Internal energy of ideal gas U = (f - 2)nRT: Worked examples

1.7.10 · D3 · Physics › Thermodynamics › Internal energy of ideal gas U = (f - 2)nRT

Everything here rests on the parent: the parent topic — internal energy of an ideal gas. Prerequisites jinka hum sahara lete hain: Kinetic theory of gases, Equipartition theorem, Degrees of freedom, First law of thermodynamics, Molar specific heats Cv and Cp, Ideal gas law.

Scenario matrix

Kisi bhi symbol ka matlab samajhne se pehle, unhe naam dete hain, phir woh knobs list karte hain jo ek problem ghuma sakti hai.

Do quantities $U$ decide karti hain: energy-storing modes ki sankhya $f$ (gas se set hoti hai) aur temperature $T$ (process se set hoti hai). Toh har problem "kaun sa gas" × " $T$ ke baare mein kaun sa sawaal" ki grid mein hoti hai.

Cell	Case class	Ise alag kya banata hai	Worked in
A	Monatomic, absolute $U$	$f = 3$ , $U$ khud chahiye	Ex 1
B	Diatomic, change $Δ U$	$f = 5$ , $Δ T$ se $Δ U$ chahiye	Ex 2
C	Polyatomic (nonlinear)	$f = 6$ , zyada $U$	Ex 3
D	Isothermal (degenerate $Δ T = 0$ )	$Δ U = 0$ chahe $P, V$ kuch bhi ho	Ex 4
E	Path-independence check	do alag paths se same $Δ U$	Ex 5
F	Cooling / negative $Δ T$	$Δ U$ ka sign negative hoga	Ex 6
G	Do gases ka mixture	parts ka $U$ add karo, alag $f$ ke saath	Ex 7
H	Limiting: high- $T$ vibration "unfreezes"	$f$ jumps from 5 to 7	Ex 8
I	Real-world word problem + first law	$Δ U = Q - W$ combine karo	Ex 9
J	Exam twist: $U$ diya, $T$ nikalo	formula ko invert karo	Ex 10
K	Adiabatic ( $Q = 0$ )	$Δ U$ poori tarah work se aata hai	Ex 11

Neecha wala grid usi table ko ek picture ki tarah draw karta hai: vertical axis kaun sa gas hai (yeh $f$ fix karta hai), horizontal axis problem $T$ ke baare mein kya poochh rahi hai. Har shaded box ek lettered cell (A–K) hai jo upar wali table aur use solve karne wale example se match karti hai — kisi bhi exam problem ko ek box tak trace karo.

Figure — Internal energy of ideal gas U = (f - 2)nRT

Worked examples

Cell A — Monatomic gas ki absolute internal energy

Worked example Example 1 — Ek tank mein Argon

$3 mol$ of argon (monatomic, so $f = 3$ ) $T = 350 K$ par hai. Total internal energy $U$ nikalo.

Forecast: Compute karne se pehle order of magnitude guess karo — kilojoules? tens of kJ? Likh lo.

Step 1. $f$ identify karo. Argon ek single atom hai → sirf 3 translational modes. Yeh step kyun? $f$ molecular shape se set hota hai (dekho Degrees of freedom); ek akela atom rotation store nahi kar sakta (spin karne ke liye "arms" nahi hain) ya vibration (koi bond nahi hai).

Step 2. $U = \frac{f}{2} n R T$ apply karo. Yeh step kyun? Yeh master formula hai; absolute energy ke liye hum $T$ khud use karte hain (change nahi). $U = \frac{3}{2} (3) (8.314) (350)$

Step 3. Multiply karo. Yeh step kyun? Sirf arithmetic; constants ko group karo taaki Joules mein ek single number mile. $U = 1.5 \times 3 \times 8.314 \times 350 \approx 13, 095 J \approx 13.1 kJ$

Verify: Units: $mol \cdot (J mol^{- 1} K^{- 1}) \cdot K = J$ . ✓ Aur $13 kJ$ kuch moles ke liye room-ish temperature par sahi ballpark hai (energy per mole $\approx 4.4 kJ$ ).

Cell B — Diatomic gas ki internal energy mein change

Worked example Example 2 — Nitrogen ko garam karna

$2 mol$ of N₂ (diatomic, $f = 5$ ) ko $280 K$ se $380 K$ tak warm kiya jaata hai. $Δ U$ nikalo.

Forecast: Kyunki hum sirf heat add kar rahe hain, expect karo $Δ U > 0$ .

Step 1. Change form $Δ U = \frac{f}{2} n R Δ T$ use karo. Yeh step kyun? $U$ sirf $T$ par depend karta hai, toh do states subtract karne par sirf $Δ T$ bachta hai, baaki sab cancel ho jaata hai.

Step 2. $Δ T = 380 - 280 = 100 K$ compute karo, $f = 5$ ke saath plug in karo. Yeh step kyun? Moderate $T$ par diatomic: 3 translation + 2 rotation = 5 (bond axis ke around rotation frozen hai). $Δ U = \frac{5}{2} (2) (8.314) (100)$

Step 3. Multiply karo. Yeh step kyun? Sirf arithmetic — chaar factors ko collapse karke Joules mein ek energy value banao. $Δ U = 2.5 \times 2 \times 8.314 \times 100 = 4157 J \approx 4.16 kJ$

Verify: Positive, jaise forecast kiya tha. ✓ Dhyan do ki $Δ U$ ko sirf $Δ T$ chahiye tha — change nikalne ke liye starting temperature kabhi use nahi ki.

Cell C — Polyatomic gas (zyada storage bins)

Worked example Example 3 — Water vapour

$0.5 mol$ of steam (H₂O nonlinear polyatomic hai, $f = 6$ ) $500 K$ par. $U$ nikalo.

Forecast: Diatomic se zyada degrees of freedom → same $T$ par per mole zyada energy.

Step 1. $f = 6$ set karo: 3 translation + 3 rotation (ek bent, non-linear molecule teeno axes ke around spin kar sakta hai aur har ek energy store karta hai). Yeh step kyun? Ek seedhi dumbbell ke unlike, ek bent molecule ka teeno axes ke around real moment of inertia hota hai → teesra rotation frozen nahi hai.

Step 2. Master formula $U = \frac{f}{2} n R T$ apply karo $f = 6$ ke saath. Yeh step kyun? Hum absolute $U$ chahte hain, toh $T$ khud feed karte hain; substitute karne se har symbol ek number ban jaata hai. $U = \frac{6}{2} (0.5) (8.314) (500) = 3 \times 0.5 \times 8.314 \times 500$

Step 3. Multiply karo. Yeh step kyun? Sirf arithmetic — product ko Joules mein ek single energy mein reduce karo. $U = 6235.5 J \approx 6.24 kJ$

Verify: $f /2 = 3$ , toh $U = 3 n R T$ — parent table mein polyatomic case. Per mole $= 12, 471 J$ , argon ke per-mole $\frac{3}{2} R T$ se same $T$ par zyada hai. ✓

Cell D — Degenerate case: isothermal ( $Δ T = 0$ )

Worked example Example 4 — Isothermal expansion trap

Ek ideal gas isothermally expand hoti hai; uska volume triple ho jaata hai aur pressure drop ho jaata hai. $Δ U$ nikalo.

Forecast: Gas kaam karti hai aur uska $P, V$ dramatically change hota hai — surely $U$ change hoga? (Yahi trap hai.)

Step 1. Word isothermal padho: temperature constant, toh $Δ T = 0$ . Yeh step kyun? "Iso-thermal" ka literally matlab = same temperature; yahi ek quantity hai jis par $U$ dhyan deta hai.

Step 2. $Δ U = \frac{f}{2} n R Δ T$ apply karo jahan $Δ T = 0$ hai. Yeh step kyun? Jo ek fact humein pata hai ( $Δ T = 0$ ) substitute karne se poora product vanish ho jaata hai — koi aur quantity ise revive nahi kar sakti, kyunki $U$ ko $P$ aur $V$ ki parwaah nahi. $Δ U = \frac{f}{2} n R (0) = 0$

Verify: First law of thermodynamics se, $Δ U = Q - W$ . Yahan gas heat $Q$ absorb karti hai aur equal work $W$ karti hai, toh $Q = W$ aur $Δ U = 0$ . ✓ $P$ aur $V$ change hue, lekin $U$ nahi — kyunki intermolecular forces nahi hain.

Cell E — Do alag paths se same $Δ U$

Worked example Example 5 — Path independence

$1 mol$ of a monatomic gas ( $f = 3$ ) $300 K$ se $400 K$ tak (a) constant volume, phir (b) constant pressure se jaata hai. Dikhao ki $Δ U$ identical hai.

Forecast: Supplied heat alag hogi (constant- $P$ ko zyada chahiye, work karne ke liye bhi), lekin $Δ U$ match karni chahiye.

Step 1. Dono paths ke same endpoints share hain $Δ T = 100 K$ . Yeh step kyun? Kyunki $U$ ek state function hai (abhi upar define ki gayi), sirf endpoints matter karte hain, route nahi.

Step 2. $Δ U$ ek baar compute karo. Yeh step kyun? Kyunki $Δ U$ sirf $Δ T$ dekhta hai, ek calculation dono paths ko ek saath cover karti hai. $Δ U = \frac{3}{2} (1) (8.314) (100) = 1247.1 J$

Step 3. Specific heats se cross-check karo. Constant volume par heat $Q_{V} = n C_{V} Δ T$ hai jahan $C_{V} = \frac{f}{2} R = \frac{3}{2} R$ (dekho Molar specific heats Cv and Cp). Yeh step kyun? $Q_{V}$ equals $Δ U$ hota hai (koi work nahi), toh same number dena chahiye. $Q_{V} = (1) (\frac{3}{2} \cdot 8.314) (100) = 1247.1 J$

Verify: Dono routes $Δ U = 1247.1 J$ dete hain. ✓ Constant-pressure path ko zyada heat chahiye ( $Q_{P} = n C_{P} Δ T$ jahan $C_{P} = \frac{5}{2} R$ ), lekin extra work mein jaata hai, $U$ mein nahi.

Neecha $P$ – $V$ diagram in do paths ko draw karta hai taaki tum path independence dekh sako: blue route (fixed volume par pressure badhao, phir fixed pressure par expand karo) aur yellow route (pehle expand karo, phir pressure badhao) red dot (State 1, $300 K$ ) se shuru hote hain aur green dot (State 2, $400 K$ ) par khatam hote hain. Do coloured tracks alag areas enclose karte hain — toh unmein alag work aur alag heat involve hoti hai — phir bhi dono same do temperatures ko connect karte hain, toh $Δ U$ dono ke liye same $1247 J$ hai.

Cell F — Negative $Δ T$ (cooling)

Worked example Example 6 — Refrigerated oxygen

$4 mol$ of O₂ (diatomic, $f = 5$ ) ko $320 K$ se $250 K$ tak cool kiya jaata hai. $Δ U$ nikalo.

Forecast: Cooling molecular jiggle remove karti hai → expect karo $Δ U < 0$ .

Step 1. $Δ T = 250 - 320 = - 70 K$ (sign rakho!). Yeh step kyun? $Δ T$ ka sign energy flow ki direction carry karta hai; isse drop karna ek classic error hai.

Step 2. Change formula $Δ U = \frac{f}{2} n R Δ T$ apply karo. Yeh step kyun? Same master relation; negative $Δ T$ seedha pass through hoga aur $Δ U$ ko negative banayega. $Δ U = \frac{5}{2} (4) (8.314) (- 70)$

Step 3. Multiply karo. Yeh step kyun? Sirf arithmetic — negative factor final energy ka sign set karta hai. $Δ U = 2.5 \times 4 \times 8.314 \times (- 70) = - 5819.8 J \approx - 5.82 kJ$

Verify: Negative, jaise forecast kiya tha. ✓ Magnitude wahi hai jo $70 K$ se heating add karti — energy $\pm Δ T$ mein symmetric hai.

Cell G — Do gases ka mixture

Worked example Example 7 — Helium aur nitrogen ka mix

Ek vessel mein $2 mol$ He ( $f = 3$ ) aur $3 mol$ N₂ ( $f = 5$ ) thermal equilibrium mein $300 K$ par hain. Total internal energy nikalo.

Forecast: Har gas apna $f$ rakhti hai; total $U$ = parts ka sum.

Step 1. Internal energy additive hai: $U_{tot} = U_{He} + U_{N_{2}}$ . Yeh step kyun? Energy saare molecules par ek bulk sum hai; do species degrees of freedom share nahi karti.

Step 2. Same $T = 300 K$ par har part compute karo. Yeh step kyun? Woh thermal equilibrium mein hain, toh ek single $T$ dono terms ko feed karti hai — hum master formula har gas par apne $f$ ke saath apply karte hain. $U_{He} = \frac{3}{2} (2) (8.314) (300) = 7482.6 J$ $U_{N_{2}} = \frac{5}{2} (3) (8.314) (300) = 18706.5 J$

Step 3. Add karo. Yeh step kyun? Step 1 ki additivity ka matlab hai ki total sirf do energies ka sum hai. $U_{tot} = 7482.6 + 18706.5 = 26189.1 J \approx 26.2 kJ$

Verify: Units sab Joules hain; dono terms positive hain. ✓ Ek "effective" mixture $f$ hoga $\frac{2 ( 3 ) + 3 ( 5 )}{5} = 4.2$ , aur $\frac{4.2}{2} (5) (8.314) (300) = 26189 J$ match karta hai. ✓

Cell H — Limiting behaviour: high $T$ par vibration unfreeze hoti hai

Worked example Example 8 — Bahut high temperature par diatomic gas

$1 mol$ of a diatomic gas itni zyada garam hoti hai ki uski vibrational mode activate ho jaati hai, $f$ ko $5$ se $7$ kar deti hai. $T = 2000 K$ par $f = 5$ vs $f = 7$ ke liye $U$ compare karo.

Forecast: Zyada active modes → same $T$ par zyada stored energy.

Step 1. $f = 7$ kyun? Ek vibrating bond do tareeqon se energy store karta hai: kinetic (atoms ka movement) aur potential (spring stretch). Har ek ek quadratic term hai → $+ 2$ degrees of freedom. Yeh step kyun? Equipartition theorem har quadratic energy term count karta hai, aur vibration unmein se do contribute karta hai.

Step 2. Master formula $U = \frac{f}{2} n R T$ se dono compute karo. Yeh step kyun? Do scenarios mein sirf $f$ alag hai; dono values ko same formula mein feed karna extra modes ke effect ko isolate karta hai. $U_{f = 5} = \frac{5}{2} (1) (8.314) (2000) = 41570 J$ $U_{f = 7} = \frac{7}{2} (1) (8.314) (2000) = 58198 J$

Step 3. Difference nikalo. Yeh step kyun? Subtract karne se exactly woh energy isolate hoti hai jo do nayi vibrational bins carry karti hain. $Δ U = 58198 - 41570 = 16628 J \approx 16.6 kJ$

Verify: Extra $= \frac{2}{2} n R T = n R T = (1) (8.314) (2000) = 16628 J$ — exactly do vibrational bins ki worth. ✓ Low $T$ par yeh bins frozen hoti hain (quantum gap bahut bada), toh $f = 5$ ; parent ka " $f$ at moderate $T$ " caveat exactly yahi limiting case hai.

Cell I — First law ke saath real-world word problem

Worked example Example 9 — Piston kaam karta hai heating ke dauran

$1 mol$ monatomic gas ( $f = 3$ ) ka ek cylinder $Q = + 2000 J$ heat absorb karta hai aur $W = + 500 J$ kaam karta hai piston push karte hue (positive kyunki gas apne surroundings par kaam karti hai). $Δ U$ aur temperature rise $Δ T$ nikalo.

Forecast: Saari heat internal energy nahi banegi — kuch work ke roop mein bahar jaayegi, toh $Δ U < 2000 J$ .

Step 1. First law: $Δ U = Q - W$ . Yeh step kyun? First law of thermodynamics — heat in minus work done by the gas = woh energy jo andar rehti hai (upar boxed sign convention use karke). $Δ U = 2000 - 500 = 1500 J$

Step 2. $Δ T$ pane ke liye $Δ U = \frac{f}{2} n R Δ T$ invert karo. Yeh step kyun? Hume $Δ U$ pata hai; sirf unknown $Δ T$ ke liye solve karo. $Δ T = \frac{2 Δ U}{f n R} = \frac{2 ( 1500 )}{3 ( 1 ) ( 8.314 )}$

Step 3. Compute karo. Yeh step kyun? Sirf arithmetic — divide karke kelvin mein temperature rise pao. $Δ T = \frac{3000}{24.942} \approx 120.3 K$

Verify: Plug back: $\frac{3}{2} (1) (8.314) (120.3) = 1500 J = Δ U$ . ✓ Aur $Δ U < Q$ jaise forecast kiya tha, kyunki $500 J$ work mein gaya. ✓

Cell J — Exam twist: $U$ diya, $T$ nikalo

Worked example Example 10 — Temperature reverse-engineer karna

$0.25 mol$ diatomic gas ( $f = 5$ ) ka sample internal energy $U = 1300 J$ hai. Uski temperature $T$ nikalo.

Forecast: Formula rearrange karna — expect karo few hundred kelvin.

Step 1. $U = \frac{f}{2} n R T$ ko $T$ ke liye solve karo. Yeh step kyun? Same formula, unknown ko subject banao; pure algebra, koi nayi physics nahi. $T = \frac{2 U}{f n R}$

Step 2. Jaane hue numbers substitute karo. Yeh step kyun? Har symbol ki jagah uski value rakhne se algebra ek single arithmetic fraction ban jaata hai. $T = \frac{2 ( 1300 )}{5 ( 0.25 ) ( 8.314 )} = \frac{2600}{10.3925}$

Step 3. Compute karo. Yeh step kyun? Sirf arithmetic — kelvin mein temperature pane ke liye divide karo. $T \approx 250.2 K$

Verify: Forward-check: $\frac{5}{2} (0.25) (8.314) (250.2) = 1300 J$ . ✓ Units: $\frac{J}{mol \cdot J mol ^{- 1} K ^{- 1}} = K$ . ✓ Answer sensible few-hundred kelvin hai, jaise forecast kiya tha. ✓

Cell K — Adiabatic process ( $Q = 0$ )

Worked example Example 11 — Insulated compression

Ek perfectly insulated cylinder mein $2 mol$ monatomic gas ( $f = 3$ ) hai. Ek piston gas compress karta hai, toh work uske par ki jaati hai: hamare sign convention mein yeh $W = - 900 J$ hai (work by the gas negative hai). Koi heat enter ya leave nahi karti ( $Q = 0$ ). $Δ U$ aur temperature change $Δ T$ nikalo.

Forecast: Heat flow ke bina, saari compression work internal energy ban jaati hai → $T$ zaroor badhegi, toh $Δ U > 0$ .

Step 1. First law $Q = 0$ ke saath: $Δ U = Q - W = 0 - (- 900) = + 900 J$ . Yeh step kyun? First law of thermodynamics — "adiabatic" matlab insulation saari heat block karti hai ( $Q = 0$ ), toh energy change ka ek maatra source work hai. Yahan $W = - 900 J$ (gas par kiya kaam), jo $U$ badhata hai.

Step 2. $Δ U = \frac{f}{2} n R Δ T$ se $Δ U$ ko $Δ T$ mein convert karo, $Δ T$ ke liye solve karo. Yeh step kyun? $Δ U$ ab pata hai; master formula ise us temperature change se link karta hai jo hum chahte hain. $Δ T = \frac{2 Δ U}{f n R} = \frac{2 ( 900 )}{3 ( 2 ) ( 8.314 )}$

Step 3. Compute karo. Yeh step kyun? Sirf arithmetic — divide karke kelvin mein rise pao. $Δ T = \frac{1800}{49.884} \approx 36.1 K$

Verify: Positive $Δ T$ , jaise forecast kiya tha (compression gas ko garam karta hai). ✓ Back-check: $\frac{3}{2} (2) (8.314) (36.1) = 900 J = Δ U$ . ✓ Cell D (isothermal) se contrast: wahan $Q$ aur $W$ balance hue; yahan $Q = 0$ toh work akela $U$ drive karta hai.

Recall

Recall Which cell does each cue belong to?

"Volume triple hota hai, $T$ fixed hai, $Δ U$ nikalo" ::: Cell D — isothermal, $Δ U = 0$ . "He + N₂ ek box mein, total $U$ nikalo" ::: Cell G — parts ka sum karo apne $f$ ke saath. "Bond $2000 K$ par vibrate karne lagta hai" ::: Cell H — $f$ jumps $5 \to 7$ . " $U$ diya, $T$ poochha" ::: Cell J — $T = 2 U / (f n R)$ se invert karo. " $320 K$ se $250 K$ tak cool kiya" ::: Cell F — negative $Δ T$ rakho. "Insulated, koi heat nahi, piston gas compress karta hai" ::: Cell K — adiabatic, $Δ U = - W$ .

Connections

Parent topic — master formula ki derivation
Degrees of freedom — har cell ka $f$ kaise choose hota hai
Equipartition theorem — vibration 2 kyun add karta hai (Cell H)
First law of thermodynamics — $Δ U = Q - W$ (Cells I, D, K)
Molar specific heats Cv and Cp — Cell E mein cross-check
Ideal gas law — path independence ko underpin karta hai
Kinetic theory of gases — microscopic root of it all