Exercises — γ = Cp - Cv — for monatomic, diatomic, polyatomic

KYA: Ek monatomic gas ek single ball-jaisa atom hota hai. Woh sirf 3 directions mein move (translate) kar sakta hai: $x,y,z$. Use spin karte hue feel nahi kiya ja sakta (ek point ka koi measurable moment of inertia nahi hota). KYUN: Energy rakhne ka har independent tarika ek degree of freedom hai. Toh $f=3$ (sirf translation). $\gamma =1+\frac{2}{f}=1+\frac{2}{3}=\frac{5}{3}\approx \mathrm{1.67.}$ Yeh figure mein sabse left waala (pale-yellow) dot hai — sabse chota $f$, sabse bada γ.

YEH DOF KYUN: ek diatomic molecule ek dumbbell hai — woh 3 directions mein translate karta hai aur 2 perpendicular axes ke around tumble karta hai, toh room temperature par $f=5$ hota hai (bond-axis spin aur vibration quantum-frozen hain, toh woh koi energy store nahi karte). $C_p=C_v+R$ KYUN: extra $R$ woh expansion work hai jo gas ko constant pressure par har kelvin mein per mole karna padta hai (Mayer's Relation). $C_v=\frac{f}{2}R=\frac{5}{2}R,C_p=C_v+R=\frac{5}{2}R+R=\frac{7}{2}R.$

Jhooth. Kyunki $C_p=C_v+R$ aur dono $C_v>0$ aur $R>0$ hain, toh $C_p>C_v$ milta hai, isliye $\gamma =C_p/C_v>1$ hamesha. Wajah: constant pressure par gas expansion work bhi karta hai, jisme hamesha extra heat lagti hai (dekho First Law of Thermodynamics).

$f=5$ KYUN: nitrogen ek diatomic dumbbell hai, toh 3 translation + 2 rotation se $f=5$ milta hai (vibration room temperature par frozen hai). Yeh figure mein chalk-blue dot hai. KYUN equipartition $C_v$ set karta hai: har active degree of freedom $\frac{1}{2}RT$ per mole store karta hai, toh $U=\frac{f}{2}RT$ aur $C_v=dU/dT=\frac{f}{2}R$ (Degrees of Freedom & Equipartition Theorem). $C_v=\frac{5}{2}R=2.5\times 8.314=20.785{\text{J mol}}^{-1}{\text{K}}^{-1}.$ $C_p$ ke liye $R$ kyun add karte hain: Mayer's relation kehta hai ki constant-pressure capacity, constant-volume capacity plus expansion-work cost $R$ hoti hai. $C_p=C_v+R=20.785+8.314=29.099{\text{J mol}}^{-1}{\text{K}}^{-1}.$ Sanity check: $\gamma =29.099/20.785=1.40$ ✓ — $7/5$ se match karta hai.

INVERT KYUN: γ, $f$ ka ek one-to-one function hai (figure mein curve kabhi ek height repeat nahi karta), toh measured γ exactly ek $f$ pin karta hai. $f$ ke liye $\gamma =1+2/f$ solve karo: $f=\frac{2}{\gamma -1}=\frac{2}{1.40-1}=\frac{2}{0.40}=5.$ YEH GAS KYO IDENTIFY KARTA HAI: $f=5$ matlab 3 translation + 2 rotation, jo ek dumbbell ki pehchaan hai → diatomic (jaise ${\text{O}}_2$, ${\text{N}}_2$). Figure par yeh padhna hai ki "height 1.40 par kaun sa dot baitha hai?" — woh chalk-blue wala $f=5$ par.

KYA: Constant volume par, $Q=n{C}_v\Delta T$ (koi kaam nahi, saari heat → internal energy). $C_v$ KYUN aur $C_p$ nahi: volume fixed hai, toh koi expansion work nahi — First Law of Thermodynamics deta hai $dQ=dU$. Argon monatomic: $C_v=\frac{3}{2}R=1.5\times 8.314=12.471$. $Q=n{C}_v\Delta T=2\times 12.471\times (400-300)=2494.2\text{J}.$

Constant pressure par: $Q=n{C}_p\Delta T$ jahan $C_p=\frac{7}{2}R=3.5\times 8.314=29.099$. $\Delta T=\frac{Q}{n{C}_p}=\frac{5000}{3\times 29.099}=57.27\text{K}.$ Internal energy part: $\Delta U=n{C}_v\Delta T=3\times 20.785\times 57.27=3571.4\text{J}$. Work part — KYUN $W=nR\Delta T$: expanding gas dwara kiya gaya work $W=P\Delta V$ hai (First Law of Thermodynamics). Constant pressure par ideal-gas law $PV=nRT$ deta hai $P\Delta V=nR\Delta T$ (temperature change ko $nR$ se multiply karo, kyunki $P$ aur $n$ fixed hain). Toh expansion work hai $W=P\Delta V=nR\Delta T=3\times 8.314\times 57.27=1428.6\text{J}.$ Check: $3571.4+1428.6=5000$ J ✓. Work fraction $R/C_p=8.314/29.099=0.286$ hai, matlab $28.6\%$ heat piston push karne mein gayi.

2 KYUN, 1 NAHI: ek vibrating bond energy dono motion (kinetic) aur stretch (potential) mein store karta hai; equipartition har quadratic term ko $\frac{1}{2}RT$ deta hai, toh ek vibration = 2 degrees of freedom. Naya $f=5+2=7$. $\gamma =1+\frac{2}{7}=\frac{9}{7}\approx \mathrm{1.286.}$ Room-T value $7/5=1.40$ tha. Toh use garam karna (zyada active DOF) gas ko figure mein curve par rightward aur downward slide karta hai — γ 1 ki taraf gir jaata hai.

Ratio se shuru karo aur upar-neeche dono ko $C_v$ se divide karo: $\frac{C_p-C_v}{C_p}=\frac{\frac{C_p}{C_v}-1}{\frac{C_p}{C_v}}=\frac{\gamma -1}{\gamma }=1-\frac{1}{\gamma }.$ Monatomic ke liye $\gamma =5/3$: $1-\frac{1}{5/3}=1-\frac{3}{5}=\frac{2}{5}=\mathrm{0.4.}$ Matlab: monatomic gas ke liye constant-pressure heat ka $40\%$ expansion work ban jaata hai.

ENERGIES KYO ADD KARTE HAIN: internal energy extensive hoti hai — total $U$ har gas ke $U$ ka sum hai (dekho Internal Energy of Ideal Gas). $C_{v,\text{mix}}=\frac{n_1\left(\frac{3}{2}R\right)+n_2\left(\frac{5}{2}R\right)}{n_1+n_2}=\frac{2(1.5)+3(2.5)}{5}R=\frac{3+7.5}{5}R=2.1R.$ Numerically $C_{v,\text{mix}}=2.1\times 8.314=17.459{\text{J mol}}^{-1}{\text{K}}^{-1}.$ $C_{p,\text{mix}}=C_{v,\text{mix}}+R=3.1R=25.773{\text{J mol}}^{-1}{\text{K}}^{-1}.$ ${\gamma }_{\text{mix}}=\frac{C_{p,\text{mix}}}{C_{v,\text{mix}}}=\frac{3.1}{2.1}=\mathrm{1.476.}$ Dhyaan do ki yeh $5/3=1.67$ (pure He) aur $7/5=1.40$ (pure O₂) ke beech mein aata hai, jaisa hona chahiye.

SIRF γ KYUN MATTER KARTA HAI YAHAN: $v=\sqrt{\gamma RT/M}$ mein $R$, $T$, $M$ dono gases ke liye identical hain, toh ratio mein cancel ho jaate hain aur sirf γ bachta hai (dekho Speed of Sound in Gas). $\frac{v_{\text{mono}}}{v_{\text{dia}}}=\sqrt{\frac{{\gamma }_{\text{mono}}}{{\gamma }_{\text{dia}}}}=\sqrt{\frac{5/3}{7/5}}=\sqrt{\frac{25}{21}}=\mathrm{1.091.}$ Sound monatomic gas mein lagbhag $9.1\%$ tez travel karta hai — ek stiffer, springier medium kyunki uski zyada heat pressure mein jaati hai (rotation mein kam chhupti hai).

YEH LAW KYUN: ek Adiabatic Process mein koi heat nahi nikalti, aur $P{V}^{\gamma }$ constant rehta hai — γ exponent ke roop mein aata hai. $P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }\Rightarrow \frac{P_2}{P_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma }={\left(\frac{1}{2}\right)}^{5/3}.$ $\frac{P_2}{P_1}=2^{-5/3}=\mathrm{0.3150.}$ Pressure apni shuruat ka lagbhag $31.5\%$ ho jaata hai — isothermal $\frac{1}{2}$ se steeper, kyunki gas thandi bhi hoti hai.

Jab $f$ badhta hai, $2/f\to 0$, toh $\gamma \to 1^{+}$ (1 ki taraf upar se approach karta hai, kabhi pahunchta nahi) — yeh figure mein curve ki flat right-hand tail hai. Physical matlab: bahut zyada storage boxes hone par, lagbhag saari incoming heat internal motions mein share ho jaati hai, temperature change ke liye bahut kam bachta hai. Expansion work ka extra $R$ huge $C_v$ ke relative negligible ho jaata hai, toh $C_p/C_v\to 1$. Gas kabhi actually γ = 1 nahi hit kar sakti kyunki $R$ expansion ka ek fixed, non-zero cost hai.

$f=2/(\gamma -1)$ use karo:

• $300$ K par: $f=2/0.400=5.00$.
• $2000$ K par: $f=2/0.350=5.714$. Effective $f$ $5$ se $7$ ki taraf badha. Non-integer $5.71$ ka matlab hai ki vibrational mode partially unfreeze hua hai — quantum mechanics use temperature ke saath gradually activate hone deta hai, toh $2000$ K par yeh apne full 2 DOF ka ek fraction contribute karta hai. Full activation ($f=7$) se $\gamma =9/7=1.286$ milta.

$\gamma =C_p/C_v$ se milta hai $C_p=\gamma C_v$. Ise Mayer's $C_p-C_v=R$ mein dalo: $\gamma C_v-C_v=R\Rightarrow C_v(\gamma -1)=R\Rightarrow \boxed{C_v=\frac{R}{\gamma -1}}.$ Phir $C_p=\gamma C_v=\frac{\gamma R}{\gamma -1}.$ Verify (diatomic, $\gamma =7/5$): $C_v=\frac{R}{7/5-1}=\frac{R}{2/5}=\frac{5}{2}R✓,C_p=\frac{(7/5)R}{2/5}=\frac{7}{2}R✓.$ Yeh parent table se exactly match karta hai — formulas sirf Mayer + definition rearranged hain, koi equipartition nahi chahiye.

L1: γ = 1+2/f padho. L2: plug karo aur invert karo. L3: heat ko $\Delta U$ aur $W$ mein split karo. L4: mixtures mein capacities add hoti hain, γ sound aur adiabatics mein kaam aata hai. L5: general $C_v=R/(\gamma -1)$ prove karo aur fractional $f$ ko partial quantum activation ke roop mein padho.