Heat and internal energy — microscopic vs macroscopic: Question bank

1.7.3 · D5 · Physics › Thermodynamics › Heat and internal energy — microscopic vs macroscopic

Definition Is page par use hone wale symbols — shuru karne se pehle ek baar padh lo

Neeche har letter yahan introduce kiya gaya hai taaki baad mein kuch bhi surprise na kare:

$U$ (internal energy) — gas ke andar stored total energy; ek state function.
$Q$ (heat) aur $W$ (work) — boundary ke paar energy transfers; process quantities.
$T$ = temperature, $P$ = pressure, $V$ = volume — macroscopic bulk measurements.
$N$ = molecules ki number; $m$ = ek molecule ki mass; $⟨ v^{2} ⟩$ = sabhi molecules par squared speed ka average.
$n$ = moles ki number (ek mole sirf molecules ki ek fixed count hai, $N_{A} = 6.02 \times 1 0^{23}$ molecules).
$f$ = degrees of freedom: ek molecule ke kinetic energy store karne ke independent tarike ki sankhya (teen $x, y, z$ along slide karne ke liye; aur agar woh rotate bhi kar sake to zyada). Ise hum neeche explain karte hain.

True or false — justify

Ye saare woh sentences hain jo tum class mein sun sakte ho. True/false decide karo, phir ek line mein kyun batao.

Ek hot cup of coffee "mein bahut saari heat hai."

False. Usme bahut badi internal energy

U

hai; heat tabhi exist karti hai jab

Δ T

ki wajah se energy boundary cross kare. "Kisi body mein heat" aise hai jaise "nadi mein baarish."

Agar do bodies ka temperature same hai, unki internal energy bhi same hai.

False. Same

T

ka matlab hai same average molecular KE, lekin

U

kitne molecules hain uske saath bhi scale karta hai (

U = \frac{f}{2} n R T

, jahan

n

mole count hai) —

4 0^{\circ}

C par ek bathtub aur ek teaspoon

U

mein bahut alag hote hain.

Kisi system mein heat add karna hamesha uska temperature badhata hai.

False. Isothermal expansion mein heat andar aati hai phir bhi

T

unchanged rehta hai kyunki gas equal work karta hai; phase change ke dौraan heat bonds todne mein jaati hai,

T

badhane mein nahi.

Ideal gas ki internal energy change ho sakti hai even if temperature fixed rahe.

False. Ideal gas ke liye

U

sirf

T

par depend karta hai, isliye

Δ T = 0 \Rightarrow Δ U = 0

, chahe

P

V

kuch bhi kare.

Δ U

do states ke beech liye gayi path par depend karta hai.

False.

U

ek state function hai, isliye

Δ U

sirf start aur end states par depend karta hai.

Q

aur

W

individually path-dependent hote hain.

Gas par work karna aur use heat karna alag physical processes hain, lekin dono

U

ko same amount se badha sakte hain.

True.

Q

(

Δ T

se disordered energy) aur

W

(distance par ordered force) alag transfers hain, phir bhi

Δ U = Q - W

sirf stored energy par unka net effect count karta hai.

Real gas mein intermolecular forces ke saath bhi,

U

sirf temperature par depend karta hai.

False. Real molecules attract/repel karte hain, isliye

U

mein ek potential-energy term hota hai jo separation ke saath, hence

V

ke saath change karta hai. Sirf ideal gas is term ko drop karta hai.

Heat hamesha zyada internal energy wali body se kam internal energy wali body mein flow karti hai.

False. Heat higher se lower temperature ki taraf flow karti hai, higher

U

se nahi. Ek bada lukewarm object ek chote hot object se zyada

U

rakh sakta hai, phir bhi hot object se heat receive kar sakta hai.

Spot the error

Har line mein ek specific mistake hai. Use naam do aur correct karo.

Δ U = Q + W

jahan

W

gas dwara ki gayi work hai."

Humare convention ke khilaf sign error hai.

W

= work done by the gas ke saath, first law hai

Δ U = Q - W

; plus sign tabhi apply hota hai jab symbol work done on the gas mean kare.

"Vacuum mein free expansion mein, gas work karta hai, isliye uska temperature girta hai."

Gas kisi cheez ke against push nahi karta, isliye

W = 0

; saath mein

Q = 0

bhi, hence

Δ U = 0

aur (ideal gas)

T

unchanged rehta hai. Koi piston nahi, koi work nahi.

"Kyunki

C_{V} = \frac{3}{2} R

, kisi bhi process mein add ki gayi saari heat internal energy mein jaati hai."

Ye sirf constant volume par sach hai (jahan

W = 0

hota hai). Generally kuch heat work ke roop mein nikl jaati hai, isliye

Q \neq = Δ U

. Dekho Specific Heats Cv and Cp.

"Temperature molecules ki total kinetic energy hai."

Ye average translational KE per molecule hai:

\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T

(yahan

m

ek molecule ki mass hai). Total KE

U

hai, jo molecule count par bhi depend karta hai.

"Ek diatomic aur ek monatomic gas same

T

par per mole same internal energy rakhte hain."

False — alag degrees of freedom

f

hote hain. Diatomic mein

f = 5

hai (rotation add hoti hai), isliye

U = \frac{5}{2} n R T

versus

\frac{3}{2} n R T

"Kyunki

P V = \frac{1}{3} N m ⟨ v^{2} ⟩

, volume double karne par

⟨ v^{2} ⟩

double ho jaata hai."

Nahi —

⟨ v^{2} ⟩

sirf

T

se fix hota hai. Fixed

T

par,

V

double karne par

P

half ho jaata hai, isliye product

P V

(aur hence

⟨ v^{2} ⟩

) constant rehta hai.

"Heat ek tarah ka matter hai jo bodies ke beech flow karta hai."

Heat energy in transit hai, koi substance nahi. Ye discredited caloric picture hai jiske against topic warn karta hai.

Why questions

Apne aap ko mechanism dene ke liye force karo, sirf label nahi.

Ideal-gas internal energy sirf

T

par kyun depend karta hai,

V

par kyun nahi?

Ideal molecules mein koi intermolecular forces nahi hote, isliye koi potential-energy term nahi hota jo unke spread out hone par change kare — sirf KE bachti hai, aur KE

T

se fix hoti hai.

First law mein

W

ke aage minus sign kyun hota hai?

W

= work done by the gas ke saath: agar gas ek piston ko bahar push karta hai, woh apni stored energy kharch karta hai, isliye

U

girta hai. Work ke roop mein nikl rahi energy subtract karni padti hai.

Hum

⟨ v^{2} ⟩

jaise averages kyun use kar sakte hain har molecule ko track karne ki jagah?

\sim 1 0^{23}

molecules hote hain; individual paths hoplessly chaotic hain, lekin unke statistical averages stable aur reproducible hote hain — macroscopic variables exactly yahi measure karte hain. Dekho Kinetic Theory of Gases.

Har degree of freedom ko

\frac{1}{2} k_{B} T

energy kyun milti hai?

Equipartition theorem: thermal equilibrium mein energy independent quadratic modes of motion mein equally share hoti hai. Teen translational directions se

3 \times \frac{1}{2} k_{B} T = \frac{3}{2} k_{B} T

per molecule milta hai. Dekho Equipartition Theorem.

Heat ek state function kyun nahi hai jaise internal energy hai?

Heat sirf

Δ T

se driven transfer ke dौraan exist karti hai; andar aa jaane ke baad, woh energy sirf

U

hai. Tum kisi state ko point karke nahi keh sakte "itni heat isme hai," isliye iska koi state value nahi hota.

Same temperature wali do bodies ke beech net heat exchange kyun nahi hoti?

Koi temperature difference nahi hone par koi thermal driving nahi hota; molecular collisions dono taraf barabar energy transfer karti hain. Yahi Temperature and the Zeroth Law ka basis hai.

Isothermal expansion mein heat input kyun chahiye even though

T

nahi badhlta?

Gas piston par work karta hai, jo uski KE drain karke use cool kar deta; heat exactly us rate par andar aati hai jo

⟨ v^{2} ⟩

(hence

T

) ko constant rakhne ke liye zaroori hai, isliye

Q = W

Edge cases

Woh scenarios jo naive rules tod dete hain — inhe sab cover karo.

Constant-volume heating:

W

Q

Δ U

kya hain?

W = 0

(koi volume change nahi, koi piston motion nahi), isliye

Q = Δ U = \frac{f}{2} n R Δ T

. Saari heat internal energy ban jaati hai.

Isothermal ideal-gas process:

Δ U

aur

Q

kya hain?

Δ U = 0

(kyunki

T

fixed hai), isliye first law deta hai

Q = W

— andar aane wala har joule work ke roop mein nikal jaata hai.

Adiabatic process (

Q = 0

): jab gas expand karta hai toh

U

kaise change hota hai?

Δ U = - W

: koi heat exchange nahi hone par gas sirf apni energy se work kar sakta hai, isliye

U

(aur

T

) girta hai. Compression ulta karta hai. Dekho Isothermal and Adiabatic Processes.

Vacuum mein ideal gas ki free expansion:

Q

W

Δ U

Δ T

Chaaro zero hain: koi heat cross nahi karta, vacuum ke against koi work nahi hoti, isliye

Δ U = 0

aur

T

unchanged rehta hai volume double hone ke bawajood.

Zero-Kelvin limit: ideal gas ke liye

⟨ v^{2} ⟩

aur

U

ka kya hota hai?

\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T

se, jaise

T \to 0

average KE

\to 0

, isliye ideal-gas

U \to 0

. (Real matter is se pehle quantum limits tak pahunch jaata hai.)

Ek single molecule ek box mein bounce kar raha hai — kya "temperature" ka koi matlab hai?

Itna nahi: temperature bahut saare molecules par ek statistical average hai. Ek molecule ke paas ek KE hoti hai lekin koi well-defined thermodynamic

T

nahi hota; concept tabhi sharpen hota hai jab

N

bada hota hai.

Do identical gas samples, ek constant

P

par heat kiya gaya, ek constant

V

par, same

Δ T

tak: kaun zyada heat absorb karta hai?

Constant-

P

wala, kyunki woh expansion work bhi karta hai, isliye

Q_{P} = Δ U + W > Q_{V} = Δ U

. Exactly isliye

C_{P} > C_{V}

hota hai. Dekho Specific Heats Cv and Cp.

Recall Lock karne ke liye ek-line summary

Store vs transfer, path-independence, aur ideal gas ke liye sirf $U (T)$ — yahan almost har trap in teeno mein se kisi ek ki confusion hai.

Woh trap jo zyatatar logon ko pakad leta hai

"Heat added" aur "internal energy raised" ko confuse karna — ye equal hote hain sirf tab jab koi work nahi hoti (

V

constant ho).

Connections

Heat and internal energy — microscopic vs macroscopic
First Law of Thermodynamics
Kinetic Theory of Gases
Equipartition Theorem
Specific Heats Cv and Cp
Isothermal and Adiabatic Processes
Temperature and the Zeroth Law
Degrees of Freedom and Molecular Structure