Maxwell-Boltzmann speed distribution — derivation (key for propulsion): Worked examples

1.7.12 · D3 · Physics › Thermodynamics › Maxwell-Boltzmann speed distribution — derivation (key for p

Neeche sab kuch sirf do constants aur ek formula se hai, toh inhe ek baar pin kar lete hain.

Scenario matrix

Har MB problem in case classes mein se ek (ya unka blend) hoti hai. Last column batata hai kaun sa worked example use cover karta hai.

#	Case class	Tricky kyon hai	Covered by
C1	Plain numeric speed	$m$ ko molar mass se sahi nikalna	Ex 1
C2	Ratio / universal	$k_{B} T / m$ cancel ho jaata hai — answer pure number	Ex 2
C3	Temperature scaling	Saari speeds $\propto T$	Ex 3
C4	Mass scaling (propulsion)	Saari speeds $\propto 1/ m$	Ex 4
C5	Degenerate input $v = 0$	$f (0) = 0$ , peak nahi	Ex 5
C6	Limiting $T \to 0$ aur $T \to \infty$	Curve collapse / spread hoti hai	Ex 5
C7	Peak height vs peak position	Height $T$ badhne pe girta hai	Ex 6
C8	Speed window mein fraction (integral)	$f (v) d v$ ka meaning samajhna	Ex 7
C9	Real-world word problem (nozzle)	Physics ko $v_{r m s}$ mein convert karna	Ex 8
C10	Exam twist: mixture / two gases	Har gas apna $m$ rakhti hai	Ex 9

Ex 1 — C1: plain numeric speed

Worked example Oxygen O₂ ka RMS aur mean speed 300 K par

O₂ ka molar mass $M = 32 g/mol = 0.032 kg/mol$ hai. Avogadro $N_{A} = 6.022 \times 1 0^{23}$ . $T = 300 K$ par $v_{r m s}$ aur $\overset{v}{ˉ}$ nikalo.

Forecast: Parent mein N₂ $\approx 517$ m/s tha. O₂ thoda bhaari hai (32 vs 28). Guess: 500 m/s se thoda neeche.

Step 1 — Ek molecule ka mass nikalo. $m = \frac{M}{N _{A}} = \frac{0.032}{6.022 \times 1 0 ^{23}} = 5.31 \times 1 0^{- 26} kg .$ Yeh step kyon? Formula ko per molecule mass chahiye, per mole nahi — yahi #1 jagah hai jahan log galti karte hain.

Step 2 — $v_{r m s} = 3 k_{B} T / m$ mein plug karo. $v_{r m s} = \frac{3 ( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{5.31 \times 1 0 ^{- 26}} = 2.339 \times 1 0^{5} \approx 484 m/s .$ Yeh step kyon? "rms" fast molecules ko weight karta hai — yeh speed pressure aur energy ke liye relevant hai.

Step 3 — Fixed ratio $\overset{v}{ˉ} / v_{r m s} = 8/ π / 3 = 8/ (3 π)$ se mean speed nikalo. $\overset{v}{ˉ} = v_{r m s} \frac{8}{3 π} = 484 \times 0.9213 \approx 446 m/s .$ Yeh step kyon? Ek baar ek speed mil jaaye toh baaki sirf constant multiples hain — dobara integrate karne ki zaroorat nahi.

Verify: $484$ m/s $500$ se thoda kam aur N₂ ke $517$ se bhi kam hai (O₂ bhaara ⟹ slow) ✓. Units: $J / kg = m^{2} / s^{2} =$ m/s ✓.

Ex 2 — C2: universal ratio

Worked example Kisi bhi gas ke liye fraction

$v_{p} / v_{r m s}$ Dikhao ki most-probable aur rms speed ka ratio ek pure number hai, aur uski value nikalo.

Forecast: $v_{p}$ peak par hai, $v_{r m s}$ tail ki wajah se right mein khicha hai. Toh ratio 1 se kam hona chahiye — guess around 0.8.

Step 1 — Dono likho. $\frac{v _{p}}{v _{r m s}} = \frac{2 k _{B} T / m}{3 k _{B} T / m} .$ Yeh step kyon? Dono mein identical factor $k_{B} T / m$ hai — inhe side by side likhne se cancellation dikh jaata hai.

Step 2 — $k_{B} T / m$ cancel karo. $\frac{v _{p}}{v _{r m s}} = \frac{2}{3} = 0.8165.$ Yeh step kyon? Kyunki physics ( $k_{B}, T, m$ ) drop ho jaati hai, yeh number har gas ke liye har temperature par same hai — curve ki shape ka signature hai.

Verify: $0.8165 < 1$ ✓ (peak rms se left mein hai). Parent ke ratio $1.000 : 1.128 : 1.225$ se cross-check: $1/1.225 = 0.816$ ✓.

Ex 3 — C3: temperature scaling

Ex 4 — C4: mass scaling (propulsion punchline)

Worked example Thruster propellant ke roop mein Helium vs Xenon

He ( $M = 4 g/mol$ ) aur Xe ( $M = 131 g/mol$ ) ke $v_{r m s}$ ko same temperature par compare karo. Cold-gas thruster ke liye kaun better hai (zyada exhaust speed = zyada specific impulse)?

Forecast: Xe $\approx 33 \times$ bhaari hai. Speed $\propto 1/ m$ , toh He roughly $33 \approx 5.7 \times$ faster hona chahiye.

Step 1 — Same $T$ hai, toh $v_{r m s} \propto 1/ m$ , aur molar mass ratio molecular mass ratio ke barabar hai. $\frac{v _{He}}{v _{Xe}} = \frac{M _{Xe}}{M _{He}} = \frac{131}{4} = 32.75 .$ Yeh step kyon? Bhaari mass denominator mein square root ke neeche hai, isliye $1/ m$ ke roop mein aata hai; ratio $k_{B} T$ ko wipe out kar deta hai.

Step 2 — Evaluate karo. $\frac{v _{He}}{v _{Xe}} = 5.72.$ Yeh step kyon? Yahi reason hai ki light gases zyada thrust per kilogram deti hain — exhaust bahut zyada speed se nikalta hai.

Verify: He, Xe se faster ✓, aur magnitude forecast $5.7$ ke paas ✓. (Graham's law ke saath consistent, rate $\propto 1/ m$ .)

Ex 5 — C5 & C6: degenerate aur limiting inputs

Worked example

$v = 0$ par kya hota hai, aur jab $T \to 0$ ya $T \to \infty$ ? $f (0)$ evaluate karo, aur dono temperature limits mein curve describe karo.

Forecast: $v = 0$ par kuch bhi move nahi ho raha — rare hona chahiye, toh $f (0)$ chhota. $T \to 0$ par sab freeze; $T \to \infty$ par sab spread out.

Step 1 — Formula mein $v = 0$ plug karo. $f (0) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} \cdot 0^{2} \cdot e^{0} = 0.$ Yeh step kyon? $v^{2}$ shell factor value ko kill kar deta hai: radius $0$ ke sphere ka zero surface area hota hai, isliye koi bhi velocity vector exactly zero speed nahi rakhta. Curve origin se start hoti hai, peak se nahi.

Step 2 — Limit $T \to 0$ . Tab $v_{p} = 2 k_{B} T / m \to 0$ aur Gaussian width shrink hoti hai — poori curve $v = 0$ par ek spike mein collapse ho jaati hai. Yeh step kyon? Koi thermal energy nahi matlab koi motion nahi; single value $0$ ki distribution ek spike hoti hai.

Step 3 — Limit $T \to \infty$ . Tab $v_{p} \to \infty$ , peak right mein chala jaata hai, curve flatten aur widen hoti hai (area $1$ rehta hai). Yeh step kyon? Infinite heat molecules ko saari speeds par spread kar deti hai; probability normalized rehni chahiye, isliye width badhne par height girta hai.

Verify (figure): Neeche teen curves exactly yahi dikhati hain — cold = tall narrow spike near zero, hot = short wide curve far right, aur teeno $v = 0$ par $0$ ko touch karti hain ✓.

Ex 6 — C7: peak height vs peak position

Worked example Kya zyada garam gas ka peak taller hota hai?

Jab $T$ badhta hai, toh $f (v)$ ka peak height upar jaata hai ya neeche? Nikalo ki peak height $T$ ke saath kaise scale karti hai.

Forecast: Peak right move karta hai (faster). Kyunki total area $1$ fixed hai, ek wider curve shorter honi chahiye. Guess: height girta hai.

Step 1 — Peak $v_{p} = 2 k_{B} T / m$ par hota hai. $v = v_{p}$ ko $f (v)$ mein substitute karo. Note karo $m v_{p}^{2} / (2 k_{B} T) = 1$ , toh exponential $e^{- 1}$ ban jaata hai. $f (v_{p}) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} v_{p}^{2} e^{- 1} .$ Yeh step kyon? Peak par evaluate karne se messy exponent constant $e^{- 1}$ ban jaata hai, $T$ -dependence isolate ho jaati hai.

Step 2 — $T$ ke powers track karo. Prefactor $T^{- 3/2}$ deta hai; $v_{p}^{2} = 2 k_{B} T / m$ se $T^{+ 1}$ aata hai. $f (v_{p}) \propto T^{- 3/2} \cdot T^{+ 1} = T^{- 1/2} .$ Yeh step kyon? Saare $T$ 's collect karne se pata chalta hai height $1/ T$ scale karti hai — gas garm hone par girta hai.

Verify: 300 K se 1200 K tak ( $\times 4$ ), peak height $\to (4)^{- 1/2} = 1/2$ of its value. Ex 5 ke figure mein hottest curve sabse chhoti hai ✓. Har curve ke neeche area $1$ rehta hai (spread up, height down) ✓.

Ex 7 — C8: speed window mein fraction of molecules

Worked example

$v_{p}$ ke $\pm 1%$ mein speed wale fraction $f (v) d v$ use karke estimate karo ki kitne fraction molecules ki speed $v_{p}$ aur $1.01 v_{p}$ ke beech hai (peak par ek thin window).

Forecast: 1% window thin hai. Roughly (fraction) $\approx f (v_{p}) \times$ (window width). Guess: lagbhag 1%.

Step 1 — Definition use karo: fraction $\approx f (v_{p}) Δ v$ jahan $Δ v = 0.01 v_{p}$ . Yeh step kyon? $f (v) d v$ hi hai ek narrow slice mein fraction; peak ke paas $f$ almost flat hota hai, isliye rectangle ek achha approximation hai.

Step 2 — $f (v_{p})$ ko $v_{p}$ ke units mein compute karo. $β = 1/ v_{p}^{2}$ ke saath formula deta hai $f (v_{p}) = \frac{4}{π} \frac{1}{v _{p}} e^{- 1} = \frac{4}{π} e^{- 1} \frac{1}{v _{p}} = 0.8302 \frac{1}{v _{p}} .$ Yeh step kyon? Sab kuch $v_{p}$ ke terms mein likhne se $v_{p}$ 's, $Δ v$ ke saath cancel ho jaate hain, ek pure number bacht jaata hai.

Step 3 — Window se multiply karo. $fraction \approx 0.8302 \frac{1}{v _{p}} \times 0.01 v_{p} = 8.30 \times 1 0^{- 3} \approx 0.83%.$ Yeh step kyon? Yahi hai "fraction of molecules" ka concrete meaning — ek actual chhota number jo tum count kar sakte ho.

Verify: Result dimensionless hai ✓ ( $v_{p}$ ke units cancel ho gaye), positive hai, aur forecast ke anusaar 1% se kam hai ✓.

Ex 8 — C9: real-world word problem (nozzle)

Worked example Hot-hydrogen thruster ki exhaust speed

Ek nuclear-thermal rocket, hydrogen gas (H₂, $M = 2 g/mol$ ) ko nozzle se nikalne se pehle $2800 K$ tak garam karta hai. H₂ molecules ki characteristic thermal speed $v_{r m s}$ estimate karo. Yeh high specific impulse kyun suggest karta hai?

Forecast: Bahut light gas, bahut hot — several thousand m/s expect karo.

Step 1 — Ek H₂ molecule ka mass. $m = \frac{0.002}{6.022 \times 1 0 ^{23}} = 3.32 \times 1 0^{- 27} kg .$ Yeh step kyon? Phir se molar → per-molecule convert karo $k_{B}$ use karne se pehle.

Step 2 — $v_{r m s} = 3 k_{B} T / m$ compute karo. $v_{r m s} = \frac{3 ( 1.38 \times 1 0 ^{- 23} ) ( 2800 )}{3.32 \times 1 0 ^{- 27}} = 3.491 \times 1 0^{7} \approx 5908 m/s .$ Yeh step kyon? $v_{r m s}$ exhaust ka energy scale set karta hai; nearly 6 km/s woh thermal speed hai jo directed jet ko feed karti hai.

Step 3 — Interpret karo. Specific impulse $\propto$ exhaust velocity $\propto v_{r m s} \propto 1/ m$ . Hydrogen ka chhota $m$ hi reason hai ki uske molecules ek given $T$ par itne fast hote hain. Yeh step kyon? Number ko parent note ke propulsion payoff se connect karta hai.

Verify: $\sim 5900$ m/s expected thousands mein hai ✓. O₂ at 300 K ( $484$ m/s) se compare karo: H₂ lighter hai ( $\times 16$ ) aur hotter ( $\times 9.3$ ), factor $16 \times 9.3 = 12.2$ , aur $484 \times 12.2 = 5905$ ✓.

Ex 9 — C10: exam twist — do-gas mixture

Worked example Same box, do gases — kya woh ek speed share karte hain?

Ek sealed box mein He aur Ar ek temperature $T$ par thermal equilibrium mein hain. Ek student claim karta hai "equilibrium mein dono gases ka $v_{r m s}$ same hota hai." Sach ya jhooth? Ratio $v_{r m s, He} / v_{r m s, Ar}$ nikalo ( $M_{He} = 4$ , $M_{Ar} = 40$ g/mol).

Forecast: Equilibrium ka matlab equal temperature hai (equal mean energy), equal speed nahi. Lighter He faster move karega. Guess ratio $\approx 10 \approx 3.2$ .

Step 1 — Equilibrium, per molecule average kinetic energy ko equalize karta hai (equipartition ke zariye): $\frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$ har gas ke liye. Yeh step kyon? Temperature — speed nahi — woh cheez hai jo do gases mix hone par share karti hain. Student ne "same $T$ " ko "same $v$ " samajh liya.

Step 2 — Equal energy lekin different mass ⟹ different speed. Ratio lo. $\frac{v _{r m s, He}}{v _{r m s, Ar}} = \frac{M _{Ar}}{M _{He}} = \frac{40}{4} = 10 = 3.162.$ Yeh step kyon? $k_{B} T$ fixed hai (same box), sirf mass differ karta hai, $1/ m$ ke roop mein aata hai.

Step 3 — Verdict: claim galat hai; sirf $v_{r m s, He} = 10 v_{r m s, Ar}$ sahi hai. Yeh step kyon? Specific misconception name karta hai taaki woh dobara na ho.

Verify: He faster ✓, ratio $3.16 > 1$ ✓; equal energies check: $\frac{1}{2} m_{He} v_{He}^{2}$ vs $\frac{1}{2} m_{Ar} v_{Ar}^{2}$ ka ratio $(1/10) (10) = 1$ ✓.

Recall Kaun sa cell sabse mushkil tha? Self-test

$f (v)$ ka peak height scale karta hai ::: $T^{- 1/2}$ ke saath (gas garm hone par girta hai). $f (0) = ?$ aur kyon ::: $0$ , kyunki $v^{2}$ shell factor zero speed par vanish ho jaata hai. Equilibrium par mix hoti do gases share karti hain ::: temperature (mean energy), speed nahi. Kisi bhi gas ke liye $v_{p} / v_{r m s}$ ::: $2/3 = 0.8165$ , universal.

Connections

Parent: MB derivation
Kinetic Theory of Gases — pressure $v_{r m s}$ se link karta hai
Equipartition Theorem — mixture mein equal energy (Ex 9)
Boltzmann Distribution — energy-space parent law
Mean Free Path — $\overset{v}{ˉ}$ use karta hai
Rocket Propulsion & Specific Impulse — Ex 4 aur Ex 8 payoff
Effusion and Graham's Law — $1/ m$ scaling reused