Carnot cycle — full derivation, efficiency = 1 − T_C - T_H: Visual walkthrough

1.7.21 · D2 · Physics › Thermodynamics › Carnot cycle — full derivation, efficiency = 1 − T_C - T_H

Jis stage pe hum rehte hain: P–V plane

Kisi bhi cycle se pehle, humein ek stage chahiye. Socho ek gas cylinder mein band hai, piston ke neeche.

Piston ko neeche push karo → gas ke paas kam jagah bachti hai → uska volume $V$ chhota ho jaata hai.
Gas ko kam jagah mein squeeze karo → woh zyada force se push karta hai wapas → uska pressure $P$ (piston par force per unit area) badh jaata hai.

Toh gas ki state ek flat map par ek dot hai: horizontal axis = ==volume $V$ == (kitni jagah hai), vertical axis = ==pressure $P$ == (kitni zyada push karti hai). Jaise gas change hoti hai, dot ek curve par slide karta hai. Us curve ke neeche ka area mechanical work hai. Area kyun? Kyunki work = force × distance jitna piston move karta hai, aur is map par "force" $P$ axis par hai jabki "distance moved" $V$ axis par — inhe multiply karo toh area milta hai.

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

Step 1 — Ideal-gas law se isothermal curve milti hai

KYA. Ek mole ideal gas ke liye humein ek fact diya gaya hai: $P V = R T$ , jahan $R$ ek fixed number hai (gas constant) aur $T$ absolute temperature hai kelvin mein (dekho Absolute temperature scale).

KYUN. Humein yeh jaanna hai ki isothermal move ka map par shape kaisi hogi taaki hum area (work) measure kar sakein. $T$ ko fixed rakho aur pressure ke liye solve karo: $P = pressure as room shrinks \frac{R T}{V}$

Ise term by term padho: upar $R T$ ek constant hai jab $T$ fixed hai; neeche $V$ ka matlab hai pressure aur volume inversely linked hain. Room double karo, push half ho jaati hai. Yeh ek hyperbola hai — ek curve jo neeche aur daayein swoops karta hai.

PICTURE. Zyada temperature ek hyperbola deti hai jo upar baithti hai (upar $R T$ bada). Toh hot curve ( $T_{H}$ ) cold curve ( $T_{C}$ ) ke upar float karti hai.

Step 2 — $T_{H}$ par isothermal expansion: heat $Q_{H}$ grab karna

KYA. Gas ko hot reservoir par $T_{H}$ par rakho aur use slowly expand karne do volume $V_{1}$ se bade $V_{2}$ tak, hot hyperbola par rehte hue.

KYUN. Kyunki temperature fixed hai aur (ideal gas ke liye) internal energy sirf temperature par depend karti hai, gas ki stored energy $U$ change nahi hoti: $Δ U = 0$ . First law of thermodynamics ke hisaab se, $Δ U = Q - W$ , toh $Δ U = 0$ ke saath andar aaya har joule heat work ke roop mein bahar jaata hai: $Q_{H} = W_{1 \to 2} .$

Work hot curve ke neeche ka area hai $V_{1}$ se $V_{2}$ tak. Use paane ke liye hum add up karte hain $P d V$ ko volume ke har chhote sliver par — yahi integral sign ka matlab hai, "thin strips ka sum": $W_{1 \to 2} = \int_{V_{1}}^{V_{2}} P d V = \int_{V_{1}}^{V_{2}} \frac{R T _{H}}{V} d V = R T_{H} ln \frac{V _{2}}{V _{1}} .$

Term by term: $\int \frac{1}{V} d V$ exactly woh operation hai jo ek logarithm produce karta hai — isliye $ln$ aata hai (yeh woh ek function hai jiska slope $1/ V$ hai). Uske andar, $\frac{V _{2}}{V _{1}} > 1$ kyunki humne expand kiya, toh uska $ln$ positive hai → heat genuinely andar flow karti hai.

$Q_{H} = R T_{H} ln \frac{V _{2}}{V _{1}} (> 0)$

Step 3 — Adiabatic expansion: $T_{H}$ se $T_{C}$ tak coast karna

KYA. Cylinder ko ek perfect blanket mein wrap karo ( $Q = 0$ ) aur use expand karte rehne do, $V_{2} \to V_{3}$ . Koi heat andar nahi aati, toh gas apni baahri push ki payment apni khud ki energy se karta hai, isliye woh $T_{H}$ se $T_{C}$ tak cool ho jaata hai.

KYUN. Hum gas ko bich mein kisi reservoir ko touch kiye bina cold temperature tak le jaana chahte hain — kisi mid-temperature body ko touch karna heat ko temperature gap cross karne dega, jo wasteful aur irreversible hai. Blanketing ise reversible rakhta hai. Ab hum actually is "blanketed" move ki shape derive karte hain — hum ise quote nahi kar sakte.

Definition Do heat capacities, aur

$γ$ ka matlab

== $C_{V}$ == = molar heat capacity at constant volume: woh joules ki tadaad jo tumhe ek mole mein daalni padengi uski temperature ek kelvin badhane ke liye jabki volume pinned ho (taaki koi heat work ke roop mein bahar na nikle). Volume fixed hone ki wajah se, sab kuch internal energy mein jaata hai: $d U = C_{V} d T$ .
== $C_{P}$ == = molar heat capacity at constant pressure: same idea lekin gas expand ho sakti hai, toh woh kuch heat surroundings ke against work karte hue bhi kharach karti hai. Isliye $C_{P} > C_{V}$ ; extra exactly $R$ hai, giving $C_{P} - C_{V} = R$ .
== $γ = C_{P} / C_{V}$ == unka ratio hai. Kyunki $C_{P}$ bada wala hai, $γ$ hamesha 1 se bada hota hai — isliye $γ > 1$ hai, decree se nahi balki isliye ki expanding-while-heating mein hamesha extra cost lagti hai.

Formula Adiabatic law

$T V^{γ - 1} = const$ scratch se derive karna Shuru karo blanketed move ke liye first law se ( $Q = 0$ ): internal energy mein change minus work done ke barabar hai, $change in U C_{V} d T = work leaving - P d V .$ Substitute karo ideal-gas pressure $P = R T / V$ (Step 1) right side mein: $C_{V} d T = - \frac{R T}{V} d V .$ Separate karo dono variables ko ( $T$ wali cheez left par, $V$ wali cheez right par), $C_{V} T$ se divide karke: $\frac{d T}{T} = - \frac{R}{C _{V}} \frac{d V}{V} .$ Integrate karo dono sides — har side ek $\int \frac{1}{x} d x$ hai, jo ek logarithm deta hai (same tool as Step 2): $ln T = - \frac{R}{C _{V}} ln V + const .$ Use karo $R = C_{P} - C_{V}$ , toh $\frac{R}{C _{V}} = \frac{C _{P} - C _{V}}{C _{V}} = γ - 1$ . Substitute karo: $ln T = - (γ - 1) ln V + const \Rightarrow ln T + (γ - 1) ln V = const .$ Exponentiate karo (logs ko undo karo): $ln (T V^{γ - 1}) = const$ , isliye $T V^{γ - 1} = const .$

Result padho: jaise $V$ badhta hai, product fixed rakhne ke liye $T$ zaroor ghatta hai — yeh exactly woh cooling hai $T_{H}$ se $T_{C}$ tak jo blanketed expansion mein hoti hai.

PICTURE. Adiabatic curve dono hyperbolas se steeper hoti hai — woh high hot curve se neeche low cold curve tak dive karti hai.

Is law ko is stroke par apply karne se ( $T_{H}$ at $V_{2}$ down to $T_{C}$ at $V_{3}$ ) humein pehla bridge equation milta hai: $T_H\,V_2^{\gamma-1} = T_C\,V_3^{\gamma-1}. \tag{A}$

Step 4 — $T_{C}$ par isothermal compression: heat $Q_{C}$ dump karna

KYA. Gas ko cold reservoir par $T_{C}$ par rakho aur slowly squeeze karo, $V_{3} \to V_{4}$ (with $V_{4} < V_{3}$ ), cold hyperbola par rehte hue.

KYUN. Same isothermal logic as Step 2, lekin ulta: hum gas par work karte hain, aur kyunki $Δ U = 0$ again hai, woh energy heat ke roop mein cold reservoir mein bahar jaani chahiye. Hum $Q_{C}$ ko positive amount of heat rejected define karte hain: $Q_{C} = R T_{C} ln \frac{V _{3}}{V _{4}} .$

Term by term: $\frac{V _{3}}{V _{4}} > 1$ (humne compress kiya, toh hum fraction ko flip karte hain ise 1 se upar rakhne ke liye), isliye $ln$ positive hai aur $Q_{C} > 0$ — heat sach mein bahar jaati hai. Yahaan Second law of thermodynamics kaata hai: tum yeh step skip nahi kar sakte. Kuch heat zaroor dump karni padti hai.

$Q_{C} = R T_{C} ln \frac{V _{3}}{V _{4}} (> 0)$

Step 5 — Adiabatic compression: $T_{H}$ tak wapas climb karna

KYA. Gas ko phir se blanket karo aur $V_{4} \to V_{1}$ squeeze karo, loop band karo. Koi heat exchange nahi hone se, compression work use $T_{C}$ se $T_{H}$ tak wapas heat karta hai.

KYUN. Hum exact starting dot par wapas jaana chahte hain ise ek cycle banana ke liye (hamesha ke liye repeatable). Wahi adiabatic law $T V^{γ - 1} = const$ (Step 3 mein derive kiya) is stroke par apply karne se ( $T_{C}$ at $V_{4}$ up to $T_{H}$ at $V_{1}$ ) humein doosra bridge equation milta hai: $T_H\,V_1^{\gamma-1} = T_C\,V_4^{\gamma-1}. \tag{B}$

PICTURE. Loop ab band hai: do hyperbola arcs (hot top, cold bottom) do steep adiabatic arcs se joined. Enclosed area net work per cycle hai.

Step 6 — Woh magic cancellation: volumes kyun gayab ho jaate hain

KYA. Bridge equation (A) ko bridge equation (B) se divide karo: $\frac{T _{H} V _{2}^{γ - 1}}{T _{H} V _{1}^{γ - 1}} = \frac{T _{C} V _{3}^{γ - 1}}{T _{C} V _{4}^{γ - 1}} .$

KYUN. Left par $T_{H}$ cancel ho jaate hain, right par $T_{C}$ cancel ho jaate hain, aur dono sides ka $(γ - 1)$ -th root lene se sab collapse ho jaata hai: $\frac{V _{2}}{V _{1}} = \frac{V _{3}}{V _{4}}$

Yahi poora secret hai. $Q_{H}$ ke andar chhupa volume ratio exactly equal hai $Q_{C}$ ke andar wale se. Socho: do adiabats parallel "slides" hain, toh woh same stretch factor carry karte hain hot curve se cold curve tak loop ke left aur right dono sides par.

Step 7 — Efficiency assemble karo

KYA. Efficiency $η$ poochhti hai: jis heat ke liye tumne pay kiya, uska kitna fraction work bana? Ek poore cycle mein gas wapas start par aa jaati hai, toh $Δ U = 0$ aur net work net heat ke barabar hai: $η = \frac{W _{net}}{Q _{H}} = \frac{Q _{H} - Q _{C}}{Q _{H}} = 1 - \frac{Q _{C}}{Q _{H}} .$

KYUN. Ab dono boxed heats substitute karo aur Step 6 se cancellation use karo: $\frac{Q _{C}}{Q _{H}} = \frac{R T _{C} l n ( V _{3} / V _{4} )}{R T _{H} l n ( V _{2} / V _{1} )} = \frac{T _{C}}{T _{H}} \cdot = 1 by Step 6 \frac{ln ( V _{3} / V _{4} )}{ln ( V _{2} / V _{1} )} = \frac{T _{C}}{T _{H}} .$

$R$ cancel ho jaate hain; dono logs equal hain toh unka ratio $1$ hai; sirf temperature ratio bachta hai. Isliye answer ko fark nahi padta ki tumne kaunsi gas use ki.

Step 8 — Reversible identity $Q_{H} / T_{H} = Q_{C} / T_{C}$ kahaan se aati hai

KYA. Humne yeh assume nahi kiya — yeh un do boxed heats se nikal ke aata hai jo hum pehle hi earn kar chuke hain. $Q_{H}$ ko $T_{H}$ se aur $Q_{C}$ ko $T_{C}$ se divide karo:

\qquad \frac{Q_C}{T_C} = \frac{R\,T_C\ln(V_3/V_4)}{T_C} = R\ln\frac{V_3}{V_4}.$$ **KYUN.** Har case mein neeche ka temperature heat ke andar wale temperature ko **cancel** kar deta hai, sirf volume-ratio log bachta hai. Lekin Step 6 ne prove kiya $V_2/V_1 = V_3/V_4$, toh yeh dono right-hand sides *identical* hain: $$\boxed{\dfrac{Q_H}{T_H} = \dfrac{Q_C}{T_C}}$$ **ISKA MATLAB.** Rearrange karo, quantity $Q/T$ jo tum hot stroke par *gain* karte ho woh cold stroke par *exactly return* ho jaati hai — loop ke around kuch bhi lost nahi hota. Ise poore reversible cycle par loop-sum ke roop mein likhne par, $$\oint \frac{dQ_{\text{rev}}}{T} = 0,$$ jo kehta hai ek hidden bookkeeping quantity hai jo har cycle mein apni starting value par wapas aa jaati hai. Woh quantity ==[[Entropy]]== hai, aur yeh identity uska birth certificate hai. --- ## Step 9 — Edge aur degenerate cases (kabhi uncovered mat chhodna) **KYA / KYUN / PICTURE har corner ke liye:** - **$T_C = T_H$ (koi temperature difference nahi).** Toh $\eta = 1 - 1 = 0$. Hot aur cold hyperbolas ek line mein merge ho jaati hain; loop ka **zero enclosed area** hai — koi work nahi. Tum do same-temperature cheezeon ke beech engine nahi chala sakte. - **$T_C \to 0$ (perfectly cold sink).** Toh $\eta \to 1 - 0 = 1$. Lekin absolute zero unreachable hai, toh $\eta = 1$ ek limit hai jiske paas tum jaate ho, chhute nahi. Cold hyperbola $V$-axis ki taraf sink karti hai. - **$T_C > T_H$ (mislabelled reservoirs).** Toh $\eta < 0$: heat spontaneously galat direction mein flow karti hai aur "engine" actually ek refrigerator hai jo drive ho raha hai — dekho [[Heat engines and refrigerators]]. Formula tumhe negative sign se honestly warn karta hai. - **Zero-width strokes ($V_1 = V_2$).** Koi expansion nahi matlab $\ln 1 = 0$, toh $Q_H = 0$: kuch absorb nahi, kuch done nahi. Ek degenerate flat loop. ![[deepdives/dd-physics-1.7.21-d2-s08.png]] > [!mistake] Sabse common galti > Celsius use karna. Formula ek **ratio** $T_C/T_H$ hai, aur ratios explode karte hain agar tum zero point shift karo. $300\,\text{K}/600\,\text{K} = 0.5$ lekin $27\,°\text{C}/327\,°\text{C} = 0.083$ — bilkul alag. Hamesha kelvin. --- ## Ek-picture summary Upar sab kuch ek single annotated loop mein compress kiya: hot top par heat $Q_H$ in, cold bottom par heat $Q_C$ out, do adiabatic slides jo force karte hain $V_2/V_1 = V_3/V_4$, aur enclosed area jo tumhara net work hai. ![[deepdives/dd-physics-1.7.21-d2-s09.png]] > [!recall]- Feynman ki poore walkthrough ki retelling > Humne ek map draw kiya jahan left–right hai *gas ke paas kitni jagah hai* aur up–down hai *woh kitna hard push karta hai*. Hot stove par gas phool jaati hai aur heat grab karti hai — woh heat pushing ban jaati hai, aur amount top curve ke neeche ka area hai (ek logarithm, kyunki humne $1/V$ ke thin strips ka sum kiya). Phir hum ise blanket karte hain taaki koi heat sneaks in na ho aur use coast karne dete hain, hot se cold tak cool hote hue jab woh push karta rehta hai. Ice block par hum use slowly squeeze karte hain aur woh apni leftover heat dump karta hai — ek doosra logarithm. Aakhir mein blanket karo aur exact start tak squeeze karo. Ab yeh trick hai: do blanketed slides parallel hain, toh left par stretch factor right ke stretch factor ke barabar hai — dono logarithms *identical* hain aur cancel ho jaate hain. Poori story mein sirf dono temperatures ka ratio bachta hai, isliye $\eta = 1 - T_C/T_H$ aur isliye koi fark nahi padta tumne kaunsi gas use ki. Aur tum kabhi 100% nahi pahunch sakte, kyunki uske liye infinitely cold ice block chahiye hoga. > [!recall]- Quick self-check > Volume logs kyun cancel hote hain? ::: Do adiabats force karte hain $V_2/V_1 = V_3/V_4$, toh $\ln(V_3/V_4)/\ln(V_2/V_1) = 1$. > Cancellation ke baad kya bachta hai? ::: Sirf $T_C/T_H$ — isliye efficiency working substance ko ignore karti hai. > Enclosed loop area kya represent karta hai? ::: Cycle per net work done. > Step 4 ($Q_C$ reject karna) kyun exist karna chahiye? ::: Second law forbid karta hai sari heat ko work mein convert karna; kuch dump karna hi padega. > $Q_H/T_H = Q_C/T_C$ kahaan se aata hai? ::: Har heat ko apne temperature se divide karo; $T$ cancel ho jaate hain aur Step 6 dono logs ko equal banata hai. > $\gamma > 1$ kyun hai? ::: Kyunki $\gamma = C_P/C_V$ aur $C_P = C_V + R > C_V$ (expanding while heating mein hamesha extra work cost lagti hai). --- > [!mnemonic] Shape, ek breath mein > **Top curve = heat IN. Bottom curve = heat OUT. Steep sides = free coasting. Area = work. Answer = one minus cold-over-hot.**