Work done in each process — derivation: Question bank

1.7.15 · D5 · Physics › Thermodynamics › Work done in each process — derivation

Prerequisites parent note mein aur First Law of Thermodynamics, P-V Diagrams, Adiabatic Process and PV^gamma, Isothermal Process and Boyle's Law, Internal Energy and Cv, aur Carnot Cycle mein milenge.

True or false — justify

Isothermal work zero hota hai kyunki temperature kabhi nahi badlta

False.

Δ T = 0

Δ U = 0

hota hai,

W = 0

nahi; piston phir bhi bahar nikalti hai aur

W = n R T ln (V_{2} / V_{1}) \neq = 0

. Constant temperature ka matlab constant volume nahi hota.

Gas ka kiya hua work sirf start aur end states par depend karta hai

False. Work path-dependent hota hai: ek hi do points ko join karne wale do alag curves

P

–

V

diagram par alag areas enclose karte hain, isliye alag

W

dete hain. Sirf state functions (jaise

U

) endpoints par depend karte hain.

Isochoric process mein gas phir bhi work kar sakti hai agar pressure change ho

False. Work ke liye moving piston chahiye, yaani

d V \neq = 0

. Agar

V

fixed hai, toh

d V = 0

har jagah hai aur

W = \int P d V = 0

hoga chahe

P

kitna bhi kyon na badle.

Same volume change ke liye, isobaric work hamesha isothermal expansion work se bada hota hai

True (same start point ke liye). Isothermal expansion mein pressure girta hai jab volume badhta hai, isliye isotherm ke neeche ka area constant (starting) pressure line ke neeche bane rectangle se chota hota hai.

Adiabatic expansion mein gas positive work karta hai

True. Yeh expand hoti hai (

V_{2} > V_{1}

), aur koi heat andar nahi aati (

Q = 0

) toh yeh work internal energy se pay hoti hai, isliye gas thandi hoti hai (

T_{2} < T_{1}

) aur

W = \frac{n R ( T _{1} - T _{2} )}{γ - 1} > 0

Jo gas compress hoti hai woh negative work karti hai

True. "Gas dwara kiya gaya work" hai

\int P d V

; compression mein

V_{2} < V_{1}

, isliye integral negative hai — surroundings gas par work karte hain.

Kisi bhi cyclic loop se enclosed area net absorbed heat ke barabar hota hai

True (poore cycle ke liye). Cycle mein

Δ U = 0

hota hai, isliye first law deta hai

Q_{n e t} = W_{n e t} =

enclosed area — lekin yeh sirf poore loop ke liye hai, kisi ek leg ke liye nahi.

Anticlockwise cycle positive net work produce karta hai

False. Clockwise loops mein

W_{n e t} > 0

(engine); anticlockwise loops mein

W_{n e t} < 0

(refrigerator/heat pump), yaani gas par work hota hai.

Spot the error

"Isothermal work

P Δ V

hai kyunki yeh isobaric case mein kaam aaya."

Isobaric shortcut isliye kaam karta hai kyunki

P

constant hai aur integral se bahar aa jaata hai. Isothermal mein

P = n R T / V

vary karta hai, isliye tumhe

\int n R T d V / V

integrate karna padega aur result ek logarithm aayega, koi product nahi.

"Adiabatic work

\frac{P _{2} V _{2} - P _{1} V _{1}}{γ - 1}

hai."

Sign ulta hai. Yeh

\frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1}

hona chahiye taaki expansion (

P_{1} V_{1} > P_{2} V_{2}

thande final state mein)

W > 0

de. Starting-state term pehle aati hai.

"Isobaric process ke liye

W = n R Δ T

sirf tab hai jab gas monatomic ho."

γ

isobaric work mein kabhi nahi aata.

W = P Δ V = n R Δ T

seedha

P V = n R T

se aata hai aur kisi bhi ideal gas ke liye valid hai chahe atomicity kuch bhi ho.

"Kyunki adiabatic process mein

Q = 0

hai, koi energy gas se nahi nikalti, isliye temperature unchanged rehta hai."

Energy nikalti zaroor hai — heat ke roop mein nahi, work ke roop mein.

Q = 0

ke saath first law deta hai

Δ U = - W

, isliye positive work karne se

U

drain hoti hai aur gas thandi ho jaati hai.

W = \int P d V

ke liye gas ka ideal hona zaroori hai."

Master formula

d W = P d V

sirf force

= P A

aur geometry se aata hai — yeh kisi bhi substance ke liye valid hai. Ideality baad mein use hoti hai jab

P

V

ke specific function ke roop mein likha jaata hai (jaise

n R T / V

P

–

V

diagram par, work

P

-axis (vertical axis) aur curve ke beech ka area hota hai."

Nahi — work curve ke neeche ==

V

-axis (horizontal)== tak ka area hota hai, kyunki hum

d V

par integrate karte hain.

P

-axis tak ka area

\int V d P

hoga, jo ek alag quantity hai.

"Cyclic process mein total work zero hota hai kyunki yeh starting point par wapas aata hai."

Sirf

Δ U

aur

Δ T

zero par wapas aate hain. Work ek process quantity hai: enclosed area generally nonzero hota hai, aur yehi exactly hai jo engines net output produce karne mein use karte hain.

Why questions

Isothermal work mein logarithm kyun aata hai jabki isobaric mein nahi

Kyunki isothermal mein

P = n R T / V

V

par depend karta hai, aur

\int d V / V = ln V

; isobaric mein

P

constant hai aur bahar aa jaata hai, sirf plain

\int d V = Δ V

bachta hai.

Hum isobaric process ke liye

P

ko integral se bahar kyon nikal sakte hain lekin isothermal ke liye nahi

Ek constant integral se bahar aa sakta hai; ek varying quantity nahi aa sakti. Isobaric

P

definition se fixed hai, lekin isothermal

P

gas ke phailne ke saath girta hai, isliye use andar hi rehna padta hai.

Adiabatically expand hoti gas thandi kyun hoti hai jabki isothermally expand hoti gas nahi hoti

Isothermal gas ek reservoir se heat absorb karta hai taaki usne jo energy work ke roop mein kharch ki use replace kare aur

T

fixed rahe. Adiabatic gas sealed off hota hai (

Q = 0

), isliye woh apni khud ki internal energy kharchta hai aur

T

girta hai.

Work ko path-dependent kyun kaha jaata hai jab energy conservation phir bhi laagu hoti hai

Conservation

U

(ek state function) par apply hoti hai:

Δ U = Q - W

. Lekin

Q

aur

W

individually us route par depend karte hain jo liya gaya; sirf unka combination

Q - W

endpoints se fix hota hai.

Expansion ke dauran adiabatic curve same point se guzarne wale isotherm se neeche steep kyun hoti hai

Kyunki

P V^{γ} =

const,

P V =

const se zyada tezi se girta hai (

γ > 1

). Isliye same expansion ke liye adiabat zyada neeche girta hai, kam area enclose karta hai aur isotherm se kam work deta hai.

Isochoric process apni saari heat internal energy mein kyun daal deta hai

d V = 0

ke saath,

W = 0

, isliye first law

Δ U = Q - W

ban jaata hai

Δ U = Q

; heat ka har joule internal energy (aur temperature) ko directly badhata hai.

Cyclic work ka sign clockwise vs anticlockwise traversal direction par kyun depend karta hai

Enclosed area expansion ke dauran sweep hota hai (loop ka bottom, right ki taraf move karna,

+ W

) versus compression (top, left ki taraf,

- W

). Clockwise mein expansion leg dominate karta hai, isliye net work positive hota hai.

Edge cases

Work kya hoga agar initial aur final volumes equal hain lekin path isochoric nahi thi (ek loop jo same

V

par wapas aata hai)

Zaroori nahi ki zero ho.

\int P d V

ek aisi curve par jo same

V

se jaati aur wapas aati hai area enclose kar sakti hai; sirf truly fixed-

V

(vertical-line) path

W = 0

deta hai.

Isothermal work ka kya hoga jab

V_{2} \to V_{1}

(vanishing expansion)

ln (V_{2} / V_{1}) \to ln 1 = 0

, isliye

W \to 0

smoothly — koi volume change nahi, koi work nahi, master formula ke consistent.

Limit

T_{2} \to T_{1}

(negligible cooling) mein adiabatic work kya hoga

W = \frac{n R ( T _{1} - T _{2} )}{γ - 1} \to 0

. Agar temperature barely change kare, toh almost koi internal energy spend nahi hoti, isliye almost koi work nahi hota.

Kya expansion ke dauran bhi work negative ho sakta hai

Bahar ki taraf push karne wali single gas ke liye, nahi —

d V > 0

aur

P > 0

force karte hain ki

\int P d V > 0

. Negative work ke liye path mein kahin compression (

d V < 0

) chahiye.

P

–

V

diagram par ek vertical line ke saath work kya hai (fixed volume par pressure change)

Zero. Vertical segment mein poori jagah

d V = 0

hai, isliye

\int P d V = 0

— yeh exactly ek isochoric leg hai, jaise kai cycles ki ek side.

Kya vacuum mein gas ka free expansion

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

ka ek example hai

Nahi. Free expansion mein koi opposing pressure nahi hota (

P_{e x t} = 0

), isliye moving boundary ke through koi force act nahi karta aur

W = 0

hota hai, chahe gas ka volume badhta ho.

P

–

V

diagram par ek horizontal line ke liye

W

kya hai

Horizontal segment constant pressure hota hai, isliye

W = P (V_{2} - V_{1})

— us line ke neeche plain isobaric rectangle area.

Recall Quick self-test

Path-dependent ka matlab hai work depend karta hai... ::: li gayi curve par, sirf endpoints par nahi. $Δ U = 0$ ka matlab hai... ::: $Q = W$ , $W = 0$ nahi. Vacuum mein free expansion mein work hota hai... ::: zero (koi opposing pressure nahi). Anticlockwise cyclic loop net work deta hai jo hota hai... ::: negative (gas par kiya jaata hai).