perfect refrigerator ko ordinary engine ke saath jodo to perfect engine ban jaayega, aur ek perfect engine ordinary refrigerator ko chala kar use perfect bana dega. Yeh ek hi law ke do chehre hain. 'Bina kisi aur effect ke' shabdon par dhyaan do. Heat ko poori tarah work mein badalna allowed hai agar saath mein kuch aur bhi badle. Isothermally expand hoti gas absorb ki gayi saari heat ko work mein badal deti hai, lekin aakhir mein uska volume bada ho jaata hai, isliye woh process cycle nahi hai.

Quick check

3 mein se 1

Kaun sa proposed device Kelvin–Planck statement ko todega?

ek electric heater jo apni saari electrical energy ko heat mein badal deta haielectricity se chalne wala refrigeratorek ship engine jo samudra se heat leta hai aur poori ko work mein badal deta haiek car engine jo garam exhaust gas reject karta hai

Answer dekho

Answer (C) ek ship engine jo samudra se heat leta hai aur poori ko work mein badal deta hai

Ek hi reservoir (samudra) se heat le kar use poora work mein badalna, cycle dar cycle, theek wahi hai jise Kelvin–Planck mana karta hai. Work ko poori tarah heat mein badalna, jaise heater karta hai, bilkul allowed hai.

11.10 Reversible aur irreversible processes

Lagbhag kisi bhi roz ke event ki film banao aur ulta chalao, aur tum turant bata doge ki film ulti chal rahi hai (Fig. 11.15). Ulta version mechanics ka koi law nahi todta aur energy ko bilkul conserve karta, phir bhi aisa kabhi nahi hota. Jo processes apne aap hote hain woh sirf ek taraf chalte hain; physicists unhe irreversible kehte hain. Kuch examples:

Ink ki ek boond paani ke glass mein tab tak failti hai jab tak rang ek jaisa na ho jaaye; ink kabhi khud wapas boond mein ikattha nahi hoti.
Agarbatti ka dhuaan poore room mein fail jaata hai; woh kabhi wapas stick mein nahi samaata.
Stand par rakhi garam iron room temperature tak thandi ho jaati hai; thanda room use kabhi phir se garam nahi karta.
Jali hui maachis ki teeli dhuein aur raakh se kabhi khud ko dobara nahi jodti.
Cycle rokte waqt garam hue brakes use kabhi phir aage nahi dhakelte. Unki extra internal energy ko poori tarah wapas motion mein badalna Kelvin–Planck statement ko tod dega.

2/4Ink apne aap fail jaati hai.

Fig. 11.15Mixing ek hi taraf chalti hai: uski film ulti chalao to impossible lagti hai.

Side se dikhta paani ka ek glass. Step 1: surface ke paas purple ink dots ka ek chhota, ghana jhund. Step 2: dots upar-beech mein ek dheele baadal ki tarah fail gaye hain. Step 3: dots poore paani mein barabar bikhre hain. Step 4: dots wapas upar boond mein jama ho gaye hain, unke upar ek bada cross hai aur label hai film ulti: kabhi nahi dikhti.

Irreversibility ke do main sources hain:

Non-equilibrium states. Free expansion, explosions, mixing aur finite pressure ya temperature difference se chalne wali har cheez system ko equilibrium se bahut door ki states se guzarti hai.
Dissipation. Friction, viscosity aur electrical resistance ordered energy ko internal energy mein badal dete hain: train dheemi hote waqt brake blocks garam ho jaate hain, aur charging cable mein current behne par woh garam hoti hai.

Koi bhi machine friction ya resistance se poori tarah free nahi hoti, isliye practice mein har process kam se kam thoda irreversible hota hai.

Definition

Reversible process

Aisa process jise ulta chalaya ja sake taaki system aur uske surroundings dono apni initial states mein laut aayein, aur kahin aur koi change na ho.

ExampleEk ideal gas jo frictionless piston ke against infinitely slowly, isothermally expand hoti hai (ek idealisation).

Kisi process ke reversible (उत्क्रमणीय) hone ke liye do conditions dono poori honi chahiye: woh kabhi equilibrium na chhode (yaani quasi-static ho), aur friction ya kisi aur dissipation mein kuch bhi na khoye. Socho ek bath mein rakhi ideal gas ek bilkul smooth piston ko infinitely slowly bahar dhakel rahi hai: bahar ka pressure zara sa badhao aur gas har state ko wapas follow karti hai, aur theek utni hi heat lauta deti hai jitni li thi. Aise ideal ki parwaah kyun karein jo kabhi poori tarah hota hi nahi? Kyunki yeh benchmark set karta hai. Second law aise engine ko mana karta hai jo apni saari heat ko work mein badal de, isliye engineers jaanna chahte hain ki do temperatures ke beech sabse achha kya possible hai. Woh best sirf reversible steps se bane engine ka hota hai; har irreversibility use neeche kheenchti hai.

Quick check

2 mein se 1

Inmein se kaun se processes irreversible hain?

paani mein failti inkbrake laga kar rukti cyclefrictionless piston ke against quasi-static isothermal expansionjalti hui maachis ki teeli

Saare sahi options chuno

Answer dekho

Answer

(A) paani mein failti ink
(B) brake laga kar rukti cycle
(D) jalti hui maachis ki teeli

Mixing, friction aur combustion sab irreversible hain. Ideal quasi-static, frictionless expansion reversible process ka textbook example hai.

11.11 Carnot engine

1824 mein ek young French engineer, Sadi Carnot, ne notice kiya ki steam engines zyaadatar trial and error se behtar kiye ja rahe the. Unhone ek zyada tez sawaal poocha: $T_{1}$ par ek hot reservoir aur $T_{2}$ par ek cold reservoir diye hon, to koi bhi engine zyada se zyada kya kar sakta hai, aur kis cycle ke saath? Kamaal ki baat yeh hai ki heat ko caloric samajhte hue bhi woh sahi jawab tak pahunch gaye.

Ideal engine reversible hona chahiye, kyunki irreversibility work barbaad karti hai. Isliye heat ko kabhi bhi finite temperature difference paar nahi karna chahiye. Isse do steps turant fix ho jaate hain: heat $Q_{1}$ ko $T_{1}$ par isothermally lena hoga, aur heat $Q_{2}$ ko $T_{2}$ par isothermally dena hoga. Cycle band karne ke liye working substance ko $T_{1}$ se neeche $T_{2}$ tak aur wapas upar bhi jaana hoga. Koi bhi leg jis par woh heat exchange kare, jaise isochoric leg, ko har beech wale temperature par reservoirs ki ek seedhi chahiye hogi, aur hamare engine ko sirf do reservoirs allowed hain. Isliye in legs mein bilkul bhi heat nahi honi chahiye: yeh reversible adiabatic steps hone chahiye. Do temperatures ke beech kaam karne wala reversible engine Carnot engine hai, aur do adiabats se jude do isotherms wali uski cycle Carnot cycle hai (Fig. 11.16).

1/5Do isotherms (solid) do adiabats (dashed) se jude hain: $1 \to 2 \to 3 \to 4 \to 1$ .

Fig. 11.16Carnot cycle: do adiabats se jude do isotherms (monatomic gas ke liye banaya gaya).NCERT Fig. 11.9 · p. 237

Left mein pressure aur volume ka graph jismein chaar states hain, 1 se 4: 1 se 2 tak T1 par ek solid vermillion isotherm, 2 se 3 tak ek dashed adiabat, 3 se 4 tak T2 par ek solid sky-blue isotherm aur 4 se wapas 1 tak ek dashed adiabat. Right mein piston wala ek cylinder ek base par khada hai. Step 1 poora loop dikhata hai. Step 2 1 se 2 ko highlight karta hai: cylinder hot reservoir par khada hai, Q1 label wale heat arrows andar jaate hain aur piston upar uthta hai. Step 3 2 se 3 ko highlight karta hai: insulating stand par piston aur upar uthta hai. Step 4 3 se 4 ko highlight karta hai: cold reservoir par heat Q2 bahar flow karti hai aur piston neeche aata hai. Step 5 4 se 1 ko highlight karta hai: insulating stand par piston apni shuruaati height par laut aata hai.

DerivationCarnot engine ki efficiency

Working substance ke roop mein ideal gas ke $μ$ moles lo. Step 1 → 2, $T_{1}$ par $V_{1}$ se $V_{2}$ tak isothermal expansion: Eq. (11.12) ke hisaab se gas utna work karti hai jitni heat absorb karti hai,

\begin{aligned} Q_{1} & = W_{1 \to 2} \\ = μ R T_{1} \ln \frac{V_{2}}{V_{1}} \end{aligned}

(11.18)

Step 2 → 3, $T_{1}$ se $T_{2}$ tak adiabatic expansion: Eq. (11.16) ke hisaab se gas work karti hai

\begin{aligned} W_{2 \to 3} \\ = \frac{μ R (T_{1} - T_{2})}{γ - 1} \end{aligned}

(11.19)

Step 3 → 4, $T_{2}$ par $V_{3}$ se $V_{4}$ tak isothermal compression: gas par work $W_{3 \to 4}$ kiya jaata hai, aur utni hi maatra heat $Q_{2}$ ban kar usse bahar jaati hai,

\begin{aligned} Q_{2} & = W_{3 \to 4} \\ = μ R T_{2} \ln \frac{V_{3}}{V_{4}} \end{aligned}

(11.20)

Step 4 → 1, $T_{2}$ se wapas $T_{1}$ tak adiabatic compression: gas par kiya gaya work hai

\begin{aligned} W_{4 \to 1} \\ = \frac{μ R (T_{1} - T_{2})}{γ - 1} \end{aligned}

(11.21)

Ek cycle mein gas dwara kiya gaya net work hai $W_{1 \to 2} + W_{2 \to 3} - W_{3 \to 4} - W_{4 \to 1} .$ Dono adiabatic works barabar hain, isliye cancel ho jaate hain, aur bachta hai

\begin{aligned} W & = W_{1 \to 2} - W_{3 \to 4} \\ = μ R T_{1} \ln \frac{V_{2}}{V_{1}} \\ - μ R T_{2} \ln \frac{V_{3}}{V_{4}} \end{aligned}

(11.22)

aur efficiency hai

\begin{aligned} η & = \frac{W}{Q_{1}} \\ = 1 - \frac{Q_{2}}{Q_{1}} \\ = 1 - \frac{T_{2} \ln \frac{V_{3}}{V_{4}}}{T_{1} \ln \frac{V_{2}}{V_{1}}} \end{aligned}

(11.23)

Adiabats volumes ko jodte hain. Step 2 → 3 par $T V^{γ - 1} = constant$ apply karne par,

\begin{array}{c} T_{1} V_{2}^{γ - 1} = T_{2} V_{3}^{γ - 1} \\ {(\frac{V_{2}}{V_{3}})}^{γ - 1} = \frac{T_{2}}{T_{1}} \end{array}

(11.24)

aur step 4 → 1 par,

\begin{array}{c} T_{2} V_{4}^{γ - 1} = T_{1} V_{1}^{γ - 1} \\ {(\frac{V_{1}}{V_{4}})}^{γ - 1} = \frac{T_{2}}{T_{1}} \end{array}

(11.25)

Right-hand sides barabar hain, isliye $V_{2} / V_{3} = V_{1} / V_{4}$ , jise rearrange karne par milta hai

\frac{V_{3}}{V_{4}} = \frac{V_{2}}{V_{1}}

(11.26)

Isliye Eq. (11.23) ke dono logarithms barabar hain aur cancel ho jaate hain, aur bachta hai

Carnot engine ki efficiency

η = 1 - \frac{T_{2}}{T_{1}}

(11.27)

Efficiency sirf do reservoir temperatures par depend karti hai, jo kelvin mein hone chahiye. Yeh 1 tak tabhi pahunchegi jab $T_{2}$ absolute zero ho, aur jab $T_{1} = T_{2}$ ho to yeh zero hai: engine ko chalne ke liye temperature difference chahiye hi.

Example 11.7

Thermal power station ki limit

Ek coal-fired power station $540^{\circ} C$ par steam banata hai aur use $35^{\circ} C$ ke nadi ke paani se condense karta hai. (a) In temperatures ke liye Carnot efficiency kya hai? (b) Station 38% overall efficiency par $500 M W$ electrical power deta hai. Woh kis rate se heat reject karta hai? (c) Same output wale Carnot engine ke liye yeh rate kya hota?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Kyunki Carnot cycle mein kuch bhi irreversible nahi hai, ise ulta bhi chalaya ja sakta hai, $1 \to 4 \to 3 \to 2 \to 1$ . Tab gas thande end par $Q_{2}$ sokh leti hai, use bahar se work $W$ chahiye hota hai aur garam end par $Q_{1}$ daal deti hai: engine refrigerator ban gaya hai, aur same do temperatures ke beech koi bhi refrigerator isse behtar nahi kar sakta. Asal mein Carnot engine hi akela reversible engine hai jo sirf do reservoirs ke beech kaam kar sakta hai.

3/3Saath mein: source par koi change nahi, aur sink se heat $W^{'} - W$ poori tarah work mein badal gayi. Forbidden.

Fig. 11.17Agar koi engine Carnot engine ko hara de, to yeh jodi ek reservoir ki heat ko poori tarah work mein badal degi.NCERT Fig. 11.10 · p. 239

Upar hot reservoir aur neeche cold reservoir. Step 1: left mein engine I hot reservoir se heat Q1 leta hai, work W prime right ki taraf bhejta hai aur Q1 minus W prime cold reservoir ko reject karta hai. Step 2: right mein engine R, ulta chalta hua, cold reservoir se Q1 minus W leta hai aur hot reservoir ko Q1 lautata hai, I ke work W se chalte hue; bacha hua W prime minus W upar point karta hai. Step 3: ek dashed box dono ko gherta hai; sirf do flows hain, cold reservoir se box mein heat W prime minus W aur utna hi work bahar, aur hot reservoir ka label kehta hai ki koi net change nahi.

TheoremCarnot's theorem

(a) Do diye gaye reservoirs ke beech kaam karne wala koi bhi engine, unhi reservoirs ke beech kaam karne wale Carnot engine se zyada efficient nahi ho sakta. (b) Carnot engine ki efficiency working substance par depend nahi karti.

Maan lo koi engine I, shayad irreversible, $T_{1}$ wale same source aur $T_{2}$ wale sink ke beech kaam karne wale Carnot engine R se zyada efficient hai. I ko seedha chalao: woh source se heat $Q_{1}$ leta hai, work $W^{'}$ deta hai aur sink ko $Q_{1} - W^{'}$ reject karta hai (Fig. 11.17).

Ab R ko refrigerator ki tarah ulta chalao, is size ka ki woh source ko theek $Q_{1}$ lautaye. Engine ki tarah $Q_{1}$ lene par woh work $W$ deta; ulta chalne par use utna hi work $W$ chahiye aur woh sink se $Q_{1} - W$ leta hai. Agar I zyada efficient hai, to $W^{'} > W$ , isliye I R ko chala sakta hai aur phir bhi uske paas $W^{'} - W$ bachega.

Is jodi ko ek machine maano. Source I ko $Q_{1}$ deta hai aur R se $Q_{1}$ wapas paata hai: koi net change nahi. Sink $Q_{1} - W^{'}$ paata hai lekin $Q_{1} - W$ deta hai, yaani net loss $W^{'} - W$ . Yeh combination bas itna karta hai ki ek hi reservoir se heat $W^{'} - W$ le kar use poora work mein badal deta hai, jise Kelvin–Planck statement mana karta hai. Isliye koi bhi engine R ko hara nahi sakta. Part (b) bhi isi tarah nikalta hai: agar ek gas use karne wala Carnot engine doosri gas wale Carnot engine ko hara de, to dono ko jodne se phir second law tootega.

Kyunki $T_{1}$ aur $T_{2}$ ke beech har Carnot engine ki efficiency same hai, working substance kuch bhi ho, $1 - Q_{2} / Q_{1} = 1 - T_{2} / T_{1}$ . Isliye

Carnot cycle

\frac{Q_{2}}{Q_{1}} = \frac{T_{2}}{T_{1}}

(11.28)

Yahan $Q_{1}$ aur $Q_{2}$ hot aur cold reservoirs ke saath isothermally exchange hui heats hain. Kyunki Eq. (11.28) mein kisi substance ka zikr hi nahi hai, yeh temperature scale ki definition ka kaam kar sakti hai: do temperatures ka ratio un heats ka ratio hai jo Carnot engine un temperatures par exchange karta hai. Yeh thermodynamic temperature scale mercury, gas ya kisi aur thermometric material ki properties par depend nahi karta, aur ideal gas ke liye yeh ideal gas temperature se match karta hai.

Example 11.8

Ek ideal refrigerator

Ek refrigerator $35^{\circ} C$ wali kitchen mein apne andar ka temperature $5^{\circ} C$ rakhta hai. Maan lo woh ulte Carnot engine ki tarah kaam karta. (a) Uska coefficient of performance kya hota? (b) Khaane se $1.0 k J$ heat hatane ke liye use kitna work chahiye hota, aur woh kitchen mein kitni heat chhodta?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Khud Carnot engine explore karo. Gas aur expansion ratio badlo aur dekho ki $Q_{1}$ , $Q_{2}$ aur $W$ kaise badalte hain, jabki efficiency $1 - T_{2} / T_{1}$ par tiki rehti hai.

InteractiveCarnot engine lab

Ideal gas ka ek mole 10 litres se shuru ho kar Carnot cycle chalata hai. Sliders hot reservoir ka temperature 400 se 800 kelvin, cold reservoir ka temperature 250 se 390 kelvin aur isothermal expansion ratio set karte hain; buttons se monatomic ya diatomic gas chuni jaati hai. Play state ko pressure aur volume graph par loop ke chakkar lagwata hai jabki ek cylinder hot reservoir, insulating stand ya cold reservoir par baithta hai. Cards Q1, Q2, work, efficiency aur one minus T2 over T1 dikhate hain, jo hamesha match karte hain.

Fig. 11.18Do reservoir temperatures, expansion ratio aur gas set karo, phir Play one cycle dabao.

Quick check

4 mein se 1

527^{\circ} C

aur

127^{\circ} C

ke beech kaam karne wale Carnot engine ki efficiency percentage mein kya hai?

Answer dekho

Answer 50 %

T_{1} = 800 K

T_{2} = 400 K

, isliye

η = 1 - 400 / 800 = 0.50.

Celsius values use karna (

1 - 127 / 527 = 0.76

) classic galti hai.

Recap

Summary aur key points

Thermal equilibrium mein bulk variables badalna band kar dete hain. Zeroth law (do systems jo har ek kisi teesre ke saath equilibrium mein hain, woh aapas mein bhi equilibrium mein hain) temperature ko define karta hai (Section 11.3).
Internal energy $U$ centre-of-mass frame mein molecular kinetic aur potential energy hai, aur yeh ek state variable hai. Heat, jo temperature difference se chalti hai, aur work transit mein energy hain (Section 11.4).
First law: $Δ Q = Δ U + Δ W$ , jahan $Δ Q$ system ko di gayi heat hai aur $Δ W$ uske dwara kiya gaya work; constant pressure par $Δ W = P Δ V$ (Section 11.5).
$s = Δ Q / (m Δ T)$ aur $C = Δ Q / (μ Δ T)$ . Equipartition solids ke liye $C = 3 R$ deta hai, jo ordinary temperatures par zyaadatar solids par fit hota hai (Section 11.6).
Calorie ko ${14.5}^{\circ} C$ aur ${15.5}^{\circ} C$ ke beech ki degree tak garam kiye gaye paani se define kiya jaata hai; $1 c a l = 4.186 J$ (Paani ki specific heat capacity).
Ideal gas ke liye $C_{p} - C_{v} = R$ (Ideal gas ki specific heats).
State variables equilibrium states ko describe karte hain aur $P V = μ R T$ jaise equation of state se jude hote hain. System ko aadha karne par intensive variables same rehte hain; extensive variables aadhe ho jaate hain (Section 11.7).
Quasi-static process infinitely slow hota hai aur sirf equilibrium states se guzarta hai (Section 11.8.1).
Isothermal ideal gas: $Q = W = μ R T \ln (V_{2} / V_{1})$ (Section 11.8.2). Adiabatic: $P V^{γ}$ constant aur $W = μ R (T_{1} - T_{2}) / (γ - 1)$ (Section 11.8.3).
Isochoric: $W = 0$ . Isobaric: $W = P Δ V = μ R Δ T$ . Cyclic: $Δ U = 0$ , aur net work loop ka area hai (Section 11.8.6).
Second law: koi bhi engine apni saari heat ko work mein nahi badal sakta (Kelvin–Planck), aur heat apne aap thande se garam ki taraf flow nahi kar sakti (Clausius); isliye $η < 1$ aur $α$ finite hai (Section 11.9).
Spontaneous processes irreversible hote hain. Reversible process quasi-static aur dissipation se free hona chahiye (Section 11.10).
Carnot engine (do isotherms, do adiabats) ki $η = 1 - T_{2} / T_{1}$ hoti hai, jo kisi bhi working substance ke liye do temperatures ke beech sabse zyada possible hai; saath hi $Q_{2} / Q_{1} = T_{2} / T_{1}$ (Section 11.11).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 11.1	Practice Q1
NCERT Exercise 11.2	Practice Q2
NCERT Exercise 11.3	Practice Q3
NCERT Exercise 11.4	Practice Q4
NCERT Exercise 11.5	Practice Q5
NCERT Exercise 11.6	Practice Q6
NCERT Exercise 11.7	Practice Q7
NCERT Exercise 11.8	Practice Q8

Quick check: MCQs

11.1 Introduction2 questions
11.2 Thermal equilibrium2 questions
11.3 Zeroth law of thermodynamics3 questions
11.4 Heat, internal energy aur work3 questions
11.5 First law of thermodynamics3 questions
Ideal gas ki specific heats3 questions
11.7 Thermodynamic state variables aur equation of state3 questions
11.8.3 Adiabatic process3 questions
11.8.6 Cyclic process3 questions
11.9 Second law of thermodynamics3 questions
11.10 Reversible aur irreversible processes2 questions
11.11 Carnot engine4 questions

Glossary · 15

Key terms (English + Hindi)

adiabatic processरुद्धोष्म प्रक्रमruddhoshm prakram: Bina heat exchange wala process; quasi-static ideal gas ke liye $P V^{γ}$ constant rehta hai.Chapter mein pehli baar yahan aaya →

Carnot engineकार्नो इंजनkaarno injan: Do reservoirs ke beech do isothermal aur do adiabatic steps se kaam karne wala reversible engine; iski efficiency $1 - T_{2} / T_{1}$ hai.Chapter mein pehli baar yahan aaya →

diathermic wall: Aisi wall jo heat ko jaane deti hai, jaise copper ki patli sheet. Heat ko rokne wali wall ko adiabatic kehte hain.Chapter mein pehli baar yahan aaya →

first law of thermodynamics: System ke liye energy conservation: di gayi heat internal energy mein badhotri plus system dwara kiye gaye work ke barabar hai, $Δ Q = Δ U + Δ W$ .Chapter mein pehli baar yahan aaya →

heatऊष्माushma: Woh energy jo temperature difference ki wajah se system aur uske surroundings ke beech flow karti hai.Chapter mein pehli baar yahan aaya →

internal energyआंतरिक ऊर्जाaantarik oorja: Kisi system ke molecules ki kul kinetic aur potential energy, us frame mein maapi gayi jismein poora system rest par ho; symbol $U$ .Chapter mein pehli baar yahan aaya →
isothermal processसमतापीय प्रक्रमsamtaapiya prakram: Constant temperature par hone wala process; ideal gas ke liye $P V$ constant rehta hai aur $Δ U = 0$ .Chapter mein pehli baar yahan aaya →

molar specific heat capacity: Har mole ka temperature ek kelvin badhane ke liye zaroori heat, $C = Δ Q / (μ Δ T)$ , $J m o l^{- 1} K^{- 1}$ mein.Chapter mein pehli baar yahan aaya →

quasi-static processस्थैतिककल्प प्रक्रमsthaitikkalp prakram: Ek idealised, infinitely slow process jismein system sirf equilibrium states se guzarta hai.Chapter mein pehli baar yahan aaya →

reversible processउत्क्रमणीय प्रक्रमutkramaniya prakram: Aisa process jise ulta chalaya ja sake taaki system aur surroundings dono apni shuruaati states mein laut aayein aur kahin aur koi change na ho.Chapter mein pehli baar yahan aaya →

specific heat capacityविशिष्ट ऊष्मा धारिताvishisht ushma dhaarita: Har unit mass ka temperature ek kelvin badhane ke liye zaroori heat, $s = Δ Q / (m Δ T)$ , $J k g^{- 1} K^{- 1}$ mein.Chapter mein pehli baar yahan aaya →
state variableअवस्था चरavastha char: $P$ , $V$ , $T$ ya $U$ jaisi quantity jiski value sirf equilibrium state se tay hoti hai, is baat se nahi ki state tak kaise pahunche.Chapter mein pehli baar yahan aaya →

thermal equilibriumतापीय साम्यtaapiya saamya: Thermal contact mein rakhe do systems ki woh state jab unke beech koi net heat flow nahi hota, isliye unke macroscopic variables badalna band kar dete hain.Chapter mein pehli baar yahan aaya →
thermodynamicsऊष्मागतिकीushmagatiki: Physics ki woh branch jo kuch maapne laayak bulk variables ki madad se heat, work, temperature aur energy ko aapas mein jodti hai.Chapter mein pehli baar yahan aaya →

zeroth law of thermodynamics: Agar do systems mein se har ek kisi teesre ke saath thermal equilibrium mein hai, to woh aapas mein bhi thermal equilibrium mein hain.Chapter mein pehli baar yahan aaya →

Thermodynamics: Class 11 Physics Chapter 11 Hinglish mein

Is chapter mein aap kya seekhenge

11.1 Introduction

11.2 Thermal equilibrium

11.3 Zeroth law of thermodynamics

11.4 Heat, internal energy aur work

11.5 First law of thermodynamics

11.6 Specific heat capacity

Paani ki specific heat capacity

Ideal gas ki specific heats

11.7 Thermodynamic state variables aur equation of state

11.8 Thermodynamic processes

11.8.1 Quasi-static process

11.8.2 Isothermal process

11.8.3 Adiabatic process

11.8.4 Isochoric process

11.8.5 Isobaric process

11.8.6 Cyclic process

11.9 Second law of thermodynamics

11.10 Reversible aur irreversible processes

11.11 Carnot engine

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye