Work done in each process — derivation: Visual walkthrough

1.7.15 · D2 · Physics › Thermodynamics › Work done in each process — derivation

Yeh page parent derivation ka picture-companion hai. Har symbol use se pehle earn kiya jaata hai.

Step 1 — "Work" ka matlab kya hota hai ek pushing gas ke liye

KYA HAI. Ek gas ek cylinder mein band hai jo ek piston (ek sliding lid) se seal hai. Gas bahar ki taraf press karti hai; agar piston thoda slide kare, toh gas ne work kiya hai.

YAHAN SE KYUN SHURU KAREIN. Physics mein, work = force $\times$ distance moved force ki direction mein. Symbol $W$ . Yeh measure karta hai "kitna pushing effort actually motion mein convert hua." Hume force se shuru karna hoga kyunki yeh work ki ek ek definition hai jo hamesha sahi hoti hai — baaki sab kuch hum isi se build karte hain.

PICTURE. Neeche cylinder dekho. Gas left mein bhari hai; piston amber bar hai. White arrow woh force $F$ hai jo gas piston par lagaati hai. Agar piston move nahi karta, distance $= 0$ , toh $W = 0$ — ek aisi wall ko dhakkelna jo nahi hilegi koi work nahi karta.

Figure — Work done in each process — derivation

Step 2 — Force ko pressure mein badlo (pressure kyun natural variable hai)

KYA HAI. Force $F$ ko kisi aise cheez se replace karo jo gas actually "har jagah feel" karti hai: pressure.

FORCE KI JAGAH PRESSURE KYUN? Force depend karta hai ki piston kitna bada hai; ek wider piston usi gas se zyada total force feel karta hai. Pressure $P$ = force area par spread, $P = F / A$ , toh yeh gas ki akeli property hai, piston size se independent. Isliye har gas law $P$ mein likhi hoti hai, $F$ mein nahi. Rearrange karne par gas jo force produce karti hai woh milti hai:

F = pressure of gas P \times piston area A

$P$ ::: force per unit area (pascals). Gas ki intrinsic "push strength."
$A$ ::: piston ka flat area jis par gas push karti hai.
$P A$ ::: total force, kyunki pressure poore face par act karta hai.

PICTURE. Piston ka face shaded hai; bohot saare chhote cyan arrows pressure ko har patch par act karte dikhate hain. Unhe sab area $A$ par add karo aur single bada force arrow $F = P A$ wapas mil jaata hai.

Step 3 — Ek tiny push: $d W = P d V$

KYA HAI. Piston ko ek tiny distance bahar slide karne do jise hum $d x$ kehte hain (" $d$ " ka matlab hai "bahut chhoti matra"). Kiya gaya tiny work hai

d W = F d x = (P A) d x = P = d V (A d x) = P d V

$d x$ ::: ek sliver of distance jo piston move karta hai — itna chhota ki $P$ abhi nahi badla.
$A d x$ ::: area $\times$ length ka sliver = ek thin slab of volume jo sweep out hua. Hum ise $d V$ rename karte hain.
$d V$ ::: woh tiny extra volume jo gas ne gain kiya.
$d W = P d V$ ::: ek tiny expansion ke liye tiny work.

TINY STEPS MEIN KYUN KAREIN? Kyunki $P$ constant nahi hota jab gas expand karti hai — jaise gas phailti hai woh zyada weakly push karti hai. Agar hum $W = F \times (poori distance)$ ek shot mein try karte, toh hum zyaadatar trip ke liye galat $P$ use kar rahe hote. Journey ko itne thin slivers mein kaatna ki $P$ effectively constant ho har sliver ke andar — yehi hai jo tiny " $d$ " hamein deta hai.

PICTURE. Amber slab swept volume $d V$ hai; uski width $d x$ hai, uska area $A$ hai. Work mein uski "cost" $P$ times uski width hai — yeh ek thin strip ki height-times-width hai.

Step 4 — Tiny pushes add karo = integral = curve ke neeche area

KYA HAI. $V_{1}$ se $V_{2}$ tak poora expansion millions of these slivers hai jo side by side rakhe gaye hain. "Infinitely many infinitely thin pieces ko add karo" ka ek naam aur ek symbol hai: integral $\int$ .

W = \int_{V_{1}}^{V_{2}} P d V

$\int$ ::: Sum ke liye ek stretched "S" — yeh har sliver $P d V$ add karta hai.
$V_{1}$ (bottom) ::: jahan piston start karta hai.
$V_{2}$ (top) ::: jahan piston rukta hai.
Andar $P$ ::: har strip ki height; yeh strip se strip badal sakti hai.

INTEGRAL KYUN AUR MULTIPLICATION KYUN NAHI? Kyunki ek plain multiplication $P Δ V$ tabhi kaam karta hai jab height $P$ kabhi nahi badalti. Integral precisely woh tool hai jo changing height ko handle karta hai. Agar $P$ constant hota, toh integral ek rectangle mein wapas collapse ho jaata — koi contradiction nahi.

KAISA DIKHTA HAI. $P$ upar, $V$ across plot karo — ek $P$ – $V$ diagram (dekho P-V Diagrams). Har strip height $P$ , width $d V$ ki thin rectangle hai. Unhe stack karne se curve ke neeche ka region fill hota hai. Toh:

W = area under the P - V curve

Step 5 — Case A: Isobaric (flat curve → ek rectangle)

KYA HAI. "Iso-baric" = same pressure. $P$ constant hai, toh use sum se bahar nikalo:

W = \int_{V_{1}}^{V_{2}} P d V = P \int_{V_{1}}^{V_{2}} d V = height P width (V_{2} - V_{1})

YEH EASY WALA KYUN HAI. Constant height ka matlab hai curve ke neeche area bas ek plain rectangle hai: height $P$ , width $Δ V = V_{2} - V_{1}$ . Koi summing tricks ki zaroorat nahi.

PICTURE. Height $P$ par ek horizontal line; uske neeche shaded area ek clean rectangle hai.

Ideal-gas law $P V = n R T$ use karte hue, yeh $W = n R (T_{2} - T_{1})$ bhi equal hota hai.

Step 6 — Case B: Isochoric (koi width nahi → zero area)

KYA HAI. "Iso-choric" = same volume. Piston pinned hai; $V$ kabhi nahi badalta toh $d V = 0$ .

W = \int_{V_{1}}^{V_{1}} P d V = 0

ZERO KYUN. Koi distance travel nahi hua matlab kisi bhi strip ke liye koi width nahi. Diagram par "path" ek vertical line hai, aur ek vertical line ka uske neeche zero area hai — tum kitni bhi tall rectangle nahi bana sakte agar width zero ho.

PICTURE. Fixed $V$ par ek vertical segment; shaded region ek line mein collapse ho jaata hai — bharne ke liye kuch nahi.

Step 7 — Case C: Isothermal (ek curved roof → ek logarithm)

KYA HAI. "Iso-thermal" = same temperature $T$ . Ab $P$ genuinely change karta hai. $P V = n R T$ se hum har volume par height solve karte hain:

P = \frac{n R T}{V}

Is changing height ko sum mein feed karo:

W = \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V = n R T \int_{V_{1}}^{V_{2}} \frac{d V}{V} = n R T [ln V]_{V_{1}}^{V_{2}}

W = n R T ln \frac{V _{2}}{V _{1}}

$\frac{n R T}{V}$ ::: curve ki height — yeh sink karti hai jaise $V$ badhta hai (weaker push).
$\int \frac{d V}{V}$ ::: woh ek integral jiska answer ek logarithm hai, $ln$ .
$ln (V_{2} / V_{1})$ ::: "volume kitni baar badha, multiply-scale par measure karke."

LOGARITHM KYUN AATA HAI. Strips chhoti hoti jaati hain jaise curve dive karti hai (ek hyperbola $P \propto 1/ V$ ). $1/ V$ curve ke neeche area add karna woh unique kaam hai jo $ln$ produce karta hai. Isliye yahan tool logarithm hai aur power nahi — shape ne yahi maanga tha. Dekho Isothermal Process and Boyle's Law.

PICTURE. Ek neeche ki taraf curve karti hyperbola; uske neeche shaded area start ke paas mota hai (tall strips) aur end mein pattla (short strips).

Step 8 — Case D: Adiabatic (ek steeper curve → ek power law)

KYA HAI. "A-diabatic" = wall ke paas se koi heat cross nahi karti, $Q = 0$ . Yahan gas $P V^{γ} = K$ (ek constant) obey karti hai, jahan $γ$ (gamma) 1 se bada ek number hai jo gas type se set hota hai. Height solve karo: $P = K V^{- γ}$ .

W = \int_{V_{1}}^{V_{2}} K V^{- γ} d V = K [\frac{V ^{1 - γ}}{1 - γ}]_{V_{1}}^{V_{2}} = \frac{K}{1 - γ} (V_{2}^{1 - γ} - V_{1}^{1 - γ})

Kyunki $K = P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$ , ugly powers pressures mein snap shut ho jaate hain:

W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = \frac{n R ( T _{1} - T _{2} )}{γ - 1}

$K V^{- γ}$ ::: height — yeh isothermal $1/ V$ se tezi se drop karta hai, kyunki $γ > 1$ fall ko steep karta hai.
$\int V^{- γ} d V = \frac{V ^{1 - γ}}{1 - γ}$ ::: integrate karne ka power rule; yahan use kiya kyunki height $V$ ki ek power hai, $1/ V$ nahi.
$γ - 1$ denominator mein ::: $1 - γ$ ke sign flip hone ke baad jo bachta hai expansion ko positive rakhne ke liye.

POWER LAW KYUN, LOG NAHI. Ek adiabatic curve $P \propto V^{- γ}$ hai — $V$ ki ek power. Ek power integrate karna (other than $- 1$ ) hamesha power rule se ek aur power deta hai. Sirf special exponent $- 1$ log deta hai; adiabatic $- γ \neq = - 1$ use karta hai, toh hume power answer milta hai. $P V^{γ} = K$ ka source: Adiabatic Process and PV^gamma.

PICTURE. Same start se do curves: flatter cyan isothermal hai, steeper amber adiabatic hai. Adiabatic tezi se dive karta hai, toh woh kam area enclose karta hai — same volume change ke liye kam work.

Step 9 — Case E: Cyclic (loop → enclosed area)

KYA HAI. Ek cycle apne start par wapas aata hai, ek closed loop trace karte hue. Net work loop ke andar enclosed area hai (likha $\oint$ , circle-integral):

W_{net} = \oint P d V = area inside the loop

ENCLOSED AREA KYUN AATA HAI. Top ke saath daayein jaana (expansion) ek bada positive area add karta hai; neeche se waapas left jaana (compression) ek chhota negative area subtract karta hai. Jo bachta hai woh difference hai — woh region jo do paths ke beech trapped hai.

Clockwise loop ::: top path higher-pressure wali hai ⇒ $W_{net} > 0$ (engine).
Anticlockwise loop ::: $W_{net} < 0$ (refrigerator).

PICTURE. Ek closed loop; outgoing top arc (amber) aur returning bottom arc (cyan) ek shaded region enclose karte hain — woh shaded region hi net work hai. Yeh Carnot Cycle ka dil hai.

Ek picture summary

Ek hi $P$ – $V$ diagram par char processes, same start point share karte hue. Shape padho → work padho:

Flat line = isobaric = rectangle = $P Δ V$ .
Vertical line = isochoric = zero width = $0$ .
Gentle curve = isothermal = $1/ V$ ke neeche area = $n R T ln (V_{2} / V_{1})$ .
Steep curve = adiabatic = $V^{- γ}$ ke neeche area = $\frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1}$ .

Recall Feynman retelling — poora walkthrough plain words mein

Ek gas ek lid ko push karti hai. Work hai push strength × lid kitni door gayi. Humne "force" ko "pressure" se swap kiya kyunki pressure gas ki hoti hai chahe lid kitni bhi badi ho. Lekin gas phailtey phailtey kamzor padti jaati hai, toh hum ek baar push nahi kar sakte aur multiply nahi kar sakte — hum trip ko itne thin slivers mein kaatate hain ki strength har ek ke andar steady ho, har sliver ke liye chhota work $P d V$ dhundhte hain, aur unhe sab stack karte hain. Stacking = integral = pressure-versus-volume curve ke neeche area. Ab char cases bas char shapes hain: ek flat roof ek plain rectangle deti hai (isobaric); ek wall koi width nahi deti toh koi work nahi (isochoric); ek gently sagging roof ek logarithm deti hai (isothermal, kyunki shape $1/ V$ hai); ek steeper sagging roof ek power answer deti hai (adiabatic, kyunki heat trapped hai aur curve $V^{- γ}$ hai); aur ek closed loop bas apne do arcs ke beech ek area trap karta hai (cyclic). Ek idea — curve ke neeche area — char costumes mein.

Recall

Har work formula secretly ek equation kya hai? ::: $W = \int_{V_{1}}^{V_{2}} P d V$ — P–V curve ke neeche area. Isochoric work kyun vanish ho jaata hai? ::: Path ek vertical line hai — zero width, zero area, piston move nahi karta. Isothermal log kyun deta hai lekin adiabatic power kyun? ::: Isothermal height $1/ V$ hai ( $1/ V$ ka integral $ln$ hai); adiabatic height $V^{- γ}$ hai jahan $γ \neq = 1$ (power rule ek power deta hai). Net cyclic work geometrically kya equal hai? ::: Loop ke andar enclosed area; clockwise positive hai.

Connections

Parent: Work done in each process
First Law of Thermodynamics — $Δ U = Q - W$ in $W$ values consume karta hai
P-V Diagrams — work = curve ke neeche area
Adiabatic Process and PV^gamma — $P V^{γ} = K$ ka origin
Isothermal Process and Boyle's Law — $1/ V$ curve ka origin
Internal Energy and Cv — adiabatic $W = - Δ U$
Carnot Cycle — cyclic enclosed-area idea