Heat and internal energy — microscopic vs macroscopic: Worked examples

1.7.3 · D3 · Physics › Thermodynamics › Heat and internal energy — microscopic vs macroscopic

Yeh Heat and Internal Energy — Microscopic vs Macroscopic ka "sleeves chadha ke kaam karo" wala companion note hai. Parent note ne ideas build kiye; yahan hum har tarah ke scenario ko tackle karte hain jo yeh topic throw kar sakta hai, ek worked example per cell.

Shuru karne se pehle, ek promise: neeche har symbol parent note mein define kiya gaya tha, sivaaye kuch jo hum yahan add kar rahe hain. Hum use karte hain:

$U$ = internal energy (energy jo andar stored hai), $Δ U$ = uska change,
$Q$ = heat (energy jo boundary cross karti hai temperature difference ki wajah se),
$W$ = work done by the gas (energy jo ordered pushing ke roop mein bahar jaati hai),
$n$ = number of moles, $R = 8.314 J mol^{- 1} K^{- 1}$ (gas constant), $T$ = temperature in kelvin,
$f$ = degrees of freedom, aur do master formulas $Δ U = Q - W and U = \frac{f}{2} n R T .$

Neeche har problem inhi lines ke saath sirf bookkeeping hai.

The scenario matrix

"Har tarah ke scenario" ko teen knobs ki values choose karne aur kuch special cases ke roop mein soch sakte ho. Table har cell aur usse match karne wala example list karta hai.

Cell	Isme kya khaas hai	$Q$	$W$	Sign of $Δ U$	Covered by
A Heat in, gas does work	classic first-law bookkeeping	$> 0$	$> 0$	$+$	Ex 1
B Constant volume ( $W = 0$ ), heating	saari heat $\to$ internal energy	$> 0$	$0$	$+$	Ex 2
C Isothermal ( $Δ U = 0$ )	heat in = work out exactly	$> 0$	$> 0$	$0$	Ex 3
D Adiabatic ( $Q = 0$ )	koi heat nahi; work $U$ ko change karta hai	$0$	$> 0$	$-$ (expansion)	Ex 4
E Sign flip: work done on gas, heat leaks out	dono negative	$< 0$	$< 0$	$+$	Ex 5
F Heat out, gas compressed	dono negative	$< 0$	$< 0$	either sign	Ex 6
G Degenerate: free expansion into vacuum	$Q = 0, W = 0$	$0$	$0$	$0$	Ex 7
H Degrees of freedom change ( $f$ : 3 vs 5)	diatomic gas	$> 0$	$0$	$+$	Ex 8
I Real-world word problem	tyre / pump framing	context	context	context	Ex 9
J Exam twist: cyclic process	$Δ U = 0$ over a full loop	net $> 0$	net $> 0$	$0$	Ex 10
K Heat in and gas compressed	dono $U$ ko badhate hain	$> 0$	$< 0$	$+$	Ex 11
L Heat out while gas still does work	opposite signs	$< 0$	$> 0$	$-$	Ex 12
M Constant-volume cooling	heat removed, no work	$< 0$	$0$	$-$	Ex 13
N Heat in but too much work out	net cooling despite heating	$> 0$	$> 0$ (with $W > Q$ )	$-$	Ex 14

Ex 1 — Cell A: heat in, gas does work

Ex 2 — Cell B: constant volume, saari heat internal energy ban jaati hai

Worked example Statement

Ek rigid sealed can ke andar $2 mol$ monatomic gas ko $300 K$ se $330 K$ tak heat karo. Kitni heat chahiye thi?

Forecast: ek rigid can move nahi kar sakta — toh heat kahan ja sakti hai?

Rigid $\Rightarrow$ volume fixed $\Rightarrow W = 0$ . Yeh step kyun? Work force times distance hai; jo wall move nahi kar sakti uska matlab zero distance, zero work.
Monatomic $\Rightarrow f = 3 \Rightarrow U = \frac{3}{2} n R T$ , toh $Δ U = \frac{3}{2} n R Δ T = \frac{3}{2} (2) (8.314) (30) = 748.26 J .$ Yeh step kyun? $Δ U$ ke liye sirf start aur end temperatures chahiye — yeh ek state function hai, toh path irrelevant hai.
First law with $W = 0$ : $Q = Δ U = 748.26 J$ . Yeh step kyun? Work term gone hone par, $Δ U = Q - 0$ collapse ho jaata hai $Q = Δ U$ mein — heat ke paas storage ke alawa jaane ki koi jagah nahi.

Verify: Saari heat internal energy mein gayi kyunki koi work escape nahi hui — isi liye $C_{V} = \frac{3}{2} R$ hai (upar define kiya; dekho Specific Heats Cv and Cp). Sanity: heating ne $T$ badhaya, toh $Q > 0$ . ✓

Ex 3 — Cell C: isothermal, $Δ U = 0$

Worked example Statement

$1 mol$ ideal gas $T = 350 K$ par isothermally expand karta hai volume $V_{1}$ se $V_{2} = 3 V_{1}$ tak. $Δ U$ , $W$ , aur $Q$ nikalo.

Forecast: heat clearly flow ho rahi hai andar — toh surely $U$ badh raha hai? (Nahi badh raha. Dekho.)

Isothermal ka matlab $T$ constant. Ideal gas ke liye $U = \frac{f}{2} n R T$ sirf $T$ par depend karta hai, toh $Δ U = 0$ . Yeh step kyun? Temperature change nahi $\Rightarrow$ average molecular speed unchanged $\Rightarrow$ stored energy unchanged.
Ideal gas ka isothermal work (from Isothermal and Adiabatic Processes): $W = n R T ln \frac{V _{2}}{V _{1}} = (1) (8.314) (350) ln 3 = 3196.87 J .$ Yeh step kyun? Constant $T$ par pressure $P$ girta jaata hai jaise gas phailta hai; chote-chote pushes $P Δ V$ ko add karo $P = n R T / V$ ke saath ( $P V = n R T$ se) toh logarithm milta hai.
First law: $Q = Δ U + W = 0 + 3196.87 = 3196.87 J$ . Yeh step kyun? $Δ U = Q - W$ ko rearrange karne par $Q = Δ U + W$ milta hai; $Δ U = 0$ ke saath zaruri heat exactly performed work ke barabar ho jaati hai.

Verify: $Q = W$ exactly — heat ka har joule turant work ke roop mein nikalta jaata hai, $⟨ v^{2} ⟩$ (aur isliye $T$ ) ko fixed rakhte hue. Yeh parent note ka "heat in ≠ internal energy up" concrete form mein hai. ✓

Ex 4 — Cell D: adiabatic, $Q = 0$

Figure dekho: ek gas perfect insulation mein wrapped piston ko bahar push kar rahi hai. Koi heat wall cross nahi kar sakti, phir bhi gas thandi ho jaati hai. Kyun? Kyunki bacha hua energy account sirf $U$ hai, aur gas use work karne mein spend kar deta hai.

Figure — Heat and internal energy — microscopic vs macroscopic

Worked example Statement

$1 mol$ monatomic gas adiabatically ( $Q = 0$ ) expand karta hai aur $500 J$ work karta hai. $Δ U$ aur temperature drop $Δ T$ nikalo.

Forecast: koi heat andar allowed nahi, aur gas energy work par spend kar raha hai — $T$ kis direction mein jaayega?

Adiabatic $\Rightarrow Q = 0$ . Yeh step kyun? "Insulated / fast" processes koi heat exchange nahi karte; yeh $Q$ term ko khatam kar deta hai.
First law: $Δ U = Q - W = 0 - 500 = - 500 J$ . Yeh step kyun? $500 J$ work ka ek hi source ho sakta hai woh stored energy hai, toh $U$ drop karna chahiye.
$Δ T$ mein convert karo $Δ U = \frac{3}{2} n R Δ T$ se: $Δ T = \frac{Δ U}{\frac{3}{2} n R} = \frac{- 500}{\frac{3}{2} ( 1 ) ( 8.314 )} = - 40.09 K .$ Yeh step kyun? $U$ ideal gas ke liye one-to-one $T$ se map hota hai, toh known $Δ U$ se known $Δ T$ milta hai.

Verify: $Δ T < 0$ — gas bina koi heat khoye thanda ho gaya, purely work karne se. Isi liye bike pumps compression par warm hote hain aur aerosol cans release par chill hote hain. ✓

Ex 5 — Cell E: sign flip, work done ON the gas

Ex 6 — Cell F: heat out AUR compression saath mein

Worked example Statement

Ek gas compress kiya jaata hai, uspar $70 J$ work karte hue, jabki $70 J$ heat bahar nikali jaati hai. $Δ U$ nikalo. Phir sirf $50 J$ heat nikalne par repeat karo.

Forecast: jab dono effects equal hoon, $U$ ka kya hona chahiye?

Case (i): $Q = - 70 J$ , $W = - 70 J$ (work done on gas). $Δ U = (- 70) - (- 70) = 0 J$ . Yeh step kyun? Compression se inject ki gayi energy exactly match karti hai heat ke roop mein bled off energy se — perfect cancellation, $T$ unchanged.
Case (ii): $Q = - 50 J$ , $W = - 70 J$ . $Δ U = (- 50) - (- 70) = + 20 J$ . Yeh step kyun? Ab kam heat escape karti hai, toh net $20 J$ stored rehta hai aur $T$ badh jaata hai.

Verify: Case (i) mein $Δ U = 0$ — isi tarah compressed gas ko isothermal rakha jaata hai: exactly compression work ke barabar heat nikalo. Case (ii) $> 0$ confirm karta hai "kam heat nikalo $\Rightarrow$ zyada garam." ✓

Ex 7 — Cell G: degenerate case, free expansion into vacuum

Figure degenerate scenario dikhata hai: gas left mein, vacuum right mein, beech mein valve. Valve kholo aur gas khaali half mein rush karta hai. Kuch bhi push-back nahi kar raha, aur box insulated hai.

Worked example Statement

Ek thermally insulated box ko ek valve se split kiya gaya hai. Gas left half mein hai; right half perfect vacuum hai. Valve khulta hai. $Q$ , $W$ , $Δ U$ , aur $Δ T$ nikalo.

Forecast: gas double volume mein expand karta hai — surely thanda hoga, jaise Ex 4 mein?

Insulated $\Rightarrow Q = 0$ . Yeh step kyun? Koi heat wall cross nahi kar sakti.
Gas vacuum ke against push karta hai — koi opposing force nahi, koi piston nahi, toh $W = 0$ . Yeh step kyun? Work force times distance hai; kisi bhi cheez mein expand karne ka matlab zero opposing force, isliye zero work. Yeh degenerate cell hai — dono terms vanish ho jaate hain.
$Δ U = Q - W = 0 - 0 = 0$ , toh $Δ T = 0$ : temperature change nahi hota. Yeh step kyun? $U$ unchanged hone par aur $U$ sirf $T$ set karne par, $T$ frozen rehta hai chahe $V$ double ho jaaye.

Verify: Ex 4 se compare karo (adiabatic ek piston ke against, jo thanda hua). Same $Q = 0$ , lekin yahan $W = 0$ bhi — koi work nahi matlab koi cooling nahi. Akela difference yeh hai ki koi push-back karta hai ya nahi. ✓

Ex 8 — Cell H: degrees of freedom change (diatomic gas)

Worked example Statement

$1 mol$ diatomic gas ( $f = 5$ : teen translational + do rotational) ko constant volume par $400 K$ se $420 K$ tak heat karo. Iske liye zaruri heat ko monatomic gas ke liye same rise par compare karo.

Forecast: energy store karne ke zyada tarike — kya diatomic ko same $Δ T$ ke liye zyada ya kam heat chahiye?

Constant volume $\Rightarrow W = 0 \Rightarrow Q = Δ U$ . Yeh step kyun? Fixed volume koi work nahi karta (koi distance move nahi), toh first law $Δ U = Q - W$ collapse ho jaata hai $Q = Δ U$ mein — heat ka har joule store ho jaata hai.
Diatomic: $U = \frac{5}{2} n R T$ , toh $Q_{di} = \frac{5}{2} n R Δ T = \frac{5}{2} (1) (8.314) (20) = 415.70 J .$ Yeh step kyun? $5$ degrees of freedom mein se har ek $\frac{1}{2} k_{B} T$ carry karta hai per molecule (with $k_{B}$ upar defined; equipartition, dekho Equipartition Theorem), toh zyada channels $\Rightarrow$ zyada energy per kelvin.
Monatomic same rise: $Q_{mono} = \frac{3}{2} (1) (8.314) (20) = 249.42 J$ . Yeh step kyun? $f = 3$ ke saath repeat karo $5$ ki jagah taaki like for like compare kar sakein — same $n$ , $Δ T$ , sirf storage channels ki number alag.

Verify: Ratio $415.70/249.42 = 5/3 \approx 1.667$ — exactly $f_{di} / f_{mono} = 5/3$ . Extra rotational storage proportionally zyada heat demand karta hai. Dekho Degrees of Freedom and Molecular Structure. ✓

Ex 9 — Cell I: real-world word problem

Worked example Statement

Ek hand pump $0.02 mol$ air ko seal karke (diatomic treat karo, $f = 5$ ) itni tezi se compress karta hai ki koi heat escape nahi hoti. Tum air par $18 J$ work karte ho. Temperature kitna rise karta hai?

Forecast: tumne pump barrel ko garam hote feel kiya hoga. Rise kuch degrees hoga ya ek sau?

"Koi heat escape nahi hoti" $\Rightarrow Q = 0$ ; tum gas par work karte ho $\Rightarrow W = - 18 J$ . Yeh step kyun? Fast pumping adiabatic hai; tumhari mehnat "gas dwara work" mein negative hai.
$Δ U = Q - W = 0 - (- 18) = + 18 J$ . Yeh step kyun? Tumhari saari mehnat store ho jaati hai, kyunki kuch bhi heat ke roop mein leak nahi hota.
$Δ U = \frac{5}{2} n R Δ T \Rightarrow Δ T = \frac{18}{\frac{5}{2} ( 0.02 ) ( 8.314 )} = 43.30 K$ . Yeh step kyun? Stored energy ko temperature rise mein convert karo diatomic $U$ – $T$ link use karke.

Verify: Sirf $18 J$ se $0.02 mol$ gas mein $\sim 43 K$ ka jump — isi liye barrel genuinely hot feel hota hai. Units: J / (J K $^{- 1}$ ) $=$ K. ✓

Ex 10 — Cell J: exam twist, ek full cycle

Figure ek closed loop sketch karta hai pressure–volume diagram par: gas ko kai steps se le jaaya jaata hai aur wahi exact starting point par wapas laaya jaata hai (red dot). Key fact yeh hai ki yeh kahan end hota hai.

Worked example Statement

Ek gas ek poore thermodynamic cycle se guzara jaata hai, apni starting state par wapas aata hai. Poore cycle mein woh $600 J$ heat absorb karta hai aur $370 J$ reject karta hai. $Δ U_{cycle}$ aur gas dwara kiya gaya net work nikalo.

Forecast: heat andar aayi aur bahar gayi — kya round trip ke baad koi leftover stored energy hai?

Cycle same state par wapas aata hai, aur $U$ ek state function hai, toh $Δ U_{cycle} = 0$ . Yeh step kyun? Same start aur end state $\Rightarrow$ same $U$ $\Rightarrow$ zero net change, chahe path kitna bhi loopy ho.
Net heat: $Q_{net} = 600 - 370 = 230 J$ . Yeh step kyun? Absorbed heat positive count hoti hai, rejected heat negative; unka sum boundary cross karne wali net heat hai loop ke upar.
First law over the cycle: $0 = Q_{net} - W_{net} \Rightarrow W_{net} = 230 J$ . Yeh step kyun? $Δ U = 0$ ke saath, jo bhi net heat andar aayi woh poori net work ke roop mein bahar chali gayi — isi tarah ek engine cycle se work produce karta hai.

Verify: $W_{net} = 230 J > 0$ : gas ne net work kiya, toh $P$ – $V$ loop par path clockwise gaya, $230 J$ ka area enclose karte hue. Round-trip stored change zero hai, jaise koi bhi state function demand karta hai. ✓

Ex 11 — Cell K: heat in AUR gas compressed (dono $U$ ko badhate hain)

Worked example Statement

Ek gas ko $150 J$ heat andar flow karke warm kiya jaata hai aur saath hi piston andar push kiya jaata hai, gas par $60 J$ work karte hue. $Δ U$ nikalo.

Forecast: dono effects jo energy deposit karte hain — kya $Δ U$ akeli heat se zyada hona chahiye?

Signs: heat in $\Rightarrow Q = + 150 J$ . Work done on the gas $\Rightarrow$ gas dwara work hai $W = - 60 J$ . Yeh step kyun? Dono surroundings-actions energy inward push karte hain, toh dono $U$ raise karni chahiye; signs isse reflect karni chahiye.
$Δ U = Q - W = 150 - (- 60) = 150 + 60 = + 210 J$ . Yeh step kyun? Heat $150 J$ deposit karti hai; compression aur $60 J$ deposit karta hai; kuch bahar nahi jaata, toh saara $210 J$ store ho jaata hai.

Verify: $Δ U = 210 > 150$ — akeli heat se bada, kyunki compression ne isko add kiya rather than spend kiya. Yeh Ex 1 ka mirror image hai (wahan work subtract kiya; yahan add karta hai). ✓

Ex 12 — Cell L: heat OUT jabki gas abhi bhi work karta hai

Worked example Statement

Ek gas expand karta hai, apne piston par $90 J$ work karte hue, jabki saath mein $30 J$ heat us mein se drain ho jaati hai. $Δ U$ nikalo.

Forecast: gas dono heat kho raha hai aur work par energy spend kar raha hai — $U$ kitna neeche jaayega?

Signs: heat out $\Rightarrow Q = - 30 J$ . Gas expand karta hai aur work karta hai $\Rightarrow W = + 90 J$ . Yeh step kyun? Yahan dono signs Ex 11 se opposite direction mein point karte hain — dono effects ab storage se energy withdraw karte hain.
$Δ U = Q - W = (- 30) - (90) = - 120 J$ . Yeh step kyun? $30 J$ heat ke roop mein nikla aur $90 J$ work ke roop mein nikla; dono losses $U$ se aate hain, toh dono drop ke roop mein add ho jaate hain.

Verify: $Δ U = - 120 < 0$ — adiabatic Ex 4 se zyada steep cooling, kyunki heat aur work dono nikl rahe hain. Sign combination $Q < 0, W > 0$ (woh cell jo pehle ki matrix mein miss tha) ab cover ho gaya. ✓

Ex 13 — Cell M: constant-volume cooling

Worked example Statement

$2 mol$ monatomic gas ek rigid sealed can mein $330 K$ se $300 K$ tak thanda kiya jaata hai. Gas se kitni heat bahar jaati hai?

Forecast: yeh Ex 2 ulta chala raha hai — $Q$ ka sign ab kya hona chahiye?

Rigid $\Rightarrow V$ fixed $\Rightarrow W = 0$ . Yeh step kyun? Jo wall move nahi kar sakti woh koi work nahi karti; first law reduce ho jaata hai $Q = Δ U$ mein.
$Δ U = \frac{3}{2} n R Δ T = \frac{3}{2} (2) (8.314) (300 - 330) = - 748.26 J$ . Yeh step kyun? Temperature gira, toh $Δ T < 0$ aur stored energy utni hi giri jitni Ex 2 mein badhi thi.
$Q = Δ U = - 748.26 J$ . Yeh step kyun? Koi work nahi hone par, saari lost internal energy heat ke roop mein bahar gayi — ek negative $Q$ matlab heat bahar flow ki.

Verify: $Q < 0$ (heat removed), size mein same lekin sign mein Ex 2 ke $+ 748.26 J$ se opposite — cooling heating ka ulta hai. Yeh $W = 0, Q < 0$ edge case hai. ✓

Ex 14 — Cell N: heat in, lekin itna zyada work out ki gas thandi ho jaati hai

Worked example Statement

Ek gas $80 J$ heat absorb karta hai expand hote hue aur apne piston par $200 J$ work karta hai. $Δ U$ nikalo aur batao gas warm hua ya cool.

Forecast: heat andar flow ho rahi hai — lekin gas hard work kar raha hai. Kya $T$ phir bhi gir sakta hai?

Signs: heat in $\Rightarrow Q = + 80 J$ ; gas work karta hai $\Rightarrow W = + 200 J$ . Yeh step kyun? Dono positive, lekin is baar withdrawal ( $W$ ) deposit ( $Q$ ) se bada hai.
$Δ U = Q - W = 80 - 200 = - 120 J$ . Yeh step kyun? Gas ne $200 J$ work pay kiya lekin sirf $80 J$ heat li, toh use apne store se $120 J$ nikalne pade.
Kyunki $Δ U < 0$ aur $U = \frac{f}{2} n R T$ , temperature girta hai. Yeh step kyun? Kam internal energy ka matlab kam average molecular KE, isliye kam $T$ — chahe heat andar aa rahi ho.

Verify: $Δ U = - 120 < 0$ : heat add ki gayi phir bhi gas thanda hua, kyunki work out zyada tha heat in se. Yeh intuition "heat in $\Rightarrow$ warmer" ko tod deta hai aur $Q > 0, W > 0, W > Q$ cell complete karta hai. ✓

Recall Quick self-test across the matrix

Constant volume ka matlab $W = ?$ ::: $W = 0$ , toh $Q = Δ U$ . Isothermal (ideal gas) ka matlab $Δ U = ?$ ::: $Δ U = 0$ , toh $Q = W$ . Adiabatic ka matlab $Q = ?$ ::: $Q = 0$ , toh $Δ U = - W$ . Vacuum mein free expansion: $Q, W, Δ T$ ? ::: Sab zero — koi heat nahi, koi work nahi, koi temperature change nahi. Heat in aur gas compressed: $Δ U$ ka sign? ::: Positive aur bada — dono $Q > 0$ aur $- W > 0$ energy deposit karte hain. Heat out jabki gas work karta hai: $Δ U$ ka sign? ::: Negative — dono effects energy withdraw karte hain ( $Q < 0, W > 0$ ). Heat in lekin work out isse zyada ( $W > Q > 0$ ): warm ya cool? ::: Cool — $Δ U = Q - W < 0$ chahe heat enter kar rahi ho. Ek full cycle mein, $Δ U = ?$ ::: $0$ , kyunki $U$ ek state function hai; phir $W_{net} = Q_{net}$ . Work done on the gas first law mein enter hota hai ::: negative $W$ ke roop mein (kyunki $W$ = work done by gas). $k_{B}$ aur $R$ ke beech relation ::: $k_{B} = R / N_{A}$ — $R$ per mole equals $k_{B}$ per molecule.

Connections

Heat and Internal Energy — Microscopic vs Macroscopic
First Law of Thermodynamics
Isothermal and Adiabatic Processes
Specific Heats Cv and Cp
Equipartition Theorem
Degrees of Freedom and Molecular Structure

Worked examples — Heat and internal energy — microscopic vs macroscopic

The scenario matrix

Ex 1 — Cell A: heat in, gas does work

Ex 2 — Cell B: constant volume, saari heat internal energy ban jaati hai

Ex 3 — Cell C: isothermal, $Δ U = 0$

Ex 4 — Cell D: adiabatic, $Q = 0$

Ex 5 — Cell E: sign flip, work done ON the gas

Ex 6 — Cell F: heat out AUR compression saath mein

Ex 7 — Cell G: degenerate case, free expansion into vacuum

Ex 8 — Cell H: degrees of freedom change (diatomic gas)

Ex 9 — Cell I: real-world word problem

Ex 10 — Cell J: exam twist, ek full cycle

Ex 11 — Cell K: heat in AUR gas compressed (dono $U$ ko badhate hain)

Ex 12 — Cell L: heat OUT jabki gas abhi bhi work karta hai

Ex 13 — Cell M: constant-volume cooling

Ex 14 — Cell N: heat in, lekin itna zyada work out ki gas thandi ho jaati hai

Connections

Concept Map