γ = Cp - Cv — for monatomic, diatomic, polyatomic: Question bank

1.7.17 · D5 · Physics › Thermodynamics › γ = Cp - Cv — for monatomic, diatomic, polyatomic

Intuition

Yeh page ek minefield walk hai. Neeche har item mein ek aisi jagah chupi hai jahan students γ = Cp/Cv par phisal jaate hain. Prompt padho, answer cover karo, apna jawab ZOR SE bolo, phir reveal karo. Answer side hamesha kyun batata hai — kyunki "haan/nahi" se kuch nahi seekhte.

Pehle, paanch symbols jinhe hum poori page pe use karte hain, seedhe shabdon mein define kiye hain taaki koi cheez use hone se pehle samajh aa jaaye:

$C_{v}$ = gas ke ek mole ko ek kelvin badhane ke liye chahiye heat (joules mein) jab volume lock ho — piston pina hua, koi expansion nahi.
$C_{p}$ = wohi ek-mole, ek-kelvin heat lekin pressure lock ho — piston freely slide kar sakta hai bahar ki taraf.
$γ$ (Greek "gamma") = ratio $C_{p} / C_{v}$ , ek seedha unit-less number.
$f$ = degrees of freedom = un independent "storage boxes" ki ginti jo ek molecule mein energy jaati hai ( $x$ , $y$ , $z$ mein move karna; kisi axis ke baare mein ghoomna; har ek ek box count hota hai). Har active box mein $\frac{1}{2} R T$ per mole hota hai.
$R$ = ideal-gas constant, $R \approx 8.314 J/(mol\cdotK)$ . Yeh $P V = R T$ (ek mole ke liye) mein fixed conversion factor hai: yeh batata hai ki gas ka ek mole per kelvin kitna "push" karta hai. Yeh har ideal gas ke liye same number hai, isliye yeh baad mein ratios mein cancel ho jaata hai.

Definition Chaar reference relations (sirf quote nahi, banaye gaye hain)

Neeche har trap inhi par tikha hai. Yahan har ek kahan se aata hai — taaki kabhi blindly use na karo: $C_{p} - C_{v} = R, C_{v} = \frac{f}{2} R, γ = 1 + \frac{2}{f}, f = \frac{2}{γ - 1} .$

(1) $C_{p} - C_{v} = R$ (Mayer's relation). First Law of Thermodynamics se shuru karo: heat in = internal-energy rise + work done, yaani $d Q = d U + P d V$ . Volume lock karo ( $d V = 0$ ) aur koi work nahi hoga, toh $C_{v} = d U / d T$ . Ab pressure lock karo: gas expand karti hai, aur ek mole ideal gas ke liye $P V = R T$ se $P d V = R d T$ milta hai. Toh constant-pressure heat wohi $d U$ hai plus extra $R d T$ — isse milta hai $C_{p} = C_{v} + R$ , yaani $C_{p} - C_{v} = R$ . $R$ kyun aata hai: yeh exactly expansion work per kelvin per mole hai. Dekho Mayer's Relation.

(2) $C_{v} = \frac{f}{2} R$ . Equipartition theorem har active box ko $\frac{1}{2} R T$ per mole deta hai, toh $f$ boxes ki internal energy hai $U = \frac{f}{2} R T$ . Kyunki $C_{v} = d U / d T$ hai, differentiate karne se $T$ khatam ho jaata hai aur $C_{v} = \frac{f}{2} R$ milta hai. Kyun: heat capacity bas "energy per kelvin" hai, aur har box $\frac{1}{2} R$ contribute karta hai.

(3) $γ = 1 + \frac{2}{f}$ . (1) ko rearrange karke divide karo, $γ = C_{p} / C_{v} = (C_{v} + R) / C_{v} = 1 + R / C_{v}$ , phir (2) substitute karo: $R / C_{v} = R / (\frac{f}{2} R) = 2/ f$ . Toh $γ = 1 + 2/ f$ — aur gaur karo $R$ cancel ho jaata hai, isliye γ ek pure number hai jo gas constant se independent hai.

(4) $f = \frac{2}{γ - 1}$ . Bas algebra: (3) ko $f$ ke liye solve karo. Yeh "fingerprint ulta padhna" formula hai — γ measure karo, $f$ recover karo.

In relations ko dekhne ke liye, neeche ka map wohi logic trace karta hai jo traps test karte hain:

True or false — justify karo

γ kisi bhi gas ke liye kabhi 1 se kam ya barabar ho sakta hai.

False. Mayer se,

C_{p} = C_{v} + R

jahan

R > 0

, toh

C_{p} > C_{v}

hamesha hota hai aur

γ = C_{p} / C_{v} > 1

; equivalently

γ = 1 + 2/ f

finite

f

ke saath

2/ f

strictly positive rakhta hai. 1 tak pahunchne ke liye

f \to \infty

chahiye, jo kisi molecule mein nahi hota.

Ek ideal monatomic gas ke liye,

γ = 5/3

exactly hota hai, temperature ya pressure se independent.

True. Ek atom

f = 3

(sirf translation) har ordinary condition mein rakhta hai, toh

γ = 1 + 2/3 = 5/3

ek pure number hai jo molecular shape se set hota hai, state variables se nahi.

Gas ko constant pressure aur constant volume par same

Δ T

se heat karne mein same heat lagti hai.

False. Constant-pressure heating ko expansion work

R Δ T

per mole bhi supply karna padta hai, toh zyada heat chahiye — yahi extra

R Δ T

exactly hai kyun

C_{p} > C_{v}

hota hai.

Mayer's relation

C_{p} - C_{v} = R

water vapour ke liye hold karta hai agar use ideal gas treat karein.

True jab tak ideal model ho, kyunki derivation ne sirf

P V = R T

P d V = R d T

use kiya; yeh fail hota hai real/condensed water ke liye jahan

P V = R T

break down hoti hai.

Jitna heavy gas molecule hoga, utna bada γ hoga.

False. γ sirf shape (degrees of freedom) par depend karta hai, mass par nahi; gas constant

R

toh

γ = 1 + 2/ f

mein cancel bhi ho jaata hai. Argon aur radon dono monatomic hain aur

γ = 5/3

share karte hain, bahut alag masses ke bawajood.

Diatomic gas mein hamesha

f = 5

hota hai.

Generally False.

f = 5

room-temperature value hai jab vibration frozen ho; bahut high

T

par vibrational box unfreeze hota hai aur

f \to 7

, γ

9/7

ki taraf girta hai.

γ same hota hai chahe molar heat capacities use karo ya specific (per-kg) heat capacities.

True. γ ek ratio hai, toh molar mass (ya koi bhi per-kg conversion factor) top aur bottom cancel ho jaata hai, number unchanged rehta hai.

Error dhundho

"Diatomic gas ke liye

f = 3 + 2 + 2 = 7

: teen translation, do rotation, do vibration, toh room temperature par

γ = 9/7

Vibration count galat hai. Room temperature par vibrational modes quantum-frozen hote hain, toh woh 2 hata dete hain,

f = 5

aur

γ = 7/5

milta hai.

9/7

answer sirf high

T

par apply hota hai.

"Kyunki

C_{p} - C_{v} = R

hai, toh

γ = C_{p} / C_{v} = R

Subtraction difference deta hai, ratio nahi. Mayer ko divide karne se milta hai

γ = C_{p} / C_{v} = 1 + R / C_{v}

; ratio ko numerator ke difference se replace nahi kar sakte.

"Ek diatomic molecule 3 axes ke around ghoomta hai, toh uske 3 rotational degrees of freedom hain."

Sirf 2 rotational DOF active hote hain. Bond axis ke around ghoomne ka moment of inertia negligible hota hai (mass axis par baith jaata hai), toh woh mode frozen hai — isliye 3 nahi, 2 hote hain.

"Polyatomic non-linear gas ke liye

f = 3

sirf translation, toh

γ = 5/3

Yeh rotation ignore karta hai. Non-linear molecule teeno axes ke around tumble karta hai, 3 rotational DOF add hote hain, toh

f = 6

aur

γ = 4/3

C_{v} = \frac{3}{2} R

har gas ke liye hota hai kyunki

x, y, z

mein se har ek

\frac{1}{2} R T

store karta hai."

Yeh sirf translation hai. Jo molecules rotate karte hain unke zyada boxes hote hain, toh

C_{v} = \frac{f}{2} R

jahan

f \geq 3

; sirf monatomic gases

\frac{3}{2} R

par rukti hain.

"γ badhta hai jab molecule zyada complex hoti hai, kyunki complex cheezein zyada energy store karti hain."

Ulta hai. Zyada storage boxes matlab

f

bada, aur

γ = 1 + 2/ f

chhota hota jaata hai 1 ki taraf. Complex molecules ka γ sabse chhota hota hai (

4/3

), simple ones ka sabse bada (

5/3

C_{p} = C_{v} + R

ka matlab hai extra

R

heat surroundings ko lost ho jaati hai."

Extra

R Δ T

woh work hai jo gas piston bahar dhakelte waqt karti hai, heat leak away nahi hoti. Tumhe yeh heat supply karni padti hai, aur woh mechanical work of expansion mein convert ho jaati hai.

Why questions

Physically,

C_{p}

hamesha

C_{v}

se zyada kyun hota hai?

Constant pressure par gas expand hoti hai aur work karti hai

P d V = R d T

per mole; woh work internal-energy rise ke upar extra heat se pay karni padti hai, toh zyada heat per kelvin chahiye — yahi Mayer's relation ka poora content hai.

γ molecular shape ka "fingerprint" kyun kaam karta hai?

Kyunki

γ = 1 + 2/ f

f = 2/ (γ - 1)

mein invert kiya ja sakta hai: γ measure karo aur

f

pin ho jaata hai, aur

f

(translation hamesha 3, plus rotation 0/2/3) batata hai ki molecule point hai, rod hai, ya 3-D blob hai.

\frac{1}{2} R T

per degree of freedom kyun use karte hain, koi aur fraction kyun nahi?

Equipartition theorem kehta hai ki har quadratic energy term (jaise

\frac{1}{2} m v_{x}^{2}

) thermal equilibrium mein equally share hoti hai, aur woh equal share exactly

\frac{1}{2} R T

per mole per box nikalta hai.

γ dimensionless kyun hai jab

C_{p}

aur

C_{v}

dono J/(mol·K) mein hote hain?

Kyunki γ unka ratio hai; identical units cancel ho jaate hain, aur gas constant

R

bhi

γ = 1 + R / C_{v}

mein cancel ho jaata hai jab

C_{v} = \frac{f}{2} R

substitute karte hain — ek pure number milta hai jo cleanly

P V^{γ} = const

mein appear karta hai.

Sound ki speed γ par depend karti hai, sirf temperature par kyun nahi?

Sound compressions itni fast hoti hain ki adiabatic hoti hain, toh gas ki stiffness adiabatic relation se govern hoti hai (jo γ use karta hai); stiffer response — bada γ —

v = γ R T / M

mein tez sound deta hai.

Mayer's relation real gas ke liye fail kyun ho sakta hai?

Uski derivation ne

P V = R T

P d V = R d T

assume kiya tha. Real gases alag equation of state follow karte hain, toh true difference

T (\partial P / \partial T)_{V} (\partial V / \partial T)_{P}

ban jaata hai, simply

R

nahi.

Marble jaisa atom gases mein sabse bada γ kyun rakhta hai?

Sirf 3 translational boxes aur koi rotation nahi,

f = 3

sabse chhota possible hai, aur jitna chhota

f

, utna bada

2/ f

—

γ = 1 + 2/ f

apne maximum

5/3

tak push ho jaata hai.

Edge cases

Imagined limit

f \to \infty

mein γ kya hoga?

γ = 1 + 2/ f \to 1

. Infinitely many storage boxes gas ko heat swallow karne denge almost bina temperature rise ke,

C_{p}

aur

C_{v}

almost equal ho jaayenge — lekin koi real molecule yahan tak nahi pahunchti.

Temperature badhane par diatomic gas ka γ vibrational threshold ki taraf kya hoga?

Vibrational box unfreeze hota hai,

f

5 se 7 ki taraf badhta hai, toh γ

7/5

9/7

ki taraf girta hai — ek smooth downward step jab woh mode "switch on" hota hai.

Bahut low temperature par diatomic gas

f = 3

jaisi behave kar sakti hai — kyun, aur γ kya hoga tab?

Rotation khud bahut low

T

par quantum-frozen ho jaata hai, sirf translation bachti hai, toh

f = 3

aur

γ = 5/3

— diatomic gas monatomic jaisi behave karti hai.

Monatomic aur diatomic gases ke mixture mein, kya γ bas

5/3

aur

7/5

ka average hai?

Nahi. Internal energies (extensive) add karke

C_{v, mix}

lena padta hai, phir

γ = 1 + R / C_{v, mix}

use karo; mole-weighted average of

C_{v}

lena hoga, do γ values ka naive average nahi.

Kya γ change hota hai agar same gas ke moles double kar do?

Nahi. γ intensive hai:

C_{p}

aur

C_{v}

dono amount ke saath scale hote hain, toh unka ratio untouched rehta hai — ek room bhar argon aur ek flask argon dono

γ = 5/3

share karte hain.

Kya koi gas hai jiska γ exactly

2

ho?

Uske liye

2/ f = 1

, yaani

f = 2

chahiye — har gas ke 3 translational boxes se bhi kam. Toh

γ = 2

kisi real 3-D gas ke liye impossible hai; maximum

5/3

hai

f = 3

par.

Recall Ek-line survival summary

γ ek pure number hai jo 1 se bada hota hai, molecular shape se fix hota hai ( $f$ ke zariye), aur 1 ki taraf shrink karta hai jab molecules zyada storage boxes gain karti hain — aur upar ke har "trap" mein ya toh expansion work ( $R$ ) bhool jaana hai, rotation/vibration galat count karna hai, ya ratio aur difference confuse karna hai.

Connections

Parent: γ = Cp/Cv
Mayer's Relation — woh $C_{p} - C_{v} = R$ jis par aadhe traps depend karte hain
Degrees of Freedom & Equipartition Theorem — har case ke peechhe $f$ counting
First Law of Thermodynamics — kyun constant-P heating zyada cost karti hai
Adiabatic Process — jahan γ $P V^{γ}$ ke roop mein dikhta hai
Speed of Sound in Gas — ek "why question" γ se answer hota hai
Internal Energy of Ideal Gas — woh $U = \frac{f}{2} R T$ jise hum differentiate karte hain