Work done in each process — derivation: Worked examples

1.7.15 · D3 · Physics › Thermodynamics › Work done in each process — derivation

Yeh ek drill page hai us derivation note ke liye. Wahan humne ek master idea banaya tha $W = \int_{V_{1}}^{V_{2}} P d V$ Yahan hum isme har tarah ke cases daalenge: har sign, har degenerate input, har limit, ek word problem, aur ek exam trap. Compute karne se pehle guess karo — yeh "Forecast" ki aadat hi hai jo nonsense answers pakadwati hai.

Recall Har symbol ka matlab (ek baar banao, har jagah use karo)

$P$ = pressure = force per area jo gas push karta hai (unit: pascal, $Pa = N/m^{2}$ ).
$V$ = volume jo gas fill karta hai (unit: $m^{3}$ ; note karo $1 L = 1 0^{- 3} m^{3}$ ).
$W = \int P d V$ = tiny pushes $P d V$ ka running sum jab volume change hoti hai. $P$ – $V$ picture pe yeh path ke neeche ka area hai.
$n$ = moles of gas, $R = 8.314 J mol^{- 1} K^{- 1}$ (gas constant), $T$ = absolute temperature in kelvin.
$γ$ (gamma) = ratio of heat capacities, $= 5/3$ monatomic gas ke liye, $= 7/5$ diatomic gas ke liye. Yeh sirf adiabatic work mein aata hai.
$ln$ = natural logarithm — woh function jo $e^{(\cdot)}$ ko undo karta hai; yeh isliye aata hai kyunki $\int \frac{d V}{V} = ln V$ .

The scenario matrix

Is topic se jo bhi question aa sakta hai woh is table ka ek cell hai. Neeche ke worked examples [Cell n] se tag kiye hain taaki tum dekh sako ki poora space cover ho raha hai.

Cell	Case class	Kya cheez tricky banati hai
1	Isobaric expansion ( $V_{2} > V_{1}$ )	positive work, $P$ ko bahar nikalo
2	Isobaric compression ( $V_{2} < V_{1}$ )	work negative aati hai
3	Isochoric ( $d V = 0$ ) — degenerate	area zero hai, $W = 0$ chahe kitni bhi heat ho
4	Isothermal expansion	$P$ vary karta hai → log aata hai, $W > 0$
5	Isothermal compression	same log, $V_{2} < V_{1}$ → $ln < 0$ → $W < 0$
6	Adiabatic expansion (gas cools)	power-law integral, $W > 0$ , sign trap
7	Limiting behaviour	$V_{2} \to V_{1}$ (sab $W \to 0$ ); $γ \to 1$ (adiabatic $\to$ isothermal)
8	Cyclic loop — sign by direction	net $W$ = enclosed area; clockwise $> 0$
9	Real-world word problem	words ko translate karo → kaunsa formula
10	Exam twist	mixed path / first-law combination

Figure — Work done in each process — derivation

Upar ki picture poori matrix ek nazar mein hai: same do endpoints, alag-alag coloured paths, alag-alag shaded areas. Work area hai, aur area path pe depend karta hai.

Example A — Isobaric expansion [Cell 1]

Example B — Isobaric compression [Cell 2]

Example C — Isochoric, the degenerate case [Cell 3]

Example D — Isothermal expansion [Cell 4]

Example E — Isothermal compression [Cell 5]

Example F — Adiabatic expansion [Cell 6] + sign trap

Worked example

$1$ mol diatomic gas ( $γ = 7/5$ ) adiabatically $T_{1} = 500 K$ se $T_{2} = 380 K$ tak expand karta hai. $W$ nikalo.

Forecast: Adiabatic = blanket mein lapaita, $Q = 0$ ; expansion matlab woh apni khud ki internal energy kharch karta hai, isliye cool hota hai ( $T$ girta hai, given se match) aur $W > 0$ .

Step 1 — temperature form choose karo. Yeh step kyun? Hume temperatures diye hain, isliye $W = \frac{n R ( T _{1} - T _{2} )}{γ - 1}$ use karo (derived from $\int K V^{- γ} d V$ ). Order note karo $T_{1} - T_{2}$ — initial minus final.

Step 2 — plug in karo. Yeh step kyun? $γ - 1 = 7/5 - 1 = 2/5 = 0.4$ . $W = \frac{1 \times 8.314 \times ( 500 - 380 )}{0.4} = \frac{997.68}{0.4} = 2494.2 J$

Verify: Positive ✓ (cooling expansion se $W > 0$ milta hai). Agar tumne $\frac{n R ( T _{2} - T _{1} )}{γ - 1}$ likha hota toh $- 2494.2 J$ aata, jo galat claim karta ki expanding gas negative work karti hai — yahi [Cell 6] ka sign trap hai.

Example G — Limiting behaviour [Cell 7]

Worked example

Do limits ek saath. (i) Ek isothermal process jisme $V_{2} \to V_{1}$ (barely koi expansion). (ii) Ek adiabatic curve jab $γ \to 1$ : dikhao ki yeh isothermal answer ke paas aata hai.

Forecast: (i) Koi volume change nahi → $W \to 0$ . (ii) Jab $γ = 1$ , $P V^{γ} = K$ ban jaata hai $P V = K =$ constant, jo isothermal condition hi hai — isliye dono formulas merge ho jaane chahiye.

Step 1 — limit (i) numerically. Yeh step kyun? $1$ mol $300 K$ pe lo, $V_{2} / V_{1} = 1.001$ . $W = 1 \times 8.314 \times 300 \times ln (1.001) = 2494.2 \times 0.0009995 = 2.493 J$ Jab $V_{2} / V_{1} \to 1$ , $ln (1) = 0$ , isliye $W \to 0$ ✓ — smoothly, abruptly nahi.

Step 2 — limit (ii) conceptually. Yeh step kyun? Adiabatic work $\frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1}$ mein denominator mein $γ - 1 \to 0$ hai, lekin numerator bhi $\to 0$ jaata hai kyunki jab $γ = 1$ toh curve $P V =$ const hai isliye $P_{1} V_{1} = P_{2} V_{2}$ . Yeh $0/0$ resolve hota hai (L'Hôpital se) exactly $n R T ln (V_{2} / V_{1})$ mein — yahi isothermal result hai.

Verify: Dono limits theek behave karte hain. Adiabatic Process and PV^gamma curve $γ > 1$ ke liye Isothermal Process and Boyle's Law curve se steeper hoti hai; $γ = 1$ pe dono coincide karte hain (upar figure).

Example H — Cyclic loop, sign by direction [Cell 8]

Worked example

Gas rectangular loop $A \to B \to C \to D \to A$ chalata hai jaise upar dikhaya: $A = (V = 1 m^{3}, P = 1 \times 1 0^{5} Pa)$ , $B = (3, 1 \times 1 0^{5})$ , $C = (3, 3 \times 1 0^{5})$ , $D = (1, 3 \times 1 0^{5})$ , usi order mein. $W_{net}$ nikalo.

Forecast: Order $A \to B \to C \to D \to A$ trace karo: yeh yahan anticlockwise hai (low pressure pe expand, high pressure pe compress). Anticlockwise ⇒ $W_{net} < 0$ . Magnitude = box area.

Step 1 — leg by leg. Yeh step kyun? Sirf horizontal legs mein $d V \neq = 0$ hota hai; vertical legs ( $V$ fixed) $0$ dete hain.

$A \to B$ (expand at $P = 1 0^{5}$ ): $W = 1 0^{5} \times (3 - 1) = + 2 \times 1 0^{5} J$ .
$B \to C$ (isochoric): $0$ .
$C \to D$ (compress at $P = 3 \times 1 0^{5}$ ): $W = 3 \times 1 0^{5} \times (1 - 3) = - 6 \times 1 0^{5} J$ .
$D \to A$ (isochoric): $0$ .

Step 2 — sum karo. Yeh step kyun? Net work loop ke around algebraic sum hai. $W_{net} = 2 \times 1 0^{5} + 0 - 6 \times 1 0^{5} + 0 = - 4 \times 1 0^{5} J$

Verify: Enclosed area $= Δ P \cdot Δ V = (2 \times 1 0^{5}) (2) = 4 \times 1 0^{5} J$ ; sign negative kyunki anticlockwise ✓. Carnot Cycle logic se match: loop direction sign set karta hai.

Example I — Real-world word problem [Cell 9]

Example J — Exam twist: mixed with first law [Cell 10]

Worked example

Ek isothermal expansion mein, gas $Q = 5000 J$ heat absorb karta hai. Ek student claim karta hai "work done zero hai kyunki temperature constant hai, isliye $Δ U = 0$ ." Use correct karo aur actual work nikalo.

Forecast: Claim " $Δ U = 0$ " ko " $W = 0$ " se confuse kar raha hai. Constant $T$ pe bhi piston move karta hai, isliye $W \neq = 0$ . Actually first law force karta hai ki $W = Q$ .

Step 1 — first law apply karo. Yeh step kyun? $Δ U = Q - W$ . Isothermal ⇒ $T$ constant ⇒ $Δ U = 0$ (internal energy sirf $T$ pe depend karti hai). Isliye $0 = Q - W \Rightarrow W = Q = 5000 J .$

Step 2 — student kahan galat gaya. Yeh step kyun? $Δ U = 0$ kehta hai heat aur work equal hain, na ki koi bhi zero hai. Gas quietly saare $5000 J$ absorbed heat ko work mein convert kar deta hai.

Verify: $W = 5000 J$ , positive (expansion) ✓. Log formula se consistent: jo bhi $n R T ln (V_{2} / V_{1})$ equal ho, woh yahan $5000 J$ absorbed ke barabar hona chahiye.

Recall Rapid self-test

Isobaric compression $0.05 \to 0.02 m^{3}$ at $2 \times 1 0^{5}$ Pa, work by gas ::: $- 6000 J$ Isothermal, $2$ mol, $400$ K, volume $10 \to 30$ L ::: $+ 7307 J$ Same but compressed $30 \to 10$ L ::: $- 7307 J$ Adiabatic diatomic, $500 \to 380$ K, $1$ mol ::: $+ 2494 J$ Anticlockwise rectangular loop, $Δ P = 2 \times 1 0^{5}$ , $Δ V = 2$ ::: $- 4 \times 1 0^{5} J$ Isothermal expansion absorbing $5000$ J of heat, work ::: $5000 J$ (since $Δ U = 0 \Rightarrow W = Q$ )

Connections

Parent: the derivation
First Law of Thermodynamics — Examples C aur J mein use hua
P-V Diagrams — har figure yahan rehta hai
Adiabatic Process and PV^gamma — Examples F, G, I
Isothermal Process and Boyle's Law — Examples D, E, J
Internal Energy and Cv — kyun adiabatic gas cool hoti hai
Carnot Cycle — cyclic sign rule (Example H)