Mean free path, mean speed, RMS speed — derivations: Worked examples

1.7.11 · D3 · Physics › Thermodynamics › Mean free path, mean speed, RMS speed — derivations

Ye page parent derivations ka drill floor hai. Formulas wahan already banaye ja chuke hain; yahan hum har tarah ke questions tackle karte hain jo exam ya real world throw kar sakta hai. Pehle hum cases ka poora map rakhenge, phir har ek ko work out karenge.

Agar koi symbol unfamiliar lage, toh wo parent note mein define kiya gaya tha — lekin hum har ek ko usi waqt re-state karte hain jab wo appear ho, taaki tumhe kabhi flip back na karna pade.

The scenario matrix

Har possible question ko is grid mein ek cell ki tarah socho. Hamara goal: koi empty cell nahi. Neeche har worked example us cell ke saath tagged hai jo wo fill karta hai.

#	Cell class	Tricky kya hai	Example
A	Direct plug — $v_{r m s}$	SI unit trap ( $M$ kg/mol mein)	Ex 1
B	Direct plug — $\overset{v}{ˉ}$ and $v_{p}$	teen speeds, ek gas	Ex 2
C	Ratio / scaling (T changes)	speeds $\propto T$	Ex 3
D	Ratio / scaling (different gas, same T)	speeds $\propto 1/ M$	Ex 4
E	Mean free path — direct	$n = P / k_{B} T$ form, woh $2$	Ex 5
F	Limiting / degenerate ( $P \to 0$ , $T \to 0$ )	kya blow up hota hai, kya vanish hota hai	Ex 6
G	Collision frequency ( $\overset{v}{ˉ}$ aur $λ$ combine karo)	rate $= \overset{v}{ˉ} / λ$	Ex 7
H	Real-world word problem	vacuum chamber design	Ex 8
I	Exam-style twist (reverse: $T$ ya $d$ nikalo)	formula invert karo	Ex 9

Constants jo poore mein use honge (inhein memorise karo):

$R = 8.314 J mol^{- 1} K^{- 1}$ (gas constant)
$k_{B} = 1.38 \times 1 0^{- 23} J K^{- 1}$ (Boltzmann constant, $= R / N_{A}$ )
$N_{A} = 6.022 \times 1 0^{23} mol^{- 1}$

Cell A — Direct RMS speed

Worked example

Ex 1. Oxygen ( $O_{2}$ ) ka $v_{r m s}$ nikalo $T = 300 K$ par. Molar mass $M = 32 g/mol$ .

Forecast: compute karne se pehle guess karo — kya ye tens, hundreds, ya thousands of m/s hoga? (Room-temperature gas molecules hundreds of m/s ki hoti hain.)

$M$ ko SI mein convert karo. $M = 32 g/mol = 0.032 kg/mol$ . Ye step kyun? Formula $v_{r m s} = 3 R T / M$ joules mein $R$ use karta hai, jo $kg m^{2} s^{- 2}$ hote hain. Agar $M$ grams mein chhod do toh tum ek number se divide karte ho jo 1000× too small hai, isliye speed $1000 \approx 32 \times$ too big aa jaati hai. Ye single most common exam error hai.
Plug in karo. $v_{r m s} = \frac{3 R T}{M} = \frac{3 ( 8.314 ) ( 300 )}{0.032} = 2.338 \times 1 0^{5}$ Ye step kyun? $v_{r m s}$ "energy-carrying" speed hai kyunki kinetic energy $\propto v^{2}$ hai; ye seedha $\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k_{B} T$ se aata hai.
Root lo. $v_{r m s} \approx 483 m/s$ .

Verify karo: Units — $J/(kg/mol) = (kg m^{2} s^{- 2} / mol) / (kg/mol) = m^{2} s^{- 2} = m/s$ . ✓ Hundreds of m/s, jaise forecast kiya, aur lighter than… well, hum Ex 4 mein compare karenge.

Cell B — Ek gas ki teeno speeds

Worked example

Ex 2. Usi $O_{2}$ ke liye $300 K$ par, mean speed $\overset{v}{ˉ}$ aur most probable speed $v_{p}$ bhi nikalo.

Forecast: parent note ne ordering $v_{p} < \overset{v}{ˉ} < v_{r m s}$ di hai. Toh dono naye numbers $483 m/s$ se neeche hone chahiye, $v_{p}$ sabse neeche.

Mean speed — speed ka ordinary arithmetic average, jo control karta hai ki cheezein kitni baar hoti hain (collisions, effusion): $\overset{v}{ˉ} = \frac{8 R T}{π M} = \frac{8 ( 8.314 ) ( 300 )}{π ( 0.032 )} \approx 445 m/s$ Ye step kyun? $3$ ki jagah extra $8/ π$ isliye aata hai kyunki hum $v^{2} f (v)$ ki jagah $v f (v)$ ko Maxwell-Boltzmann distribution par integrate karte hain.
Most probable speed — woh jagah jahan distribution $f (v)$ peak karti hai, single most common speed: $v_{p} = \frac{2 R T}{M} = \frac{2 ( 8.314 ) ( 300 )}{0.032} \approx 395 m/s$ Ye step kyun? $df / d v = 0$ set karne se root ke neeche factor $2$ milta hai.

Verify karo: Ratios hone chahiye $v_{p} : \overset{v}{ˉ} : v_{r m s} = 2 : 8/ π : 3 \approx 1.414 : 1.596 : 1.732$ . Check karo: $445/395 = 1.127$ aur $8/ π / 2 = 1.128$ . ✓ Ordering $395 < 445 < 483$ hold karti hai, jaise forecast kiya.

Figure — Mean free path, mean speed, RMS speed — derivations

Cell C — Temperature scaling (same gas)

Cell D — Alag gas, same temperature

Worked example

Ex 4. Same temperature par, helium ( $M = 4 g/mol$ ) aur oxygen ( $M = 32 g/mol$ ) ke $v_{r m s}$ compare karo.

Forecast: helium 8× lighter hai. Speed $1/ M$ ke saath jaati hai, aur $8 \approx 2.83$ — toh helium roughly 2.8× faster hona chahiye.

Fixed $T$ par ratio. $\frac{v _{He}}{v _{O_{2}}} = \frac{M _{O_{2}}}{M _{He}} = \frac{32}{4} = 8 \approx 2.83$ Ye step kyun? Same $T$ par har molecule ki same average kinetic energy $\frac{3}{2} k_{B} T$ hoti hai (ye Equipartition theorem hai). Same energy $\frac{1}{2} m v^{2}$ ke saath chhota mass matlab badi speed — lighter gas tezi se udti hai.
Ex 1 ki oxygen value use karo $v_{O_{2}} = 483 m/s$ at $300 K$ : $v_{He} = 2.83 \times 483 \approx 1367 m/s$

Verify karo: Direct plug: $3 (8.314) (300) /0.004 = 1.871 \times 1 0^{6} \approx 1368 m/s$ . ✓ (rounding). Isliye hi helium leak karta hai aur bhaari gases se zyada tezi se Earth ki gravity escape karta hai — Diffusion and effusion se link.

Cell E — Mean free path, direct

Worked example

Ex 5. STP par nitrogen ka mean free path $λ$ nikalo: $T = 273 K$ , $P = 1.01 \times 1 0^{5} Pa$ , molecular diameter $d = 3.7 \times 1 0^{- 10} m$ .

Forecast: parent note ne air ke liye ~68 nm quote kiya — nitrogen basically air hi hai, toh same ballpark expect karo.

Pressure form seedha use karo. $λ = \frac{k _{B} T}{2 π d ^{2} P}$ Ye step kyun? Number density $n = P / k_{B} T$ (Ideal gas law se) substitute karne se hum $n$ ko alag messy number ki tarah compute karna skip kar sakte hain.
Numbers plug karo. $λ = \frac{( 1.38 \times 1 0 ^{- 23} ) ( 273 )}{2 π ( 3.7 \times 1 0 ^{- 10} ) ^{2} ( 1.01 \times 1 0 ^{5} )}$ Numerator $= 3.767 \times 1 0^{- 21}$ . Denominator $= 2 π (1.369 \times 1 0^{- 19}) (1.01 \times 1 0^{5}) = 6.14 \times 1 0^{- 14}$ . $λ \approx 6.1 \times 1 0^{- 8} m = 61 nm$ $2$ kyun? Targets bhi move kar rahe hain. Do molecules ke beech average relative speed $2 \overset{v}{ˉ}$ hoti hai, jo collision rate ko $2$ se badhata hai aur flight ko $2$ se chhota karta hai.

Verify karo: Units — $\frac{J}{m ^{2} \cdot Pa} = \frac{J}{m ^{2} \cdot N/m ^{2}} = \frac{J}{N} = m$ . ✓ 61 nm ≈ molecular diameter ka 165× — gas mostly empty space hai. ✓ (Parent ke ~68 nm se match karta hai air ke thoda alag $d$ ke saath.)

Cell F — Limiting & degenerate cases

Worked example

Ex 6. $v_{r m s}$ aur $λ$ in extreme limits mein kya hote hain? (a) $T \to 0$ ; (b) $P \to 0$ (perfect vacuum) fixed $T$ par; (c) $P \to \infty$ .

Forecast: thandi gas freeze hoti hai → speeds khatam. Empty space → hit karne ko kuch nahi → flights badi ho jaati hain. Gas crush karo → udne ki jagah nahi.

(a) $T \to 0$ . $v_{r m s} = 3 R T / M \to 0 = 0$ . Kyun? Teeno speeds mein $T$ factor share hota hai, isliye $v_{p}, \overset{v}{ˉ}, v_{r m s} \to 0$ saath mein jaate hain. Zero thermal energy matlab (classically) koi motion nahi. Saath hi $λ = \frac{k _{B} T}{2 π d ^{2} P} \to 0$ agar $P$ fixed rakha jaaye — lekin real gas pehle liquefy ho jaayegi, isliye ye ek idealised limit hai.
(b) $P \to 0$ fixed $T$ par. $λ = \frac{k _{B} T}{2 π d ^{2} P} \to \infty$ . Kyun? Volume per molecule kam ( $n = P / k_{B} T \to 0$ ) matlab collide karne ko almost kuch nahi — ek molecule poori chamber bina kisi se takraaye cross kar sakta hai. Ye exactly "molecular flow" regime hai jo vacuum technology mein use hota hai (Ex 8). Note karo $v_{r m s}$ unchanged rehta hai — ye sirf $T$ par depend karta hai, $P$ par nahi.
(c) $P \to \infty$ . $λ \to 0$ . Kyun? Infinitely squeeze karo → infinite crowding → hits ke beech zero flight.

Verify karo: Limits $λ \propto 1/ P$ aur $v_{r m s} \propto T$ ( $P$ se independent) ke saath consistent hain. Sanity: $P$ double karo toh $λ$ aadha ho jaata hai lekin $v_{r m s}$ akela reh jaata hai. ✓

Cell G — Collision frequency ( $\overset{v}{ˉ}$ aur $λ$ combine karna)

Worked example

Ex 7. STP par ek nitrogen molecule ek second mein kitne collisions karta hai? Use karo $\overset{v}{ˉ} = 445 m/s$ (Ex 2-style se, $N_{2}$ ke liye kaafi close) aur $λ = 6.1 \times 1 0^{- 8} m$ (Ex 5).

Forecast: ye har second 445 m fly karta hai aur har flight sirf ~61 nm lambi hai — isliye flights (collisions) ki count enormous hogi, billions per second.

Collision frequency = distance per second ÷ distance per flight. $Z = \frac{v ˉ}{λ} = \frac{445}{6.1 \times 1 0 ^{- 8}}$ Ye step kyun? Ek second mein ek molecule $\overset{v}{ˉ}$ metres cover karta hai; har collision ek $λ$ lambi flight khatam karta hai, isliye count seedha unka ratio hai. Hum $\overset{v}{ˉ}$ use karte hain ( $v_{r m s}$ nahi) kyunki collision rate ordinary average speed se govern hoti hai.
Compute karo. $Z \approx 7.3 \times 1 0^{9} s^{- 1}$ .

Verify karo: $7.3 \times 1 0^{9}$ collisions per second — lagbhag 7 billion. ✓ Enormous, jaise forecast kiya. Collisions ke beech time $\approx 1/ Z \approx 1.4 \times 1 0^{- 10} s$ . Units: $\frac{m/s}{m} = s^{- 1}$ . ✓

Cell H — Real-world word problem

Worked example

Ex 8. Ek vacuum chamber 0.5 m wide hai. Engineers chahte hain ki mean free path chamber size se zyada ho (taaki molecules walls se takraayen, ek doosre se nahi — "molecular flow"). $T = 300 K$ , $d = 3.7 \times 1 0^{- 10} m$ , kaun sa pressure chahiye?

Forecast: normal air mein $λ \sim 60$ nm hai. Humein $λ \geq 0.5$ m chahiye — lagbhag $1 0^{7}$ × bada. Kyunki $λ \propto 1/ P$ , humein pressure atmospheric se lagbhag $1 0^{7}$ × chhota chahiye, yaani deep vacuum.

Condition $λ = 0.5 m$ set karo aur $P$ solve karo. $λ = \frac{k _{B} T}{2 π d ^{2} P} \Rightarrow P = \frac{k _{B} T}{2 π d ^{2} λ}$ Ye step kyun? Hum mean-free-path formula invert karte hain kyunki unknown ab $P$ hai, $λ$ nahi.
Plug in karo. $P = \frac{( 1.38 \times 1 0 ^{- 23} ) ( 300 )}{2 π ( 3.7 \times 1 0 ^{- 10} ) ^{2} ( 0.5 )}$ Numerator $= 4.14 \times 1 0^{- 21}$ . Denominator $= 2 π (1.369 \times 1 0^{- 19}) (0.5) = 3.04 \times 1 0^{- 19}$ . $P \approx 1.36 \times 1 0^{- 2} Pa$

Verify karo: Atmospheric se ratio: $1.36 \times 1 0^{- 2} /1.01 \times 1 0^{5} \approx 1.3 \times 1 0^{- 7}$ — lagbhag ten-million-fold lower, jaise forecast kiya. ✓ Isliye high-vacuum equipment (~ $1 0^{- 2}$ Pa aur neeche) molecular-flow regime mein enter karta hai.

Cell I — Exam twist (formula reverse karna)

Worked example

Ex 9. Kis temperature par hydrogen ( $M = 2 g/mol$ ) ka $v_{r m s}$ Earth ki escape speed, $11.2 km/s$ , ke barabar hoga? (Isliye Earth hydrogen ko space mein lose karta hai.)

Forecast: escape speed ~11 200 m/s hai — typical room-temperature $v_{r m s}$ se 20× zyada. Kyunki $v_{r m s} \propto T$ , humein $T$ room temperature se kaafi zyada chahiye (thousands of K), lekin absurdly zyada nahi, kyunki high-speed tail mein even a fraction of molecules kam $T$ par escape kar jaati hain.

$v_{r m s} = 3 R T / M$ ko $T$ ke liye invert karo. $T = \frac{M v _{r m s}^{2}}{3 R}$ Ye step kyun? Unknown ab $T$ hai; dono sides square karne se root hat jaata hai, phir rearrange karo.
Plug in karo ( $M = 0.002 kg/mol$ , $v_{r m s} = 11200 m/s$ ): $T = \frac{( 0.002 ) ( 11200 ) ^{2}}{3 ( 8.314 )} = \frac{( 0.002 ) ( 1.2544 \times 1 0 ^{8} )}{24.94}$ $T \approx 1.006 \times 1 0^{4} K \approx 10, 000 K$

Verify karo: Reverse-check: $3 (8.314) (10060) /0.002 = 1.254 \times 1 0^{8} \approx 11, 200 m/s$ . ✓ Thousands of K, jaise forecast kiya. (Reality mein hydrogen bahut kam $T$ par escape karta hai kyunki Maxwell-Boltzmann distribution ki fast tail continuously leak hoti rehti hai.)

Recall

Recall Kaun si speed kahan use hogi?

Energy / KE calculations ::: $v_{r m s}$ use karo Collision rate, effusion, "kitni baar" ::: $\overset{v}{ˉ}$ use karo Distribution ki peak ::: $v_{p}$ Collisions ke beech distance ::: $λ = 1/ (2 nπ d^{2})$ Collisions per second ::: $Z = \overset{v}{ˉ} / λ$

Recall Scaling shortcuts (re-plugging se bachao)

Same gas, $T$ change hota hai ::: $v \propto T$ (ratio se saare constants cancel) Same $T$ , alag gas ::: $v \propto 1/ M$ Fixed $T$ , $P$ change hota hai ::: $λ \propto 1/ P$ ; $v_{r m s}$ unchanged