Internal energy of ideal gas U = (f - 2)nRT — Hinglish notes

1.7.10 · Physics › Thermodynamics

WHAT is internal energy?

Key word hai microscopic. Isme pure container ki kinetic energy shamil nahi hai jo move kar raha ho, ya uski gravitational PE. Sirf random internal jiggaling count hoti hai.

WHY sirf temperature? (first-principles)

Ideal gas ke liye hum assume karte hain:

Molecules point particles hain → koi internal PE nahi.
Molecules ke beech koi forces nahi (sirf instantaneous collisions chhod ke) → koi inter-molecular PE nahi.

Toh saari energy kinetic hai. Aur random motion ki kinetic energy exactly wahi hai jo temperature measure karta hai. Isliye $U = U (T)$ — sirf temperature ka function.

HOW to derive $U = \frac{f}{2} n R T$

Step 1 — Translational energy ke liye kinetic theory

Kinetic theory se, gas ka pressure deta hai: $P V = \frac{1}{3} N m ⟨ v^{2} ⟩$

Yeh step kyun? Yeh kinetic-theory ka result hai jo macroscopic $P, V$ ko microscopic speeds se connect karta hai. $N$ = molecules ki sankhya, $m$ = ek ki mass, $⟨ v^{2} ⟩$ = mean-square speed.

Ideal gas law $P V = n R T = N k_{B} T$ se compare karo (using $n R = N k_{B}$ , $k_{B}$ = Boltzmann constant): $\frac{1}{3} N m ⟨ v^{2} ⟩ = N k_{B} T$

Yeh step kyun? Humare paas same $P V$ ke liye do expressions hain. Inhe equal set karne se speed aur temperature ka link milta hai.

Total translational KE solve karo: $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$

Yeh step kyun? Dono sides ko $\frac{3}{2}$ se multiply karo aur $N$ se divide karo taaki average KE per molecule mile.

Toh average translational KE per molecule $= \frac{3}{2} k_{B} T$ .

Step 2 — Equipartition theorem

Translation mein 3 degrees of freedom hote hain ( $v_{x}, v_{y}, v_{z}$ ), jo deta hai $3 \times \frac{1}{2} k_{B} T = \frac{3}{2} k_{B} T$ — Step 1 se match karta hai.

Step 3 — $f$ degrees of freedom tak generalize karo

Agar ek molecule mein $f$ total active degrees of freedom hain (translation + rotation + vibration), toh: $average energy per molecule = \frac{f}{2} k_{B} T$

Yeh step kyun? Equipartition simply har degree of freedom ke liye $\frac{1}{2} k_{B} T$ add karta hai.

Total internal energy = (energy per molecule) × (molecules ki sankhya): $U = N \cdot \frac{f}{2} k_{B} T = \frac{f}{2} (N k_{B}) T$

$N k_{B} = n R$ use karte hue: $U = \frac{f}{2} n R T$

Yeh step kyun? $n$ = moles, $R = N_{A} k_{B}$ gas constant hai. Yeh molecule count ko moles mein convert karta hai.

Degrees of freedom $f$ by gas type

Gas type	Translation	Rotation	$f$ (moderate $T$ par)	$U$
Monatomic (He, Ar)	3	0	$3$	$\frac{3}{2} n R T$
Diatomic (O₂, N₂)	3	2	$5$	$\frac{5}{2} n R T$
Linear triatomic (CO₂)	3	2	$5$ (rot.)	—
Nonlinear polyatomic (H₂O)	3	3	$6$	$3 n R T$

Figure — Internal energy of ideal gas U = (f - 2)nRT

Active recall

Recall Quick self-test (answers chhupa lo!)

Ideal gas ke liye $U$ sirf $T$ par kyun depend karta hai? → Koi intermolecular PE nahi; energy purely kinetic hai, aur $T$ kinetic energy measure karta hai.
Har quadratic DOF mein energy kitni hoti hai? → $\frac{1}{2} k_{B} T$ per molecule.
Monatomic / diatomic ke liye $f$ ? → 3 / 5.
Isothermal process mein $Δ U$ ? → 0.
$N k_{B}$ ko molar form mein convert karo? → $n R$ .

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek dabbe mein chhoti-chhoti bouncy balls bhari hain jo tez-tez ud rahi hain. Dabba jitna garam hoga, woh utni tez udenge. "Internal energy" bas yeh hai ki total mein kitni udaan ho rahi hai. Zyada balls hon (zyada gas) ya unhe tez karo (zyada heat), toh zyada udaan energy hogi. Har direction jismein ek ball move ya spin kar sakti hai woh ek alag tarika hai udaan-energy store karne ka, aur nature us energy ko un sabhi tarike se equally share karta hai. Yahi sharing rule hai jis wajah se hum " $f$ " count karte hain — hilne ke independent tarike — aur usse multiply karte hain.

Flashcards

Ideal gas ki internal energy kis cheez ka function hai?

Temperature only (koi V ya P dependence nahi), kyunki koi intermolecular forces nahi hain.

Equipartition theorem state karo.

Har quadratic degree of freedom average energy

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

per molecule carry karta hai.

f

degrees of freedom wale ideal gas ki internal energy ka formula?

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

Average translational KE per molecule?

\frac{3}{2} k_{B} T

Monatomic gas ke degrees of freedom?

3 (sab translational), isliye

U = \frac{3}{2} n R T

Moderate T par diatomic gas ke degrees of freedom?

5 (3 translational + 2 rotational), isliye

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

Diatomic molecule mein sirf 2 rotational DOF kyun hote hain?

Bond axis ke around rotation ka moment of inertia negligible hota hai, isliye woh mode frozen out ho jaata hai.

Isothermal process mein ideal gas ka

Δ U

Zero, kyunki

Δ T = 0

N k_{B}

n R

se kya relation hai?

N k_{B} = n R

kyunki

R = N_{A} k_{B}

aur

N = n N_{A}

Kis process mein

Q = Δ U

hota hai?

Constant-volume (isochoric) process mein, jahan work

W = 0

Connections

Kinetic theory of gases — $P V = \frac{1}{3} N m ⟨ v^{2} ⟩$ ka source
Equipartition theorem — $\frac{1}{2} k_{B} T$ rule
First law of thermodynamics — $Δ U = Q - W$
Degrees of freedom — $f$ kaise count hota hai
Molar specific heats Cv and Cp — $C_{V} = \frac{f}{2} R$ directly aata hai
Ideal gas law — $P V = n R T$

Internal energy of ideal gas U = (f - 2)nRT