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

$k=1.38\times 10^{-23}\text{J/K}$ (joules per kelvin). $W=1$ ke liye: $S=k\ln 1=k\cdot 0=0\text{J/K}$. Ek unique state (banane ka sirf ek tarika) zero entropy carry karta hai — yeh Third Law of Thermodynamics ka seed hai.

$S=k\ln 6=(1.38\times 10^{-23})(1.7918)=2.47\times 10^{-23}\text{J/K}.$ Yahan $\ln 6=1.7918$ kyunki $e^{1.7918}\approx 6$. Yeh parent note wala "4 coins mein se 2 heads" macrostate hai.

$W=100$ wala zyada hai (zyada $\ln W$). Fark hai: $\Delta S=k\ln 100-k\ln 10=k\ln \frac{100}{10}=k\ln 10=1.38\times 10^{-23}\times 2.3026=3.18\times 10^{-23}\text{J/K}.$ Note: $W$ ek factor of $10$ se badha, lekin $S$ sirf $k\ln 10$ se badha — ek chhota sa additive bump, tenfold nahi.

$W=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=\frac{5!}{3!2!}=\frac{120}{6\cdot 2}=10.$ $S=k\ln 10=(1.38\times 10^{-23})(2.3026)=3.18\times 10^{-23}\text{J/K}.$ Humne $\left(\genfrac{}{}{0ex}{}{5}{3}\right)$ use kiya kyunki hum choose kar rahe hain ki 5 coins mein se kaun se 3 heads dikhayenge — ek unordered selection hai.

Har molecule independently do gune position-states pata hai, isliye $W_f=2^N{W}_i$ (position choices independent molecules mein multiply hoti hain — Free Expansion of an Ideal Gas dekho). $\Delta S=k\ln \frac{W_f}{W_i}=k\ln 2^{100}=100k\ln 2.$ Numerically $\Delta S=100\times 1.38\times 10^{-23}\times 0.6931=9.57\times 10^{-22}\text{J/K}$.

$\frac{\Delta S}{N}=\frac{Nk\ln 2}{N}=k\ln 2=9.57\times 10^{-24}\text{J/K per molecule}.$ Yeh constant hai kyunki har molecule $W$ mein same factor of 2 contribute karta hai, aur $\ln$ us product ko equal terms $k\ln 2$ ki sum mein badal deta hai. Isliye entropy extensive hai (size $N$ ke saath scale karta hai).

$\left(\genfrac{}{}{0ex}{}{4}{0}\right)=1,\left(\genfrac{}{}{0ex}{}{4}{1}\right)=4,\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6,\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4,\left(\genfrac{}{}{0ex}{}{4}{4}\right)=1.$ Total microstates $=1+4+6+4+1=16=2^4$. ✓ Sabse probable macrostate: $n=2$ (do heads), jisme $W=6$. Fraction $=6/16=0.375=37.5\%$. Peaked middle jeet jaata hai kyunki uske paas sabse zyada tarike hain — Second Law ka seed jo "sabse bade pile ki taraf drift" ke roop mein re-read hota hai.

Multiplicities multiply hoti hain: $W_{AB}=W_A{W}_B=8\times 32=256$. $S_{AB}=k\ln 256.$ Alag alag: $S_A+S_B=k\ln 8+k\ln 32=k(\ln 8+\ln 32)=k\ln (8\cdot 32)=k\ln 256.$ ✓ Numerically $\ln 256=5.545$, isliye $S_{AB}=7.65\times 10^{-23}\text{J/K}$. Additivity exactly $\ln$ property $\ln a+\ln b=\ln (ab)$ hai — isliye logarithm choose kiya gaya tha (dekho Microstates and Macrostates).

$W_f/W_i=50^{N_A}$, isliye $\Delta S=k\ln (50^{N_A})=N_{A}k\ln 50=R\ln 50,$ $R=N_{A}k=8.314\text{J/(mol⋅K)}$ use karke. $\Delta S=8.314\times \ln 50=8.314\times 3.912=32.5\text{J/(mol⋅K)}.$ Positive hoga hi: liquid mein locked-in crystal ki tulna mein bahut zyada accessible microstates hote hain.

Statistical: $\Delta S=N_{A}k\ln 2=R\ln 2$. $\Delta S=8.314\times 0.6931=5.76\text{J/(mol⋅K)}.$ Yeh exactly wahi value hai jo Clausius Entropy dS = dQ_rev / T se milti hai, jab $V$ se $2V$ tak reversible isothermal expansion ke liye $dS=d{Q}_{\text{rev}}/T$ integrate karo ($\Delta S=nR\ln (V_f/V_i)=R\ln 2$). Statistical aur thermodynamic entropy agree karte hain — $S=k\ln W$ ka poora point yehi hai.

All heads: $W=\left(\genfrac{}{}{0ex}{}{10}{\mathrm{0...}}\right)$ — dhyan se, "all heads" matlab $n=10$ hai: $W=\left(\genfrac{}{}{0ex}{}{10}{10}\right)=1$, isliye $S=k\ln 1=0$. Five heads: $W=\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$, isliye $S=k\ln 252=(1.38\times 10^{-23})(5.529)=7.63\times 10^{-23}\text{J/K}$. System apna almost sara time "5 heads" ke paas spend karta hai: woh macrostate $2^{10}=1024$ microstates mein se $252$ ka owner hai ($24.6\%$), jabki "all heads" sirf $1$ ka ($0.098\%$). Zyada $W$ ⇒ zyada $S$ ⇒ system wahan milta hai — Second Law of Thermodynamics ka counting statement.

$\Delta H={\log }_2{2}^{100}=100\text{bits}.$ Har molecule ka left/right choice exactly 1 bit of missing information hai, aur 100 molecules hain. Thermodynamic entropy same count hai jo $k\ln 2$ per bit se re-scale kiya gaya hai: $\Delta S=(k\ln 2)\times 100=9.57\times 10^{-22}\text{J/K},$ jo L2.2 se match karta hai. Same $\ln (\text{count})$ idea — joules vs. bits sirf front mein constant ki choice hai.

Ek per cell: particles distinguishable hain aur cells distinct hain, isliye yeh 3 labelled particles ko 3 labelled cells mein one-to-one assign karne ke tareekon ki sankhya hai $=3!=6$. Entropy $S=k\ln 6=2.47\times 10^{-23}\text{J/K}$. Sab ek cell mein: exactly $1$ arrangement hai, isliye $W=1$ aur $S=k\ln 1=0$. Spread-out macrostate ki entropy strictly zyada hai — zyada distinct orderings. (Yeh Boltzmann Distribution ke peeche ki microscopic logic hai: energy bahut saare accessible cells mein populate hoti hai kyunki aisa karna $W$ maximize karta hai.)

Ground state unique hai ⇒ $W=1$ ⇒ $S=k\ln 1=0$. Yeh Third Law of Thermodynamics hai: ek perfectly ordered pure crystal ke liye $T\to 0$ par entropy zero ho jaati hai. Caveat: agar crystal mein randomly-placed isotopes hain, toh $T=0$ par bhi kaun sa lattice site kaun sa isotope rakhta hai iske bahut saare arrangements hote hain. Tab $W>1$ aur ek residual entropy $S=k\ln W_{\text{isotope}}>0$ bach jaati hai. "Jo order tum dekh sakte ho" woh "$W=1$" ke barabar nahi hai — hamesha microstates count karo, kabhi tidiness mat dekho.

(a) Do independent free expansions, har ek mein $50$ molecules ka volume double ho raha hai: $\Delta S=2\times (50k\ln 2)=100k\ln 2=9.57\times 10^{-22}\text{J/K}.$ (b) Saare $100$ molecules mein se har ek apne position-states double karta hai: $W_f=2^{100}{W}_i$, isliye $\Delta S=100k\ln 2=9.57\times 10^{-22}\text{J/K}.$ Same answer. Kyun: entropy additive hai (parent ke derivation ka Step 2). Chahe tum do 50-molecule contributions alag count karke add karo ($50k\ln 2+50k\ln 2$), ya saare 100 ek saath count karo ($100k\ln 2$), $\ln$ multiplied microstate counts ko same sum mein badal deta hai. Additivity guarantee karti hai ki bookkeeping consistent hai.