Work done in each process — derivation — Hinglish notes

1.7.15 · Physics › Thermodynamics

The Master Formula (ise ek baar derive karo, har jagah use karo)

KYU force se shuru karein? Kyunki work fundamentally $d W = F \cdot d x$ hota hai.

KAISE pressure laayein:

Area $A$ ka piston, gas pressure $P$ force $F = P A$ se push karta hai.
Piston $d x$ bahar nikalti hai. Tiny work $d W = F d x = P A d x$ .
Lekin $A d x = d V$ (swept volume).

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

Yahi 80/20 core hai: $W = \int P d V$ master karo aur har case bas "V ke function ke roop mein $P$ kya hai?" ka sawaal hai.

Figure — Work done in each process — derivation

1. Isobaric (constant pressure)

KYA: $P$ = constant, ise integral se bahar nikalo.

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

$P V = n R T$ use karke: $W = n R (T_{2} - T_{1})$ .

2. Isochoric (constant volume)

KYA: $V$ = constant $\Rightarrow d V = 0$ .

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

3. Isothermal (constant temperature)

KYU mushkil hai: $V$ ke saath $P$ bhi change hota hai, toh pehle $P (V)$ substitute karna padega.

KAISE: Constant $T$ par ideal gas: $P V = n R T \Rightarrow P = \frac{n R T}{V}$ .

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

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

(Boyle's law se $\frac{V _{2}}{V _{1}} = \frac{P _{1}}{P _{2}}$ use kiya.)

4. Adiabatic (koi heat exchange nahi, $Q = 0$ )

KYU naya formula: yahan $P V^{γ} = K$ (constant) hai, toh $P = K V^{- γ}$ .

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

Ab $K = P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$ , toh $K V_{2}^{1 - γ} = P_{2} V_{2}$ aur $K V_{1}^{1 - γ} = P_{1} V_{1}$ :

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

Cyclic process

Closed loop ke liye, $W_{n e t} = \oint P d V =$ loop se enclosed area. Clockwise loop → $W > 0$ ; anticlockwise → $W < 0$ .

Recall Feynman: 12-saal ke bacche ko explain karo

Socho tum ek swing ko push kar rahe ho. Agar tum poore raste same strength se push karo, toh work bas strength × distance hai (isobaric). Lekin ek gas phailte-phailte kamzor hoti jaati hai (pressure drop hota hai), toh tumhe bahut saari chhoti-chhoti pushes add karni padti hain jo shrink hoti rehti hain — yahi integral hai. Agar gas ko ek blanket mein wrap kar do taaki koi heat na nikle (adiabatic), toh woh push karne ke liye apni hi garmahat use kar leti hai, toh thandi ho jaati hai. Agar ise same garmahat par rakho (isothermal), toh woh zyada der push kar sakti hai aur total ek "log" mein nikalti hai.

Flashcards

Saare work calculations ka starting point

d W = P d V

, toh

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

(P–V curve ke neeche area)

Isobaric process mein work

W = P (V_{2} - V_{1}) = n R Δ T

Isochoric process mein work

W = 0

(kyunki

d V = 0

)

Isothermal process mein work

W = n R T ln (V_{2} / V_{1}) = n R T ln (P_{1} / P_{2})

Isothermal mein log kyun aata hai

P = n R T / V

vary karta hai, aur

\int d V / V = ln V

Adiabatic process mein work

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

Adiabatic expansion mein work kahan se aata hai

Internal energy se, isliye gas thandi ho jaati hai (

W = - Δ U

)

Cyclic process mein net work

P–V diagram par loop se enclosed area (clockwise = positive)

Kya work ek state function hai

Nahi, yeh path-dependent hai (curve par depend karta hai, sirf endpoints par nahi)

Connections

First Law of Thermodynamics — $Δ U = Q - W$ inheen $W$ values use karta hai
P-V Diagrams — work = curve ke neeche area
Adiabatic Process and PV^gamma — $P V^{γ} = K$ ka source
Isothermal Process and Boyle's Law
Internal Energy and Cv — adiabatic $W = - Δ U$ ko link karta hai
Carnot Cycle — net cyclic work