Third law of thermodynamics — S → 0 as T → 0: Exercises

1.7.25 · D4 · Physics › Thermodynamics › Third law of thermodynamics — S → 0 as T → 0

Do constants hain jo tum har jagah reuse karoge:

Level 1 — Recognition

Kya tum law state kar sakte ho aur uske symbols padh sakte ho?

Exercise 1.1

Third Law ka Planck form ek sentence mein state karo, phir equation likho.

Recall Solution 1.1

Statement: Kisi perfect crystalline substance ke liye, entropy absolute zero temperature approach karne par zero approach karti hai. $lim_{T \to 0} S = 0 (perfect crystal)$ "Perfect" word yahan saara kaam kar raha hai — yeh guarantee karta hai ki ek hi single ground arrangement hogi.

Exercise 1.2

$S = k_{B} ln W$ mein, $S$ , $k_{B}$ , aur $W$ mein se har ek ka plain words mein matlab batao.

Recall Solution 1.2

$S$ = entropy, ek aisa number jo measure karta hai "system kitne tarike se secretly arranged ho sakta hai jabki bahar se same dikhta rahe."
$k_{B}$ = conversion factor — "arrangements ki count" ko J/K ke physical units mein convert karta hai.
$W$ = microstates ki count — distinct arrangements ki number jo hum jo measure karte hain uske saath consistent hain. Dekho Boltzmann entropy S = k ln W.

Exercise 1.3

Ek perfect crystal ka non-degenerate ground state hota hai. $0 K$ par uska $W$ kya hoga, aur phir $S$ ?

Recall Solution 1.3

"Non-degenerate" ka matlab hai exactly ek lowest-energy arrangement, toh $W = 1$ . $S = k_{B} ln 1 = k_{B} \cdot 0 = 0.$ $ln 1 = 0$ kyunki $e^{0} = 1$ : $1$ tak pahunchne ke liye $e$ ka koi factor nahi chahiye, toh "logarithm-power" zero hai.

Level 2 — Application

Formula mein numbers sahi se plug karo.

Exercise 2.1

$N = 5 \times 1 0^{22}$ atoms ka ek perfectly ordered crystal $0 K$ tak cool kiya jaata hai. Uski entropy nikalo.

Recall Solution 2.1

Perfect order $\Rightarrow W = 1$ , chahe kitne bhi atoms hon. $S = k_{B} ln 1 = 0.$ Answer: $S = 0$ , aur yeh $N$ par depend nahi karta. Atoms ki count kabhi enter nahi hoti kyunki poore crystal ka sirf ek hi arrangement hai.

Exercise 2.2

Solid CO mein har molecule random tarike se CO ya OC orient hota hai. 1 mole ki residual entropy nikalo (2 orientations per molecule).

Recall Solution 2.2

$N_{A}$ molecules mein se har ek ke $2$ independent choices hain, toh poore crystal ke liye $W = 2^{N_{A}} .$ $S = k_{B} ln (2^{N_{A}}) = N_{A} k_{B} ln 2 = R ln 2.$ $S = 8.314 \times 0.6931 \approx 5.76 J mo l^{- 1} K^{- 1} .$ Answer: $\approx 5.76 J mol^{- 1} K^{- 1}$ .

Exercise 2.3

$0 K$ ke paas ek solid 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 $S$ nikalo.

Recall Solution 2.3

Kyunki Third Law $S (0) = 0$ fix karta hai, zero se calorimetric ladder integrate karo (dekho Absolute entropy and standard molar entropy): $S (T) = \int_{0}^{T} \frac{C _{p}}{T ^{'}} d T^{'} = \int_{0}^{T} \frac{a T ^{'3}}{T ^{'}} d T^{'} = a \int_{0}^{T} T^{'2} d T^{'} = \frac{a T ^{3}}{3} = \frac{C _{p} ( T )}{3} .$ $T^{'}$ ki ek power denominator ke $T^{'}$ se cancel ho jaati hai — yahi cancellation kyun integral converge karta hai (integrand $\to 0$ ). $S (10) = \frac{4.8 \times 1 0 ^{- 4} \times 1 0 ^{3}}{3} = \frac{0.48}{3} = 0.16 J mo l^{- 1} K^{- 1} .$

Level 3 — Analysis

Explain karo kyun machinery self-consistent hai.

Exercise 3.1

Dikhao ki general (degenerate) form $S (0) = k_{B} ln g$ Planck form mein reduce ho jaati hai, aur $S (0) = 0$ ke liye physical condition state karo.

Recall Solution 3.1

Maano ground state $g$ -fold degenerate hai: $g$ distinct lowest-energy arrangements, toh $0 K$ par $W = g$ . $S (0) = k_{B} ln g .$ Yeh zero hoga iff $ln g = 0$ , yaani $g = 1$ . Perfect crystal precisely woh case hai jab $g = 1$ . Toh Planck ka " $S_{0} = 0$ " special case hai, aur $k_{B} ln g$ general truth hai. Degeneracy $g > 1$ residual entropy hai.

Exercise 3.2

Absolute-entropy integral $S (T) = \int_{0}^{T} \frac{C _{p}}{T ^{'}} d T^{'}$ hai. Explain karo ki yeh lower limit par blow up kyun nahi karta, Debye behaviour use karke.

Recall Solution 3.2

Agar $C_{p}$ zero ke paas ek nonzero constant $c$ hota, toh integrand $c / T^{'}$ hota, jiska integral $c ln T^{'}$ hai — jo $T^{'} \to 0$ hone par $- \infty$ $\to$ hai. Diverges. Lekin Third Law force karta hai ki $C_{p} \to 0$ ; Debye deta hai $C_{p} = a T^{'3}$ , toh integrand $a T^{'2}$ hota hai, jo smoothly $\to 0$ hota hai. Tab $\int_{0}^{T} a T^{'2} d T^{'} = \frac{a T ^{3}}{3}, finite.$ Convergence luck nahi hai — joh law $S (0) = 0$ pin karta hai, wahi law $C_{p}$ ko bhi khatam karta hai. Dekho Heat capacity and Debye T-cubed law.

Exercise 3.3

Ek hypothetical solid $T = 0$ tak $C_{p} = c$ (constant, nonzero) rakhta hai. Prove karo ki yeh Third Law violate karta hai.

Recall Solution 3.3

$S (T) - S (ϵ) = \int_{ϵ}^{T} \frac{c}{T ^{'}} d T^{'} = c [ln T]_{ϵ}^{T} = c ln \frac{T}{ϵ} .$ Jaise lower limit $ϵ \to 0$ , $ln (T / ϵ) \to + \infty$ , toh $S (ϵ) \to - \infty$ . Entropy ek finite constant $0$ ki taraf jaane ki bajay $- \infty$ ki taraf bhag jaayegi. Third Law ek finite limit demand karta hai; isliye constant $C_{p}$ $T \to 0$ par forbidden hai. Isliye $C_{p}$ must vanish karna chahiye.

Level 4 — Synthesis

Kai ideas ya unattainability picture ko combine karo.

Exercise 4.1

Figure dekho. Do entropy curves $S_{A} (T)$ aur $S_{B} (T)$ (ek magnetic salt ki do states) dono $T = 0$ par same value ki taraf girte hain. Picture use karke explain karo ki "isothermal magnetize / adiabatic demagnetize" steps ka koi finite sequence $T = 0$ reach kyun nahi kar sakta.

Figure — Third law of thermodynamics — S → 0 as T → 0

Recall Solution 4.1

Adiabatic step (horizontal, green): heat exchange nahi hota, toh $S$ constant rehta hai, isliye tum curve $B$ se flat line pe left move karte ho aur curve $A$ par lower $T$ par land karte ho.
Isothermal step (vertical, orange): fixed $T$ par tum re-magnetize karte ho, curve $A$ se curve $B$ par jump karte ho, surroundings ko entropy dump karte ho.
Kyunki dono curves $T = 0$ par merge ho jaati hain (yeh $S \to 0$ ka direct consequence hai dono states ke liye), har horizontal step ek aisi $T$ par land karta hai jo zero ke aur karib hai lekin ek ever-smaller amount se — curves ke beech ka gap zero ho jaata hai.
Tumhe ek infinite staircase milti hai jiske steps geometrically shrink karte hain; tum $0$ approach karte ho lekin finite steps mein kabhi pahuncho nahi. Yeh Third Law ka unattainability form hai. Dekho Adiabatic demagnetization and reaching low temperatures.
Crucial link: agar curves merge na karti (yaani agar $S (0)$ states ke beech different hota), toh ek finite step gap bridge kar sakta tha — toh unattainability equivalent hai $S \to$ same limit ke.

Exercise 4.2

Ice mein residual entropy hoti hai kyunki har water molecule ke paas approximately $W_{molecule} = 3/2$ effective proton arrangements hote hain (Pauling). Ek mole ke liye $S (0)$ estimate karo aur CO ke $R ln 2$ se compare karo.

Recall Solution 4.2

Per mole $W = (3/2)^{N_{A}}$ , toh $S (0) = k_{B} ln [(3/2)^{N_{A}}] = R ln \frac{3}{2} .$ $S (0) = 8.314 \times ln 1.5 = 8.314 \times 0.4055 \approx 3.37 J mo l^{- 1} K^{- 1} .$ Yeh CO ke $R ln 2 \approx 5.76$ se kam hai kyunki ice ke paas per molecule fewer effective orientations hain ( $1.5 < 2$ ). Dono naive " $S = 0$ " ke genuine violations hain lekin $S (0) = k_{B} ln g$ ke saath perfectly consistent hain. Dekho Residual entropy of ice and CO.

Level 5 — Mastery

Ek aisa result banao jo parent note ne tumhe directly nahi diya.

Exercise 5.1

Ek two-level paramagnet mein $N$ spins mein se har ek energy $0$ (down) ya $ε$ (up) mein hota hai. Temperature $T$ par upper level mein spins ka fraction $p = \frac{e ^{- ε / k_{B} T}}{1 + e ^{- ε / k_{B} T}}$ hai. (a) $T \to 0$ par $p$ nikalo. (b) Argue karo ki ground state non-degenerate hai aur isliye $S \to 0$ , is model ke liye Third Law confirm karo.

Recall Solution 5.1

(a) Jaise $T \to 0^{+}$ , exponent $- ε / k_{B} T \to - \infty$ (kyunki $ε > 0$ ), toh $e^{- ε / k_{B} T} \to 0$ . Tab $p \to \frac{0}{1 + 0} = 0.$ Har spin lower level mein baithta hai. Exponential kyun? Boltzmann factor $e^{- ε / k_{B} T}$ measure karta hai ki thermal energy $k_{B} T$ spin ko kitni likely ek spin ko upar kick karega; jab $k_{B} T$ khatam hota hai, woh likelihood bhi khatam hoti hai. (b) Saare spins down ek single arrangement hai: $W = 1$ , toh $S = k_{B} ln 1 = 0$ . System Third Law obey karta hai $g = 1$ ke saath. (Agar $ε = 0$ ho — degenerate levels — toh dono orientations $T = 0$ par equally allowed hain, jo $W = 2^{N}$ deta hai aur residual entropy $N k_{B} ln 2$ deta hai; $g > 1$ exception ka ek clean illustration.)

Exercise 5.2

Usi paramagnet ke liye low $T$ ke paas heat capacity $C \approx N k_{B} (\frac{ε}{k _{B} T})^{2} e^{- ε / k_{B} T}$ ki tarah behave karta hai (Schottky form). Dikhao ki $T \to 0$ par $C \to 0$ hota hai, Third-Law requirement ke saath consistent, aur identify karo kaunsa factor jeet ta hai.

Recall Solution 5.2

Maano $x = ε / (k_{B} T)$ . Jaise $T \to 0$ , $x \to + \infty$ , aur $C \propto x^{2} e^{- x} .$ Exponential decay $e^{- x}$ polynomial growth $x^{2}$ ko beat karta hai: large $x$ ke liye, $e^{- x}$ kisi bhi power of $x$ se tezi se shrink karta hai, toh $x^{2} e^{- x} \to 0$ . Isliye $C \to 0$ . Isliye $C / T \to 0$ bhi, entropy integral ko finite rakhta hai — exactly woh self-consistency jo Third Law promise karta hai. Exponential polynomial ko kyun beat karta hai? $x$ mein har unit increase $e^{- x}$ ko fixed factor $1/ e \approx 0.37$ se multiply karta hai jabki $x^{2}$ mein sirf ek bounded relative amount add hota hai; $< 1$ se repeated multiplication kisi bhi power law ko crush kar deta hai.

Recall Tail ka Numeric check

$x = 20$ par: $x^{2} e^{- x} = 400 \times e^{- 20} \approx 400 \times 2.06 \times 1 0^{- 9} \approx 8.2 \times 1 0^{- 7}$ — already essentially zero, confirm karta hai ki $C \to 0$ .

Connections

Boltzmann entropy S = k ln W — har count-of-states step yahan.
Second law of thermodynamics — entropy integral ke liye $d S = δ Q_{r e v} / T$ supply karta hai.
Heat capacity and Debye T-cubed law — L2–L3 mein use hone wala $C_{p} \to 0$ .
Absolute entropy and standard molar entropy — $S (0) = 0$ ka payoff.
Adiabatic demagnetization and reaching low temperatures — L4 staircase figure.
Residual entropy of ice and CO — L2 aur L4 mein $g > 1$ exceptions.