Entropy — Clausius definition dS = dQ_rev - T

1.7.22 · Physics › Thermodynamics

Entropy KYA hai?

Subscript rev definition ka dil hai. Do states ke beech $Δ S$ compute karne ke liye aapko koi ek reversible path imagine karna hoga jo unhe connect kare — chahe real process irreversible hi kyun na rahi ho. Kyunki $S$ ek state function hai, koi bhi reversible path same answer deta hai.

$d Q / T$ KYU kaam karta hai lekin $d Q$ nahi karta?

HUM KAISE JAANTE HAIN ki yeh path-independent hai — Carnot link:

Ek Carnot cycle ke liye jo reversibly run kar raha hai hot reservoir $T_{H}$ aur cold reservoir $T_{C}$ ke beech: $\frac{Q _{H}}{T _{H}} = \frac{Q _{C}}{T _{C}}$ Yeh directly aata hai Carnot efficiency $η = 1 - \frac{Q _{C}}{Q _{H}} = 1 - \frac{T _{C}}{T _{H}}$ se, jo rearrange hoke upar wala equation deta hai. Heat in ko positive aur heat out ko negative count karte hue: $\frac{Q _{H}}{T _{H}} - \frac{Q _{C}}{T _{C}} = 0 \Rightarrow \oint \frac{d Q _{r e v}}{T} = 0 (for a Carnot cycle) .$

HUM KAISE GENERALISE KARTE HAIN (Clausius theorem): koi bhi reversible cycle infinitely many tiny Carnot cycles se tile ki ja sakti hai. Internal boundaries cancel ho jaate hain, isliye $\oint_{r e v} \frac{d Q _{r e v}}{T} = 0$ Ek quantity jiska integral har closed loop mein zero ho, woh zaroor kisi state function ka differential hai. Us function ko $S$ kahte hain. Yahi entropy ke existence ka proof hai.

Figure — Entropy — Clausius definition dS = dQ_rev - T

Ideal gas ke liye $Δ S$ ka Derivation (scratch se)

First Law se start karte hain ek reversible process ke liye: $d U = d Q_{r e v} - d W_{r e v} = d Q_{r e v} - P d V \Rightarrow d Q_{r e v} = d U + P d V .$

Yeh step kyun? First Law energy conservation hai; reversible work ke liye $d W = P d V$ exactly hota hai.

Ideal gas ke liye $d U = n C_{V} d T$ aur $P = \frac{n R T}{V}$ . Sab kuch $T$ se divide karo: $d S = \frac{d Q _{r e v}}{T} = \frac{n C _{V} d T}{T} + \frac{n R d V}{V} .$

$T$ se divide kyun kiya? Kyunki yahi literally Clausius definition hai — yeh inexact $d Q$ ko exact $d S$ mein convert karta hai.

$(T_{1}, V_{1})$ se $(T_{2}, V_{2})$ tak integrate karo: $Δ S = n C_{V} ln \frac{T _{2}}{T _{1}} + n R ln \frac{V _{2}}{V _{1}}$

Dono terms clean logs kyun hain? Kyunki $T$ se divide karne ke baad har term already ek exact differential tha — yeh confirm karta hai ki $1/ T$ ne apna kaam integrating factor ke taur par kar diya.

Worked examples

Common mistakes (steel-manned)

Active recall

Recall Quick self-test (answers chupaao)

$1/ T$ special kyun hai? ::: Yeh woh integrating factor hai jo inexact $d Q_{r e v}$ ko exact differential $d S$ bana deta hai.
$S$ state function kyun hai? ::: Kyunki $\oint d Q_{r e v} / T = 0$ (Clausius theorem, Carnot cycles tile karne se).
Free expansion ka entropy change? ::: $n R ln (V_{2} / V_{1}) > 0$ , ek reversible isothermal substitute path ke zariye.

Recall Feynman: 12-saal ke bacche ko samjhao

Socho heat paani ki tarah ek tank mein daali ja rahi hai. Yeh kitna spread out hota hai sirf is par depend nahi karta ki kitna daala, balki is par bhi ki tank pehle se kitna "crowded" hai — aur temperature batata hai crowding. Same heat ek thandi cheez mein daalo toh bada mess hoga (badi entropy jump); kisi already hot cheez mein daalo toh woh barely notice karti hai (choti jump). Entropy bas "mess made" ka running total hai, aur yeh tabhi sahi se count hoti hai jab aap bahut slowly (reversibly) daalo. Ek baar mess ho jaaye, universe kabhi theek nahi hota — isliye time aage ki taraf chalta hai.

Entropy change ki Clausius definition kya hai?

d S = d Q_{r e v} / T

;

Δ S = \int_{A}^{B} d Q_{r e v} / T

, path-independent.

d S = d Q_{r e v} / T

mein heat reversible kyun honi chahiye?

Sirf reversible path

d Q / T

ko ek exact (path-independent) differential banata hai jo true state change ke barabar ho.

Heat ka integrating factor kya hai?

1/ T

— yeh inexact

d Q_{r e v}

ko exact

d S

mein convert karta hai.

Clausius theorem (reversible cycle) state karo.

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

Clausius inequality (koi bhi cycle) state karo.

\oint d Q / T \leq 0

, equality sirf reversible cycles ke liye.

Ideal gas ke do states ke beech

Δ S

n C_{V} ln (T_{2} / T_{1}) + n R ln (V_{2} / V_{1})

Reversible isothermal expansion ke liye

Δ S

n R ln (V_{2} / V_{1})

Mass

m

T_{1}

T_{2}

tak heat karne par

Δ S

m c ln (T_{2} / T_{1})

Vacuum mein free expansion mein gas ka

Δ S

n R ln (V_{2} / V_{1}) > 0

(reversible isothermal substitute use karo;

Q_{a c t u a l} = 0

apply nahi hota).

Entropy ki units?

J K^{- 1}

Reversible adiabatic process ke liye

Δ S = ?

0

(isentropic).

Connections

Carnot Cycle and Efficiency — $Q_{H} / T_{H} = Q_{C} / T_{C}$ ka source
First Law of Thermodynamics — $d Q_{r e v} = d U + P d V$ deta hai
Second Law of Thermodynamics — $Δ S_{u ni v er se} \geq 0$
Reversible vs Irreversible Processes
Exact and Inexact Differentials — kyun $1/ T$ ek integrating factor hai
Statistical Entropy — Boltzmann S = k ln W — microscopic meaning
Clausius Inequality