Worked examples — γ = Cp - Cv — for monatomic, diatomic, polyatomic

1.7.17 · D3 · Physics › Thermodynamics › γ = Cp - Cv — for monatomic, diatomic, polyatomic

Ye page ek workout hai. Parent note mein do master formulas banaye the; yahan hum unhe har tarah ke sawaalon ke against drill karenge jo exam ya lab mein aa sakte hain. Har example se pehle tumhe answer forecast karna hai — pehle guess karna hi idea ko dimag mein chipkaata hai.

Do tools jo baar baar use honge (dono parent mein prove hue hain):

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Scenario matrix

$\gamma$, $C_p$, $C_v$ ke baare mein har sawaal in cells mein se kisi ek mein aata hai. Neeche ke examples mein cell(s) label ki gayi hain, aur milaakar ye sabhi cells cover karti hain.

#	Cell class	Tricky kyon hai	Covered by
A	Forward: shape → γ	sahi $f$ chunna (rotation count)	Ex 1
B	Backward: γ → f → shape	formula invert karo, molecule type padho	Ex 2
C	Numeric $C_p,C_v$ in J/mol·K	units track karo, Mayer use karo memory nahi	Ex 3
D	Mixture of gases	energies add hoti hain, phir total moles se divide	Ex 4
E	Degenerate / limiting $f$	$f\to \infty$ aur sabse chhota $f=3$	Ex 5
F	High-T unfreezing (vibration)	$f$ jump karta hai $5\to 7$; γ shift hota hai	Ex 6
G	Real-world word problem (sound speed)	γ doosre formula mein jaata hai	Ex 7
H	Exam twist: ratio trap	per-kg vs per-mole; γ unaffected	Ex 8

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Example 1 — Cell A (forward: shape → γ)

Step 1. Neon monatomic hai. Ek point atom $x,y,z$ directions mein translate kar sakta hai → 3 translational boxes; iske rotation feel nahi hote (0 rotational) aur koi bond nahi hai jo vibrate kare. Isliye $f=3$. Ye step kyun? $f$ molecule ki shape se fix hota hai, aur ek akela atom sabse simple shape hai — sirf translation count hoti hai.

Step 2. $C_v=\frac{f}{2}R=\frac{3}{2}R$. Ye step kyun? Equipartition har box ko $\frac{1}{2}R$ deta hai, times $f=3$ boxes.

Step 3. $C_p=C_v+R=\frac{3}{2}R+R=\frac{5}{2}R$. Ye step kyun? Mayer's relation: constant-pressure heating expansion-work ka bill $R$ bhi pay karta hai.

Step 4. $\gamma =\frac{C_p}{C_v}=\frac{5/2}{3/2}=\frac{5}{3}\approx 1.67$.

Verify: Cross-check $\gamma =1+2/f=1+2/3=5/3$ se karo. ✓ Dono routes agree karte hain.

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Example 2 — Cell B (backward: γ → f → shape)

Step 1. Master formula invert karo: $f=\frac{2}{\gamma -1}$. Ye step kyun? Hume $\gamma$ diya gaya hai aur $f$ puchha gaya hai — inverse relation exactly yahi jawaab deta hai ki "kaun sa $f$ ye $\gamma$ produce karta hai?", jaise arctan batata hai "kis angle ka tan ye hai?".

Step 2. $f=\frac{2}{1.40-1}=\frac{2}{0.40}=5$.

Step 3. $f=5=\underset{\text{translation}}{\underbrace{3}}+\underset{\text{rotation}}{\underbrace{2}}$. Do rotational boxes = dumbbell = diatomic (jaise ${\text{O}}_2$, ${\text{N}}_2$) room temperature par. Ye step kyun? $f$ ko hamesha-present 3 translations plus rotations mein tod do; 2 rotations ek linear two-atom molecule ki pehchaan hai (apne bond axis ke around spin freeze hoti hai).

Verify: Forward-check: $\gamma =1+2/5=7/5=1.40$. ✓ Diye gaye value se match karta hai.

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Example 3 — Cell C (numeric heat capacities)

Step 1. Diatomic ⇒ $f=5$, isliye $C_v=\frac{5}{2}R=2.5\times 8.314=20.785\text{J/mol⋅K}$. Ye step kyun? Shape ka $f$, $C_v=\frac{f}{2}R$ mein daalo; numbers rakhne se units $\text{J/mol⋅K}$ mein aate hain.

Step 2. $C_p=C_v+R=20.785+8.314=29.099\text{J/mol⋅K}$. Ye step kyun? Phir se Mayer — $C_p$ alag se memorize karne ki zaroorat nahi; ye hamesha $C_v+R$ hota hai.

Step 3. $\gamma =\frac{C_p}{C_v}=\frac{29.099}{20.785}=1.400$.

Verify: Fark $C_p-C_v=29.099-20.785=8.314=R$. ✓ Exactly ek $R$, jaisa Mayer demand karta hai. Units: poore mein J/mol·K. ✓

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Example 4 — Cell D (mixture)

Step 0 (condition state karo). Kyunki dono gases ek hi temperature $T$ share karti hain, hum har gas ki internal energy us same $T$ ke saath likh sakte hain aur unhe simply add kar sakte hain. Agar alag-alag temperatures hote to pehle thermal equilibrium tak pahunchne dena padta. Ye step kyun? Ideal gas ki internal energy $U=\frac{f}{2}nRT$ hai — do energies ko add karna tab hi sense deta hai jab common $T$ ho.

Step 1. Internal energy extensive hoti hai — ye add hoti hai (dono ke liye same $T$): $U=n_1\left(\frac{3}{2}RT\right)+n_2\left(\frac{5}{2}RT\right).$ Ye step kyun? Har gas apni energy independently store karti hai; total energy sum hai, average nahi.

Step 2. Molar (per-mole) $C_v$ paane ke liye total moles se divide karo: $C_{v,\text{mix}}=\frac{n_1\frac{3}{2}R+n_2\frac{5}{2}R}{n_1+n_2}=\frac{2(1.5)+3(2.5)}{5}R=\frac{3+7.5}{5}R=\frac{10.5}{5}R=2.1R.$ Ye step kyun? Molar heat capacity energy per mole hai, isliye $5$ total moles se normalize karte hain.

Step 3. $C_{p,\text{mix}}=C_{v,\text{mix}}+R=2.1R+R=3.1R$ (Mayer ideal mixture ke liye bhi valid hai).

Step 4. ${\gamma }_{\text{mix}}=\frac{3.1}{2.1}=1.476$.

Verify: ${\gamma }_{\text{mix}}=1.476$, $1.40$ aur $1.67$ ke beech mein hai, jaisa kisi bhi weighted average ko hona chahiye. $1.40$ ke kareeb kyunki mole count mein oxygen dominant hai. ✓

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Example 5 — Cell E (degenerate & limiting $f$)

Figure: horizontal axis = degrees of freedom $f$; vertical axis = heat-capacity ratio $\gamma$. Red curve $\gamma =1+2/f$ hai. Teen gases coloured dots ke roop mein mark hain: monatomic (He, Ne, Ar) $(3,5/3)$ par, diatomic (O₂, N₂, H₂) $(5,7/5)$ par, aur non-linear polyatomic (H₂O, CH₄) $(6,4/3)$ par. Dashed grey line asymptote $\gamma =1$ hai, jise curve approach karta hai par kabhi touch nahi karta.

Step 1 (sabse bada γ). Sabse chhota physically allowed $f$, $f=3$ hai (ek point atom ke paas kabhi 3 translational boxes se kam nahi hote). Tab $\gamma =1+\frac{2}{3}=\frac{5}{3}\approx 1.67$. Ye step kyun? $\gamma =1+2/f$, $f$ badhne ke saath ghatta hai, isliye ye smallest $f$ par maximize hota hai. $f<3$ wala koi ideal gas nahi hai, isliye $5/3$ ek hard ceiling hai.

Step 2 (limit). Jab $f\to \infty$: $\frac{2}{f}\to 0$, isliye $\gamma \to 1+0=1$. Ye step kyun? Itne saare storage boxes wale molecule mein heat ka har joule itni saari motions mein share hota hai ki constant-P aur constant-V heating almost identical ho jaati hai — $C_p\approx C_v$, isliye $\gamma \to 1^{+}$.

Step 3 (curve ki shape). Figure dekho: red curve blue dot $(3,1.67)$ par high se shuru hoti hai, orange diatomic dot $(5,1.40)$ aur green polyatomic dot $(6,1.33)$ se guzarti hai — us green point ke concrete examples H₂O aur CH₄ hain (non-linear, 3 rotational boxes) — aur phir dashed asymptote $\gamma =1$ ki taraf flat hoti jaati hai par kabhi nahi pahunchti.

Verify: $\gamma (3)=5/3=1.6\overline{6}$ (max ✓); ${\lim }_{f\to \infty }(1+2/f)=1$ (upar se approach hota hai, kabhi equal nahi ✓). Isliye har real gas $1<\gamma \le 5/3$ follow karta hai.

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Example 6 — Cell F (high-temperature unfreezing)

Step 1. Room-T diatomic: $f=5$ (3 translation + 2 rotation), $\gamma =7/5=1.40$. Ye step kyun? Vibration jaagne se pehle ye hamara baseline hai.

Step 2. Vibration unfreeze hoti hai ⇒ 2 quadratic boxes add: $f=5+2=7$. Ye step kyun? High $T$ par quantum "freeze" lift ho jaati hai; har vibrational mode potential aur kinetic energy contribute karta hai — 2 quadratic DOF.

Step 3. ${\gamma }_{\text{hot}}=1+\frac{2}{7}=\frac{9}{7}\approx 1.286$.

Verify: $\gamma$ $1.40$ se gir ke $9/7\approx 1.286$ ho gaya. ✓ Zyada boxes ⇒ γ 1 ke kareeb, exactly Ex 5 ke curve ka trend. Isliye $\gamma$ strictly constant nahi hota — ye sirf tab shift karta hai jab naye DOF unfreeze hon.

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Example 7 — Cell G (real-world word problem: speed of sound)

Step 1 (hawa). $v=\sqrt{\frac{(1.4)(8.314)(300)}{0.029}}$. Numerator $=1.4\times 8.314\times 300=3491.88$. $M$ se divide: $3491.88/0.029=120409.7$. Root: $v\approx 347.0\text{m/s}$. Ye step kyun? Speed of Sound in Gas formula humara thermodynamic $\gamma$ directly ek measurable everyday number mein badalta hai — isliye γ textbook se bahar bhi matter karta hai.

Step 2 (argon). $v_{\text{Ar}}=\sqrt{\frac{(5/3)(8.314)(300)}{0.040}}$. Numerator $=1.6667\times 8.314\times 300=4157.0$. Divide: $4157.0/0.040=103925$. Root: $v_{\text{Ar}}\approx 322.4\text{m/s}$. Ye step kyun? Same formula, naya molecule — bhaari mass ($0.040$ vs $0.029$) aur bada $\gamma$ ($1.67$ vs $1.40$) compete karte hain.

Step 3 (compare). $322.4<347.0$, isliye sound argon mein slower travel karta hai. Ye step kyun? Bhaari molar mass jeetta hai: $v\propto \sqrt{\gamma /M}$, aur $M$ $\gamma$ se bade factor se badha.

Verify: Units: $\sqrt{\frac{\text{J/mol⋅K}\cdot \text{K}}{\text{kg/mol}}}=\sqrt{\frac{\text{J}}{\text{kg}}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$. ✓ Hawa ≈ 347 m/s textbook ke ~343 m/s se match karta hai ($M$ aur $T$ ki rounding). ✓

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Example 8 — Cell H (exam twist: per-kg vs per-mole trap)

Step 1. Per-kg ko per-mole mein convert karo: $C_v=c_v\times M=650\times 0.032=20.8\text{J/mol⋅K}$. Ye step kyun? "Molar" ka matlab per mole hai; per-kg value ko kg-per-mole se multiply karo taaki kg cancel ho jaaye: $\frac{\text{J}}{\text{kg⋅K}}\cdot \frac{\text{kg}}{\text{mol}}=\frac{\text{J}}{\text{mol⋅K}}$.

Step 2. $C_p=C_v+R=20.8+8.314=29.114\text{J/mol⋅K}$.

Step 3. $\gamma =\frac{C_p}{C_v}=\frac{29.114}{20.8}=1.400$. Ye step kyun? γ ek hi units mein do heat capacities ka ratio hai — molar mass factor upar aur neeche cancel ho jaata hai, isliye γ same hoga chahe per-kg use karo ya per-mole.

Verify: $C_v\approx 20.8=\frac{5}{2}R$ ✓ (diatomic value recover hoti hai), aur $\gamma =1.40=7/5$ ✓. Trap: γ paane ke liye tumhe $M$ ki zaroorat nahi thi — sirf molar numbers ke liye. γ mass-independent hai.

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Recall Ye kaun sa cell hai? (self-test)

"$\gamma =1.33$, kya molecule?" — kaun sa matrix cell? ::: Cell B (backward: $f=6$ tak invert karo, non-linear polyatomic). "Helium mein sound speed?" — kaun sa cell? ::: Cell G (γ $v=\sqrt{\gamma RT/M}$ formula mein jaata hai). "$c_v$ J/kg·K mein diya, γ nikalo." — kaun sa cell? ::: Cell H (per-kg trap; γ unaffected). "Sabse bada possible γ?" — kaun sa cell aur jawaab? ::: Cell E; $\gamma =5/3$ at $f=3$.

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Connections

• Parent: γ = Cp/Cv — yahan drill ki gayi formulas
• Degrees of Freedom & Equipartition Theorem — $f$ aur $C_v=\frac{f}{2}R$ ka source
• Mayer's Relation — har example mein use kiya ($C_p=C_v+R$)
• Speed of Sound in Gas — Example 7 word problem
• First Law of Thermodynamics — Mayer's relation ka aadhar
• Internal Energy of Ideal Gas — $U=\frac{f}{2}RT$, mixture example mein use kiya
• Adiabatic Process — jahan γ aage use hota hai