Third law of thermodynamics — S → 0 as T → 0: Question bank

1.7.25 · D5 · Physics › Thermodynamics › Third law of thermodynamics — S → 0 as T → 0

Shuru karne se pehle, teen words jinke baare mein tum bilkul sure hone chahiye (parent note ne inhe build kiya — yeh quick reminder hai):

Parent Third Law topic note aur uske Boltzmann entropy S = k ln W bridge ko dekhte raho jab bhi koi symbol unfamiliar lage.

Conclusion sahi hai lekin stated reason loose hai: yeh isliye hold karta hai kyunki har reactant aur product apni khud ki zero ki taraf tend karta hai, isliye difference zero ki taraf tend karta hai — Nernst form. Yeh isliye nahi ki entropy somehow conserved hoti hai.

C_{p} \to 0

, isliye

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

Galat.

C_{p} \to 0

sirf lower limit par hota hai.

0 \to T

range par integrand positive hai, isliye integral ek positive number hai (jaise Debye case ke liye

\frac{1}{3} a T^{3}

), zero nahi.

"Glass ek solid hai, aur solids 0 K par

S = 0

tak pahunchte hain, isliye glass ki 0 K par entropy zero hai."

"Solids reach

S = 0

" sirf perfect crystalline solids ke liye sach hai. Glass amorphous hai — frozen disorder — isliye woh positive residual entropy rakhta hai.

"Temperature thermal energy hai, isliye

T = 0

par atoms ki total energy zero hoti hai aur woh completely ruk jaate hain."

Do slips hain:

T

excited levels ko populate karne ke liye available energy measure karta hai, total energy nahi; aur quantum zero-point motion ground state mein bhi bani rehti hai. Jo khatam hota hai woh thermal excitation hai, ek ground state bachti hai, isliye

S \to 0

"Agar ground state 3-fold degenerate hoti, toh

S (0) = k_{B} \cdot 3

Formula hai

S = k_{B} ln W

, isliye 3-fold degenerate state

S_{0} = k_{B} ln 3

deti hai,

k_{B} \cdot 3

nahi. Logarithm essential hai — yahi woh cheez hai jo

W = 1

ko exactly zero entropy deti hai.

"Kyun" wale questions

Deep reason batao, headline nahi.

Third Law humein absolute entropy define karne deta hai, jabki First aur Second Laws sirf differences dete hain — kyun?

Second Law

d S = δ Q_{r e v} / T

define karta hai, jo sirf

S

mein changes fix karta hai. Third Law integration ka constant pin karta hai:

S (0) = 0

, isliye

C_{p} / T

ko 0 se integrate karne par ek absolute number milta hai. Dekho Absolute entropy and standard molar entropy.

Third Law self-consistent hone ke liye

C_{p} \to 0

T \to 0

kyun zaroori hai?

Agar

C_{p}

finite rehti, toh

\int_{0}^{T} C_{p} / T^{'} d T^{'}

diverge hota, isliye

S

finite nahi ho sakta aur zero par pin nahi ho sakta. Law ko

C_{p} / T \to 0

chahiye, jo require karta hai

C_{p} \to 0

— jo Heat capacity and Debye T-cubed law deliver karta hai.

"Perfect crystal" essential clause kyun hai, decoration nahi?

Sirf ek defect-free, order-locked lattice ka non-degenerate ground state hota hai (

W = 1

). Woh clause hataao aur ground state degenerate ho sakta hai (

g > 1

), residual entropy

k_{B} ln g

bachti hai. Dekho Residual entropy of ice and CO.

Finitely many cooling steps mein absolute zero kyun nahi reach kiya ja sakta?

Har step (jaise adiabatic demagnetization)

T

ko do states ke beech entropy gap exploit karke drop karta hai. Jaise

T \to 0

hota hai, dono entropy curves same value par converge hoti hain, isliye gap — aur har step ki cooling — zero ho jaati hai. Dekho Adiabatic demagnetization and reaching low temperatures.

Thermal energy

\sim k_{B} T \to 0

system ko uske ground state mein kyun force karta hai?

Jab particles ko higher-energy levels mein lift karne ke liye koi thermal energy nahi hoti, toh har excited state ka Boltzmann weight vanish ho jaata hai; sirf lowest level occupied rehta hai, isliye accessible microstate count collapse ho jaata hai.

Residual entropy

N

ke saath scale kyun karta hai lekin perfect-crystal entropy nahi?

Residual entropy har particle ke independent choices se aati hai:

W = g^{N}

, isliye

S = N k_{B} ln g

N

ke saath badhta hai. Perfect crystal mein ek global arrangement hoti hai (

W = 1

)

N

ki parwah kiye bina, isliye

S = 0

hamesha.

S = k_{B} ln W

mein log exactly woh cheez kyun hai jo

W = 1

ko zero deta hai?

ln 1 = 0

: ek arrangement matlab koi choice nahi, koi missing information nahi, koi disorder nahi. Log "one way" ko "zero entropy" mein convert karta hai —

W

ka ek linear function yeh cleanly nahi kar sakta tha.

Edge cases

Woh scenarios jo headline statement quietly skip kar jaati hai.

S (0)

kya hogi agar ground state

g

-fold degenerate ho?

S (0) = k_{B} ln g > 0

— residual entropy. Strict "

S \to 0

" sirf

g = 1

ke liye hold karta hai; general law hai "

S \to

ek constant."

Ideal gas ke liye Third Law ka kya hoga jaise

T \to 0

Ek true ideal gas mein koi interactions nahi hote, isliye woh crystallize nahi kar sakta; classical entropy formula

- \infty

tak jaata. Reality intervene karti hai: real gases 0 K se pehle condense/solidify ho jaate hain, aur quantum statistics le lete hain,

S

ko finite rakhte hain aur ordered solid ke liye

\to 0

Ek substance ki 0 K entropy kya hogi jiska ek unique lowest state ho lekin bahut kareeb upar low-lying excited states bhi hon?

Phir bhi

S (0) = 0

— degeneracy sirf ground state ke baare mein hai only. Nearby excited states small nonzero

T

par entropy affect karte hain, lekin jaise

T \to 0

hota hai unki occupation vanish ho jaati hai aur

S \to k_{B} ln 1 = 0

Do different elements ke perfect crystals — kya unki 0 K par reaction ke liye

Δ S \to 0

hoga?

Haan. Har pure perfect crystal

S = 0

ki taraf tend karta hai, isliye reactants aur products dono zero approach karte hain aur

Δ S \to 0

hota hai (Nernst form) — low-temperature reaction-entropy calculations ka basis.

Ek single frozen-in point defect wala crystal vs.

1 0^{20}

random defects wala crystal — 0 K entropy?

Single fixed defect (ek specified arrangement) phir bhi

W = 1

S = 0

deta hai. Randomly-placed defects kaafi arrangements dete hain,

W ≫ 1

, aur ek positive residual entropy — disorder, sirf imperfection nahi, matter karta hai.

Kya 0 K par zero entropy ka matlab hai ki atoms bilkul still hain?

Nahi. Quantum zero-point motion ground state mein bhi persist karti hai.

S = 0

sirf yeh matlab hai ki ek single quantum state occupied hai (

W = 1

), motion band nahi hui.

Kya entropy exactly 0 K par zero hai, ya sirf zero approach karti hai?

Law ek limit hai:

lim_{T \to 0} S = 0

. Kyunki

T = 0

unattainable hai, crystal literally kabhi

S = 0

par nahi baithti — woh jitna cool karein utna close approach karti hai.

Recall Traps ka one-line summary

Teen killers hain: "perfect crystal" clause bhool jaana (residual entropy exist karti hai), yeh sochna ki $T = 0$ reachable hai (nahi hai), aur entropy integral ke diverge hone ka dar (yeh converge karta hai kyunki $C_{p} \to 0$ ).

Connections

Boltzmann entropy S = k ln W — woh counting engine jis par har jawab depend karta hai.
Second law of thermodynamics — $d S = δ Q_{r e v} / T$ supply karta hai.
Heat capacity and Debye T-cubed law — kyun $C_{p} \to 0$ aur integral converge karta hai.
Absolute entropy and standard molar entropy — $S (0) = 0$ ka payoff.
Adiabatic demagnetization and reaching low temperatures — practice mein unattainability.
Residual entropy of ice and CO — in half traps ke peechhe ke exceptions.

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

True ya false — justify karo

Error pakdo

"Kyun" wale questions

Edge cases

Connections