Entropy — Clausius definition dS = dQ_rev - T: Worked examples

1.7.22 · D3 · Physics › Thermodynamics › Entropy — Clausius definition dS = dQ_rev - T

Yeh page Clausius definition $d S = \frac{d Q _{r e v}}{T}$ ke liye "flight simulator" hai, jo parent note Entropy — Clausius definition se aata hai. Exam problem touch karne se pehle, tumhe har tarah ke case se guzarna chahiye jo entropy tumhare saamne rakh sakti hai: heat andar, heat bahar, temperature constant, temperature change hoti hui, adiabatic-aur-reversible, adiabatic-aur-irreversible, do bodies ka milna, aur woh sneaky "answer zero hai" wale traps.

Yahan use kiye gaye har symbol ko parent note mein build kiya gaya hai. Ek line ka refresher:

Scenario matrix

Har entropy problem jo tum miloge woh neeche diye gaye cells mein se ek (ya combination) hogi. Aane wale worked examples mein har example ke saath uska cell tag kiya gaya hai. Note karo ki Cell J (system vs surroundings vs universe) ek bookkeeping layer hai jo kisi bhi process ke upar baitha ho sakta hai — Example 5 ise spontaneous two-body flow ke liye dikhata hai aur Example 8 ise gas-plus-reservoir expansion ke liye, toh ye dono alag applications hain, repeat nahi.

Cell	Situation	$Δ S_{sy s}$ ka Sign	Key move
A	Constant $T$ par Heat andar (isothermal expansion)	$+$	$Q_{r e v} / T$ , $T$ bahar aa jaata hai
B	Constant $T$ par Heat bahar (isothermal compression)	$-$	same, $Q < 0$
C	Temperature badh rahi hai (heating, variable $T$ )	$+$	$\int m c d T / T$ integrate karo
D	Temperature gir rahi hai (cooling)	$-$	same, $T_{2} < T_{1}$
E	Reversible adiabatic ( $Q = 0$ , reversible)	$0$	isentropic — trap-buster
F	Irreversible adiabatic (free expansion, $Q = 0$ )	$+$	reversible substitute path
G	Do bodies thermal equilibrium tak (irreversible mixing)	overall $+$	do variable- $T$ integrals ka sum
H	General ideal-gas state change (dono $T$ aur $V$ change)	koi bhi sign	two-log master formula
I	Real-world word problem (ice melting)	$+$	constant $T$ par latent heat
J	Bookkeeping layer: system vs surroundings vs universe	har ek check karo	$Δ S_{u ni v}$ compute karo (Ex 5 = spontaneous flow, Ex 8 = gas + reservoir)

Ab hum saare das cells cover karenge aath examples ke saath (kuch examples ek saath do cells hit karte hain).

Example 1 — Isothermal expansion aur compression (Cells A + B)

Forecast: padhne se pehle signs guess karo. Expansion → gas ko zyada jagah milti hai → molecules ke liye zyada jagah → zyada mess. Compression ise exactly undo karna chahiye.

Har leg par $Δ U = 0$ . Yeh step kyun? Ideal gas ki internal energy sirf $T$ par depend karti hai, aur $T$ fixed hai, isliye $Δ U = 0$ .
First Law → saari heat work ban jaati hai: $Δ U = 0$ ke saath First Law $d U = d Q_{r e v} - P d V$ deta hai $d Q_{r e v} = P d V$ . Integral log kyun ban jaata hai? Ideal-gas law $P = \frac{n R T}{V}$ substitute karo, toh $\int P d V = \int \frac{n R T}{V} d V = n R T ln \frac{V _{2}}{V _{1}}$ ( $T$ constant ke saath), jisse $Q_{r e v} = W = n R T ln \frac{V _{2}}{V _{1}}$ milta hai.
Entropy: $Δ S = \frac{Q _{r e v}}{T} = n R ln \frac{V _{2}}{V _{1}}$ . $T$ kyun bahar aa sakta hai? Yeh constant hai, isliye integral se seedha factor out ho jaata hai.
(a) expansion: $Δ S_{a} = 2 (8.314) ln 3 = + 18.26 J K^{- 1}$ .
(b) compression: $V_{2} / V_{1} = 1/3$ , toh $Δ S_{b} = 2 (8.314) ln \frac{1}{3} = - 18.26 J K^{- 1}$ .

Verify: signs forecast se match karte hain (expand $+$ , compress $-$ ). Round trip deta hai $Δ S_{a} + Δ S_{b} = 0$ — sahi hai, kyunki $S$ ek state function hai aur hum start par wapas aa gaye. Units: $mol \cdot J K^{- 1} mol^{- 1} = J K^{- 1}$ . ✔

Example 2 — Paani ko garam karna aur thanda karna (Cells C + D)

Forecast: garam karne se $Δ S > 0$ hona chahiye, thanda karne se $Δ S < 0$ , aur magnitudes equal.

$d Q_{r e v} = m c d T$ . "Rev" kyun? Reservoirs ki ek infinite ladder imagine karo, jahan har reservoir paani se sirf $d T$ zyada garam ho, toh har slice reversibly move hota hai.
$d S = \frac{m c d T}{T}$ . Sirf $Q / T$ kyun nahi? Kyunki $T$ constant nahi hai — tumhe har slice ka local temperature use karna hoga, isliye integrate karo.
$Δ S = m c ln \frac{T _{2}}{T _{1}}$ . Log kyun? $\int d T / T = ln T$ ; ratio $T_{2} / T_{1}$ bachta hai.
(a): $Δ S_{a} = 0.5 (4186) ln \frac{350}{290} = + 394.9 J K^{- 1}$ . Rounding par note: $c = 4186 J kg^{- 1} K^{- 1}$ ke saath $350/290$ ka log hai $ln (1.2069) = 0.18814$ , toh $0.5 (4186) (0.18814) = 394.9 J K^{- 1}$ (full log carry karo, $Δ T / T$ approximate mat karo).
(b): $Δ S_{b} = 0.5 (4186) ln \frac{290}{350} = - 394.9 J K^{- 1}$ .

Verify: garam karna $+$ , thanda karna $-$ , magnitudes equal (state function, same endpoints reversed). Units $kg \cdot J kg^{- 1} K^{- 1} = J K^{- 1}$ . ✔

Example 3 — Reversible adiabatic (Cell E, "answer zero hai" wala case)

Forecast: $Q = 0$ dekh ke bahut log kehte hain " $Δ S = 0$ trivially." Dhyan se — kya yeh zero sahi wajah se hai?

Adiabatic ka matlab har instant par $d Q_{r e v} = 0$ . Yeh kyun matter karta hai: $d Q_{r e v} / T$ ka numerator poore path par zero hai.
$Δ S = \int \frac{0}{T} = 0$ . Yeh yahan legitimate kyun hai: process reversible hai, isliye definition seedhe actual path par apply hoti hai — koi substitute path ki zaroorat nahi.

Verify: master ideal-gas formula (Example 6) se cross-check karo. Upar ke $γ$ ke baare mein do facts use karke aur $\frac{T _{2}}{T _{1}} = (\frac{V _{1}}{V _{2}})^{γ - 1}$ ko $Δ S = n C_{V} ln \frac{T _{2}}{T _{1}} + n R ln \frac{V _{2}}{V _{1}}$ mein plug karne par exactly $0$ milta hai (numerically checks mein verified). Zero earned hai, assume nahi kiya. ✔

Example 4 — Vacuum mein free expansion (Cell F)

Forecast: $Q = 0$ , $W = 0$ (kuch push nahi kiya), isliye $Δ U = 0$ aur $T$ constant rehta hai. $Δ S = 0$ likhne ki temptation hogi. Kya yeh sahi hai?

Endpoints identify karo: same $T$ , volume $V_{1} \to 2 V_{1}$ . Sirf endpoints kyun? $S$ ek state function hai — actual violent path irrelevant hai.
Ek reversible substitute banao: same endpoints ke beech ek reversible isothermal expansion. Kyun allowed hai? Same states ke beech koi bhi reversible path true $Δ S$ deta hai.
Example 1 ka result apply karo: $Δ S = n R ln \frac{V _{2}}{V _{1}} = (1) (8.314) ln 2 = + 5.76 J K^{- 1}$ .

Verify: $Δ S > 0$ hona chahiye kisi isolated system mein irreversible process ke liye. Actual heat $Q_{a c t u a l} = 0$ enter nahi hoti — yahi parent note ka headline trap hai. ✔

Example 5 — Do bodies equilibrium tak pahunchti hain (Cells G + J, spontaneous flow)

Figure — Entropy — Clausius definition dS = dQ_rev - T

Forecast: metal thanda hoga ( $Δ S < 0$ ), paani garam hoga ( $Δ S > 0$ ). Kyunki process (finite temperature gap ke across heat flow) irreversible hai, guess karo $Δ S_{u ni v} > 0$ . Upar ki figure dikhati hai ki final temperature paani ki value ke paas kyun baithta hai: do curves follow karo — aluminium (orange) bahut neeche girta hai jabki paani (teal) muskil se utha kyunki paani ki heat capacity metal ki heat capacity se kaafi zyada hai, aur woh dashed plum line $T_{f}$ par milte hain.

$T_{f}$ ke liye energy balance: Al ne jo heat khoyi = paani ne jo heat gain ki. $m_{A l} c_{A l} (400 - T_{f}) = m_{w} c_{w} (T_{f} - 300) .$ Kyun? Insulated system se koi heat nahi nikalti, isliye First Law $\Rightarrow$ conserved internal energy. Figure par ise "har curve vertically kitna travel karta hai" ke roop mein padho.
Solve karo: $0.30 (900) (400 - T_{f}) = 1.0 (4186) (T_{f} - 300)$ . Numerically $T_{f} = 306.06 K$ . Kyun sensible hai? Paani ki heat capacity ( $4186 J K^{- 1}$ ) metal ki ( $270 J K^{- 1}$ ) se bahut zyada hai, isliye equilibrium paani ke start ke paas baithta hai — figure mein near-flat teal curve exactly yehi dikhata hai.
Har body ka $Δ S$ — dono ek range par cool/warm hote hain, isliye Cell-C/D integral use karo: $Δ S_{A l} = m_{A l} c_{A l} ln \frac{T _{f}}{400}, Δ S_{w} = m_{w} c_{w} ln \frac{T _{f}}{300} .$ Local $T$ kyun? Dono bodies temperature change karti hain, isliye imagined ladder of reservoirs ke along $m c d T / T$ integrate karo.
Numbers: $Δ S_{A l} = 270 ln \frac{306.06}{400} = - 72.42 J K^{- 1},$ $Δ S_{w} = 4186 ln \frac{306.06}{300} = + 83.73 J K^{- 1} .$ Yeh signs aur sizes kyun follow karte hain: $T_{f} < 400$ isliye aluminium ka log negative hai (yeh thanda hua); $T_{f} > 300$ isliye paani ka log positive hai (yeh garam hua). Paani ka magnitude zyada hai kyunki uska $m c$ (log ke aage multiplier) metal ka roughly 15 times hai, jo uske chhote temperature swing ki kami ko kaafi zyada compensate kar deta hai.
Universe: $Δ S_{u ni v} = Δ S_{A l} + Δ S_{w} = + 11.31 J K^{- 1}$ .

Verify: metal negative, paani positive (forecast ✔). Sum positive hai, jaisa ki Second Law ek spontaneous, irreversible heat flow ke liye demand karta hai finite gap ke across. Agar hum heat ko infinitely many intermediate reservoirs ke through flow hone dete (reversible), toh sum $0$ ki taraf tend karta. ✔

Example 6 — General ideal-gas state change (Cell H, master formula)

Forecast: temperature double hoti hai ( $T$ term se $+$ ) aur volume double hota hai ( $V$ term se $+$ ). Dono entropy ko upar push karte hain, isliye solidly positive answer expect karo.

Master formula (parent mein derive ki gayi): $Δ S = n C_{V} ln \frac{T _{2}}{T _{1}} + n R ln \frac{V _{2}}{V _{1}}$ . Do terms kyun? First Law ko $T$ se divide karne ke baad, ek exact piece $d U = n C_{V} d T$ se aaya aur ek $P d V = n R T d V / V$ se.
Plug in karo $C_{V} = \frac{3}{2} (8.314) = 12.471$ ke saath: $Δ S = 0.5 (12.471) ln 2 + 0.5 (8.314) ln 2.$
Compute karo: $= 4.322 + 2.882 = + 7.20 J K^{- 1}$ .

Verify: dono terms positive (forecast ✔). Tool par sanity check: $T_{2} = T_{1}$ set karne par Example 1 ka isothermal result milta hai; $V_{2} = V_{1}$ set karne par Example 2 ka heating result milta hai. Reversible-adiabatic (Example 3) do terms ko exactly cancel kar deta hai. ✔

Example 7 — Ice melting, real-world word problem (Cell I)

Forecast: solid → liquid matlab molecules ko ghoomne ki freedom milti hai, isliye entropy badhni chahiye.

Heat absorbed: $Q_{r e v} = m L_{f}$ . "Rev" kyun? Melting fixed $273 K$ par hoti hai surroundings ke saath jo sirf infinitesimally zyada garam hote hain — genuinely reversible.
Constant $T$ → $T$ bahar aa jaata hai: $Δ S = \frac{Q _{r e v}}{T} = \frac{m L _{f}}{T}$ . Integrate kyun nahi? Phase change ke dauran temperature change nahi hoti.
Compute karo: $Δ S = \frac{0.20 ( 3.34 \times 1 0 ^{5} )}{273} = + 244.7 J K^{- 1}$ .

Verify: positive (solid → liquid, zyada disorder ✔). Units: $\frac{kg \cdot J kg ^{- 1}}{K} = J K^{- 1}$ . ✔

Example 8 — Exam twist: gas plus reservoir (Cell J, reversible vs irreversible)

Forecast: reversible mein $Δ S_{u ni v} = 0$ exactly hona chahiye; irreversible mein $Δ S_{u ni v} > 0$ hona chahiye.

Reversible case:

System: $Δ S_{sy s} = n R ln 2 = 8.314 ln 2 = + 5.76 J K^{- 1}$ . Kyun? Cell A result.
Surroundings: reservoir same $T$ par same heat $Q = n R T ln 2$ khota hai, isliye $Δ S_{s u r r} = - \frac{Q}{T} = - 5.76 J K^{- 1}$ . Negative kyun? Heat nikalna reservoir ki entropy ko kam karta hai.
Universe: $Δ S_{u ni v} = 5.76 - 5.76 = 0$ . Zero kyun? Reversible ⇒ koi entropy generate nahi hoti.

Irreversible (free) case: 4. System: same endpoints ⇒ same $Δ S_{sy s} = + 5.76 J K^{- 1}$ . 5. Surroundings: $Q_{a c t u a l} = 0$ , reservoir untouched ⇒ $Δ S_{s u r r} = 0$ . Zero kyun? Free expansion mein gas kisi cheez par push nahi karta aur insulated walls koi heat pass nahi karti, isliye reservoir ke saath koi heat exchange nahi hoti — temperature $T$ par zero heat cross hone ke saath, $Δ S_{s u r r} = Q / T = 0$ . 6. Universe: $Δ S_{u ni v} = + 5.76 J K^{- 1} > 0$ .

Verify: reversible → $Δ S_{u ni v} = 0$ ; irreversible → $Δ S_{u ni v} > 0$ (dono forecasts ✔). Dono cases mein same $Δ S_{sy s}$ confirm karta hai ki $S$ ek state function hai; sirf surroundings' entropy dono processes ko alag karti hai. Yahi Clausius Inequality aur Second Law of Thermodynamics ka operational content hai. ✔

Active recall

Recall Kaun sa cell? Scenario ko uske move se match karo (answers chhupao)

"Ice 0 °C par pighalti hai" ::: Cell I — constant- $T$ latent heat, $Δ S = m L_{f} / T$ .
"Gas vacuum mein volume double karta hai, insulated" ::: Cell F — reversible isothermal substitute, $Δ S = n R ln 2 > 0$ .
"Reversible adiabatic compression" ::: Cell E — $Δ S = 0$ (isentropic).
"Garam metal thande paani mein" ::: Cell G — do $m c ln (T_{f} / T_{i})$ terms ka sum, universe $> 0$ .
"Gas ke liye dono $T$ aur $V$ change hote hain" ::: Cell H — master formula $n C_{V} ln (T_{2} / T_{1}) + n R ln (V_{2} / V_{1})$ .

Connections

Parent · Clausius definition
Carnot Cycle and Efficiency — kyun $Q_{H} / T_{H} = Q_{C} / T_{C}$ in sums ko underpin karta hai
First Law of Thermodynamics — Examples 5 aur 6 mein energy balance
Second Law of Thermodynamics — Examples 5 aur 8 mein $Δ S_{u ni v} \geq 0$
Reversible vs Irreversible Processes — Cells E vs F, reversible vs free expansion
Exact and Inexact Differentials — substitute-path trick kyun kaam karta hai
Statistical Entropy — Boltzmann S = k ln W — "zyada jagah ⇒ zyada microstates"
Clausius Inequality — Example 8 ka irreversible universe increase