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

1.7.21 · D1 · Physics › Thermodynamics › Carnot cycle — full derivation, efficiency = 1 − T_C - T_H

Yeh page har symbol build karta hai jo parent note use karta hai, bilkul zero assumed knowledge se shuru karke. Ise upar se neeche padho: har idea woh ground hai jis par agla khada hai.

1. Pressure $P$ , Volume $V$ , Temperature $T$ — gas ke teen dials

Picture. Ek syringe imagine karo jiska nozzle sealed hai. Plunger andar push karo: $V$ neeche jaata hai, band hawa pushback karti hai zyada zor se ( $P$ upar jaata hai). Woh pushback pressure hai — ek real force jo aap apne anguthe mein feel karte ho.

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

Topic ko teeno ki zaroorat kyun hai. Carnot cycle gas ki alag-alag $(P, V, T)$ states ke through ek safar hai. Us safar ko draw aur reason karne ke liye hume yeh batana hoga ki gas kahan hai — aur $P$ , $V$ , $T$ uske coordinates hain. Jab gas cycle mein chaar corner states visit karta hai, hum unke volumes ko $V_{1}, V_{2}, V_{3}, V_{4}$ label karte hain — sirf "corner 1 par volume" ke naam, aur aise hi aage.

2. Absolute temperature $T$ — kelvin kyun, Celsius kyun nahin

Picture. Ek thermometer socho jisme tube ka bottom absolute zero par nailed hai, na ki "barf pighalti hai" par. Paani $273.15 K$ par freeze hota hai, $373.15 K$ par boil karta hai.

Topic ko iski zaroorat kyun hai — yeh optional nahin hai. Efficiency formula ek ratio $T_{C} / T_{H}$ use karta hai. Ratio tabhi sense banta hai jab zero kuch nahin matlab ho. Celsius mein, $0° C$ ek arbitrary point hai (pighalti barf), isliye $\frac{20° C}{40° C}$ meaningless bakwaas hai. Kelvin mein, $\frac{300 K}{600 K} = \frac{1}{2}$ genuinely matlab hai "aadhi jiggle-energy". Dekho Absolute temperature scale.

3. Ideal gas law $P V = R T$ — teen dials ko link karne wala rule

Picture. Agar aap $T$ fixed rakho aur $V$ badhne do, toh $P = R T / V$ zaroor ghutna chahiye — curve $P = const / V$ ek smooth neeche jaane wala swoop hai jise hyperbola kehte hain. Woh swoop ek Isothermal process line hai.

Topic ko iski zaroorat kyun hai. Derivation mein har integral $P$ ko is law se $R T / V$ se replace karta hai. Yeh woh bridge hai jo hume sirf temperatures aur volumes se work compute karne deta hai.

4. Change ke do sizes: $Δ$ (finite) vs $d$ (infinitesimal)

5. Internal energy $U$ — aur exactly yeh $T$ par kaise depend karta hai

Picture. $U$ ko bouncing balls ke ek swarm ki total kinetic energy socho. Unhe garam karo ( $T$ badhao) aur woh tez bounce karte hain ( $U$ badhta hai). Walls ko bina garam kiye move karo (akele $V$ change karo) aur swarm ki total energy untouched rehti hai.

Topic ko iski zaroorat kyun hai. Kyunki $U = C_{V} T$ , koi bhi process jo same $T$ par return karta hai uska $Δ U = 0$ hoga. Ek poore cycle ke baad gas apne start par return karta hai, isliye cycle par $Δ U = 0$ — woh single zero first law ko "net work = net heat" mein collapse kar deta hai (§7).

6. Work $W$ aur heat $Q$ — energy move karne ke do tarike

Picture — area = work kyun. Jab gas ek sliver $d V$ (§4) se pressure $P$ par expand karta hai, toh woh work ka ek sliver $P d V$ karta hai (force × distance, neatly packaged). Un saare thin slivers ko integral $\int P d V$ se stack karo aur aapko $P$ – $V$ curve ke neeche area milta hai.

Accounting ke liye First law of thermodynamics dekho, aur "engine" ka matlab Heat engines and refrigerators mein dekho.

7. First law — energy bookkeeping

Picture. Ek bank account socho. $Q$ deposit ki gayi raqam hai, $W$ withdraw ki gayi raqam hai, $Δ U$ aapke balance mein change hai. Kuch bhi appear ya vanish nahin hota.

Topic ko iski zaroorat kyun hai. Do special cases sara kaam karte hain:

Isothermal ( $Δ U = 0$ ): saari heat work ban jaati hai, $Q = W$ .
Adiabatic ( $Q = 0$ ): work purely internal energy se aati hai, $W = - Δ U$ .

8. $γ$ aur $C_{V}$ — aur adiabat par $T V^{γ - 1}$ constant kyun hai

Picture. Gas ko sealed box mein garam karo (constant $V$ ): har joule jiggle banta hai. Gas ko free lid wale box mein garam karo (constant $P$ ): kuch joules lid uthate hain — isliye aapko same temperature rise ke liye zyada heat chahiye. Hence $C_{P} > C_{V}$ aur $γ > 1$ .

Topic ko iski zaroorat kyun hai. Dono adiabatic steps par apply karne par, yeh relation $V_{2} / V_{1} = V_{3} / V_{4}$ force karta hai — woh magic cancellation jo efficiency ko sirf temperature par dependent chhod deta hai.

Dono move-types ke liye poori tarah Isothermal process aur Adiabatic process dekho.

9. Reversibility, $d Q_{rev}$ , aur loop integral $\oint$

Definition Reversible heat aur loop integral

$\oint$

Reversible matlab hai process seedha ulta chal sakti hai har state retrace karke, kuch bhi waste kiye bina. Iske liye infinitely slow changes aur koi finite temperature gaps nahin chahiye.
== $d Q_{rev}$ == sirf heat $d Q$ (§4, §6) ka ek infinitesimal sliver hai jo reversibly exchange hua — subscript "rev" flag karta hai ki yeh particular heat koi temperature gap cross nahin ki, isliye exchange undo kiya ja sakta tha. Yeh wahi heat $Q$ hai jaise pehle, sirf reversible tagged aur sliver mein kaata gaya.
== $\oint$ (loop integral)== matlab hai "slivers ko ek closed cycle ke around, wapas start tak add karo". Toh $\oint d Q_{rev} / T = 0$ kehta hai: agar aap har sliver $d Q_{rev}$ ko us temperature $T$ se divide karke add karo jis par woh exchange hua, poore loop ke around, aap zero par return karte ho.

Topic ko iski zaroorat kyun hai. Reversibility hi kyun Carnot sabse best possible hai — yeh Second law of thermodynamics ka theme hai. Aur $\oint d Q_{rev} / T = 0$ woh discovery hai ki ek naya quantity, Entropy, loop ke around conserved hai.

Prerequisite map

Equipment checklist

Khud test karo — aap parent derivation ke liye ready ho jab aap har ek ka answer bina dekhe de sako.

Δ V

aur

d V

mein kya fark hai?

Δ V

ek finite, measurable change hai jis par aap ruler rakh sako;

d V

ek infinitely small sliver hai jise aap integral se add karte ho.

T_{C} / T_{H}

ke liye temperatures kelvin mein kyun honi chahiye, Celsius mein kyun nahin?

Kyunki yeh ek ratio hai, aur ratios tabhi kuch matlab rakhte hain jab zero ka matlab "bilkul jiggle nahin" ho — sirf kelvin mein woh true zero hai.

P

–

V

curve ke neeche area kya represent karta hai?

Gas ka kiya gaya work,

\int P d V

\int d V / V

logarithm kyun deta hai?

ln

woh function hai jiska slope

1/ V

hai, isliye yeh exactly

1/ V

ka antiderivative hai.

Ideal gas ke liye

U

kya hai aur yeh sirf

T

par kyun depend karta hai?

U = C_{V} T

; particles attract nahin karte, isliye akele

V

change karne par koi energy nahi lagti — sirf

T

U

ko move karta hai. Hence same

T

⇒

Δ U = 0

V_{1}, V_{2}

aur

Q_{H}

ka kya matlab hai?

V_{1}, V_{2}

cycle corners 1 aur 2 par volumes hain;

Q_{H}

woh heat hai jo hot isotherm par absorb hoti hai jab volume

V_{1}

V_{2}

tak badhta hai.

First law words mein batao.

Internal energy mein change = heat in minus work out,

Δ U = Q - W

(energy conservation).

T V^{γ - 1} = const

kahan se aata hai?

Adiabat par

d U = - P d V

se,

U = C_{V} T

aur

P = R T / V

ke saath,

R = C_{P} - C_{V}

use karke integrate karke.

γ = C_{P} / C_{V} > 1

kyun hai?

Constant pressure par kuch heat piston uthati hai, isliye

C_{P} > C_{V}

, ratio 1 se zyada ho jaata hai.

d Q_{rev}

kya hai aur

\oint d Q_{rev} / T = 0

kya hint karta hai?

Reversibly exchange ki gayi heat ka ek sliver (koi temperature gap nahin);

d Q_{rev} / T

ka loop sum zero hona ek conserved quantity — entropy — ki taraf hint karta hai.

"Reversible" physically kya chahiye?

Infinitely slow changes aur koi finite temperature gaps nahin, taaki kuch bhi waste na ho.

Recall Ek saans mein summary

$P, V, T$ gas ko locate karte hain; kelvin ratios ko meaningful banata hai; $P V = R T$ teeno ko link karta hai; $Δ$ ek finite change hai jabki $d$ ek infinitesimal sliver hai jise hum integrate karte hain; $P$ – $V$ ke neeche area work hai (isotherms par logs deta hai); $U = C_{V} T$ sirf $T$ par depend karta hai; first law energy bookkeeping hai; $γ$ adiabat ko $T V^{γ - 1} = const$ ke zariye shape karta hai; reversibility Carnot ko champion banata hai aur entropy ko janam deta hai.