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

1.7.25 · Physics › Thermodynamics

WHAT hai Third Law?

DO forms kyun hain? Nernst ne pehle sirf $Δ S \to 0$ claim kiya tha (jo measurable hai). Planck ne ise strengthen karke " $S_{0} = 0$ khud" bana diya, jisse entropy ka absolute zero point fix ho jaata hai — kuch jo First aur Second Laws kabhi nahi dete.

YEH SACH KYUN HAI? (Statistical mechanics, first principles se)

$S = k_{B} ln W$ se Third Law ka Derivation:

Step 1 — $T \to 0$ par kya hota hai? Thermal energy $\sim k_{B} T \to 0$ . System ab excited nahi ho sakta; woh apni ground state (lowest energy) mein aa jaata hai. Yeh step kyun? Temperature ek measure hai available thermal energy ka jo higher levels ko populate kare; use hatao aur sirf bottom level bachti hai.
Step 2 — Ground states gino. Ek perfect crystal ke liye ground state non-degenerate hoti hai: exactly ek microstate hota hai, $W = 1$ . Yeh step kyun? "Perfect crystal" = har atom apni unique lattice site par locked, koi defect nahi, koi ambiguity nahi → sirf ek arrangement.
Step 3 — Plug in karo. $S = k_{B} ln (1) = k_{B} \cdot 0 = 0.$ Yeh step kyun? $ln 1 = 0$ . Zero disorder ⇒ zero entropy. QED.

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

ABSOLUTE ENTROPY KAISE CALCULATE KARTE HAIN? (Calorimetric ladder)

Kyunki Third Law $S (0) = 0$ pin karta hai, hum heat capacity ko zero se integrate karke ek absolute entropy nikal sakte hain:

Integral $T = 0$ par converge kyun karta hai? Integrand $C_{p} / T$ hai, jo blow up kar deta agar $C_{p}$ constant hoti. Lekin Third Law force karta hai $C_{p} \to 0$ as $T \to 0$ (neeche dekho), jisse $C_{p} / T$ finite hota hai aur integral well-defined. Yeh law self-consistent hai.

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

State the Third Law (Planck form).

Ek perfect crystalline substance ke liye,

S \to 0

T \to 0

Boltzmann's entropy formula aur har symbol ka matlab.

S = k_{B} ln W

;

W

=microstates ki sankhya,

k_{B} = 1.38 \times 1 0^{- 23}

J/K.

Perfect crystal ke liye 0 K par

S = 0

kyun?

Non-degenerate ground state

\Rightarrow W = 1 \Rightarrow S = k_{B} ln 1 = 0

Residual entropy kya hai aur ek example do.

Non-zero

S (0) = k_{B} ln g

ek degenerate frozen-in state se; e.g. CO mein

S_{0} = R ln 2

T \to 0

par heat capacities ke liye Third Law ka consequence.

C_{p}, C_{V} \to 0

(e.g. Debye

C_{V} \propto T^{3}

Third Law use karke absolute entropy ka formula.

S (T) = \int_{0}^{T} \frac{C _{p}}{T ^{'}} d T^{'}

kyunki

S (0) = 0

Entropy integral

T = 0

par converge kyun karta hai?

Kyunki

C_{p} \to 0

T^{3}

, toh

C_{p} / T \to 0

, finite integrand.

Third Law ka Unattainability statement.

Absolute zero finite steps/operations mein reach nahi kiya ja sakta.

Nernst aur Planck forms mein fark.

Nernst:

Δ S \to 0

. Planck (stronger):

S_{0} = 0

khud, entropy ka absolute zero fix karta hai.

1 mole CO ki residual entropy (2 orientations).

S_{0} = R ln 2 \approx 5.76

J mol⁻¹ K⁻¹.

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek box mein magnetic dice hain jo tab tak tumble karte rehte hain jab tak table shake karti hai (woh shaking heat hai). Shake karte waqt dice hazaron alag patterns mein land karte hain — woh messiness entropy hai. Ab dheere-dheere shaking band karo (absolute zero ki taraf cool karo). Dice ek lowest-energy pattern mein settle ho jaate hain. Sirf ek possible pattern hone se koi messiness nahi bachti — entropy zero ho jaati hai. Lekin agar dice sticky hain aur fully settle hone se pehle kuch alag patterns mein freeze ho jaayein, toh thodi si messiness trapped reh jaati hai — woh leftover residual entropy hai.

Connections

Boltzmann entropy S = k ln W — law ke peeche statistical engine.
Second law of thermodynamics — $d S = δ Q_{r e v} / T$ define karta hai jo $S$ integrate karne mein use hota hai.
Heat capacity and Debye T-cubed law — explain karta hai $C_{p} \to 0$ kyun.
Absolute entropy and standard molar entropy — $S (0) = 0$ ka practical payoff.
Adiabatic demagnetization and reaching low temperatures — unattainability action mein.
Residual entropy of ice and CO — exceptions jo "perfect crystal" clause prove karte hain.