Carnot cycle — full derivation, efficiency = 1 − T_C - T_H: Question bank

1.7.21 · D5 · Physics › Thermodynamics › Carnot cycle — full derivation, efficiency = 1 − T_C - T_H

Neeche use hone wale symbols ki reminders (sab parent note mein build kiye gaye hain, yahan restate kiye gaye hain taaki ye page akela bhi kaam kare):

$T_{H}, T_{C}$ — hot aur cold reservoir temperatures, kelvin mein (kabhi Celsius nahi).
$Q_{H}$ — hot side se absorb ki gayi heat (ek positive number); $Q_{C}$ — cold side ko reject ki gayi heat (ye bhi positive number ke roop mein stored hai).
$W$ — net work out per cycle; $η = W / Q_{H} = 1 - T_{C} / T_{H}$ — the efficiency.
$V_{1} \to V_{2}$ — hot isothermal ke dauran volumes; $V_{3} \to V_{4}$ — cold isothermal ke dauran volumes (neeche figures dekho).

Jis cycle ke baare mein tum reason kar rahe ho

Traps se pehle, actual loop dekho. Pehla figure char strokes ka $P$ – $V$ diagram hai; neeche har sawaal isi picture ki ek leg ko refer karta hai.

Figure — Carnot cycle — full derivation, efficiency = 1 − T_C - T_H

Net work per cycle woh area hai jo loop ke andar enclosed hai. Jis direction mein tum travel karte ho woh matter karta hai, isliye agla figure sign convention dikhata hai: ek clockwise loop ek engine hai (work out), ek counterclockwise loop ek refrigerator/heat pump hai (work in).

True ya false — justify karo

Carnot engine ki efficiency is baat par depend karti hai ki tum kaunsi gas use karte ho

False — adiabatic links force karte hain ki

V_{2} / V_{1} = V_{3} / V_{4}

, isliye volume logs cancel ho jaate hain aur sirf ratio

T_{C} / T_{H}

bachta hai, working substance se independent.

Dono

T_{H}

aur

T_{C}

ko double karne se efficiency unchanged rehti hai

True —

η = 1 - T_{C} / T_{H}

sirf ratio par depend karta hai, aur dono ko double karne se

2 T_{C} /2 T_{H} = T_{C} / T_{H}

same rehta hai.

Carnot engine

η = 1

reach kar sakta hai agar

T_{H}

kaafi bada kar diya jaaye

False —

η = 1 - T_{C} / T_{H}

sirf

T_{H} \to \infty

par

1

ke paas aata hai, kabhi equal nahi hota, aur

T_{C} > 0

hamesha ek nonzero fraction reject karne ke liye chodta hai.

Ek full cycle mein internal energy change

Δ U

zero hota hai

True — internal energy ek state function hai aur gas apne exact starting state mein wapas aa jaati hai, isliye

Δ U = 0

regardless of the path taken.

Heat sirf isothermal expansion ke dauran absorb hoti hai, kahi aur nahi

True — dono adiabatics by definition koi heat exchange nahi karte, aur cold isothermal heat reject karta hai, isliye absorption sirf step 1 mein

T_{H}

par hoti hai.

Carnot engine ko backwards chalane se woh ideal refrigerator ban jaata hai

True — kyunki har step reversible hai, cycle ko reverse karna (counterclockwise loop) work

W

use karke heat

Q_{C}

ko cold se hot mein pump karta hai, jo ek Carnot refrigerator ki definition hai.

Ek Carnot cycle mein universe ka total entropy change positive hota hai

False — ek Carnot cycle fully reversible hai, isliye

\oint d Q_{r e v} / T = 0

aur gas + reservoirs + surroundings ka total entropy change exactly zero hota hai.

Error dhundo

"Kyunki

η = 1 - T_{C} / T_{H}

jahan

T_{C} = 2 7^{\circ}

C aur

T_{H} = 22 7^{\circ}

η = 1 - 27/227 = 0.88

Error yeh hai ki Celsius ko ek ratio mein use kiya gaya; pehle convert karo —

T_{C} = 300

T_{H} = 500

K, jo deta hai

η = 1 - 300/500 = 0.40

0.88

se bahut alag.

"Gas adiabatic expansion ke dauran

Q_{H}

absorb karta hai kyunki woh expand kar raha hai aur work kar raha hai."

Ek adiabatic by definition koi heat exchange nahi karta (

d Q = 0

); expansion mein kiya gaya work gas ki apni internal energy se aata hai, isliye woh cool hoti hai.

Q_{C} / Q_{H} = (T_{C} / T_{H}) \cdot ln (V_{3} / V_{4}) / ln (V_{2} / V_{1})

, isliye efficiency volumes par depend karti hai."

Ek ideal gas ki adiabatic

T V^{γ - 1} = const

obey karti hai (

C_{V} d T = - \frac{R T}{V} d V

ko integrate karne se), isliye step 2 deta hai

T_{H} V_{2}^{γ - 1} = T_{C} V_{3}^{γ - 1}

aur step 4 deta hai

T_{H} V_{1}^{γ - 1} = T_{C} V_{4}^{γ - 1}

; inhein divide karne par

V_{2} / V_{1} = V_{3} / V_{4}

milta hai, jo log ratio ko exactly

1

banata hai isliye volumes cancel ho jaate hain aur

Q_{C} / Q_{H} = T_{C} / T_{H}

"Ek cleverly engineered irreversible engine usi temperature ke dono reservoirs ke beech Carnot ko beat karta hai,

η = 0.6

reach karta hai jahan Carnot

0.5

deta hai."

Yeh Carnot's theorem violate karta hai — koi bhi engine do reservoirs ke beech reversible value se zyada nahi ja sakta, kyunki ise ek reversed Carnot engine ke saath chain karna ek perpetual-motion machine of the second kind bana dega.

"Heat ko isothermally aur reversibly gas aur reservoir ke beech

10 K

ke gap par transfer kiya ja sakta hai."

Reversible heat transfer ke liye gas aur reservoir ko same temperature par hona chahiye (infinitesimal gap); ek finite

10 K

gap step ko irreversible banata hai aur available work waste karta hai.

"Kyunki cycle mein

Δ U = 0

hai, engine koi net work nahi karta."

Δ U = 0

ka matlab hai net heat equals net work (

W = Q_{H} - Q_{C}

), zero work nahi; work

P

–

V

diagram par clockwise loop ke enclosed area ke equal hai.

Why questions

Heat-exchange steps isothermal kyun hone chahiye, naaki isobaric?

Reversible rehne ke liye, gas ko reservoir ki temperature par baithe rehna chahiye jab bhi woh heat leta hai, jo exchange ke dauran

T

constant force karta hai; ek isobaric step

T

ko drift karne deta aur ek temperature gap kholta.

Intermediate temperature changes ke liye adiabatics use kyun karte hain sirf ek warm reservoir ko touch karne ki jagah?

Ek intermediate-temperature reservoir ko touch karna heat ko finite gap se transfer karta (irreversible); gas ko insulate karna use work ke zariye apna temperature khud change karne deta hai, har step ko reversible rakhta hai.

T_{C}

ko ek degree kam karna

T_{H}

ko ek degree badhane se zyada help kyun karta hai (typical values ke paas)?

η = 1 - T_{C} / T_{H}

ke partial derivatives compare karo:

\partial η / \partial T_{C} = - 1/ T_{H}

aur

\partial η / \partial T_{H} = + T_{C} / T_{H}^{2}

; unke magnitudes ka ratio hai

(1/ T_{H}) / (T_{C} / T_{H}^{2}) = T_{H} / T_{C} > 1

jab bhi

T_{H} > T_{C}

, isliye

T_{C}

mein ek-degree drop hamesha

η

zyada change karta hai — e.g.

T_{H} = 600, T_{C} = 300

par cold side har degree ke liye do guna zyada effective hai.

Identity

Q_{H} / T_{H} = Q_{C} / T_{C}

ko "the birth of entropy" kyun kaha jaata hai?

Q_{H} / T_{H} - Q_{C} / T_{C} = 0

ki tarah rearrange karne par, yeh kehta hai ki

\oint d Q_{r e v} / T = 0

reversible loop ke liye, matlab

d Q_{r e v} / T

ek state function tak integrate hota hai — woh function entropy hai.

Koi bhi real engine practice mein Carnot efficiency kyun nahi reach kar sakta?

Real processes mein friction, finite temperature gaps, aur finite speeds hote hain — sab irreversible — isliye woh entropy generate karte hain aur ideal reversible bound se short rehte hain.

Efficiency formula absolute temperature scale kyun assume karta hai?

Kyunki

T_{C} / T_{H}

ek ratio hai, sirf woh scale jiska zero true absence of thermal energy hai (kelvin) ek physically meaningful fraction deta hai; zero point shift karna (Celsius) ratio ko corrupt kar deta hai.

Loop ko counterclockwise traverse karne se net work ka sign flip kyun hota hai?

Clockwise path par tum higher pressure par expand karte ho jitne par compress karte ho, isliye

\oint P d V > 0

(work out); direction reverse karna swap karta hai ki kaun se legs expansion vs compression hain, isliye

\oint P d V < 0

aur tumhe work supply karna padta hai — loop refrigerator ban jaata hai.

Edge cases

T_{C} \to 0

limit mein

η

ka kya hota hai?

η \to 1 - 0 = 1

, ek perfect engine — lekin

T_{C} = 0

K third law ke according unreachable hai, isliye yeh ek limit hai, kabhi achievable value nahi.

Jab

T_{H} \to T_{C}

(reservoirs equal) hota hai toh

η

ka kya hota hai?

η \to 1 - T_{C} / T_{C} = 0

; koi temperature difference nahi hai toh koi driving gap nahi hai, loop zero enclosed area mein collapse ho jaata hai, aur koi net work extract nahi kiya ja sakta.

Agar tum "engine" ko

T_{C} > T_{H}

ke saath chalane ki koshish karo (cold source hot se zyada garam), toh formula kya kehta hai?

η = 1 - T_{C} / T_{H}

negative ho jaata hai, yeh signal karta hai ki heat galat direction mein flow karegi — tumhe work input karna padega, yani yeh ab engine nahi raha balki heat pump ban gaya.

Kya reversible identity

Q_{H} / T_{H} = Q_{C} / T_{C}

ek irreversible engine ke liye valid hai?

Nahi — irreversible operation ke liye

Q_{H} / T_{H} < Q_{C} / T_{C}

(entropy badhti hai), isliye equality woh special reversible boundary hai jo sirf Carnot achieve karta hai.

Jab

T_{H} = T_{C}

ho toh enclosed

P

–

V

area kya hai?

Zero — equal temperatures ke saath dono isotherms merge ho jaate hain aur dono adiabatics ka koi temperature span nahi hota, isliye loop ek zero enclosed area aur zero work ki line mein pinch ho jaata hai.

Recall Ek-line self-test

Har answer cover karo aur reason re-derive karo, verdict nahi. Agar kisi "why" mein ek se zyada sentence ki struggle lagi, toh parent note ka woh section dobara padho.