Entropy — Clausius definition dS = dQ_rev - T: Question bank

1.7.22 · D5 · Physics › Thermodynamics › Entropy — Clausius definition dS = dQ_rev - T

Neeche use hone wale notation ke reminders, taaki kuch bhi unclear na rahe:

$Δ S$ = kisi chosen body ka entropy change do states ke beech, unit $J K^{- 1}$ .
$Q$ = woh heat jo actually flow hui; $Q_{r e v}$ = heat ek imagined reversible path ke saath same endpoints par.
$T$ = absolute temperature (kelvin, hamesha positive).
"System" = woh cheez jo hum track karte hain; "surroundings/reservoir" = woh sab cheez jiske saath heat exchange hoti hai; "universe" = system + surroundings.
$\oint$ = ek closed cycle ke around poora integral (wapas starting state tak).

True or false — justify

Reversible adiabatic process mein

Δ S = 0

hota hai.

True. Adiabatic ka matlab hai

d Q = 0

aur reversible ka matlab hai hum

d S = d Q_{r e v} / T = 0

use kar sakte hain, toh koi bhi entropy create nahi hoti — yahi exactly ek isentropic process hai.

Har adiabatic process mein

Δ S = 0

hota hai.

False. Free expansion adiabatic hai (

Q = 0

) phir bhi

Δ S = n R ln (V_{2} / V_{1}) > 0

;

d Q = 0

sirf tab entropy change ko zero karta hai jab path bhi reversible ho, isliye irreversible process mein

Q = 0

plug karna galat hai.

Entropy ek body ke andar stored heat ka ek form hai.

False. Heat stored nahi hoti (yeh ek inexact differential hai — tum yeh nahi keh sakte ki "is gas mein 5 J heat baith rahi hai"); entropy ek state function hai jo measure karti hai ki energy kitni spread-out hai,

J K^{- 1}

mein, joules mein nahi.

Kisi bhi real (irreversible) process ke liye universe ki entropy badhti hai.

True. Second Law of Thermodynamics kehta hai

Δ S_{u ni v er se} \geq 0

, irreversible processes ke liye strict inequality ke saath — woh inequality hi "irreversible" ka matlab hai.

Agar gas compress ki jaaye, toh uski entropy zaroor giregi.

False. Compression

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

term ko kam karta hai, lekin agar temperature badhta hai toh

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

term usse zyada compensate kar sakta hai;

Δ S

ka sign dono endpoints par depend karta hai, sirf volume par nahi.

Ek reversible cycle ke along,

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

hota hai lekin

\oint d Q_{r e v}

zero hona zaroori nahi.

True.

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

isliye kyunki

S

ek state function hai, phir bhi

\oint d Q_{r e v} =

net heat

=

net work

\neq = 0

generally — woh non-zero loop integral exactly wahi reason hai kyun raw heat inexact hai aur

1/ T

integrating factor ki zaroorat padti hai.

System ki entropy ek real process ke dauran kam ho sakti hai.

True. Sirf universe ki entropy force se upar jaati hai; ek system apni entropy surroundings mein dump kar sakta hai (jaise ek fridge paani thanda karta hai), jab tak surroundings zyada gain karein.

Same states ke beech do alag reversible paths alag

Δ S

dete hain.

False. Kyunki

S

ek state function hai, fixed endpoints ke beech har reversible path identical

Δ S

deta hai — woh path-independence hi

T

se divide karne ka pura point hai.

Spot the error

"Gas freely vacuum mein expand hoti hai,

Q = 0

, isliye

Δ S = 0

Error yeh hai ki irreversible path ki actual heat ko

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

mein use kar rahe hain. Path ko same endpoints ke saath ek reversible isothermal se replace karo:

Δ S = n R ln (V_{2} / V_{1}) > 0

"Ek irreversible engine ka entropy change

\oint d Q / T

hai, jise hum zero set karte hain."

Ek irreversible cycle ke liye Clausius Inequality deta hai

\oint d Q / T < 0

, naa ki

= 0

; sirf reversible cycles equality tak pahunche hain. Ise zero set karna chupke se reversibility assume karta hai.

"Paani garam karte waqt, main

Δ S = Q / T

use karunga final temperature

T

ke saath."

System temperature poore process mein change hoti hai, isliye ek single

T

galat hai; tumhe local

T

use karke

d S = m c d T / T

integrate karna padega, jo deta hai

Δ S = m c ln (T_{2} / T_{1})

d Q_{r e v} / T

mein

T

reservoir temperature hai, isliye system entropy ke liye main reservoir ka

T

use karta hoon."

System entropy ke liye system ki temperature use karo. Reversible exchange mein dono temperatures sirf infinitesimally alag hoti hain isliye fark nahi padta, lekin system bookkeeping ke liye hamesha system ka

T

lo.

"Kyunki

Δ S_{u ni v er se} \geq 0

hai, koi bhi process kabhi bhi kisi bhi entropy ko kam nahi kar sakta."

Bound universe par hai, parts par nahi. Ek local region (system) thanda ho sakta hai aur entropy kho sakta hai, bas surroundings ko kam se kam utna gain karna chahiye.

"Carnot cycle ke liye

Q_{H} = Q_{C}

hai kyunki woh apne start par wapas aata hai."

Jo start par wapas aata hai woh

U

hai (isliye net

Q =

net

W

), individual heats nahi. Jo actually balance hota hai woh hai

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

, yaani entropy exchange, jo deta hai

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

"Entropy ki units joules hain kyunki yeh heat se aati hai."

Entropy heat divided by temperature hai, isliye iska unit

J K^{- 1}

hai, J nahi —

T

se division exactly wahi hai jo dimensions change karta hai.

Why questions

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

mein heat reversible kyun honi chahiye?

Sirf ek reversible path ke along hi

d Q / T

ek exact (path-independent) differential banta hai jo true state change ke barabar hota hai; irreversible

Q / T

Δ S

se kam padta hai.

1/ T

heat ka "integrating factor" kyun hai aur koi aur function kyun nahi?

Kyunki Carnot Cycle and Efficiency force karta hai

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

, toh

d Q_{r e v}

1/ T

se multiply karna har reversible loop ko zero par integrate karta hai — yeh exact differential ki defining property hai. Dekho Exact and Inexact Differentials.

T

se divide karna ek path-dependent quantity ko path-independent kyun bana deta hai?

Raw heat

\oint d Q_{r e v} \neq = 0

ek loop ke around, lekin

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

(Clausius theorem, tiny Carnot cycles se tiling karke); ek quantity jiska zero loop integral ho woh ek state function ka differential hai.

Wahi heat ek thande body mein zyada bada entropy jump kyun create karti hai ek garam body ki tulna mein?

Kyunki

d S = d Q / T

mein denominator mein

T

hai: chhota

T

d Q / T

bada karta hai. Physically, kisi cheez mein jo already "quiet" hai energy add karna proportionally zyada naye microstates kholata hai.

Agar definition mein reversibility chahiye toh hum kisi irreversible process ka

Δ S

compute kar hi kyun sakte hain?

Kyunki

S

ek state function hai, iska change sirf endpoints par depend karta hai. Hum koi bhi reversible path same endpoints ke saath invent karte hain aur uspe integrate karte hain; answer wahi true

Δ S

hai, chahe actually kuch bhi hua ho.

Entropy time ko ek direction kyun deti hai ("arrow of time")?

Real processes sirf

Δ S_{u ni v er se}

badhate hain; reverse (spontaneously un-mixing, un-spreading) use kam karta aur kabhi nahi hota, isliye increasing-entropy direction "future" label karta hai. Microscopically yeh Statistical Entropy — Boltzmann S = k ln W hai.

Surroundings ka

Δ S

sirf

- Δ S_{sy s t e m}

kyun nahi hai?

Sirf reversible processes mein hi woh exactly cancel karte hain. Irreversible processes mein surroundings zyada gain karte hain jitna system khoata hai, ek positive

Δ S_{u ni v er se}

chodke.

Edge cases

Ek reversible cycle: ek poore loop mein system ka total

Δ S

kya hai?

Zero. System apne start state par wapas aata hai aur

S

ek state function hai, isliye

\oint d S = 0

— chahe boundary ke across kitni bhi heat gayi ho.

T = 0 K

par isothermal expansion: kya

Δ S = n R ln (V_{2} / V_{1})

abhi bhi finite hai?

Formula finite hai, lekin

T = 0

unreachable hai (third law); zyada zaroori baat yeh hai ki

T \to 0

par koi bhi real heat exchange

d Q / T

blow up karta hai, yeh signal deta hai ki finite steps mein absolute zero kyun nahi paaya ja sakta.

Same

T

aur

P

par do identical gases mix karo:

Δ S

kya hai?

Zero. Koi bhi distinguishable state change nahi hai ("before" aur "after" physically identical hain), isliye koi naye microstates nahi khulte — yeh Gibbs paradox ka resolution hai.

Q > 0

(heat absorbed) lekin

Δ S_{sy s t e m} < 0

wala process: possible hai?

Ek single reversible input ke liye nahi, kyunki

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

d Q_{r e v}

ka sign share karta hai. Yeh sirf tab ho sakta hai jab saath mein work/other paths state ko reshape karein; pure reversible heat exchange ke liye,

Q > 0 \Rightarrow Δ S_{sy s t e m} > 0

Reversible isothermal compression (

V_{2} < V_{1}

Δ S

ka sign?

Negative, kyunki

ln (V_{2} / V_{1}) < 0

. Squeeze karna accessible microstates hatata hai — lekin surroundings expelled heat absorb karte hain aur kam se kam utna gain karte hain,

Δ S_{u ni v er se} \geq 0

rakhte hue.

Perfect insulation aur perfectly reversible:

Δ S_{sy s t e m}

kya hai?

Zero — yeh ek reversible adiabat (isentropic) hai. Dono conditions zaroori hain: sirf reversibility ya sirf insulation

Δ S = 0

force nahi karta.

Equal temperature par do bodies ke beech heat flow: entropy ka kya hota hai?

Kuch net nahi — equal

T

par koi spontaneous flow nahi hota, aur koi bhi infinitesimal reversible exchange mein ek side ka

+ d Q / T

, doosre side ke

- d Q / T

se cancel hota hai, isliye

Δ S_{u ni v er se} = 0

. Unequal

T

hi flow ko irreversible aur

Δ S_{u ni v er se} > 0

banata hai.

Connections

Clausius Inequality — woh $\oint d Q / T \leq 0$ jo zyatar "spot the error" items ko power karta hai
Reversible vs Irreversible Processes — woh line jis par yahan har trap khada hai
Second Law of Thermodynamics — $Δ S_{u ni v er se} \geq 0$ ka source
Carnot Cycle and Efficiency — jahan se $Q_{H} / T_{H} = Q_{C} / T_{C}$ aata hai
Exact and Inexact Differentials — kyun $1/ T$ heat ko integrate karta hai
Statistical Entropy — Boltzmann S = k ln W — "more spread out" ke peeche microstate meaning
First Law of Thermodynamics — $Q$ aur $W$ ke neeche energy accounting