γ = Cp - Cv — for monatomic, diatomic, polyatomic: Visual walkthrough

1.7.17 · D2 · Physics › Thermodynamics › γ = Cp - Cv — for monatomic, diatomic, polyatomic

Yeh parent topic hai jo poori tarah ek visual walkthrough ke roop mein bataya gaya hai. Har symbol ko use karne se pehle earn kiya gaya hai.

Step 1 — Heat, aur yeh kahan ja sakti hai

WHAT. Ek sealed box of gas imagine karo. Hum usme thodi si heat dalte hain, jise $d Q$ likhte hain ("heat Q ki ek chhoti si matra"). Woh heat gayab nahi ho sakti — use kahin na kahin reappear karna hoga. Woh sirf do jagah ja sakti hai:

WHY. Yeh bookkeeping rule First Law of Thermodynamics hai. Yeh kehta hai ki heat do hisson mein batti hai: $d Q = molecules ke andar store hoti hai d U + bahar dhakelte waqt kharch hoti hai d W$

Term by term:

$d Q$ — woh chhoti heat jo hum daal rahe hain (input).
$d U$ — internal energy mein rise: molecules ki tez, zyada wild motion. Energy jo gas ke andar rehti hai.
$d W$ — gas dwara apne surroundings par kiya gaya work, yaani piston ko bahar dhakelnaa.

PICTURE. Heat arrow do arrows mein split hoti hai — ek "andar jaati hai" aur ek "bahar dhakelta hai."

Step 2 — "No-piston" experiment: $C_{v}$ define karo

WHAT. Piston ko lock karo taaki volume change na ho sake. Gas kuch bhi push nahi kar sakti, isliye $d W = 0$ . Ab heat ka har joule sirf molecules ke andar jaane ke liye majboor hai.

WHY. Hum "energy stored inside" ko "energy spent working" se alag karna chahte hain. Piston bolt karne se work term cleanly khatam ho jaata hai: $d V = 0 \Rightarrow d W = P d V = 0 \Rightarrow d Q = d U$

Term by term:

$d V$ — volume mein chhota change; hum isse zero force karte hain.
$P d V$ — expansion ka work (pressure $\times$ swept volume); zero jab $d V = 0$ ho.
To $d Q$ ka sara hissa $d U$ ban jaata hai.

Molar heat capacity at constant volume ka matlab hai "har degree, har mole ke liye chahiye heat, volume fixed rakh ke": $C_{v} \equiv (\frac{d Q}{d T})_{V} = \frac{d U}{d T}$

$C_{v}$ — kitne joules ek mole ko ek kelvin badhate hain, piston bolted.
$()_{V}$ — chhota $V$ matlab "volume constant rakha."

PICTURE. Bolted piston; heat arrow poori tarah "hotter molecules" thermometer mein flow karti hai.

Step 3 — "Free-piston" experiment: $C_{p}$ define karo

WHAT. Ab piston ko freely slide karne do, jo constant pressure $P$ par dabaya hua hai (maano atmosphere ka bojh). Utni hi heat daalo. Gas garm hoti hai — lekin jaise garm hoti hai, woh expand hoti hai, piston ko utha leti hai. Woh uthana work hai, aur yeh tumhari heat ka ek hissa chura leta hai.

WHY. Temperature badhte waqt pressure fix rakhne ke liye, volume ko grow karna hi hoga. Ek ideal gas ke liye, $P V = R T$ (ek mole ke liye), to constant $P$ par: $P d V = R d T$

Term by term:

$P$ — constant pressure jo neeche dab raha hai.
$R$ — universal gas constant, $8.314 J/mol\cdotK$ , temperature aur gas energy ke beech fixed conversion.
$R d T$ — expansion work jo tum warming ke har degree ke liye zaroor pay karte ho.

Ise First Law $d Q = d U + P d V$ mein daalo aur $d T$ se divide karo: $C_{p} = \frac{d U}{d T} + R = C_{v} + R$

$C_{p}$ — constant pressure par molar heat capacity.
Extra $+ R$ exactly woh piston-lifting tax hai.

PICTURE. Free piston $d V$ se upar uthta hai; heat arrow split hoti hai — zyada thermometer mein, ek hissa side mein jaata hai jis par likha hai "lifting work $R d T$ ."

Step 4 — Mayer's relation: do experiments compare karte hain

WHAT. Dono experiments ko side by side rakho. Sirf ek fark tha — piston-lifting tax $R$ . Subtract karo: $C_{p} - C_{v} = R$

Yeh Mayer's Relation hai. Yeh kehta hai do heat capacities ka gap hamesha exactly $R$ hota hai, chahe koi bhi gas ho — jab tak woh $P V = R T$ follow kare.

WHY. Kyunki constant pressure par tum same internal-energy rise plus ek fixed lump of expansion work pay karte ho. Woh lump $R$ per mole per kelvin hai, bas itna hi.

PICTURE. Do bar charts: $C_{p}$ woh $C_{v}$ hai jis par ek extra orange block of height $R$ upar stacked hai.

Step 5 — "Motion boxes" ginna (degrees of freedom)

WHAT. Ek mole kitni internal energy rakhta hai? Yeh depend karta hai ki uske molecules kitne independent tareekon se move kar sakte hain. Har aisa independent tarika ek degree of freedom hai, jise $f$ likhte hain.

WHY yeh idea, koi aur nahi. Hume $C_{v} = d U / d T$ chahiye, to hume $U$ chahiye. Equipartition theorem hume $U$ free mein de deta hai: nature thermal energy ko saari active motion boxes mein equally share karta hai, har box ko exactly $\frac{1}{2} R T$ per mole deta hai.

Ek single atom (ek marble): woh $x$ , $y$ , $z$ ke saath move kar sakta hai — 3 translational boxes. Ek point ko spin karna meaningless hai (pakadne ko kuch nahi) → $0$ rotational.
Ek dumbbell (diatomic): 3 translational + 2 rotational (woh do axes ke baare mein tumble karta hai; apne pattale bond ke baare mein spinning almost kuch bhi store nahi karti) → $f = 5$ .
Ek non-linear molecule: 3 translational + 3 rotational → $f = 6$ .

PICTURE. Teen molecules chote arrows ke saath: marble (3 seedhe arrows), dumbbell (3 seedhe + 2 curved), tripod-shaped molecule (3 seedhe + 3 curved).

$U = \frac{f}{2} R T \Rightarrow C_{v} = \frac{d U}{d T} = \frac{f}{2} R$

$\frac{f}{2} R T$ — $f$ boxes mein se har ek $\frac{1}{2} R T$ carry karta hai; sabko add karo.
$C_{v} = \frac{f}{2} R$ — $T$ ke respect mein differentiate karo; $T$ drop ho jaata hai, sirf ek pure count bachta hai.

Step 6 — Count se γ assemble karo

WHAT. Step 4 ke ratio mein $C_{v} = \frac{f}{2} R$ daalo: $γ = 1 + \frac{R}{C _{v}} = 1 + \frac{R}{\frac{f}{2} R} = 1 + \frac{2}{f}$

Term by term:

$R$ 's cancel ho jaate hain — γ ke koi units nahi, yeh ek bare number hai.
$\frac{2}{f}$ — molecular shape par poori dependency yahan hai. Zyada boxes $f$ → chhota fraction → γ 1 ki taraf slide karta hai.

WHY. Ek "fancier" molecule incoming heat ko zyada boxes mein spread karta hai, to uska temperature zyada slowly badhta hai, to $C_{v}$ bada hota hai, to $R$ -tax relatively kam mayne rakhta hai — aur γ shrink hota hai.

PICTURE. Curve $γ = 1 + 2/ f$ plot kiya gaya, teen real gases us par dots ki tarah marked hain.

Step 7 — Edge cases (agar koi box freeze ya unfreeze ho jaye?)

WHAT. Count $f$ patthar mein nahi likha — yeh depend karta hai ki kaun se boxes active hain.

WHY. Quantum mechanics kuch boxes band rakhta hai jab tak itni heat nahi milti ki woh khul sakein. Yahi woh jagah hai jahan γ badal sakta hai:

Bahut thanda diatomic gas. Rotation bhi freeze out ho sakta hai → sirf 3 translational boxes bachte hain → $f \to 3$ , $γ \to \frac{5}{3}$ . Ek diatomic gas thande mein monatomic jaisa pretend kar sakta hai.
Ordinary temperature diatomic. Rotation active, vibration abhi frozen → $f = 5$ , $γ = 1.40$ . (Yeh roz ka case hai.)
Bahut garam diatomic. Vibration unfreeze hoti hai, 2 aur boxes add karta hai (jiggling bond ka kinetic + potential) → $f \to 7$ , $γ \to \frac{9}{7} \approx 1.29$ .

Degenerate limit. Jaise $f \to \infty$ (socho ek molecule jiske countless wobbling parts hain), $\frac{2}{f} \to 0$ , to $γ \to 1^{+}$ . γ 1 ke paas aa sakta hai lekin kabhi reach ya neeche nahi jaata — kyunki $C_{p} > C_{v}$ hamesha hold karta hai (Step 4). γ $(1, \frac{5}{3}]$ mein trapped hai.

PICTURE. Ek diatomic gas ke liye ek staircase: jaise $T$ badhti hai, $f$ steps 3 → 5 → 7, aur γ steps down 1.67 → 1.40 → 1.29.

Ek-picture summary

Upar sab kuch ek single flow mein collapse ho jaata hai: First Law → do experiments → Mayer → equipartition → γ = 1 + 2/f → teen gases.

Recall Feynman: poori walkthrough simple words mein

Gas mein heat daalo. Piston bolt karo aur saari heat wilder motion ban jaati hai — yahi $C_{v}$ hai, "heat per degree jab kuch push karne ko nahi." Piston slide karne do aur gas ko garm hote waqt usse bhi uthana padta hai; woh ek fixed extra lump cost karta hai, exactly $R$ per mole per degree — isliye $C_{p}$ hamesha $C_{v} + R$ hota hai (Mayer). To $γ = C_{p} / C_{v} = 1 + R / C_{v}$ . Ab, $C_{v}$ kitna bada hai? Count karo ki molecule kitne tareekon se move kar sakti hai — uske "motion boxes." Heat unme khud ko equally share karti hai, $\frac{1}{2} R T$ har ek mein, to $C_{v} = \frac{f}{2} R$ . Plug in karo: $R$ 's cancel ho jaate hain aur $γ = 1 + 2/ f$ . Ek akela atom 3 boxes rakhta hai (γ = 1.67), ek dumbbell 5 rakhta hai (γ = 1.40), ek floppy molecule 6 rakhta hai (γ = 1.33). Zyada boxes, zyada sharing, slower heating, γ 1 ke paas — lekin kabhi 1 par nahi, kyunki woh piston-lifting tax hamesha dena padta hai.

Recall

$C_{p}$ hamesha exactly $C_{v} + R$ kyun hota hai? ::: Constant pressure par gas expansion work $R d T$ per mole bhi karti hai; woh fixed extra heat Mayer's relation hai. $γ = 1 + 2/ f$ kahan se aata hai? ::: $C_{v} = \frac{f}{2} R$ ko $γ = 1 + R / C_{v}$ mein substitute karo; $R$ 's cancel ho jaate hain. Kya γ kabhi 1 ke equal ya neeche ja sakta hai? ::: Nahi — kyunki $C_{p} > C_{v}$ hamesha hota hai, $γ > 1$ ; yeh sirf 1 ke paas jaata hai jab $f \to \infty$ . Ek real gas ka γ temperature ke saath kyu badalta hai? ::: Motion boxes (rotation, vibration) freeze ya unfreeze hote hain, active $f$ change karte hain.

Connections

First Law of Thermodynamics — woh $d Q = d U + P d V$ jahan se humne shuru kiya (Step 1)
Mayer's Relation — Step 4 mein build kiya gaya $C_{p} - C_{v} = R$
Degrees of Freedom & Equipartition Theorem — $C_{v} = \frac{f}{2} R$ supply karta hai (Step 5)
Internal Energy of Ideal Gas — $U = \frac{f}{2} R T$
Adiabatic Process — yahan yeh γ aage use hota hai, $P V^{γ} = const$ ke zariye
Speed of Sound in Gas — $v = γ R T / M$ directly γ par ride karta hai