Exercises — First law of thermodynamics — dU = dQ − dW, sign conventions

KYA: "Absorbs" = heat gas ke andar flow hoti hai, toh $Q=+250\text{J}$. KYUN: Physics convention mein $Q>0$ ka matlab hai heat gas mein ADD hui. PICTURE: heat bahar queue mein khadi hai aur andar step kar rahi hai — "QUWe by" mnemonic wali queue. $Q=+250\text{J}$

KYA: $W=0$. KYUN: $W=\int PdV$. Rigid box $\Rightarrow dV=0\Rightarrow W=0$. Yeh ek isochoric (constant-volume) process hai — dekho Isothermal, adiabatic, isobaric, isochoric processes. PICTURE: piston ek jagah bolted hai; gas kuch bhi push nahi kar sakti, toh koi kaam nahi karti. Tab $\Delta U=Q$: saari heat internal energy ban jaati hai.

Signs: $Q=+600$ (add hua), $W=+220$ (gas expand hoti hai → kaam by gas). Apply karo: $\Delta U=Q-W=600-220=+380\text{J}$. PICTURE: jo $600\text{J}$ andar aayi, unme se $220\text{J}$ piston ke through wapas bahar chali gayi; $380\text{J}$ ruki → gas garam hua.

Signs: gas par kaam $=180\text{J}\Rightarrow W_{\text{by}}=-180\text{J}$. Heat release hui $\Rightarrow Q=-70\text{J}$. Apply karo: $\Delta U=Q-W=(-70)-(-180)=+110\text{J}$. PICTURE: $180\text{J}$ compression se pump in hua, $70\text{J}$ heat ke roop mein leak out hua → net $+110\text{J}$ store hua, gas warm hoti hai.

KYUN constant $P$ easy hai: $W={\int }_{V_1}^{V_2}PdV$; $P$ constant hone par yeh integral se bahar aa jaata hai, aur $W=P(V_2-V_1)=P\Delta V$ milta hai. Apply karo: $W=(2.0\times 10^5)(3.0\times 10^{-3}-1.0\times 10^{-3})=(2.0\times 10^5)(2.0\times 10^{-3})=400\text{J}$. PICTURE (neeche): $P$–$V$ diagram par isobar ek flat horizontal line hoti hai; kaam uske neeche ka rectangle hota hai.

Key fact: Ideal gas ke liye $U=U(T)$ sirf (dekho Internal energy and degrees of freedom). Constant $T\Rightarrow \Delta U=0$. Apply karo: $0=Q-W\Rightarrow Q=W=520\text{J}$. PICTURE: heat ka har joule jo andar aata hai seedha kaam ke roop mein bahar chala jaata hai; internal "warmth" ka level kabhi nahi hilta.

Key facts: Adiabatic $\Rightarrow Q=0$. Gas par kaam $\Rightarrow W_{\text{by}}=-310\text{J}$. Apply karo: $\Delta U=Q-W=0-(-310)=+310\text{J}>0\Rightarrow$ gas garm hoti hai. KYUN: Koi heat channel nahi hone par, compression energy ka sirf ek hi jagah jaana mumkin hai — internal energy. Isliye bicycle pump garm ho jaata hai.

Key fact: $U$ ek state function hai. Start state = end state $\Rightarrow \Delta U_{\text{cycle}}=0$, chahe path koi bhi ho. Apply karo: $0=Q_{\text{cycle}}-W_{\text{cycle}}\Rightarrow Q_{\text{cycle}}=W_{\text{cycle}}=+150\text{J}$. PICTURE (neeche): $P$–$V$ diagram par ek cycle ek closed loop hai; enclosed area net kaam hai, aur yahan woh area net heat absorb ko supply karta hai.

Step 1 — Path 1 se $\Delta U$ nikalo. $\Delta U=Q_1-W_1=900-350=+550\text{J}$. Step 2 — ise reuse karo. $\Delta U$ sirf endpoints A aur B par depend karta hai, toh Path 2 ka same $\Delta U=+550\text{J}$ hoga. Step 3 — $Q_2$ solve karo. $550=Q_2-120\Rightarrow Q_2=670\text{J}$. KYUN yeh kaam karta hai: $U$ ek state function hai — yahi lake analogy ka poora point hai. Path badlo, toh $Q$ aur $W$ saath mein change hote hain, lekin unka difference pinned rehta hai.

$W=0$: gas vacuum ke against expand hoti hai — kuch bhi push back nahi kar raha, toh $\int P_{\text{ext}}dV=0$. Koi piston move nahi hua, koi surroundings lift nahi hua. $Q=0$: insulated walls, toh koi heat cross nahi karti. $\Delta U=Q-W=0-0=0$. $\Delta T=0$: ideal gas $U=U(T)$; $\Delta U=0\Rightarrow T$ unchanged. KYUN log surprised hote hain: gas expand karti hai phir bhi koi kaam nahi karti aur thandi nahi hoti — kyunki "expanding" tabhi energy cost karti hai jab tum kisi cheez ke against push karo.

Leg A→B (isochoric): $dV=0\Rightarrow W_{AB}=0$. Tab $\Delta U_{AB}=Q_{AB}-W_{AB}=600\text{J}$.

Leg B→C (isothermal, ideal gas): $\Delta U_{BC}=0$ (constant $T$). Toh $W_{BC}=Q_{BC}-\Delta U_{BC}=450-0=450\text{J}$.

Leg C→A (isobaric compression): gas par $250\text{J}$ kaam $\Rightarrow W_{CA}=-250\text{J}$. Tab $\Delta U_{CA}=Q_{CA}-W_{CA}=(-720)-(-250)=-470\text{J}$.

(b) Gas dwara net kaam: $W_{\text{cycle}}=W_{AB}+W_{BC}+W_{CA}=0+450-250=+200\text{J}$.

(c) $\Delta U_{\text{cycle}}$ check karo: $\Delta U_{AB}+\Delta U_{BC}+\Delta U_{CA}=600+0-470=+130\text{J}$.

Yeh ZERO NAHI hai — numbers inconsistent hain! Ek sacche cycle mein $\Delta U_{\text{cycle}}=0$ hona chahiye. Heat se cross-check karein: $Q_{\text{cycle}}=600+450-720=+330\text{J}$, lekin $W_{\text{cycle}}=200\text{J}$, aur cycle require karti hai $Q_{\text{cycle}}=W_{\text{cycle}}$. Yeh mismatch ($330\ne 200$) ek over-specified, unphysical data set expose karta hai.

Mastery ka lesson: First Law ek consistency check hai. Ek genuine cycle mein tum individual legs par heats aur works freely specify kar sakte ho, lekin unhe $\sum \Delta U=0$ aur $Q_{\text{cycle}}=W_{\text{cycle}}$ satisfy karna hoga. Numbers trust karne se pehle hamesha loop audit karo.

Requirement: $\Delta U_{AB}+\Delta U_{BC}+\Delta U_{CA}=0$. Hum $W_{CA}=-250\text{J}$ rakhte hain, toh $\Delta U_{CA}=Q_{CA}-(-250)=Q_{CA}+250$. $600+0+(Q_{CA}+250)=0\Rightarrow Q_{CA}=-850\text{J}.$ Net heat: $Q_{\text{cycle}}=600+450-850=+200\text{J}$. Net kaam: unchanged, $W_{\text{cycle}}=0+450-250=+200\text{J}$. Check: $Q_{\text{cycle}}=W_{\text{cycle}}=200\text{J}$ ✓ aur $\Delta U_{\text{cycle}}=0$ ✓. Ab yeh ek valid engine cycle hai jo har loop mein $200\text{J}$ kaam deta hai.