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

1.7.12 · D4 · Physics › Thermodynamics › Maxwell-Boltzmann speed distribution — derivation (key for p

Figure — Maxwell-Boltzmann speed distribution — derivation (key for propulsion)

Upar wali figure tumhara visual cheat-sheet hai: wahi curve jisme teen speeds mark ki gayi hain. Peak $v_{p}$ maximum par baithti hai, $\overset{v}{ˉ}$ thoda us se right mein hai, aur $v_{r m s}$ aur bhi right mein, lambi tail ki taraf.

Level 1 — Recognition

L1.1

Kaun sa formula molecules ka fraction deta hai jinka speed $v$ aur $v + d v$ ke beech hai? Likho aur uske units batao.

Recall Solution (L1.1)

Distribution function khud: $f (v) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} v^{2} e^{- \frac{m v ^{2}}{2 k _{B} T}} .$ Fraction hai $f (v) d v$ . Kyunki $f (v) d v$ ek pure number (fraction) hona chahiye aur $d v$ ki units speed ki hain, isliye $f (v)$ ki units $(speed)^{- 1}$ hain, matlab $s/m$ .

L1.2

Teen characteristic speeds ko smallest se largest tak order karo, aur $f (0)$ batao.

Recall Solution (L1.2)

Hamesha $v_{p} < \overset{v}{ˉ} < v_{r m s}$ (ratios $1.000 : 1.128 : 1.225$ ). Aur $f (0) = 0$ , kyunki $v^{2}$ factor function ko $v = 0$ par zero kar deta hai — koi bhi molecule exactly zero speed par "most likely" value ke roop mein nahi baithta.

L1.3

Product $v^{2} e^{- m v^{2} /2 k_{B} T}$ mein, kaun sa factor $v$ ke saath badhta hai aur kaun sa ghatta hai? Dono ka product kya shape banata hai?

Recall Solution (L1.3)

$v^{2}$ badhta hai (ek growing parabola), $e^{- m v^{2} /2 k_{B} T}$ ghatta hai (ek decaying exponential). Dono ka product ek skewed peak hai: yeh $0$ se chadh ke $v_{p}$ par maximum reach karta hai, phir ek lambi high-speed tail ke saath decay karta hai.

Level 2 — Application

L2.1

N₂ ke liye $T = 300$ K par $v_{r m s}$ compute karo.

Recall Solution (L2.1)

$v_{r m s} = \frac{3 k _{B} T}{m} = \frac{3 ( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{4.65 \times 1 0 ^{- 26}} = 2.671 \times 1 0^{5} \approx 517 m/s .$ $\overset{v}{ˉ}$ ki jagah $v_{r m s}$ kyun? RMS fast molecules ko zyada weight deta hai (yeh $⟨ v^{2} ⟩$ se aata hai), jo exactly wahi hai jis par energy aur pressure depend karti hain.

L2.2

Usi N₂ ke liye 300 K par $v_{p}$ aur $\overset{v}{ˉ}$ compute karo.

Recall Solution (L2.2)

$v_{p} = \frac{2 k _{B} T}{m} = \frac{2 ( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{4.65 \times 1 0 ^{- 26}} = 1.781 \times 1 0^{5} \approx 422 m/s .$ $\overset{v}{ˉ} = \frac{8 k _{B} T}{π m} = \frac{8 ( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{π ( 4.65 \times 1 0 ^{- 26} )} = 2.267 \times 1 0^{5} \approx 476 m/s .$ Ordering check karo: $422 < 476 < 517$ ✓.

L2.3

Agar N₂ ko 300 K se 600 K tak heat kiya jaye, toh $v_{p}$ kitne factor se badlega? Naya $v_{p}$ kya hoga?

Recall Solution (L2.3)

$v_{p} \propto T$ , isliye $T$ double karne par $v_{p}$ , $2 \approx 1.414$ se multiply hota hai. Naya $v_{p} = 422 \times 1.414 \approx 597 m/s$ . Square root kyun? $T$ , $v_{p}$ mein andar square root ke under aata hai ( $v_{p} = 2 k_{B} T / m$ ), isliye andar ka factor 2 bahar $2$ ban jaata hai.

Level 3 — Analysis

L3.1

Ek hi temperature par H₂ ( $m = 3.3 \times 1 0^{- 27}$ kg) aur CO₂ ( $m = 7.3 \times 1 0^{- 26}$ kg) ko compare karo. $v_{r m s, H_{2}} / v_{r m s, CO_{2}}$ nikalo.

Recall Solution (L3.1)

Kyunki fixed $T$ par $v_{r m s} \propto 1/ m$ : $\frac{v _{H_{2}}}{v _{CO_{2}}} = \frac{m _{CO_{2}}}{m _{H_{2}}} = \frac{7.3 \times 1 0 ^{- 26}}{3.3 \times 1 0 ^{- 27}} = 22.12 \approx 4.70.$ Propulsion ka punchline: halka exhaust ⟹ faster molecules ⟹ zyada exhaust velocity ⟹ zyada specific impulse. Dekho Rocket Propulsion & Specific Impulse.

L3.2

Ek gas temperature $T$ par hai. Kis temperature $T^{'}$ par ek do gune bhaari gas ka sama $v_{r m s}$ hoga?

Recall Solution (L3.2)

$v_{r m s}$ equal set karo: $\frac{3 k _{B} T}{m} = \frac{3 k _{B} T ^{'}}{2 m} \Rightarrow T^{'} = 2 T$ . Kyun: mass double karne ke baad same RMS speed rakhne ke liye temperature bhi double karni padegi — mass aur temperature $3 k_{B} T / m$ ke andar ek-ke-badle-ek trade off karte hain.

L3.3

Do speeds of interest satisfy karte hain $\overset{v}{ˉ} / v_{p} = 4/ π$ . Is ratio ko verify karo aur iska numeric value batao.

Recall Solution (L3.3)

$\frac{v ˉ}{v _{p}} = \frac{8 k _{B} T / π m}{2 k _{B} T / m} = \frac{8}{π \cdot 2} = \frac{4}{π} \approx 1.128.$ Yeh universal hai — $k_{B} T / m$ cancel ho jaata hai, isliye ratio har gas ke liye har temperature par identical hai.

Level 4 — Synthesis

L4.1

$g (v_{x}) = \frac{m}{2 π k _{B} T} e^{- m v_{x}^{2} /2 k_{B} T}$ se shuru karke, $⟨ v_{x}^{2} ⟩ = \frac{k _{B} T}{m}$ derive karo aur isse $v_{r m s}$ obtain karo.

Recall Solution (L4.1)

$β = \frac{m}{2 k _{B} T}$ leke, standard Gaussian moment hai: $⟨ v_{x}^{2} ⟩ = \frac{\int _{- \infty}^{\infty} v _{x}^{2} e ^{- β v_{x}^{2}} d v _{x}}{\int _{- \infty}^{\infty} e ^{- β v_{x}^{2}} d v _{x}} = \frac{1}{2 β} = \frac{2 k _{B} T}{2 m} = \frac{k _{B} T}{m} .$ Isotropy se $⟨ v^{2} ⟩ = ⟨ v_{x}^{2} ⟩ + ⟨ v_{y}^{2} ⟩ + ⟨ v_{z}^{2} ⟩ = 3 \cdot \frac{k _{B} T}{m} = \frac{3 k _{B} T}{m}$ , isliye $v_{r m s} = ⟨ v^{2} ⟩ = \frac{3 k _{B} T}{m} . ✓$ Yeh Equipartition Theorem result $⟨ \frac{1}{2} m v_{x}^{2} ⟩ = \frac{1}{2} k_{B} T$ har component ke liye use karta hai.

L4.2

Explicitly dikhao ki $\overset{v}{ˉ} = \frac{8 k _{B} T}{π m}$ , $\overset{v}{ˉ} = \int_{0}^{\infty} v f (v) d v$ evaluate karke.

Recall Solution (L4.2)

$β = \frac{m}{2 k _{B} T}$ aur $C = 4 π (\frac{β}{π})^{3/2}$ likho taaki $f (v) = C v^{2} e^{- β v^{2}}$ ho. Phir: $\overset{v}{ˉ} = C \int_{0}^{\infty} v^{3} e^{- β v^{2}} d v = C \cdot \frac{1}{2 β ^{2}} .$ ( $\int_{0}^{\infty} v^{3} e^{- β v^{2}} d v = \frac{1}{2 β ^{2}}$ use karke.) $C$ substitute karne par: $\overset{v}{ˉ} = 4 π (\frac{β}{π})^{3/2} \frac{1}{2 β ^{2}} = \frac{2}{π β} = \frac{4}{π β} = \frac{4 \cdot 2 k _{B} T}{π m} = \frac{8 k _{B} T}{π m} . ✓$

L4.3

Temperature $2 T$ par peak height $f (v_{p})$ aur temperature $T$ par $f (v_{p})$ ka ratio nikalo. (Scaling se argue karo, constants mat grind karo.)

Recall Solution (L4.3)

Peak par, $v_{p} \propto T$ aur $f (v_{p}) \propto v_{p}^{2} e^{- 1} \cdot (\frac{m}{2 π k _{B} T})^{3/2}$ . Kyunki $e^{- m v_{p}^{2} /2 k_{B} T} = e^{- 1}$ factor temperature-independent hai (yeh peak par evaluate hota hai), $f (v_{p}) \propto (T)^{2} \cdot T^{- 3/2} = T \cdot T^{- 3/2} = T^{- 1/2} .$ Isliye $\frac{f ( v _{p} ) at 2 T}{f ( v _{p} ) at T} = (2)^{- 1/2} = \frac{1}{2} \approx 0.707$ . Interpretation: heat karne se peak girta hai (curve phail jaata hai) jabki total area $1$ rehti hai — parent note ke Forecast-then-Verify se match karta hai.

Level 5 — Mastery

L5.1

Ek tiny hole se effusion molecules ko $\propto v f (v)$ rate se sample karti hai (fast molecules hole ko zyada baar hit karte hain). Effusing molecules ki most probable speed nikalo, yaani $v f (v) \propto v^{3} e^{- β v^{2}}$ ka peak.

Recall Solution (L5.1)

$h (v) = v^{3} e^{- β v^{2}}$ maximize karo. $h^{'} (v) = 0$ set karo: $h^{'} (v) = (3 v^{2} - 2 β v^{4}) e^{- β v^{2}} = v^{2} (3 - 2 β v^{2}) e^{- β v^{2}} = 0.$ Nonzero root: $3 - 2 β v^{2} = 0 \Rightarrow v = \frac{3}{2 β} = \frac{3 k _{B} T}{m} = v_{r m s}$ . Khoobsurat result: typical effusing molecule $v_{r m s}$ par move karti hai — bulk peak $v_{p}$ se faster, kyunki hole preferentially fast tail ko sample karta hai. Yeh Effusion and Graham's Law ki foundation hai.

L5.2

Rocket exhaust "energy speed" ko $v_{r m s} = 3 k_{B} T / m$ se best capture kiya jaata hai. Ek chamber $T = 3000$ K par H₂ ( $m = 3.3 \times 1 0^{- 27}$ kg) jalake run karta hai. $v_{r m s}$ estimate karo aur comment karo ki real exhaust velocities kyun kam hoti hain.

Recall Solution (L5.2)

$v_{r m s} = \frac{3 ( 1.38 \times 1 0 ^{- 23} ) ( 3000 )}{3.3 \times 1 0 ^{- 27}} = 3.764 \times 1 0^{7} \approx 6135 m/s .$ Toh random thermal speed ka $\sim 6.1$ km/s. Real exhaust kam hota hai kyunki nozzle is isotropic random motion ko directed flow mein imperfectly convert karta hai (finite expansion, recombination, heat loss). Phir bhi, yeh dikhata hai ki halki hot gas kyun valued hai: $v_{r m s} \propto T / m$ — dekho Rocket Propulsion & Specific Impulse.

L5.3

Dikhao ki $v_{p}$ se upar speed wale molecules ka fraction temperature- aur mass-independent hai (ek universal number), aur ise estimate karo. (Dimensionless integral set up karo; tum numeric value quote kar sakte ho.)

Recall Solution (L5.3)

$u = v / v_{p}$ substitute karo, toh $β v^{2} = v^{2} / v_{p}^{2} = u^{2}$ (kyunki $v_{p} = 1/ β$ ). Phir $f (v) d v$ ek pure function of $u$ ban jaata hai: $P (v > v_{p}) = \int_{1}^{\infty} \frac{4}{π} u^{2} e^{- u^{2}} d u .$ Saara $k_{B} T / m$ cancel ho gaya — answer ek universal constant hai. Numerically evaluate karne par: $P (v > v_{p}) \approx 0.572.$ Toh molecules ka lagbhag 57% most probable speed se faster move karta hai — yeh right-skew ka direct evidence hai (peak population ka middle nahi hai).

Connections

Kinetic Theory of Gases — distribution ke peeche collision picture
Equipartition Theorem — har DOF ke liye $\frac{1}{2} k_{B} T$ deta hai (L4.1 mein use hua)
Boltzmann Distribution — MB ka energy-space parent
Mean Free Path — $\overset{v}{ˉ}$ use karta hai
Rocket Propulsion & Specific Impulse — L5.2 payoff
Effusion and Graham's Law — L5.1 payoff