Entropy and disorder — Boltzmann S = k·ln(W): Worked examples

1.7.24 · D3 · Physics › Thermodynamics › Entropy and disorder — Boltzmann S = k·ln(W)

Tumne formula $S = k ln W$ parent note mein dekha hai. Yahan hum ise tumhari haddiyon mein utar denge — har tarah ki situation pe hit karke jo yeh face kar sakta hai — bada $W$ , chhota $W$ , degenerate $W = 1$ case, ratios, real chemistry, aur ek exam trap. Compute karne se pehle guess karo. Wahi pe learning hoti hai.

Definition Vocabulary refresh (yahan kuch bhi assumed nahi hai)

$W$ ::: number of microstates (alag-alag microscopic arrangements) jo bahar se ek jaisi dikhti hain. Ek plain counting number, hamesha $\geq 1$ .
$k = 1.38 \times 1 0^{- 23} J/K$ ::: Boltzmann constant, woh fixed number jo ek pure count ko joules-per-kelvin mein convert karta hai.
$N_{A} = 6.022 \times 1 0^{23}$ ::: Avogadro's number, ek mole mein particles ki count — ek plain (bahut bada) counting number bina kisi unit ke, jab bhi hum "per molecule" se "per mole" jaate hain tab use hota hai.
$ln$ ::: natural logarithm — sawaal "e ko kis power pe raise karein toh yeh number mile?" Hum ise use karte hain (na ki $W$ ko directly) kyunki counts multiply hote hain jab systems combine hote hain lekin entropy add honi chahiye, aur $ln (x y) = ln x + ln y$ × ko + mein badal deta hai.
$S$ ::: entropy, $J/K$ mein measure hoti hai.

Scenario matrix

Boltzmann formula jo bhi problem throw kare, woh in cells mein se kisi ek mein aati hai. Neeche har worked example us cell ke saath tagged hai jo woh fill karta hai.

Cell	Kya special hai iske baare mein	Example
A. $W = 1$ (degenerate / ordered)	bilkul unique arrangement, entropy ka "zero"	Ex 1
B. Small finite $W$ (combinatorics)	$(n N)$ se haath se count karo	Ex 2
C. Ratio $W_{f} / W_{i}$ (change, absolute nahi)	absolute $W$ unknown, sirf factor matter karta hai	Ex 3
D. Exponential $W \sim a^{N}$ (huge $N$ )	states list nahi ho sakte; $ln (a^{N}) = N ln a$ use karo	Ex 4
E. Real-world numbers (units check)	$k$ plug in karo, joules milenge, magnitude sanity-check karo	Ex 5
F. Limiting behaviour ( $T \to 0$ , $W \to 1$ )	Third-Law edge case	Ex 6
G. Do systems compare karo (kaun bada hai?)	puri numbers ke bina reasoning	Ex 7
H. Exam twist (the trap)	" $W$ doubled → $S$ doubled?" waala catch	Ex 8

Neeche wali figure tumhari is matrix ki map hai: yeh master curve $S / k = ln W$ ko ek baar plot karti hai, aur us par un teen cells ko mark karti hai jinki shape tumhe feel karni chahiye — degenerate point (Cell A), slow doubling nudge (Cell H), aur woh general slow climb jis par Cells B–G chalte hain. Baar baar isko dekhte rehna: har example us magenta curve par kahin na kahin land karta hai.

Figure — Entropy and disorder — Boltzmann S = k·ln(W)

Figure kya dikhati hai (alt text): horizontal axis multiplicity $W$ hai (number of microstates), $1$ se $60$ tak; vertical axis entropy $k$ ki units mein hai, yaani $S / k = ln W$ . Ek single magenta curve $W = 1$ ke paas steeply utha phir flat ho jaata hai — logarithm ki signature slow growth. Ek violet dot $(W = 1, S / k = 0)$ par baitha hai: degenerate zero-entropy point (Cell A). Do orange dots $W = 15$ aur $W = 30$ par hain, dotted horizontal lines se vertical axis se jude hue; un do lines ke beech vertical gap sirf $ln 2 \approx 0.69$ hai, jo dikhata hai ki $W$ double karne se $S$ barely raise hoti hai (Cell H). Har doosra example usi magenta curve par kahin ride karta hai — sirf uska $W$ -coordinate alag hota hai.

Cell A — degenerate $W = 1$ state

Worked example Example 1 — Absolute zero par ek perfect crystal

Ek pure element ka block itna thanda kar diya gaya hai ki har atom apni ek lowest-energy lattice site par frozen baitha hai, kahin koi ambiguity nahi. Use arrange karne ka exactly ek hi tarika hai: $W = 1$ . $S$ nikalo. (Yeh figure par violet dot hai.)

Forecast: padhne se pehle guess karo — kya $S$ small-but-positive hai, exactly zero, ya negative?

$W$ identify karo. Har atom ka ek fixed spot hai, toh poore block ka exactly ek microstate hai. $W = 1$ . Yeh step kyun? Entropy ek count hai; pehle count pakka karna zaroori hai.
Formula apply karo: $S = k ln W = k ln 1$ . Yeh step kyun? Direct substitution — yahan koi combinatorics ki zaroorat nahi.
$ln 1 = 0$ use karo (kyunki $e^{0} = 1$ ). Toh $S = k \cdot 0 = 0$ . Yeh step kyun? Ek ka log zero hota hai: " $e$ ko power 0 par raise karein toh 1 milta hai."

Answer: $S = 0 J/K$ .

Verify: Sirf ek arrangement ka matlab hai "koi missing information nahi" — kuch bhi uncertain nahi, toh entropy vanish honi chahiye. Yeh exactly Third Law of Thermodynamics hai: jab $T \to 0$ , $W \to 1$ , $S \to 0$ . Units: $k$ J/K hai times dimensionless $ln$ , toh $S$ J/K hai. ✓

Cell B — haath se small finite $W$

Worked example Example 2 — Chhe coins, exactly chaar heads

$N = 6$ distinguishable coins toss karo. Macrostate "exactly $n = 4$ heads" consider karo. $W$ aur $S$ nikalo. (Magenta curve par $W = 15$ pe land karta hai.)

Forecast: "all heads" macrostate se zyada arrangements ya kam? (Apni gut trust karo.)

Macrostate fix karta hai kitne heads, kaun se nahi. Count karo ki 6 coins mein se kaun se 4 heads dikhate hain: $W = (4 6)$ . Yeh step kyun? Ek subset choose karna combination hai — picking ka order matter nahi karta, toh $(n N)$ .
Compute karo $(4 6) = \frac{6 !}{4 ! 2 !} = \frac{720}{24 \cdot 2} = 15$ . Yeh step kyun? $(n N) = \frac{N !}{n ! ( N - n )!}$ distinct subsets ki sankhya hai.
Entropy: $S = k ln 15 = 1.38 \times 1 0^{- 23} \times ln 15$ . Yeh step kyun? Ab $W$ ek plain number hai; substitute karo.
$ln 15 \approx 2.708$ , toh $S \approx 1.38 \times 1 0^{- 23} \times 2.708 \approx 3.74 \times 1 0^{- 23} J/K$ . Yeh step kyun? Yeh count → log → times $k$ ka final "times $k$ " hai: hum $ln 15$ numerically evaluate karte hain, phir $k$ se multiply karte hain taaki pure count ko joules-per-kelvin mein convert kar sakein.

Answer: $W = 15$ , $S \approx 3.74 \times 1 0^{- 23} J/K$ .

Verify: "All heads" ( $n = 6$ ) mein $(6 6) = 1$ hai, entropy 0. Fifteen $>$ one, toh mixed macrostate zyada probable hai aur zyada entropy hai — intuition se match karta hai ki middling ("disordered") outcomes dominate karte hain. ✓

Cell C — ratio $W_{f} / W_{i}$

Worked example Example 3 — Partition hatana (free expansion,

$N = 3$ ) Teen distinguishable gas molecules box ke left half mein hain. Partition hata di gayi toh har ek ab poore box mein ghoom sakta hai (double volume). $W$ kitne factor se bada, aur $Δ S$ kya hai?

Forecast: compute karne se pehle multiplicity factor guess karo.

Har molecule ki position-choices ki sankhya double hoti hai (double accessible volume). Yeh step kyun? Double space → har molecule ke liye double position-states.
Independent molecules ki choices multiply hoti hain: $W_{f} = 2 \times 2 \times 2 \times W_{i} = 2^{3} W_{i} = 8 W_{i}$ . Yeh step kyun? Independent choices multiply hoti hain (yeh counting ka multiplication rule hai).
Entropy change sirf ratio use karta hai: $Δ S = k ln W_{f} - k ln W_{i} = k ln \frac{W _{f}}{W _{i}} = k ln 2^{3} = 3 k ln 2.$ Yeh step kyun? $ln W_{f} - ln W_{i} = ln (W_{f} / W_{i})$ — unknown absolute $W_{i}$ cancel ho jaata hai, isliye hum ise jaane bina solve kar sakte hain.
Numerically $3 k ln 2 = 3 (1.38 \times 1 0^{- 23}) (0.6931) \approx 2.87 \times 1 0^{- 23} J/K$ . Yeh step kyun? Phir se "times $k$ " finish: $ln 2 \approx 0.6931$ pure-count part hai, aur $3 k$ se multiply karne se woh physical J/K units mein aa jaata hai taaki measured entropies se compare kar sakein.

Answer: factor $8$ ; $Δ S = 3 k ln 2 \approx 2.87 \times 1 0^{- 23} J/K$ .

Verify: General law $Δ S = N k ln 2$ with $N = 3$ gives $3 k ln 2$ . ✓ Dekho Free Expansion of an Ideal Gas.

Cell D — exponential $W \sim a^{N}$ , huge $N$

Worked example Example 4 — Ek mole apna volume double karta hai

Ab wahi free expansion ek mole ke liye ho, $N = N_{A} = 6.022 \times 1 0^{23}$ molecules (yaad karo $N_{A}$ Avogadro's number hai, ek mole mein particles ki count). $Δ S$ nikalo.

Forecast: kya answer astronomically large hoga, ya everyday-sized number of J/K?

Ex 3 jaisi physics lekin $N = N_{A}$ : $W_{f} = 2^{N_{A}} W_{i}$ . Hum $2^{N_{A}}$ states list nahi kar sakte — isliye hum kabhi directly $W$ ko touch nahi karte. Yeh step kyun? $ln$ exponential ko tame karta hai: $ln (a^{N}) = N ln a$ .
$Δ S = k ln 2^{N_{A}} = N_{A} k ln 2$ . Yeh step kyun? Log $N_{A}$ ko front par kheench laata hai — yahi wajah hai ki formula mein $ln$ aata hai.
Recognize karo $N_{A} k = R = 8.314 J/(mol\cdotK)$ , gas constant. Toh $Δ S = R ln 2$ . Yeh step kyun? Boltzmann-per-molecule × Avogadro = universal-per-mole; stat-mech ko classical thermo se jodata hai.
$R ln 2 = 8.314 \times 0.6931 \approx 5.76 J/K$ . Yeh step kyun? Final numeric evaluation: $ln 2$ count part hai, $R$ units carry karta hai, toh product directly ek mole ke liye J/K mein hai.

Answer: $Δ S = R ln 2 \approx 5.76 J/K$ .

Verify: $W$ ka $2^{6 \times 1 0^{23}}$ tak balloon karne ke bawajood, entropy ek humble $5.76 J/K$ hai — kyunki $ln$ bahut dheere dheere barhta hai. Yeh isothermal doubling ke liye classical Clausius result ke barabar hai. Statistical aur thermodynamic entropy agree karte hain. ✓

Cell E — real-world numbers, units audit

Worked example Example 5 — Do isotopes ek crystal mein mix hote hain

Ek crystal mein $N = 100$ lattice sites hain; $50$ isotope A se bhare hain, $50$ isotope B se, randomly arrange hue. Bhale hi woh ek tidy solid jaisa lagta ho, isotopes kai microstates dete hain. Is configurational disorder se $S$ nikalo.

Forecast: ek rigid, ordered-looking crystal — kya $S$ zero hoga ya clearly positive?

Count karo ki 100 sites mein se kaun se 50 isotope A rakhte hain: $W = (50 100)$ . Yeh step kyun? A ko sites ke subset par rakhna phir se ek combination hai.
Yeh enormous hai; $ln$ use karo. Numerically $(50 100) \approx 1.0089 \times 1 0^{29}$ , toh $ln W \approx 66.78$ . Yeh step kyun? Humein $ln W$ chahiye, $W$ nahi; log number ko manageable rakhta hai.
$S = k ln W = 1.38 \times 1 0^{- 23} \times 66.78 \approx 9.22 \times 1 0^{- 22} J/K$ . Yeh step kyun? "Times $k$ " close karta hai: $ln W = 66.78$ pure count hai; $k$ se multiply karne par physical entropy J/K mein milti hai.

Answer: $S \approx 9.2 \times 1 0^{- 22} J/K$ — chhota lekin zero nahi.

Verify: Yeh parent note ka steel-man hai: ek "perfectly ordered" crystal phir bhi isotopic arrangements se entropy carry kar sakta hai. Tidiness dekh ke mislead ho jaate ho — tumhe count karna hoga. Units: J/K. ✓

Cell F — limiting behaviour, $W \to 1$

Worked example Example 6 — Ek two-level spin ko absolute zero ki taraf thanda karna

Har spin ke paas sirf do energy levels hain $ε$ gap se alag. Hum sirf un do clean limits ke baare mein poochhte hain jahan accessible states equally likely hain, toh Boltzmann ka $S = k ln W$ exactly apply hota hai. $T \to 0$ aur $T \to \infty$ par $W$ count karo aur $S$ nikalo.

Forecast: $T \to 0$ par, kya $S$ $0$ ki taraf jaata hai, $k ln 2$ ki taraf, ya blow up karta hai?

Cold limit $T \to 0$ : upper level tak pahunchne ke liye thermal energy bilkul nahi hai (cost $ε$ hai), toh har spin apne single ground level mein baitha hai. Accessible states per spin: $W = 1$ . Yeh step kyun? Jab sirf ek state reachable ho, count unambiguous hota hai — koi partition-function subtlety nahi, bas $W = 1$ .
Phir $S = k ln 1 = 0$ . Yeh step kyun? Degenerate count → zero entropy, phir se Cell A result.
Hot limit $T \to \infty$ : thermal energy gap $ε$ se kahin zyada hai, toh dono levels equal probability se reach hote hain. Accessible, equally-likely states per spin: $W = 2$ . Yeh step kyun? Do equally-likely states exactly woh regime hai jahan $S = k ln W$ valid hai — hum honestly $W = 2$ count karte hain.
Phir $S = k ln 2 \approx 0.693 k \approx 9.57 \times 1 0^{- 24} J/K$ per spin. Yeh step kyun? "Times $k$ " close karta hai: $ln 2$ pure count part hai, $k$ ise ek spin ke liye J/K mein convert karta hai.

Answer: $S \to 0$ jab $T \to 0$ ; $S \to k ln 2$ jab $T \to \infty$ .

Verify: $T \to 0 \Rightarrow S \to 0$ exactly Third Law of Thermodynamics hai. High- $T$ ceiling $k ln 2$ kehta hai "do equally likely states = 1 bit of uncertainty," Information Entropy (Shannon) echo karta hai. (Intermediate $T$ par do states equally likely nahi hain, toh zyada general $S = - k \sum_{i} p_{i} ln p_{i}$ use karna padega Boltzmann Distribution se — $ln W$ nahi; exactly isliye humne sirf do clean limits tak restrict kiya.) ✓

Cell G — puri numbers ke bina compare karo

Worked example Example 7 — Zyada entropy kiske paas hai: solid ya liquid?

$W$ compute kiye bina, argue karo ki same substance ke liye $S_{liquid} > S_{solid}$ hai ya nahi.

Forecast: $Δ S_{melt}$ ka sign predict karo.

Solid mein har molecule lattice site par pin hota hai limited orientations ke saath → few microstates, chhota $W_{s}$ . Yeh step kyun? Kam accessible positions/orientations ⇒ chhota count.
Liquid mein molecules freely translate aur tumble karte hain → kai microstates, $W_{l} ≫ W_{s}$ . Yeh step kyun? Zyada accessible arrangements ⇒ bada count.
Kyunki $ln$ ek increasing function hai, $W_{l} > W_{s} \Rightarrow ln W_{l} > ln W_{s} \Rightarrow S_{l} > S_{s}$ . Yeh step kyun? $ln$ order preserve karta hai — bada input, bada output.
Isliye $Δ S_{melt} = k ln (W_{l} / W_{s}) > 0$ .

Answer: liquid ki entropy zyada hoti hai; melting $S$ badhata hai.

Verify: Everyday fact se match karta hai — melting par heat absorb hoti hai, aur Clausius se $Δ S = Q_{rev} / T > 0$ . Do languages, ek hi conclusion. ✓

Cell H — exam trap

Worked example Example 8 — "

$W$ double hua, toh entropy double hui." Sach ya jhooth? Ek student kehta hai: "Multiplicity $W$ se $2 W$ ho gayi, isliye entropy double ho gayi." Unhe theek karo, aur actual change compute karo agar $W$ $1 0^{20}$ se shuru hua tha. (Yeh figure par orange-dots step hai.)

Forecast: kya $S$ double hogi, ya ek fixed chhoti amount se change hogi?

Student $S \propto W$ ko $S \propto ln W$ se confuse kar raha hai. Change honestly likho: $Δ S = k ln (2 W) - k ln W = k (ln 2 + ln W - ln W) = k ln 2.$ Yeh step kyun? $ln (2 W) = ln 2 + ln W$ ; bada $ln W$ cancel ho jaata hai, sirf $ln 2$ bachta hai.
Toh $W$ double karna hamesha wahi tiny amount add karta hai, starting $W$ se independent: $k ln 2 \approx 9.57 \times 1 0^{- 24} J/K$ . Yeh step kyun? Trap reveal hota hai — answer $W$ ke saath scale hi nahi karta; yeh exactly figure par near-flat orange stretch hai.
Original entropy $S_{i} = k ln (1 0^{20}) = k \times 46.05 \approx 6.36 \times 1 0^{- 22} J/K$ se compare karo. Change ek $\sim 1.5%$ nudge hai, doubling nahi. Yeh step kyun? Additive change ko base value ke perspective mein rakhta hai — $W$ ka doubling $S$ ko sirf lagbhag dedh percent move karta hai.

Answer: Jhooth. $Δ S = k ln 2 \approx 9.57 \times 1 0^{- 24} J/K$ , tiny aur fixed — doubling nahi. Entropy lagbhag $1.5%$ barhti hai, $100%$ nahi.

Verify: $k ln 2$ wahi number hai jo Ex 6 ke high- $T$ limit mein aur (per molecule) free expansion mein aaya tha — "one bit" quantum of entropy. Kyunki $W \sim a^{N}$ , sirf system size $N$ (via $ln a^{N} = N ln a$ ) $S$ double kar sakta hai; sirf $W$ double karna kabhi nahi kar sakta. ✓

Recall Scenario checklist

Kaun sa cell sirf ratio chahta hai, kabhi absolute $W$ nahi? ::: Cell C (aur D, H) — $ln$ $W_{f} / W_{i}$ ko difference mein badal deta hai aur unknown $W_{i}$ cancel ho jaata hai. Jab bhi $W = 1$ ho toh $S$ kya hoti hai? ::: $0$ , kyunki $ln 1 = 0$ (Cells A aur F cold limit). $W$ double karne se $S$ kitni change hoti hai? ::: Ek fixed $k ln 2$ , starting $W$ chahe kuch bhi ho (Cell H). Cell F mein intermediate $T$ par $ln W$ kyun use nahi kar sakte? ::: Wahan states equally likely nahi hain, toh $S = - k \sum_{i} p_{i} ln p_{i}$ use karna padega.

Connections

Parent topic — Boltzmann entropy
Microstates and Macrostates — woh counting jo $W$ produce karta hai.
Free Expansion of an Ideal Gas — Cells C & D.
Clausius Entropy dS = dQ_rev / T — Cells D & G verify karta hai.
Third Law of Thermodynamics — Cells A & F.
Boltzmann Distribution — jahan Cell F ka equally-vs-unequally-likely sawaal settle hota hai, aur $S = - k \sum_{i} p_{i} ln p_{i}$ ka source.
Information Entropy (Shannon) — "one bit = $k ln 2$ " theme.
Second Law of Thermodynamics — kyun systems biggest- $W$ cell ki taraf drift karte hain.