Work in expansion - reversible isothermal w = −nRT ln(V₂ - V₁), irreversible w = −P_ext ΔV

2.5.4 · Chemistry › Thermodynamics (Chemical)

Core Intuition

Intuition Path kyun matter karta hai work ke liye?

Work state function nahi hai — yeh depend karta hai kaise aap state A se state B pahunche, sirf endpoints pe nahi. Socho ek heavy book ko shelf pe uthane ka kaam spring balance se karo:

Reversible (slow, careful): Tum use infinitely slowly uthate ho, toh har pal upward force jo tum lagate ho exactly gravity ko balance karta hai (sirf ek infinitesimal excess movement ke liye). Koi jhatka nahi, koi energy dissipate nahi — tum process ko ek infinitesimal change se reverse kar sakte ho aur poori energy wapas pa sakte ho. Maximum useful work exchange hota hai.
Irreversible (sudden): Tum book ko ek single constant resisting force ke against jhatke se uthate ho. Yeh jerky hai, energy dissipate hoti hai (heat/sound), aur tum ise ek infinitesimal reversal se recover nahi kar sakte.

Gases ke liye: reversible expansion ka matlab hai ki external pressure $P_{ext}$ gas pressure $P_{gas}$ ke equal hota hai har moment (sirf ek infinitesimal $d P$ ka difference) — system hamesha quasi-equilibrium mein hota hai. Irreversible expansion ka matlab hai $P_{ext}$ finite hai aur noticeably kam hai $P_{gas}$ se — gas ek "burst" mein kaam karta hai lower external pressure ke against aur difference dissipate ho jaata hai.

Definitions

Derivation: Reversible Isothermal Work

Isothermal kyun? Temperature constant $\Rightarrow$ ideal gas ke liye, $P V = n R T$ poore process mein hold karta hai.

Step 1: Work integral se shuru karo $w = - \int_{V_{1}}^{V_{2}} P_{ext} d V$

Step 2: Reversible process ke liye, $P_{ext} = P_{gas}$ Kyunki gas hamesha quasi-equilibrium mein hoti hai, external pressure gas pressure ke equal hota hai (infinitesimal $d P$ integral mein kuch contribute nahi karta): $P_{ext} = P_{gas} = \frac{n R T}{V} (ideal gas law)$

Hum $P_{gas}$ substitute kyun kar sakte hain? Sirf reversible processes mein! Irreversible mein, $P_{ext} \neq = P_{gas}$ .

Step 3: Integral mein substitute karo $w = - \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V$

Step 4: Constants bahar nikalo (T isothermal mein constant hai) $w = - n R T \int_{V_{1}}^{V_{2}} \frac{1}{V} d V$

Step 5: Integrate karo $w = - n R T [ln V]_{V_{1}}^{V_{2}} = - n R T (ln V_{2} - ln V_{1})$

Step 6: Log properties se simplify karo $w_{rev} = - n R T ln (\frac{V _{2}}{V _{1}})$

Iska matlab kya hai?

Agar $V_{2} > V_{1}$ (expansion): $ln (V_{2} / V_{1}) > 0 \Rightarrow w < 0$ (work done by gas)
Agar $V_{2} < V_{1}$ (compression): $ln (V_{2} / V_{1}) < 0 \Rightarrow w > 0$ (work done on gas)
Logarithmic dependence: volume ko 1→2 L se double karne pe utna hi work hota hai jitna 10→20 L se (same ratio).

Derivation: Irreversible Expansion

KYA ALAG HAI? External pressure $P_{ext}$ poore process mein constant hai — socho ek gas atmospheric pressure ke against suddenly expand ho rahi hai.

Step 1: Work integral $w = - \int_{V_{1}}^{V_{2}} P_{ext} d V$

Step 2: $P_{ext}$ constant hai $w = - P_{ext} \int_{V_{1}}^{V_{2}} d V$

Step 3: Integrate karo $w = - P_{ext} (V_{2} - V_{1}) = - P_{ext} Δ V$

Yeh reversible se kam work kyun hai? Reversible expansion mein, gas ek pressure ke against dhakka deta hai jo smoothly decrease karta hai jab volume badhti hai (kyunki $P_{ext} = P_{gas} = n R T / V$ gas ko track karta hai). Constant- $P_{ext}$ irreversible expansion mein, gas poore time ek lower constant pressure ke against dhakka deta hai. Kam resistance $\Rightarrow$ kam work extracted.

Visual Representation

DIAGRAM KAISE PADHEIN:

Blue curve (reversible): $V_{1}$ se $V_{2}$ tak $P V$ isotherm ke neeche ka area. Yeh $w_{rev}$ hai — maximum work.
Red rectangle (irreversible): Constant $P_{ext}$ line ke neeche ka area. Yeh $w_{irrev}$ hai — hamesha chhota.
Gap: Irreversibility ki wajah se "lost" work represent karta hai.

Worked Examples

Worked example Example 3: Two-Step Irreversible Expansion

Problem: Same gas 5 L se 10 L, 2 atm ke against expand karta hai, phir 10 L se 15 L, 1 atm ke against. Total work?

Step 1: Har step mein work $w_{1} = - P_{ext, 1} Δ V_{1} = - (2 atm) (5 L) = - 10 L\cdotatm$ $w_{2} = - P_{ext, 2} Δ V_{2} = - (1 atm) (5 L) = - 5 L\cdotatm$

Alag alag kyun? Har step mein alag $P_{ext}$ hai — yeh exactly "path-dependent" $P_{ext} (V)$ case hai; hum piecewise integrate karte hain.

Step 2: Total work $w_{total} = w_{1} + w_{2} = - 10 - 5 = - 15 L\cdotatm$

Step 3: Joules mein convert karo $1 L\cdotatm = 101.325 J$ $w_{total} = - 15 \times 101.325 = - 1520 J = - 1.52 kJ$ INSIGHT: Phir bhi reversible (−5.48 kJ) se kam hai lekin single-step irreversible (−1.01 kJ) se zyada. Zyada steps equilibrium ke kareeb $\Rightarrow$ zyada work — yeh dikhata hai ki irreversible work sach mein $P_{ext}$ path pe depend karta hai.

Common Mistakes

Active Recall Questions

#flashcards/chemistry

Ideal gas ke liye reversible isothermal work ka formula kya hai?

w_{rev} = - n R T ln (V_{2} / V_{1})

Constant external pressure ke against irreversible expansion work ka formula kya hai?

w_{irrev} = - P_{ext} Δ V

Expansion ke liye reversible work magnitude mein hamesha irreversible work se zyada kyun hota hai?

Reversible expansion ek aise pressure ke against kaam karta hai jo gas ko track karta hai (

P_{ext} = P_{gas} = n R T / V

), jabki irreversible expansion ek lower external pressure ke against kaam karta hai, toh kam total work extract hoti hai.

Equation

w = - \int P_{ext} d V

mein hum

P_{gas}

ki jagah

P_{ext}

kyun use karte hain?

Work us pressure pe depend karta hai jiske against gas dhakka deta hai (external), apne internal pressure pe nahi. Sirf reversible processes mein yeh dono har instant pe equal hote hain.

Kya irreversible work sirf endpoints aur

P_{ext}

pe depend karta hai?

Sirf tab jab

P_{ext}

constant ho. Generally, irreversible work poore

P_{ext} (V)

path pe depend karta hai, toh

- \int P_{ext} d V

integrate karna hoga.

5 L se 15 L expansion ke liye, work positive hai ya negative?

Negative (work done BY the gas on surroundings).

Reversible isothermal work formula mein

(V_{2} - V_{1})

ki jagah

ln (V_{2} / V_{1})

kyun hai?

Kyunki pressure volume ke saath

P = n R T / V

ki tarah change karta hai;

1/ V

integrate karne se natural logarithm milta hai.

w = - n R T ln (V_{2} / V_{1})

use karne ki key condition kya hai?

Process (1) reversible, (2) isothermal, aur (3) ideal gas honi chahiye.

−10 L·atm ko joules mein convert karo

−1013.25 J (1 L·atm = 101.325 J)

Agar gas vacuum ke against expand kare (

P_{ext} = 0

), toh kiya gaya work kya hoga?

w = 0

(free expansion; dhakka dene ke liye koi external pressure nahi).

Work path function kyun hai, state function kyun nahi?

Work depend karta hai kaise process hoti hai (reversible vs irreversible,

P_{ext}

path, steps ki sankhya) pe, sirf initial aur final states pe nahi.

Feynman Explanation

Recall Ek 12-saal ke bachche ko explain karo

Socho tumhare paas ek spring-loaded piston hai jo hawa se bhara hai.

Slow aur careful tarika (reversible): Tum bahut slowly piston ko bahar aane dete ho, hawa ke push ko exactly har instant pe match karte ho. Kyunki kuch bhi jerky nahi hai aur kuch bhi waste nahi hota, hawa maximum possible work karta hai — aur tum use equally tiny effort se wapas andar dhakka de sakte ho.

Fast, rough tarika (irreversible): Tum piston ko achanak ordinary room air pressure ke against release kar dete ho. Hawa POP hokar us ek constant, lower pressure ke against bahar aati hai. Kuch work hota hai, lekin kam, kyunki yeh carefully resistance match nahi kar rahi — bas ek low fixed pressure ke against dhakka deti hai, aur extra energy dissipate ho jaati hai.

Reversible formula mein logarithm ( $ln V_{2} / V_{1}$ ) $1/ V$ ki math se aata hai. Volume double karna hamesha utna hi extra work karta hai, chahe tum 1→2 cups se jaao ya 100→200 cups se (same ratio).

Mnemonic

Connections

First Law of Thermodynamics — work ek tarika hai energy transfer ka ( $Δ U = q + w$ )
Isothermal Process — temperature constant kyun rehta hai aur yeh ideal gas ke liye $Δ U = 0$ se kaise relate karta hai
PV Diagram — work ko curve ke neeche area ke roop mein visualize karna
Ideal Gas Law — reversible work derive karne ki foundation
Entropy and Irreversibility — irreversible processes entropy generate karti hain; reversible limit hai
Carnot Cycle — maximum efficiency ke liye reversible isothermal expansions/compressions use karta hai
Free Expansion — limiting case jahan $P_{ext} = 0 \Rightarrow w = 0$
Heat Capacity — alag processes mein energy $q$ aur $w$ ke beech kaise distribute hoti hai