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

1.7.21 · Physics › Thermodynamics

Carnot cycle KYA hota hai?

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

HAR step KYU? (design ki logic)

Efficiency KAISE derive karein — first principles se

Hum 1 mole ideal gas use karte hain: $P V = R T$ , internal energy $U = U (T)$ sirf.

General efficiency definition. Ek poore cycle mein $Δ U = 0$ , toh first law se net heat, net work ke barabar hoti hai: $η = \frac{W _{net}}{Q _{H}} = \frac{Q _{H} - Q _{C}}{Q _{H}} = 1 - \frac{Q _{C}}{Q _{H}}$

Ab hum $Q_{H}$ aur $Q_{C}$ compute karte hain.

Step 1 — Isothermal expansion at $T_{H}$ ( $V_{1} \to V_{2}$ )

Kyunki $T$ constant hai, $Δ U = 0$ , toh $Q_{H} = W_{1 \to 2}$ . $W_{1 \to 2} = \int_{V_{1}}^{V_{2}} P d V = \int_{V_{1}}^{V_{2}} \frac{R T _{H}}{V} d V = R T_{H} ln \frac{V _{2}}{V _{1}}$

Yeh step kyun? Ideal gas law se $P = R T_{H} / V$ use kiya aur integrate kiya; log isliye aaya kyunki $\int d V / V = ln V$ . $Q_{H} = R T_{H} ln \frac{V _{2}}{V _{1}}$

Step 3 — Isothermal compression at $T_{C}$ ( $V_{3} \to V_{4}$ )

Same logic, heat reject hoti hai (gas compress hoti hai, $V_{4} < V_{3}$ ): $Q_{C} = - W_{3 \to 4} = R T_{C} ln \frac{V _{3}}{V _{4}}$

Sign kyun? Hum $Q_{C}$ ko heat released (positive number) define karte hain, toh magnitude lete hain. Kyunki $V_{3} > V_{4}$ , $ln (V_{3} / V_{4}) > 0$ . ✓

Steps 2 & 4 — Adiabatics volumes ko link karte hain

Ek ideal gas ki reversible adiabatic ke liye, $T V^{γ - 1} = const$ .

Step 2 ( $T_{H}, V_{2} \to T_{C}, V_{3}$ ) aur step 4 ( $T_{C}, V_{4} \to T_{H}, V_{1}$ ) par apply karo: $T_{H} V_{2}^{γ - 1} = T_{C} V_{3}^{γ - 1}, T_{H} V_{1}^{γ - 1} = T_{C} V_{4}^{γ - 1}$ Dono equations divide karo: $\frac{V _{2}^{γ - 1}}{V _{1}^{γ - 1}} = \frac{V _{3}^{γ - 1}}{V _{4}^{γ - 1}} \Rightarrow \frac{V _{2}}{V _{1}} = \frac{V _{3}}{V _{4}}$

Yeh key trick kyun hai? $Q_{H}$ aur $Q_{C}$ mein aane wale dono volume ratios equal hain! Yahi wajah hai ki log terms cancel ho jaati hain.

Combine karo

$\frac{Q _{C}}{Q _{H}} = \frac{R T _{C} l n ( V _{3} / V _{4} )}{R T _{H} l n ( V _{2} / V _{1} )} = \frac{T _{C}}{T _{H}} \cdot = 1 \frac{ln ( V _{3} / V _{4} )}{ln ( V _{2} / V _{1} )} = \frac{T _{C}}{T _{H}}$

Worked examples

Common mistakes (steel-manned)

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho ek bouncy gas ek cylinder mein hai. Tum use ek hot stove se touch karne do, aur woh dhire-dhire piston push karti hai hot rehte hue (woh heat grab karti hai aur work karti hai). Phir tum use ek blanket mein wrap karte ho taaki koi heat andar-bahar na jaye, aur woh push karte-karte thandi ho jaati hai. Ab tum use ek cold ice block ke saath press karte ho, use dheere squeeze karte ho taaki woh bacha khucha heat dump kar sake. Aakhir mein tum use phir blanket karte ho aur squeeze karke start pe le aate ho, use garam karte hue. Ghoomta-ghoomta yeh heat ko useful pushing mein convert karta hai. Pakad: jitna thanda tumhara ice aur jitna garam tumhara stove, utna zyada useful work milta hai — lekin tum kabhi 100% nahi pa sakte, kyunki kuch warmth hamesha ice mein jaani hi padti hai.

Flashcards

Carnot cycle ke chaar steps order mein kya hain?

Isothermal expansion (at

T_{H}

), adiabatic expansion, isothermal compression (at

T_{C}

), adiabatic compression.

Heat transfer steps isothermal kyun hone chahiye?

Unhe reversible rakhne ke liye — gas reservoir ke temperature par rehti hai toh heat koi finite temperature gap cross nahi karti.

Carnot efficiency formula kya hai?

η = 1 - T_{C} / T_{H}

, temperatures kelvin mein.

Step 1 (isothermal at

T_{H}

) mein kitni heat absorb hoti hai?

Q_{H} = R T_{H} ln (V_{2} / V_{1})

, kyunki

Δ U = 0

toh

Q = W

Adiabats volumes ke beech kya relation force karte hain?

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

, dono adiabatics par

T V^{γ - 1} = const

se.

Efficiency working substance par kyun depend nahi karti?

Volume-ratio logs cancel ho jaate hain, sirf temperature ratio

T_{C} / T_{H}

bachta hai.

Heats ko link karne wali reversible identity kya hai?

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

, yaani

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

— entropy ka aadhar.

Carnot's theorem bolo.

Do reservoirs ke beech kaam karne wala koi bhi engine, unhi reservoirs ke beech ek reversible (Carnot) engine se zyada efficient nahi ho sakta.

η = 1

kyun nahi ho sakta?

Iske liye

T_{C} = 0

K chahiye hoga (pahuncha nahi ja sakta, 3rd law); kuch heat hamesha reject karni hi padti hai (2nd law).

Common kelvin mistake kya hai?

T_{C} / T_{H}

mein Celsius use karna — ratios zero point par depend karte hain, toh kelvin use karna zaroori hai.

Connections

Second law of thermodynamics — Carnot's theorem efficiency ki ceiling set karta hai.
Entropy — $\oint d Q_{r e v} / T = 0$ state function $S$ define karta hai.
Adiabatic process — $T V^{γ - 1} = const$ deta hai.
Isothermal process — $R T ln (V_{f} / V_{i})$ work deta hai.
Absolute temperature scale — $Q_{H} / Q_{C} = T_{H} / T_{C}$ se define hota hai.
Heat engines and refrigerators — Carnot ulta chalaya jaaye toh ideal fridge/heat pump ban jaata hai.
First law of thermodynamics — $η = 1 - Q_{C} / Q_{H}$ deta hai.