Worked examples — Equipartition theorem — ½k_BT per quadratic degree of freedom

2.4.14 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Equipartition theorem — ½k_BT per quadratic degree of freedo

Neeche sab kuch sirf do ideas pe based hai jo tumhare paas pehle se hain:

• Har quadratic energy slot (energy jo $\alpha q^2$ form mein ho) average par $\frac{1}{2}{k}_{B}T$ carry karta hai.
• $C_V={\left(\frac{\partial U}{\partial T}\right)}_V$, molar $U$ mein $N_A{k}_B=R$ use hota hai, aur $\gamma =C_P/C_V$ jahan ideal gas ke liye $C_P=C_V+R$ hai.

Constants jo main baar baar use karunga: $k_B=1.380649\times 10^{-23}\text{J/K}$, $R=8.314\text{J/(mol⋅K)}$, $N_A=6.022\times 10^{23}$.

---

Scenario matrix

Har equipartition problem inhi cells mein se ek hoti hai. Neeche ke examples mein har ek batata hai ki woh kaun si cell(s) hit kar raha hai.

Cell	Kya vary karta hai	Kaun sa danger test hota hai	Example
A — clean count	ordinary gas, sabhi modes active	kya tum T-R-V sahi count kar sakte ho?	Ex 1, Ex 2
B — vibration doubles	high-T diatomic	KE aur PE yaad rakhna	Ex 3
C — degenerate/zero	ek mode jisme koi energy cost nahi (frozen)	kuch bhi contribute nahi karta	Ex 4
D — solid (Dulong–Petit)	6 slots per atom	3 KE + 3 PE	Ex 5
E — real-world word problem	Brownian speck, actual numbers	SI mein plug karna, units	Ex 6
F — non-quadratic energy	$E\propto q^4$	the general $\frac{1}{n}{k}_{B}T$ rule	Ex 7
G — exam twist / mixture	gas mixture, derive $\gamma$	averaging over species	Ex 8

Hum all seven cells ko aath examples mein cover karte hain.

---

Example 1 — Cell A: monatomic gas energy (clean count)

Step 1 — Quadratic slots count karo. Ek helium atom ek point mass hai: woh sirf $x,y,z$ mein move kar sakta hai. Energy $=\frac{1}{2}m{v}_x^2+\frac{1}{2}m{v}_y^2+\frac{1}{2}m{v}_z^2$ = 3 quadratic slots. Yeh step kyun? Equipartition ek counting rule hai — jab tak hum nahi jaante kitne slots hain, kuch nahi hota.

Step 2 — Energy per atom. $⟨E⟩=3\times \frac{1}{2}{k}_{B}T=\frac{3}{2}{k}_{B}T=\frac{3}{2}(1.380649\times 10^{-23})(300)=6.21\times 10^{-21}\text{J}.$ Yeh step kyun? Har slot $\frac{1}{2}{k}_{B}T$ deta hai; teen slots $\frac{3}{2}{k}_{B}T$ dete hain.

Step 3 — Total internal energy. $U=n\cdot \frac{3}{2}RT=2.0\times \frac{3}{2}\times 8.314\times 300=7483\text{J}.$ Yeh step kyun? Per-mole energy $\frac{3}{2}RT$ ko moles ki sankhya se multiply karo; $N_A{k}_B=R$ "per atom" ko "per mole" mein convert karta hai.

Verify: Sanity — $U=n\cdot \frac{3}{2}RT$ ko $N\cdot \frac{3}{2}{k}_{B}T$ ke barabar bhi hona chahiye jahan $N=n{N}_A$ ho. Check: $2\times 6.022\times 10^{23}\times 6.21\times 10^{-21}\approx 7.48\times 10^3\text{J}$. ✓ Same number. Forecast answer: $\approx 7500\text{J}$.

---

Example 2 — Cell A: diatomic gas at room T (frozen 3rd rotation)

Step 1 — Kaun se modes active hain? Translation: 3 slots. Rotation: molecule ek dumbbell hai, isliye woh bond ke 2 perpendicular axes ke baare mein tumble karta hai (bond axis ke baare mein rotation ka moment of inertia almost zero hota hai — dekho Quantum freezing of degrees of freedom — isliye woh frozen hai). Vibration: $300\text{K}$ par frozen. Total $=3+2=5$ slots.

Step 2 — Heat capacities. $C_V=\frac{5}{2}R=20.79\text{J/(mol⋅K)},C_P=C_V+R=\frac{7}{2}R=29.10\text{J/(mol⋅K)}.$ Yeh step kyun? $U=\frac{5}{2}RT$ isliye $C_V=\partial U/\partial T=\frac{5}{2}R$; ideal gas ke liye $C_P-C_V=R$ (dekho Heat capacity of gases).

Step 3 — Adiabatic index. $\gamma =\frac{C_P}{C_V}=\frac{7/2}{5/2}=\frac{7}{5}=\mathrm{1.40.}$

Verify: air (mostly ${\text{N}}_2$, ${\text{O}}_2$) ka measured $\gamma \approx 1.40$ hai. ✓ Forecast: $1.40$.

---

Example 3 — Cell B: same gas, high temperature (vibration doubles)

Step 1 — Vibrational slots add karo. Ek vibrating bond ek spring hai: woh kinetic energy $\frac{1}{2}\mu {\dot{r}}^2$ aur potential energy $\frac{1}{2}\kappa r^2$ store karta hai. Yeh 2 naye quadratic slots hain, ek nahi. Yeh step kyun? Yeh classic trap hai — equipartition energy terms count karta hai, motions nahi. "Vibes are doubled."

Step 2 — Naya count aur capacities. $3(\text{trans})+2(\text{rot})+2(\text{vib})=7\text{slots}.$ $C_V=\frac{7}{2}R=29.10\text{J/(mol⋅K)},\gamma =\frac{C_P}{C_V}=\frac{9/2}{7/2}=\frac{9}{7}\approx \mathrm{1.286.}$

Verify: $C_V$ $\frac{5}{2}R$ se $\frac{7}{2}R$ tak badha, yaani exactly $R$ se (do half-slots), $\frac{1}{2}R$ se nahi. ✓ Aur $\gamma$ $1.40$ se $1$ ki taraf gira, kyunki modes add karne se hamesha $\gamma$ kam hota hai. ✓ Forecast: jump by $R$.

---

Example 4 — Cell C: the degenerate mode (a slot that pays nothing)

Step 1 — Bond axis ke baare mein moment of inertia dekho. Bond axis ke baare mein spinning ka matlab hai line par hi do tiny atomic nuclei ko spin karna. Moment of inertia $I_{\parallel }$ essentially zero hai (mass axis par hi baitha hai). Yeh step kyun? Rotational energy $\frac{1}{2}I{\omega }^2$ hai; agar $I\to 0$ toh us mode ki classical energy bhi tiny hai — lekin honest reason yeh hai ki woh quantum ki wajah se frozen hai.

Step 2 — Yeh truly zero kyun carry karta hai. Rotational energy levels $\Delta E\sim {\hslash }^2/(2I)$ ki tarah spaced hote hain. Ek vanishing $I$ gap $\Delta E\gg k_{B}T$ bana deta hai, isliye mode excite nahi ho sakta (dekho Quantum freezing of degrees of freedom). Ek frozen slot exactly $0$ contribute karta hai, $\frac{1}{2}{k}_{B}T$ nahi. $⟨E_{\text{3rd rot}}⟩=0.$

Step 3 — Sahi count barakar rehta hai. Toh room-temperature count genuinely $3+2=5$ hai aur $\gamma =1.40$ hai, $1.33$ nahi.

Verify: student ka $\gamma =(6/2+1)/(6/2)=4/3\approx 1.333$ exactly wahi milega agar mode active hota — aur experiment $1.40$ kehta hai, jo use rule out karta hai. ✓ Degenerate/zero cell $0$ deta hai.

---

Example 5 — Cell D: classical solid (Dulong–Petit)

Step 1 — Solid mein slots per atom count karo. Har atom ek 3D harmonic well mein baitha hai, neighbours se bonded. Har direction $x,y,z$ mein uske paas kinetic energy $\frac{1}{2}m{\dot{x}}^2$ aur potential energy $\frac{1}{2}\kappa x^2$ hai. Yeh $3\times 2=6$ slots per atom hain. Yeh step kyun? Ek solid atom 3 independent springs ki tarah hai — har spring, vibration ki tarah, doubled hai.

Step 2 — Energy aur heat capacity. $⟨E⟩=6\times \frac{1}{2}{k}_{B}T=3k_{B}T⟹U=3RT⟹C_V=3R=24.94\text{J/(mol⋅K)}.$ Yeh Dulong–Petit law hai.

Verify: predicted $3R=24.94$; measured $\approx 24.4$. $2\%$ ke andar agreement. ✓ Forecast: $3R$.

---

Example 6 — Cell E: Brownian speck (real-world word problem)

Step 1 — Equipartition speck ke ek direction mein translation par apply karo. Ek bada object bhi "ek direction mein ek quadratic slot" hai: $\frac{1}{2}m{v}_x^2$. Equipartition ko parwah nahi kitna bhaari hai — woh phir bhi $\frac{1}{2}{k}_{B}T$ deta hai. $\frac{1}{2}m⟨v_x^2⟩=\frac{1}{2}{k}_{B}T.$ Yeh step kyun? Yeh exactly Cell A logic hai, lekin ek visible particle par apply hota hai — parent note mein prove ki gayi mass-independence ki khoobsurti.

Step 2 — Mean-square speed ke liye solve karo. $⟨v_x^2⟩=\frac{k_{B}T}{m}=\frac{(1.380649\times 10^{-23})(293)}{1.0\times 10^{-15}}=4.045\times 10^{-6}{\text{m}}^2/{\text{s}}^2.$

Step 3 — Square root lo. $v_{\text{rms}}=\sqrt{⟨v_x^2⟩}=2.01\times 10^{-3}\text{m/s}\approx 2\text{mm/s}.$ Yeh step kyun? "rms" matlab root-mean-square — speed pane ke liye squaring undo karo.

Verify: units: $\sqrt{\text{J/kg}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$ ✓. Ek air molecule ($m\sim 5\times 10^{-26}\text{kg}$) se same $T$ par compare karo: woh $\sim 500\text{m/s}$ move karega. Speck $2\times 10^{10}$ times bhaari hai, isliye $\sqrt{2\times 10^{10}}\approx 1.4\times 10^5$ se slower, giving $\sim 3\text{mm/s}$ — same ballpark. ✓ Forecast: millimetres/s.

---

Example 7 — Cell F: non-quadratic energy ($E\propto q^4$)

Step 1 — General theorem yaad karo. Parent note ne dikhaya: $E=\alpha |q{|}^n$ ke liye, same $-\frac{d\ln Z}{d\beta }$ machinery deta hai $⟨E⟩=\frac{1}{n}{k}_{B}T.$ Yeh step kyun? $\frac{1}{2}$ universal nahi hai — yeh Gaussian integral se aata hai, jise $n=2$ chahiye. Power change karo, fraction change ho jaata hai.

Step 2 — $n=4$ plug in karo. $⟨E⟩=\frac{1}{4}{k}_{B}T.$

Step 3 — Spring se kam kyun hai. Ek quartic well ki walls parabola se steeper hoti hain, $q$ ko zyada tightly confine karti hain; isliye per mode trapped energy kam hoti hai. General rule exactly quantify karta hai kitni: $1/n$.

Verify: derivation sketch — $u={\beta }^{1/n}q$ ko $Z=\int e^{-\beta \alpha q^4}dq$ mein sub karo, milega $Z\propto {\beta }^{-1/4}$, isliye $\ln Z=\text{const}-\frac{1}{4}\ln \beta$ aur $-d\ln Z/d\beta =\frac{1}{4\beta }=\frac{1}{4}{k}_{B}T$. ✓ Forecast: $\frac{1}{4}{k}_{B}T$.

---

Example 8 — Cell G: gas mixture (exam twist)

Step 1 — Har species ka $C_V$ nikalo. Helium: $C_{V,\text{He}}=\frac{3}{2}R$. Nitrogen: $C_{V,{\text{N}}_2}=\frac{5}{2}R$ (Ex 2). Yeh step kyun? Har gas independently equipartition follow karta hai; woh same $T$ share karte hain lekin apne slots khud count karte hain.

Step 2 — Mole-weighted average $C_V$. Heat capacity extensive hai — total heat capacities add karo, phir total moles se divide karo: $C_{V,\text{mix}}=\frac{n_{\text{He}}{C}_{V,\text{He}}+n_{{\text{N}}_2}{C}_{V,{\text{N}}_2}}{n_{\text{He}}+n_{{\text{N}}_2}}=\frac{1\cdot \frac{3}{2}R+3\cdot \frac{5}{2}R}{4}=\frac{\frac{3}{2}+\frac{15}{2}}{4}R=\frac{9}{4}R.$ Toh $C_{V,\text{mix}}=2.25R=18.71\text{J/(mol⋅K)}$.

Step 3 — Mixture $\gamma$. Ideal-gas mixture ke liye $C_{P,\text{mix}}=C_{V,\text{mix}}+R=\frac{13}{4}R$, isliye ${\gamma }_{\text{mix}}=\frac{C_{P,\text{mix}}}{C_{V,\text{mix}}}=\frac{13/4}{9/4}=\frac{13}{9}\approx \mathrm{1.444.}$

Verify: ${\gamma }_{\text{mix}}=1.444$ ${\gamma }_{{\text{N}}_2}=1.40$ aur ${\gamma }_{\text{He}}=1.667$ ke beech hai, jaise kisi bhi weighted mix mein hona chahiye. ✓ Forecast: range ke andar.

---

Recall Which cell was hardest for you?

Non-quadratic ($E=\alpha q^4$) $\frac{1}{4}{k}_{B}T$ deta hai, $\frac{1}{2}{k}_{B}T$ nahi ::: kyunki $\frac{1}{2}$ sirf Gaussian ($n=2$) integral se aata hai; general rule $\frac{1}{n}{k}_{B}T$ hai. Ek vibrating bond $C_V$ mein kitna add karta hai? ::: $R$ (do slots: KE + PE), $\frac{1}{2}R$ nahi. Ek diatomic ka teesra rotation contribute karta hai ::: $0$ — yeh quantum gap se frozen out ho jaata hai. Mixture $C_V$ combine hota hai ::: har species ke $C_V$ ka mole-weighted average lekar.

---

Related: Boltzmann distribution · Partition function Z · Maxwell–Boltzmann speed distribution · Ideal gas law PV=NkT