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

WHAT is internal energy?

The key word is microscopic. It does not include the kinetic energy of the whole container moving, or its gravitational PE. Only the random internal jiggling counts.

WHY only temperature? (first-principles)

For an ideal gas we assume:

Molecules are point particles → no internal PE.
No forces between molecules (except instantaneous collisions) → no inter-molecular PE.

So all the energy is kinetic. And kinetic energy of random motion is exactly what temperature measures. Hence $U = U(T)$ — a function of temperature alone.

HOW to derive $U = \frac{f}{2}nRT$

Step 1 — Kinetic theory for translational energy

From kinetic theory, the pressure of a gas gives: $PV = \frac{1}{3}Nm\langle v^2\rangle$

Why this step? This is the kinetic-theory result connecting macroscopic $P,V$ to microscopic speeds. $N$ = number of molecules, $m$ = mass of one, $\langle v^2\rangle$ = mean-square speed.

Compare with the ideal gas law $PV = nRT = Nk_BT$ (using $nR = Nk_B$ , $k_B$ = Boltzmann constant): $\frac{1}{3}Nm\langle v^2\rangle = Nk_BT$

Why this step? We have two expressions for the same $PV$ . Setting them equal links speed to temperature.

Solve for total translational KE: $\frac{1}{2}m\langle v^2\rangle = \frac{3}{2}k_BT$

Why this step? Multiply both sides by $\frac{3}{2}$ and divide by $N$ to get the average KE per molecule.

So average translational KE per molecule $= \frac{3}{2}k_BT$ .

Step 2 — The equipartition theorem

Translation has 3 degrees of freedom ( $v_x, v_y, v_z$ ), giving $3\times\tfrac12 k_BT = \tfrac32 k_BT$ — matching Step 1.

Step 3 — Generalize to $f$ degrees of freedom

If a molecule has $f$ total active degrees of freedom (translation + rotation + vibration), then: $\text{average energy per molecule} = \frac{f}{2}k_BT$

Why this step? Equipartition simply adds $\tfrac12 k_BT$ for each degree of freedom.

Total internal energy = (energy per molecule) × (number of molecules): $U = N\cdot\frac{f}{2}k_BT = \frac{f}{2}(Nk_B)T$

Using $Nk_B = nR$ : $\boxed{U = \frac{f}{2}nRT}$

Why this step? $n$ = moles, $R = N_A k_B$ is the gas constant. This converts the molecule count into moles.

Degrees of freedom $f$ by gas type

Gas type	Translation	Rotation	$f$ (at moderate $T$ )	$U$
Monatomic (He, Ar)	3	0	$3$	$\tfrac32 nRT$
Diatomic (O₂, N₂)	3	2	$5$	$\tfrac52 nRT$
Linear triatomic (CO₂)	3	2	$5$ (rot.)	—
Nonlinear polyatomic (H₂O)	3	3	$6$	$3nRT$

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

Active recall

Recall Quick self-test (cover the answers!)

Why does $U$ depend only on $T$ for an ideal gas? → No intermolecular PE; energy is purely kinetic, and $T$ measures kinetic energy.
What is energy per quadratic DOF? → $\tfrac12 k_BT$ per molecule.
$f$ for monatomic / diatomic? → 3 / 5.
$\Delta U$ in isothermal process? → 0.
Convert $Nk_B$ to molar form? → $nR$ .

Recall Feynman: explain to a 12-year-old

Imagine a box full of tiny bouncy balls zooming around. The hotter the box, the faster they zoom. The "internal energy" is just how much zooming there is in total. If you have more balls (more gas) or you make them faster (more heat), there's more zooming energy. Each direction a ball can move or spin is a separate way to store zoom-energy, and nature shares the energy equally among all those ways. That sharing rule is why we count " $f$ " — the number of independent ways to wiggle — and multiply by it.

Flashcards

What is the internal energy of an ideal gas a function of?

Temperature only (no V or P dependence), because there are no intermolecular forces.

State the equipartition theorem.

Each quadratic degree of freedom carries average energy

\tfrac12 k_BT

per molecule.

Formula for internal energy of ideal gas with f degrees of freedom?

U = \frac{f}{2}nRT

Average translational KE per molecule?

\frac{3}{2}k_BT

Degrees of freedom of a monatomic gas?

3 (all translational), so

U=\tfrac32 nRT

Degrees of freedom of a diatomic gas at moderate T?

5 (3 translational + 2 rotational), so

U=\tfrac52 nRT

Why does a diatomic molecule have only 2 rotational DOF?

Rotation about the bond axis has negligible moment of inertia, so that mode is frozen out.

\Delta U

for an ideal gas in an isothermal process?

Zero, because

\Delta T = 0

How does

Nk_B

relate to

nR

Nk_B = nR

since

R = N_A k_B

and

N = nN_A

In which process does

Q = \Delta U

Constant-volume (isochoric), where work

W=0

Connections

Kinetic theory of gases — source of $PV=\tfrac13 Nm\langle v^2\rangle$
Equipartition theorem — the $\tfrac12 k_BT$ rule
First law of thermodynamics — $\Delta U = Q - W$
Degrees of freedom — how $f$ is counted
Molar specific heats Cv and Cp — $C_V = \tfrac{f}{2}R$ follows directly
Ideal gas law — $PV = nRT$

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, ideal gas ek box hai jisme bahut saare chhote molecules randomly zoom kar rahe hain. Internal energy $U$ ka matlab hai in sabhi molecules ki total energy. Kyunki ideal gas mein molecules ke beech koi force nahi hota (no intermolecular attraction), isliye saari energy sirf kinetic hoti hai — yaani sirf unke hilne-dulne aur ghoomne ki energy. Aur temperature toh basically yahi batata hai ki molecules kitni tezi se hil rahe hain. Isliye $U$ sirf $T$ pe depend karta hai, $P$ ya $V$ pe nahi.

Ab formula $U = \frac{f}{2}nRT$ kaise aaya? Equipartition theorem kehta hai: har independent "way of moving" (degree of freedom) ko nature equally energy deta hai — exactly $\tfrac12 k_BT$ per molecule. Monatomic gas (He, Ar) sirf 3 direction mein move karta hai, toh $f=3$ . Diatomic gas (O $_2$ , N $_2$ ) ek dumbbell jaisa hai, woh move bhi karta hai (3) aur do axes pe ghoomta bhi hai (2), toh $f=5$ . Bas $f$ ko gino aur formula mein daal do.

Sabse important trick exam ke liye: $\Delta U = \frac{f}{2}nR\Delta T$ . Yeh path par depend nahi karta — chahe gas isobaric expand kare ya kuch aur, sirf temperature change ( $\Delta T$ ) maayne rakhta hai. Isothermal process mein $\Delta T = 0$ , toh $\Delta U = 0$ — bhale hi gas kaam kare aur heat le. Yeh concept bahut questions mein aata hai.

Yaad rakhna: "Half-f, n-R-T". Aur "Mono-3, Di-5". Agar yeh do cheezein clear hain, toh internal energy ke saare numericals 30 second mein ho jayenge. Bas pehle gas type pehchaano, fir $f$ daalo.

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

WHAT is internal energy?

WHY only temperature? (first-principles)

HOW to derive $U = \frac{f}{2}nRT$

Step 1 — Kinetic theory for translational energy

Step 2 — The equipartition theorem

Step 3 — Generalize to $f$ degrees of freedom

Degrees of freedom $f$ by gas type

Worked examples

Common mistakes

Active recall

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Thermodynamics

Connections