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

1.7.13 · D2 · Physics › Thermodynamics › First law of thermodynamics — dU = dQ − dW, sign conventions

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Step 1 — System draw karo aur har arrow ko ek naam do

• Yellow arrow jo andar point kar raha hai woh heat hai. Hum isse naam dete hain $Q$ (Latin quantitas caloris se). Yeh woh energy hai jo wall cross karti hai kyunki bahar andar se zyada hot hai.
• Red arrow piston par work hai. Hum isse $W$ kehte hain. Yeh woh energy hai jo gas piston ko bahar push karke kharach karti hai.
• Blue jiggling dots molecules hain. Us saari jiggling ki total energy internal energy hai, jise $U$ kaha jaata hai.

$$\underset{\text{heat in, yellow}}{\underbrace{Q}}\underset{\text{work out, red}}{\underbrace{W}}\underset{\text{energy stored inside, blue}}{\underbrace{U}}$$

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Step 2 — Energy kabhi kho nahi jaati: ledger likho

Paani ki ek tank ki picture socho (figure). Paani ek tap se andar aata hai ($Q$). Paani store hota hai, level badhata hai ($\Delta U$). Paani ek spout se bahar jaata hai ($W$). Tank ka iron rule:

$$\underset{\text{poured in}}{\underbrace{Q}}=\underset{\text{level rise}}{\underbrace{\Delta U}}+\underset{\text{poured out}}{\underbrace{W}}$$

• $Q$ — woh sab jo andar aaya (tap).
• $\Delta U$ — jo store hai uska change. Chhota triangle $\Delta$ matlab hai "change in" — final minus initial.
• $W$ — woh sab jo bahar gaya (spout).

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Step 3 — Famous form mein rearrange karo

Ledger $Q=\Delta U+W$ se start karke, dono sides se $W$ subtract karo:

$$\boxed{\Delta U=\underset{\text{heat added}}{\underbrace{Q}}-\underset{\text{work done BY gas}}{\underbrace{W}}}$$

Process ke ek chhote se sliver ke liye hum usi law ko tiny changes ke saath likhte hain:

$$dU=dQ-dW$$

• $d$ aage matlab hai "infinitesimally small amount of." Poori journey ki jagah, yeh uska ek microscopic step hai.

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Step 4 — Work actually kahan se aata hai? $dW=PdV$

Piston close-up dekho. Maano piston ki area $A$ hai. Gas usse pressure $P$ se press karta hai (force per unit area). Toh total push hai:

$$\underset{\text{force}}{\underbrace{F}}=\underset{\text{pressure}}{\underbrace{P}}\times \underset{\text{piston area}}{\underbrace{A}}$$

Ab piston ko ek tiny distance $dx$ bahar slide hone do (red arrow). Work force times distance hai:

$$dW=Fdx=PAdx$$

Lekin woh shaded green slab dekho jise piston sweep out karta hai — uska volume area times thickness hai:

$$\underset{\text{swept slab}}{\underbrace{Adx}}=dV\text{(tiny volume change)}$$

Substitute karo aur geometry gaayab ho jaati hai, sirf pure thermodynamics bachti hai:

$$\boxed{dW=PdV}⟹W={\int }_{V_1}^{V_2}PdV$$

• $P$ — us instant par pressure.
• $dV$ — us instant mein add hua volume ka sliver.
• $\int$ (integral) ek running total hai — yeh start volume $V_1$ se end volume $V_2$ tak har sliver $PdV$ ko add karta hai. Humein iske zaroorat hai kyunki $P$ gas expand hone par change ho sakta hai, isliye hum sirf ek baar multiply nahi kar sakte.

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Step 5 — $W$ ka sign $dV$ ke sign se

Figure mein do panels:

• Expansion (left): piston bahar jaata hai, volume badhta hai, toh $dV>0$. Phir $dW=PdV>0$. Gas surroundings par work karta hai — energy bahar jaati hai. $\Delta U=Q-W$ mein yeh stored energy se subtract karta hai.
• Compression (right): piston andar push hota hai, volume shrink karta hai, toh $dV<0$. Phir $dW=PdV<0$. Surroundings gas par work karte hain — energy andar aati hai. $\Delta U=Q-W$ mein, negative $W$ subtract karna energy add karta hai.

$$dV>0\Rightarrow W>0\text{ (gas pushes out)}dV<0\Rightarrow W<0\text{ (gas gets squeezed)}$$

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Step 6 — Chaar special processes: ek ek term ko zero karo

Har panel ek $P$–$V$ path hai (volume across, pressure upar):

Process	Kya fixed hai	Jo term khatam hoti hai	Law ban jaata hai
Isochoric	volume $V$	$dV=0\Rightarrow W=0$	$\Delta U=Q$
Isobaric	pressure $P$	koi nahi	$\Delta U=Q-P\Delta V$
Isothermal	temperature $T$	$\Delta U=0$ (ideal gas)	$Q=W$
Adiabatic	koi heat cross nahi karti	$Q=0$	$\Delta U=-W$

• Isochoric (vertical line): piston bolt hai. Woh move nahi kar sakta, toh koi volume sweep nahi hota, $W=0$. Saari heat stored energy ban jaati hai — gas bas zyada hot ho jaati hai.
• Isobaric (horizontal line): pressure constant, toh $W=P\Delta V$ (ek constant ka integral bas $P$ times width hota hai).
• Isothermal (gentle curve): ek ideal gas ke liye, $U$ sirf temperature $T$ par depend karta hai (dekho Internal energy and degrees of freedom). $T$ fixed rakho aur $U$ change nahi ho sakta: $\Delta U=0$, toh heat ka har joule work ban ke bahar jaata hai.
• Adiabatic (steep curve): walls insulated hain, $Q=0$. Ab work aur internal energy directly trade karte hain — ise compress karo aur yeh garm ho jaata hai, heat ke liye koi escape nahi hota.

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Step 7 — Degenerate case: vacuum mein free expansion

Ek box ki picture socho jo ek wall se split hai: left par gas, right par vacuum (figure). Wall mein ched karo. Gas poore box mein fill hone ke liye rush karta hai — volume double ho jaata hai.

• Koi piston nahi, koi push nahi: gas empty space mein expand karta hai, toh koi external pressure nahi hai jiske against push kare. Work done $=0$, chahe volume change bhi kyu na ho. ($dW=PdV$ mein, relevant opposing pressure zero hai.)
• Insulated walls: koi heat cross nahi karti, $Q=0$.

Dono ko law mein feed karo:

$$\Delta U=Q-W=0-0=0$$

Toh $U$ unchanged rehta hai — aur ek ideal gas ke liye iska matlab hai temperature bilkul nahi badlta, chahe gas expand bhi ho gayi! Yeh woh counter-intuitive gem hai jo prove karta hai ki $\Delta U$ state track karta hai, na ki volume change ke drama ko.

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Ek picture mein poora summary

Sab kuch ek canvas par: heat $Q$ andar aata hai (yellow), work $W$ piston se bahar jaata hai (red), aur jo bacha woh internal energy $U$ badhata hai (blue). Ledger exactly balance karta hai:

$$\underset{\text{stayed}}{\underbrace{\Delta U}}=\underset{\text{came in}}{\underbrace{Q}}-\underset{\text{went out}}{\underbrace{W}},W=\int PdV$$

Recall Feynman retelling — plain words mein poora walkthrough

Gas ke ek box mein teen darwaaze hain: heat in, work out, aur andar energy stored. Kuch bhi kahin se appear nahi ho sakta (Step 1–2), isliye jo aaya = jo raha + jo gaya — bus yahi poora law hai, sirf ek tap aur spout wali paani ki tank (Step 3). "Work out" woh gas hai jo piston ko shove karta hai: force pressure times area hai, distance hai kitna door woh slide karta hai, aur multiply karne par $PdV$ milta hai — woh volume jo sweep hota hai (Step 4). Woh work plus hai ya minus yeh bas depend karta hai ki volume bada ya chota hua (Step 5). Law ko feel karne ke liye, ek term kill karo: piston bolt karo (no work), ya walls insulate karo (no heat), ya temperature hold karo taaki stored energy change hi na ho sake (Step 6). Aur woh trick jo sab prove karta hai: gas ko vacuum mein burst hone do — volume double ho jaata hai lekin woh kuch push nahi karta aur koi heat nahi khota, toh temperature bhi change nahi hota (Step 7). Jo tum khaate ho woh equals hai jo store karo plus jo kharach karo.

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Connections

• Parent topic
• Conservation of energy (mechanics)
• Work done in thermodynamic processes (PV diagrams)
• Isothermal, adiabatic, isobaric, isochoric processes
• Internal energy and degrees of freedom
• Heat capacities Cp and Cv
• Second law of thermodynamics