Third law of thermodynamics — S → 0 as T → 0: Worked examples

1.7.25 · D3 · Physics › Thermodynamics › Third law of thermodynamics — S → 0 as T → 0

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Scenario matrix

Har Third-Law question inhi cells mein se kisi ek mein aata hai. Neeche jo examples hain unhe us cell se tag kiya gaya hai jo woh cover karte hain, taaki koi cheez chhoote nahi.

Cell	Kya vary karta hai	Question type	Example
A	$W = 1$ (perfect order)	entropy bilkul zero hai	Ex 1
B	$W > 1$ frozen-in, binary	residual entropy $R ln g$	Ex 2
C	$W > 1$ non-binary ( $g = 1.5$ , ice)	fractional-log residual	Ex 3
D	$C_{p} = a T^{3}$ (Debye limit)	integration se absolute entropy	Ex 4
E	$C_{p}$ = pieces ka sum + phase change	full calorimetric ladder	Ex 5
F	$T = 0$ par do states merge ho jaate hain	unattainability / cooling steps	Ex 6
G	Real-world word problem	glass vs crystal, measured mismatch	Ex 7
H	Exam twist / trap	"kya $C_{p} / T$ divergent hai?" limiting check	Ex 8

Teen "degenerate/limiting" edges — $W = 1$ (koi disorder nahi), $C_{p} \to 0$ (integrand tame hai), aur $T \to 0$ unreachable — inhe apna-apna example diya gaya hai taaki koi bhi boundary untested naa rahe.

Cell A — Perfect order, $W = 1$

Worked example Example 1 — 0 K par ek flawless silicon crystal ki Entropy (Cell A)

Q: $N = 5.0 \times 1 0^{22}$ silicon atoms ka ek perfect crystal exactly 0 K tak thanda kiya gaya hai, har atom apni lattice site par lock hai. $S$ nikaaliye.

Forecast: Kya answer $N$ par depend karta hai? Padhne se pehle guess karein.

Arrangements count karein. Perfect crystal → har atom ke liye exactly ek site → poori cheez banane ka sirf ek tarika hai, isliye $W = 1$ . Yeh step kyun? Entropy distinguishable arrangements count karti hai; perfect order mein kahin koi choice nahi bachti.
Counting engine mein feed karein. $S = k_{B} ln W = k_{B} ln 1.$ Yeh step kyun? Hum $T = 0$ par hain, isliye counting engine (integral nahi) sahi tool hai — integrate karne ke liye koi heat hi nahi hai.
Log evaluate karein. $ln 1 = 0$ , isliye $S = k_{B} \cdot 0 = 0$ . Yeh step kyun? $ln 1 = 0$ kyunki $e^{0} = 1$ — "ek tarika" matlab zero information, zero entropy.

Answer: $S = 0 J K^{- 1}$ , $N$ se independent. Woh $5 \times 1 0^{22}$ ek distractor tha.

Verify: Count ko kisi bhi factor se multiply karo — $W = 1$ wahi $1$ rahega, $ln 1 = 0$ wahi $0$ rahega. Units: $k_{B}$ ka unit $J K^{- 1}$ hai, ek pure number se multiply hua, isliye $S$ ka unit $J K^{- 1}$ hai. Consistent hai.

Cell B — Binary frozen disorder

Worked example Example 2 — Solid CO ki Residual Entropy (Cell B)

Q: Solid carbon monoxide mein har molecule ya toh CO ya OC direction mein freeze hota hai, randomly choose karke. Ek mole ke liye $S (0)$ nikaaliye.

Forecast: Har molecule ke liye do options aur $6 \times 1 0^{23}$ molecules — kya answer astronomically bada hoga, ya ek modest number jaise $6 J K^{- 1}$ ?

Total arrangements count karein. Har molecule independently $g = 2$ choices rakhta hai; $N$ molecules ke liye $W = 2^{N}$ . Yeh step kyun? Independent choices multiply hoti hain: $2 \times 2 \times \dots = 2^{N}$ .
Counting engine apply karein. $S = k_{B} ln 2^{N} = N k_{B} ln 2.$ Yeh step kyun? $ln (x^{N}) = N ln x$ se giant power ek manageable product ban jaati hai.
Per mole mein jaao. Ek mole ke liye $N = N_{A}$ , aur $N_{A} k_{B} = R$ : $S (0) = R ln 2 = 8.314 \times 0.6931.$ Yeh step kyun? $N_{A} k_{B}$ ko $R$ se replace karne par per-mole units mil jaate hain.

Answer: $S (0) = R ln 2 \approx 5.76 J mo l^{- 1} K^{- 1}$ .

Verify: Number modest hai (naa ki $1 0^{23}$ ) kyunki log ne exponent ko crush kar diya — yeh ek achha sanity check hai. Units: $R$ ka unit $J mol^{- 1} K^{- 1}$ hai aur ek pure number se multiply hua → correct.

Cell C — Non-binary / fractional log (ice)

Worked example Example 3 — Ice ki Residual Entropy (

$g = 3/2$ ) (Cell C) Q: Water ice mein har oxygen ke 4 hydrogen bonds hote hain; Pauling ki counting ek effective $g = 3/2$ orientations per molecule deti hai. $S (0)$ per mole nikaaliye aur CO se compare karein.

Forecast: Ice ka $g = 1.5$ , CO ke $g = 2$ se chhota hai. Kya ice ki residual entropy CO ke $5.76$ se zyada hogi ya kam?

Effective count. $W = (3/2)^{N}$ — Pauling ne dikhaya ki ice-rules, average par, $3/2$ allowed configurations per molecule chhod dete hain. Yeh step kyun? Number poora integer hona zaroori nahi; yeh har molecule ki ek average multiplicity hai, aur log fractions ko fine handle karta hai.
Per mole counting engine. $S (0) = R ln (\frac{3}{2}) = 8.314 \times ln 1.5.$ Yeh step kyun? Wahi $ln (x^{N}) = N ln x$ wali move; per mole karne par $R ln (3/2)$ milta hai.
Number. $ln 1.5 = 0.4055$ , isliye $S (0) = 8.314 \times 0.4055$ . Yeh step kyun? Log ki value plug karo.

Answer: $S (0) \approx 3.37 J mo l^{- 1} K^{- 1}$ — forecast ke mutabiq CO ke $5.76$ se chhota, kyunki $3/2 < 2$ .

Verify: Monotonic check — chhota $g$ zaroor chhoti entropy dega: $ln 1.5 < ln 2$ ✓. Yeh experimentally measured ice residual entropy ( $\approx 3.4 J mol^{- 1} K^{- 1}$ ) se match karta hai. Dekho Residual entropy of ice and CO.

Cell D — Debye tail se absolute entropy

Worked example Example 4 —

$C_{p} = a T^{3}$ follow karne wale solid ki Entropy (Cell D) Q: 0 K ke paas ek crystal Debye law $C_{p} = a T^{3}$ follow karta hai jahan $a = 4.8 \times 1 0^{- 4} J mol^{- 1} K^{- 4}$ . $T = 10 K$ par $S$ nikaaliye.

Forecast: Kya $S (10)$ roughly $C_{p} (10)$ ke barabar hoga, uska ek-tehai, ya teen guna? (Pehle neeche di gayi figure dekho.)

Zero se integral set up karein. Kyunki $S (0) = 0$ (Third Law), ladder ko absolute zero se shuru karo: $S (T) = \int_{0}^{T} \frac{C _{p} ( T ^{'} )}{T ^{'}} d T^{'} = \int_{0}^{T} \frac{a T ^{'3}}{T ^{'}} d T^{'} .$ Yeh step kyun? Third Law hi woh cheez hai jo hume $0$ se shuru karne deti hai; warna hume sirf entropy ka difference milta.
$T^{'}$ ki ek power cancel karein. $\frac{T ^{'3}}{T ^{'}} = T^{'2}$ , isliye integrand $a T^{'2}$ ho gaya. Yeh step kyun? Yahi wajah hai ki integral safe hai — khatarnaak $1/ T^{'}$ ko $T^{'3}$ kha jaata hai, sirf ek tame $T^{'2}$ bacha rehta hai (Cell H ki yahi chinta thi, jo pehle hi handle ho gayi).
Integrate karein. $\int_{0}^{T} a T^{'2} d T^{'} = a \frac{T ^{3}}{3}$ . Yeh step kyun? Power rule $\int T^{'2} d T^{'} = T^{'3} /3$ ; lower limit $0$ kuch contribute nahi karta.
Shortcut pehchaano. Kyunki $a T^{3} = C_{p} (T)$ , $S (T) = \frac{a T ^{3}}{3} = \frac{C _{p} ( T )}{3} .$ Yeh step kyun? Ek neat Debye rule: $T^{3}$ region mein entropy, heat capacity ki ek-tehai hoti hai.
Number. $C_{p} (10) = 4.8 \times 1 0^{- 4} \times 1 0^{3} = 0.48 J mol^{- 1} K^{- 1}$ , isliye $S (10) = 0.48/3$ .

Answer: $S (10) = 0.16 J mo l^{- 1} K^{- 1}$ (aur $S (10) = C_{p} (10) /3$ , forecast "one-third" se match karta hai).

Verify: Units: $a [J mol^{- 1} K^{- 4}] \times T^{3} [K^{3}] /3 = J mol^{- 1} K^{- 1}$ ✓. Jab $T \to 0$ , $S \to 0$ ✓ (Third Law self-consistent hai). Dekho Heat capacity and Debye T-cubed law.

Cell E — Phase change ke saath full calorimetric ladder

Worked example Example 5 — Melting transition ke through absolute entropy (Cell E)

Q: Ek substance ko 0 K se warm kiya ja raha hai. Debye region ke neeche maano $C_{p} = a T^{3}$ (Ex 4 wala $a$ ) $T_{1} = 10 K$ tak. $10 K$ aur melting point $T_{m} = 200 K$ ke beech solid ka constant $C_{p}^{solid} = 25 J mol^{- 1} K^{- 1}$ hai. $T_{m}$ par latent heat $L_{f u s} = 5000 J mol^{- 1}$ ke saath yeh melt ho jaata hai. Melting ke theek baad total entropy $S (T_{m}^{+})$ nikaaliye.

Forecast: Teen contributions add hoti hain (Debye tail + heating + melting). Aapko kya lagta hai kaun sa dominant hoga — compute karne se pehle guess karo.

Debye piece (0→10 K). Ex 4 se, $S_{1} = C_{p} (10) /3 = 0.16 J mol^{- 1} K^{- 1}$ . Yeh step kyun? Converged Debye integral reuse karo — $T = 0$ ke through jaane ka yahi ek safe tarika hai.
Constant- $C_{p}$ heating (10→200 K). $S_{2} = \int_{10}^{200} \frac{C _{p}^{solid}}{T ^{'}} d T^{'} = C_{p}^{solid} ln \frac{200}{10} = 25 ln 20.$ Yeh step kyun? Jab $C_{p}$ constant ho, $\int d T^{'} / T^{'} = ln (T_{hi} / T_{lo})$ — " $C_{p}$ constant" segments ke liye logarithm woh tool hai.
$S_{2}$ ka number. $ln 20 = 2.996$ , isliye $S_{2} = 25 \times 2.996 = 74.9 J mol^{- 1} K^{- 1}$ .
Melting (fixed $T_{m}$ par phase change). $S_{3} = \frac{L _{f u s}}{T _{m}} = \frac{5000}{200} = 25.0 J mol^{- 1} K^{- 1} .$ Yeh step kyun? Phase change par $T$ constant rehta hai, isliye $Δ S = Q / T = L / T_{m}$ — koi integral ki zaroorat nahi.
Ladder add karo. $S (T_{m}^{+}) = S_{1} + S_{2} + S_{3} = 0.16 + 74.9 + 25.0$ . Yeh step kyun? Entropy ek state function hai — reversible path ke saath contributions seedhe sum ho jaate hain.

Answer: $S (T_{m}^{+}) \approx 100.1 J mo l^{- 1} K^{- 1}$ ; heating segment $S_{2}$ dominant hai, jo forecast se match karta hai agar aapne wide temperature span notice ki thi.

Verify: Har term ka unit $J mol^{- 1} K^{- 1}$ ✓. Debye tail tiny hai (0.16) — sensible, kyunki 0 K ke paas bahut kam heat flow hoti hai. Absolute entropy and standard molar entropy tables bilkul isi tarah build hoti hain.

Cell F — Unattainability / cooling steps

Worked example Example 6 — Har cooling step kyun kam deta hai (Cell F)

Q: Adiabatic demagnetization mein, ek magnetize–demagnetize cycle temperature ko ek fixed amount se nahi balki ek factor se girata hai. Maano har cycle temperature ko $r = 0.5$ se multiply karta hai (half karta hai), starting from $T_{0} = 1 K$ . Exactly $T = 0$ tak pahunchne ke liye kitne cycles chahiye?

Forecast: Ek finite number, ya kabhi nahi?

Steps model karo. $n$ cycles ke baad, $T_{n} = T_{0} r^{n} = (0.5)^{n} K$ . Yeh step kyun? Har step subtract nahi balki scale karta hai, kyunki $T \to 0$ ke saath do entropy curves merge ho jaate hain (Third Law), isliye gains geometrically shrink hoti hain.
$T_{n} = 0$ set karo aur solve karo. $(0.5)^{n} = 0$ ke liye $n \to \infty$ chahiye. Yeh step kyun? Kisi bhi finite power par ek positive number positive hi rehta hai — geometric decay kabhi zero nahi hit karta.
Trend numerically check karo. $T_{1} = 0.5, T_{2} = 0.25, T_{3} = 0.125, \dots$ — hamesha positive, hamesha half hota jaata hai. Yeh step kyun? Figure mein "diminishing returns" ki picture confirm karta hai: curves ek doosre ke paas aate jaate hain lekin kabhi touch nahi karte.

Answer: Infinitely many steps — kisi bhi finite number of operations mein absolute zero unreachable hai (Third Law ki unattainability form).

Verify: $lim_{n \to \infty} (0.5)^{n} = 0$ lekin koi finite $n$ , $0$ nahi deta ✓. Adiabatic demagnetization and reaching low temperatures ke saath physically consistent hai.

Cell G — Real-world word problem

Worked example Example 7 — Glassy glycerol: measured vs. crystal entropy (Cell G)

Q: Glycerol ke liye, crystal ki calorimetric entropy 298 K par ( $S (0) = 0$ ke saath) integrate hoke ek spectroscopic value deti hai. Lekin agar aap glass ko cool karke uski $C_{p}$ integrate karo, toh aapko ek value milti hai jo $\approx 19 J mol^{- 1} K^{- 1}$ se zyada kam hai. Is deficit ko Third Law se interpret karo aur har molecule ke freeze configurations ki effective sankhya $g$ estimate karo.

Forecast: Kya $g$ 1 ke paas hai (nearly ordered), 2 ke paas (binary), ya bahut zyada bada (very disordered)?

Deficit diagnose karo. Glass bahut saari random configurations mein freeze ho jaata hai, isliye uska saccha $S (0) = k_{B} ln g^{N} = R ln g$ per mole zero nahi hai. Uski $C_{p}$ ko zero se integrate karne par $S (0) = 0$ maana jaata hai, isliye yeh exactly $R ln g$ miss kar deta hai. Yeh step kyun? Sirf ek perfect crystal ka $S (0) = 0$ hota hai; glass Cell B/C ka disguised version hai — residual entropy missing $R ln g$ mein chhupi hai.
Measured deficit se match karo. $R ln g = 19 ⟹ ln g = \frac{19}{8.314} = 2.285.$ Yeh step kyun? Deficit hi residual entropy hai.
$g$ solve karo. $g = e^{2.285} = 9.8 \approx 10$ . Yeh step kyun? Log ko undo karne ke liye exponentiate karo.

Answer: $g \approx 10$ frozen configurations per molecule — "2 se bahut zyada bada," ek strongly disordered glass.

Verify: $R ln 10 = 8.314 \times 2.303 = 19.1 J mol^{- 1} K^{- 1}$ ✓ — deficit reproduce ho gaya. Bilkul isi tarah chemists lab mein residual entropy detect karte hain.

Cell H — Exam twist: "diverging integrand" trap

Worked example Example 8 — Kya

$\int_{0}^{T} C_{p} / T^{'} d T^{'}$ blow up karta hai? (Cell H) Q: Ek student panic kar raha hai: " $C_{p} / T^{'}$ mein ek $1/ T^{'}$ hai jo $T^{'} = 0$ par diverge karta hai, isliye absolute entropy undefined hai." Debye tail $C_{p} = a T^{'3}$ ke liye integrand ki limit check karke ise resolve karo.

Forecast: Kya integrand $T^{'} \to 0$ ke saath $\infty$ jaata hai, ek finite constant par rehta hai, ya $0$ jaata hai?

Integrand likho. $f (T^{'}) = \frac{C _{p}}{T ^{'}} = \frac{a T ^{'3}}{T ^{'}} = a T^{'2}$ . Yeh step kyun? Naive darr yeh ignore karta hai ki $C_{p}$ khud bhi vanish hoti hai; judge karne se pehle substitute karna zaroori hai.
Limit lo. $T^{'} \to 0 lim a T^{'2} = 0$ . Yeh step kyun? $T^{'2} \to 0$ — integrand sirf finite nahi rehta, yeh zero ho jaata hai. Koi divergence nahi.
Bura case contrast karo. Agar $C_{p}$ ek constant $C_{0}$ hoti (jo Third Law 0 K ke paas forbid karta hai), toh $f = C_{0} / T^{'} \to \infty$ aur integral truly diverge karta. Yeh step kyun? Dikhata hai ki Third Law ki requirement $C_{p} \to 0$ hi woh cheez hai jo integral ko bachati hai — yeh luck nahi hai.

Answer: Integrand $\to 0$ jaata hai; integral converge karta hai. Trap dissolve ho jaata hai jab aapko yaad aata hai ki $C_{p} \to 0$ hota hai.

Verify: $T^{'} = 0.01 K$ par numerically: $f = 4.8 \times 1 0^{- 4} \times (0.01)^{2} = 4.8 \times 1 0^{- 8}$ — vanishingly small ✓, convergence confirm karta hai.

Cell coverage recap

Recall Kya humne har cell hit ki?

A (Ex1 $W = 1$ ) ::: bilkul zero entropy B (Ex2 binary CO) ::: $R ln 2 \approx 5.76$ C (Ex3 ice $g = 3/2$ ) ::: $R ln 1.5 \approx 3.37$ D (Ex4 Debye) ::: $S = C_{p} /3 = 0.16$ at 10 K E (Ex5 ladder) ::: $\approx 100.1$ J mol⁻¹K⁻¹ ek melt ke saath F (Ex6 steps) ::: infinite steps, unattainable G (Ex7 glass) ::: $g \approx 10$ ek $19$ J deficit se H (Ex8 trap) ::: integrand $\to 0$ , converge karta hai

Connections

Boltzmann entropy S = k ln W — counting engine (Cells A, B, C, G).
Second law of thermodynamics — $d S = δ Q / T$ jo Cells D, E, H mein use hua.
Heat capacity and Debye T-cubed law — $C_{p} = a T^{3}$ tail (Cells D, E, H).
Absolute entropy and standard molar entropy — Cell E se bani payoff table.
Adiabatic demagnetization and reaching low temperatures — lab mein Cell F.
Residual entropy of ice and CO — Cells B, C, G exceptions.
Parent par wapas: Third Law topic note.

Worked examples — Third law of thermodynamics — S → 0 as T → 0