Exercises — Boltzmann's entropy S = k_B ln(Ω)

• (a) $S$ entropy hai, units J/K (joule per kelvin). Yeh ek macroscopic, thermodynamic quantity hai.
• (b) $k_B$ Boltzmann's constant hai, $1.380649\times 10^{-23}$ J/K. Iska poora kaam units carry karna hai.
• (c) $\Omega$ multiplicity hai — ek pure number, macrostate se match karne wale microstates ki count. Ek count ke units nahi hote.
• Ek pure number ka $\ln$ bhi pure number hota hai, isliye $\ln \Omega$ dimensionless hai. $S$ mein saare J/K $k_B$ se aate hain. ✔

$\Delta S=S_Y-S_X=k_B(\ln 32-\ln 8)=k_B\ln \frac{32}{8}=k_B\ln 4$. Subtraction ratio kyun ban jaati hai: $\ln a-\ln b=\ln (a/b)$ — logarithms division ko subtraction mein convert kar dete hain. Kyunki $4=2^2$ hai, $\ln 4=2\ln 2\approx 1.386$. Toh $\Delta S\approx 1.386k_B$. ✔

$4$ coins mein se $r$ coins heads hain, yeh choose karne ke tarike binomial coefficient $\left(\genfrac{}{}{0ex}{}{4}{r}\right)=\frac{4!}{r!(4-r)!}$ se mile hain (yeh bas "kitne distinct arrangements mein $r$ heads hain" hai):

• 0 heads: $\left(\genfrac{}{}{0ex}{}{4}{0}\right)=1$
• 1 head: $\left(\genfrac{}{}{0ex}{}{4}{1}\right)=4$
• 2 heads: $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$
• 3 heads: $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$
• 4 heads: $\left(\genfrac{}{}{0ex}{}{4}{4}\right)=1$

Total $=1+4+6+4+1=16=2^4$ ✔ (har coin independently H ya T). Peak "2 heads" hai jahan $\Omega =6$ hai, toh $S/k_B=\ln 6\approx 1.792$. Figure mein bell shape dikhata hai ki kyun middle macrostate dominate karta hai. ✔

Ek gas ki multiplicity ek position part aur ek momentum part mein factorize hoti hai: $\Omega \propto (\text{position volume})\times (\text{momentum-shell volume})$. Sirf position part yahaan change hota hai.

• Momentum part unchanged rehta hai. Vacuum mein free expansion mein gas koi work nahi karta ($p_{\text{ext}}=0$) aur koi heat exchange nahi hoti, toh uski total kinetic energy $E$ unchanged rehti hai. Momentum-shell volume sirf $E$ par depend karta hai (yeh surface $\sum p_i^2=2mE$ hai), toh $E$ fixed hone par momentum count pehle aur baad mein identical hai. Yahi wajah hai ki yeh process ek ideal gas ke liye isothermal bhi hai ($E$ sirf $T$ par depend karta hai).
• Position part har particle ke liye double hota hai. Accessible volume double karne par har particle ke position choices double ho jaati hain, toh $\Omega \to 2^N\Omega$ ($N$ particles mein se har ek independently factor 2 gain karta hai).

$\Delta S=k_B\ln (2^N\Omega )-k_B\ln \Omega =k_{B}N\ln 2=R\ln 2.$ Numerically $\Delta S=8.314\times 0.693147=5.76\text{J/K}$. ✔ Yeh textbook ke $nR\ln (V_f/V_i)$ ke barabar hai jahan $n=1,V_f/V_i=2$ — counting picture classical thermodynamics ko reproduce karta hai.

(a) Har spin independently up/down hai, toh kaunse $n$, $N$ mein se up hain yeh choose karne par $\Omega (n)=\left(\genfrac{}{}{0ex}{}{N}{n}\right)=\frac{N!}{n!(N-n)!}$ milta hai. (b) $\left(\genfrac{}{}{0ex}{}{N}{n}\right)$ $n=N/2=50$ par peak karta hai. Wahaan $\Omega =\left(\genfrac{}{}{0ex}{}{100}{50}\right)\approx 1.0089\times 10^{29}$ hai, toh $\frac{S}{k_B}=\ln \left(\genfrac{}{}{0ex}{}{100}{50}\right)=\mathrm{66.78.}$ (c) Stirling's approximation do grades mein aati hai. Crude form $\ln m!\approx m\ln m-m$ bade factorials handle karta hai lekin sub-leading piece khota hai. Refined form ek aur term rakhta hai: $\ln m!\approx m\ln m-m+\frac{1}{2}\ln (2\pi m).$ Is refined form ko $\ln \left(\genfrac{}{}{0ex}{}{N}{N/2}\right)=\ln N!-2\ln (N/2)!$ mein substitute karo. Leading $m\ln m-m$ pieces collapse hokar $N\ln 2$ ban jaate hain (wahi crude answer), aur teen $\frac{1}{2}\ln (2\pi m)$ terms combine hote hain: $\frac{1}{2}\ln (2\pi N)-2\cdot \frac{1}{2}\ln (2\pi \frac{N}{2})=\frac{1}{2}\ln \frac{2\pi N}{(2\pi N/2{)}^2}=\frac{1}{2}\ln \frac{2}{\pi N}=-\frac{1}{2}\ln \frac{\pi N}{2}.$ Yahi woh correction term hai — ab derive kiya gaya hai, quote nahi kiya. Toh $\ln \left(\genfrac{}{}{0ex}{}{N}{N/2}\right)\approx N\ln 2-\frac{1}{2}\ln (\pi N/2)$. $N=100$ ke saath: $100\ln 2-\frac{1}{2}\ln (50\pi )=69.315-2.531=\mathrm{66.78.}$ Dono do decimals tak agree karte hain — Stirling $N=100$ par bhi kamal ka hai. ✔ (Real gases ke liye $N\sim 10^{23}$ hota hai, jahan exact factorials impossible hain aur Stirling essential hai.)

Log lo, har factor ko explicitly rakhte hue (kuch bhi "const" mein chhupaao mat): $\frac{S}{k_B}=N\ln V-N\ln h^3-\ln N!+\frac{3N}{2}\ln (2\pi mE)-\ln \Gamma (\frac{3N}{2}).$ Ab Stirling dono par apply karo: $\ln N!\approx N\ln N-N$ aur $\ln \Gamma (\frac{3N}{2})\approx \frac{3N}{2}\ln \frac{3N}{2}-\frac{3N}{2}$. Terms group karo: $\frac{S}{k_B}=N\ln \frac{V}{N}+\frac{3N}{2}\ln \left(\frac{2\pi mE}{(3N/2)}\right)+\frac{5N}{2}-3N\ln h.$ $E=\frac{3}{2}N{k}_{B}T$ recognize karke aur tidy karne par mashoor Sackur–Tetrode form milta hai $S=N{k}_B\left[\ln \left(\frac{V}{N}{\left(\frac{4\pi mE}{3N{h}^2}\right)}^{3/2}\right)+\frac{5}{2}\right],$ jismein Planck's constant $h$ log ke andar baitha hai — yahi woh jagah hai jahan quantum mechanics ek classical gas ki entropy mein enter karti hai, aur yahi wajah hai ki $S$ ka ek absolute zero-point hai (Third Law of Thermodynamics). (a) Extensivity test: $V\to \lambda V,N\to \lambda N,E\to \lambda E$ scale karo. Bracket ke andar $\frac{V}{N}\to \frac{V}{N}$ (intensive) aur $\frac{E}{N}\to \frac{E}{N}$ (intensive) hai, toh poora bracket unchanged hai; sirf prefactor $N{k}_B\to \lambda N{k}_B$ hai. Isliye $S\to \lambda S$. Extensive ✔ — Stirling ka $-\ln N!$ hi woh cheez thi jisne $\ln V^N$ ko $N\ln (V/N)$ mein convert kiya, $V$ ko intensive density ratio $V/N$ se replace kiya. (b) $1/N!$ ke bina: Stirling ka $-N\ln N$ gayab ho jaata hai, aur $\frac{S}{k_B}$ mein $N\ln V$ reh jaata hai $N\ln (V/N)$ ki jagah. Scaling ab $\lambda N\ln (\lambda V)=\lambda N\ln V+\lambda N\ln \lambda$ deta hai, ek extra $\lambda N\ln \lambda$ term jo wahan hone ki koi wajah nahi. Do identical gas boxes mix karne par ($\lambda =2$) phir ek spurious entropy of mixing $\Delta S=2N{k}_B\ln 2\ne 0$ predict hoti hai jabki kuch physical hua hi nahi — Gibbs' paradox. $1/N!$ ise remove karta hai. ✔ Gibbs Paradox & Indistinguishability dekho.

(a) $S/k_B=\ln \left(\genfrac{}{}{0ex}{}{N}{n}\right)=\ln \frac{N!}{n!(N-n)!}$. (b) Kyun hum $n$ ko continuous treat kar sakte hain: $\Omega (n)$ discrete steps mein jump karta hai, lekin bade $N$ ke liye $n$ se $n+1$ tak fractional change tiny hota hai ($\sim 1/N$), toh staircase us scale par smooth hai jo matter karta hai aur hum differentiate kar sakte hain. Har factorial par refined Stirling form use karte hue (jaise L3.1 mein derive kiya), $\frac{d}{dn}\ln \left(\genfrac{}{}{0ex}{}{N}{n}\right)=\ln \frac{N-n}{n}(\text{valid for }1\ll n\ll N-1).$ Edge caution: $n=0$ ya $n=N$ par yeh derivative diverge karta hai ($0$ ya $\infty$ ka $\ln$) aur continuous approximation fail hoti hai — wahaan $\Omega =1$, $S=0$ exactly hai, aur ends ko discretely treat karna padega. Edges se door formula excellent hai. Kyunki $E=n\epsilon$ hai, $\frac{\partial }{\partial E}=\frac{1}{\epsilon }\frac{d}{dn}$ hai, toh $\frac{1}{T}=\frac{k_B}{\epsilon }\ln \frac{N-n}{n}.$

• Agar $n<N/2$: $\frac{N-n}{n}>1$, $\ln (\cdot )>0$, toh $T>0$ (normal). Energy add karna microstates badhata hai.
• $n=N/2$ par: $\ln 1=0$, $1/T=0$, toh $T\to +\infty$ — multiplicity maximum hai, entropy peak par hai.
• Agar $n>N/2$: $\frac{N-n}{n}<1$, $\ln (\cdot )<0$, toh $\boxed{T<0}$. Energy add karna ab microstates kam karta hai ($\partial S/\partial E<0$). Negative temperature — kisi bhi positive $T$ se zyada hot. Negative Temperature & Spin Systems dekho. ✔

(c) $N=1000,n=750$ (edges se kaafi door, toh formula apply hota hai): $\frac{N-n}{n}=\frac{250}{750}=\frac{1}{3}$, toh $\ln \frac{1}{3}=-1.0986$. $\frac{1}{T}=\frac{k_B}{\epsilon }(-1.0986)\Rightarrow T=\frac{\epsilon }{k_B}\cdot \frac{1}{-1.0986}=-0.9102\frac{\epsilon }{k_B}.$ Toh $T\approx -0.910\epsilon /k_B$ — negative, jaise predict kiya tha. ✔ Figure mein $S(E)$ rising dikhta hai, $E=N\epsilon /2$ par peak hota hai, phir girta hai; $n=750$ falling ($T<0$) side par hai.

Agar $N_A$ molecules mein se har ek ke $W=3/2$ independent orientational choices hain, toh total multiplicity $\Omega =W^{N_A}$ hai (independent choices multiply hoti hain). Tab $S_0=k_B\ln \left(W^{N_A}\right)=N_A{k}_B\ln W=R\ln \frac{3}{2}.$ Numerically $S_0=8.314\times \ln 1.5=8.314\times 0.405465=3.37\text{J/(mol⋅K)}$. ✔ Yeh experimental $\approx 3.4$ J/(mol·K) se khoobsurat tarike se match karta hai — yeh ek striking confirmation hai ki $S=k_B\ln \Omega$ literally orientational disorder count karta hai, aur Third Law of Thermodynamics ($S\to 0$) ek perfectly ordered crystal ki zaroorat hai, jo ice nahi hai.