Thermodynamic processes — isothermal (T const), isochoric (V const), isobaric (P const), adiabatic (Q = 0): Question bank

1.7.14 · D5 · Physics › Thermodynamics › Thermodynamic processes — isothermal (T const), isochoric (V

Traps mein jaane se pehle, har ek symbol ko earn karte hain — taaki neeche koi bhi reveal tumhe surprise na kare.

Definition Is page ki alphabet (har symbol, simple shabdon mein)

$P$ = gas ka pressure (yeh walls par kitna force lagata hai, per unit area). Diagram par yeh height axis hai.
$V$ = gas ka volume. Diagram par yeh width axis hai.
$T$ = temperature (kelvin mein), gas kitna hot hai.
$n$ = moles ki sankhya — literally kitna gas hai (ek mole = molecules ka ek fixed bahut bada count, $6.02 \times 1 0^{23}$ ). Zyada gas → sab cheez zyada, isliye $n$ energies ko multiply karta hai.
$R$ = universal gas constant $\approx 8.31 J/(mol\cdotK)$ — ek fixed conversion number jo "moles aur degrees" ko "joules of energy" mein convert karta hai. Yeh har ideal gas ke liye same hai; isliye ise universal kehte hain.
$U$ = internal energy — molecules ki jiggiling mein chhupi hui total kinetic energy.
$Q$ = gas mein add ki gayi heat; $W$ = gas dwara kiya gaya work.

Figure — Thermodynamic processes — isothermal (T const), isochoric (V const), isobaric (P const), adiabatic (Q = 0)

Chaar processes bas states ke beech chaar paths hain. Yeh picture is poore page ka map hai:

Recall Do woh facts jo almost har trap resolve kar dete hain

$Δ U = n C_{V} Δ T$ har process mein (ideal gas ki internal energy sirf temperature par depend karti hai — First Law of Thermodynamics dekho).
First Law $Δ U = Q - W$ hamesha sach hai; chaar named processes mein se har ek bas ek term ya ek variable ko khatam karta hai.

True or false — justify karo

Har statement ke baad honest reasoning hai, sirf verdict nahi.

Isothermal ka matlab hai koi heat flow nahi hoti.

Jhooth. Isothermal temperature fix karta hai, isliye

Δ U = 0

aur First Law force karta hai

Q = W \neq = 0

—

T

constant rakhne ke liye heat freely flow karti hai. Zero-heat process adiabatic hoti hai.

Adiabatic process mein temperature constant rehti hai.

Jhooth. Adiabatic

Q = 0

fix karta hai,

T

nahi. Koi heat andar nahi aane par, gas dwara kiya gaya koi bhi work internal energy se pay hota hai, isliye expansion par

T

girta hai aur compression par badhta hai.

Kyunki

C_{V}

"constant-volume" heat capacity hai,

Δ U = n C_{V} Δ T

sirf constant volume par hi hold karta hai.

Jhooth.

U

sirf

T

par depend karta hai, isliye yeh relation har path par hold karta hai.

C_{V}

sirf

Δ T

aur

Δ U

ke beech conversion factor hai; iske naam mein "V" uss tarike se aata hai jismein ise measure kiya jaata hai, na ki jahan yeh apply hota hai.

Isochoric process mein gas koi work nahi karta, isliye usme koi energy enter nahi hoti.

Jhooth. Work zero hai, lekin heat nahi hai:

Q = Δ U = n C_{V} Δ T

. Heat ka har joule internal energy mein jaata hai kyunki expansion par kuch kharch nahi hota.

Same point se gujarne wale isotherm aur adiabat ka shape same hota hai.

Jhooth. Isotherm hai

P V = const

; adiabat hai

P V^{γ} = const

jismein

γ > 1

hai. Adiabat steeper hai, same volume increase ke liye lower pressure tak girta hai.

Isobaric expansion ke liye,

Q = Δ U

kyunki pressure constant hai.

Jhooth. Constant

P

par gas work bhi karta hai

W = P Δ V

, isliye

Q = Δ U + W = n C_{P} Δ T

, jo strictly

Δ U

se bada hai.

Mayer's relation

C_{P} = C_{V} + R

sirf isobaric process par apply hoti hai jahan se yeh derive ki gayi thi.

Jhooth. Yeh gas ke do material properties ko relate karta hai; ek baar prove hone ke baad yeh hamesha hold karta hai. Humne ise isobaric process use karke derive kiya, lekin conclusion path-independent hai.

Agar gas apne starting

P

V

, aur

T

par wapas aa jaaye, to round trip ke liye

Q = 0

hai.

Jhooth. Closed loop mein

Δ U = 0

hota hai, isliye

Q = W

— aur

W

enclosed loop area ke barabar hota hai, jo generally nonzero hai. Sirf

Δ U

guaranteed zero hai.

P

–

V

diagram par ek vertical line isobaric process represent karti hai.

Jhooth. Vertical matlab

V

fixed → isochoric. Horizontal (constant height

P

) isobaric hai.

Gas dwara kiya gaya work hamesha positive hota hai jab gas heat paata hai.

Jhooth. Isochoric heating mein

Q > 0

lekin

W = 0

; adiabatic compression mein

W < 0

aur

Q = 0

ke saath.

W

ka sign volume change track karta hai, heat nahi.

Error dhundo

Galat line padho, phir batao kya toot raha hai.

"Adiabatic:

W = n C_{V} (T_{1} - T_{2})

; gas compress hua to

T_{2} > T_{1}

, aur mujhe

W > 0

mila."

Sign galat hai.

T_{2} > T_{1}

ke saath, factor

(T_{1} - T_{2}) < 0

hai, isliye

W < 0

— gas par work kiya gaya hai, yahi reason hai ki compression ise garam karta hai.

"Isothermal work:

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

Ratio ulta hai. Sahi hai

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

; expansion (

V_{2} > V_{1}

) se

W > 0

milna chahiye, lekin ulta version negative number dega.

"Kyunki isothermal process mein

Δ U = 0

hai, aur

Δ U = n C_{V} Δ T

, gas ki koi heat capacity nahi hai yahan."

Yeh reasoning

Δ U = 0

ko "

C_{V}

vanish ho jaata hai" ke saath confuse kar rahi hai.

Δ U = 0

sirf isliye hai kyunki

Δ T = 0

hai;

C_{V}

gas ka ek fixed property hai aur unchanged hai.

"Adiabatic ke liye main end state dhundne ke liye

P V = const

use karunga."

Galat curve. Adiabatic

P V^{γ} = const

ko follow karta hai;

P V = const

isotherm hai. Ise use karne par underestimate hoga ki pressure kitna neeche jaata hai.

"Isobaric expansion mein

W = V Δ P

(pressure times volume change)."

Variables swap ho gaye hain. Pressure constant wala hai;

W = P Δ V

(constant pressure times volume change), jo height

P

ke rectangle ka area hai.

Q = n C_{P} Δ T

kisi bhi heating ke liye kaam karta hai, kyunki

C_{P}

sirf heat capacity hai."

Q = n C_{P} Δ T

isobaric result hai. Heat path-dependent hai: isochoric ke liye yeh

n C_{V} Δ T

hai, isothermal ke liye yeh

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

hai. Tum sabhi paths ke liye ek

C

nahi chun sakte.

"Adiabat aur isotherm dono

Δ U = 0

satisfy karte hain."

Sirf isotherm karta hai (

Δ T = 0

). Adiabat mein

Q = 0

hota hai, aur

Δ U = - W \neq = 0

generally kyunki temperature change hota hai.

Why questions

"Kyun" wahi jagah hai jahan yaad kiye gaye formulas ya to survive karte hain ya collapse ho jaate hain.

C_{P}

C_{V}

se bada kyun hota hai?

Constant pressure par gas garam hone ke saath expand hota hai, isliye aane wali heat ko dono temperature rise aur expansion work ke liye pay karna hota hai; Mayer's relation is gap ko exactly

R

ke roop mein quantify karta hai.

Adiabatically expand karne wala gas kyun thanda hota hai jabki koi heat "leta" nahi?

Q = 0

ke saath, baahir push karne ka work gas ki apni internal energy se hi finance ho sakta hai;

U

kam hone se

T

kam hota hai. Isliye utha hua air clouds banata hai aur spray can thandi ho jaati hai.

Work

P

–

V

curve ke neeche ke area ke barabar kyun hota hai?

Pressure

P

ke against har chota expansion

d V

contribute karta hai

d W = P d V

; in pateele strips ko sum (integrate) karna bilkul curve ke neeche ke area ke barabar hai (upar map figure mein shaded strip dekho).

P

–

V

path ki shape work ke liye kyun matter karti hai lekin

Δ U

ke liye nahi?

W

path ke neeche ka area hai, isliye same endpoints ke beech different paths different areas sweep karte hain.

Δ U

sirf endpoints ki temperatures par depend karta hai, isliye yeh path-independent hai.

Isothermal change slowly (quasi-statically) kyun hona chahiye?

"Quasi-statically" matlab itna gently ki gas har instant par balance mein ho. Sirf ek slow process hi allow karta hai ki heat itni slowly trickle kare ki temperature poore time reservoir se pinned rahe; jaldi karo, to gas ke kuch hisse equilibrium restore hone se pehle heat ya cool ho jaayenge aur single

T

nahi rahega.

Adiabatic change ya to well-insulated ya bahut fast kyun hona chahiye?

"Well-insulated" matlab walls se physically heat cross nahi kar sakti.

Q = 0

ke liye koi heat leak nahi chahiye: insulation path block karta hai, ya extreme speed process ko itni jaldi khatam kar deti hai ki koi bhi meaningful heat boundary cross nahi kar paati.

Adiabat same point par isotherm se steeper kyun hai?

Adiabat par gas expand hone ke saath temperature bhi ghatta hai, isliye pressure do compounding reasons ke liye girta hai (zyada volume aur kam

T

); exponent

γ > 1

is extra steepness ko encode karta hai.

Δ U = n C_{V} Δ T

hume path jaane bina internal energy change kyun dhundhne deta hai?

Kyunki

U

ideal gas ke liye sirf

T

ka ek state function hai — koi bhi kalpanic path chuno same

Δ T

ke saath aur tumhe same

Δ U

milega.

Edge cases

Boundary aur degenerate scenarios jo topic quietly assume karta hai.

Isothermal process mein work kya hoga jahan volume bilkul nahi badlta?

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

, aur phir

Q = W = 0

bhi — "process" sirf gas ka wahan baitha rehna hai.

Constant temperature par isobaric process ke liye kya hoga?

Agar dono

P

aur

T

fixed hain, to

P V = n R T

force karta hai

V

bhi fixed — kuch nahi badlta; yeh ek single equilibrium point hai, process nahi.

Kya koi process simultaneously isothermal aur adiabatic ho sakti hai (real change of state ke liye)?

Sirf trivially. Isothermal ko

Q = W

chahiye; adiabatic ko

Q = 0

chahiye; saath mein yeh force karte hain

W = 0

aur koi volume change nahi, yaani koi actual process nahi.

Limit

γ \to 1

mein adiabat kya ban jaata hai?

P V^{γ} \to P V = const

, isliye adiabat isotherm par collapse ho jaata hai. Physically

γ \to 1

matlab expansion se

T

bahut kam badlta hai, do curves blur ho jaati hain.

Agar gas ko adiabatically compress kiya jaaye zero net temperature change ke saath, kya yeh possible hai?

Nahi. Adiabatic compression gas par positive work karta hai

Q = 0

ke saath, isliye

Δ U > 0

aur

T

zaroor badhega; zero

Δ T

ke liye

Δ U = 0

chahiye hoga, jo

W \neq = 0

se contradict karta hai.

Ek complete cyclic process (

P

–

V

diagram par closed loop) ke liye

Δ U

kya hai?

Exactly zero, kyunki gas apne initial temperature par wapas aa jaata hai. Net heat phir net work ke barabar hoti hai, jo enclosed area hai — engines jaise Carnot Cycle ka basis.

Isochoric process ke liye, kya added heat ka koi hissa work ke roop mein "waste" hota hai?

Kuch nahi —

d V = 0

ke saath gas zero work karta hai, isliye heat ka

100%

internal energy ban jaata hai. Yeh temperature badhane ke liye har joule par sabse "efficient" hai.