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

Energy $\frac{1}{2}m{v}^2$ ki tarah scale karti hai — yeh $v^2$ ke average par depend karta hai. Toh energy carry karne wala speed root-mean-square speed $v_{rms}=\sqrt{3k_{B}T/m}$ hai.

Collision frequency is baat par depend karti hai ki molecules actually kitni tezi se ek doosre ke paas se guzarte hain — yeh $v$ ka ordinary average hai. Woh mean speed $\bar{v}=\sqrt{8k_{B}T/\pi m}$ hai.

Dono $v_p$ (distribution ka peak) use nahi karte, jo sirf "single most common speed" hai.

$v_p<\bar{v}<v_{rms}$ Kyun: averaging se pehle square karna (jaisa $v_{rms}$ karta hai) tail mein fast molecules ko extra weight deta hai, number ko upar kheenchta hai. Peak $v_p$ tail ko bilkul ignore karta hai, isliye woh sabse neeche baith jaata hai. Numerical ratios hain $\sqrt{2}:\sqrt{8/\pi }:\sqrt{3}\approx 1.41:1.60:1.73$.

$n=N/V$ number density hai — per cubic metre kitne molecules hain. Yeh gas ki "crowdedness" hai.

$\lambda \propto 1/n$, toh $n$ ko teen guna karne se $\lambda$ ek tihaai reh jaata hai. Teen guna zyada targets ⇒ ek tihaai distance ke baad collisions hote hain.

$v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3(8.314)(300)}{0.004}}=\sqrt{1.871\times 10^6}\approx 1368\text{m/s}$ Itna fast kyun? Helium halka hai ($M$ chota, denominator mein), toh same temperature par yeh nitrogen (~517 m/s) se bahut tezi se move karta hai. Isliye helium balloons se jaldi leak ho jaata hai.

$\bar{v}=\sqrt{\frac{8RT}{\pi M}}=\sqrt{\frac{8(8.314)(273)}{\pi (0.032)}}=\sqrt{\frac{18158}{0.1005}}=\sqrt{1.806\times 10^5}\approx 425\text{m/s}$ $\pi$ kyun? Yeh directly us Gaussian integral se aata hai jo Maxwell–Boltzmann distribution par $v$ ko average karne ke liye use hota hai — parent derivation dekho.

Pressure form use karo taaki hum $n$ manually calculate na karein: $\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}=\frac{(1.38\times 10^{-23})(300)}{\sqrt{2}\pi (2.7\times 10^{-10}{)}^2(1.0\times 10^5)}$ Numerator: $4.14\times 10^{-21}$. Denominator: $\sqrt{2}(3.1416)(7.29\times 10^{-20})(1.0\times 10^5)=3.239\times 10^{-14}$. $\lambda \approx 1.28\times 10^{-7}\text{m}\approx 128\text{nm}$ Itna bada kyun? Chota molecule ($d$ chota ⇒ $d^2$ denominator mein tiny) ⇒ lambi flights.

$\frac{v_{rms}}{\bar{v}}=\frac{\sqrt{3RT/M}}{\sqrt{8RT/\pi M}}=\sqrt{\frac{3RT/M}{8RT/\pi M}}=\sqrt{\frac{3\pi }{8}}$ Har ek $R$, $T$, $M$ cancel ho jaata hai — yahi toh poora point hai. Numerically: $\sqrt{\frac{3\pi }{8}}=\sqrt{1.178}\approx 1.085$ Toh $v_{rms}$ hamesha $\bar{v}$ se lagbhag $8.5\%$ bada hota hai, chahe gas ya temperature kuch bhi ho. Isliye dono ko confuse karna itna aasaan hai.

(a) $v_{rms}\propto \sqrt{T}$, toh $T$ double karne se $v_{rms}$ $\sqrt{2}\approx 1.414$ se multiply ho jaata hai.

(b) Density form $\lambda =1/(\sqrt{2}n\pi d^2)$ use karo. Constant volume par molecules ki sankhya aur $V$ unchanged hain, isliye $n$ unchanged hai, isliye $\lambda$ bilkul change nahi hota (factor $1$).

Surprise kyun? Pressure form $\lambda =k_{B}T/(\sqrt{2}\pi d^{2}P)$ dekhne mein $\lambda \propto T$ jaisi lagti hai — lekin constant $V$ par, heating bhi $P$ ko double karti hai (kyunki $P\propto T$ fixed $n$ par). Upar ka $T$ aur neeche ka $P$ cancel ho jaate hain. $\lambda$ actually crowding ($n$) ke baare mein hai, temperature ke nahi.

Same $T$ aur $P$ par, dono ka $n=P/k_{B}T$ same hai. Toh $\frac{{\lambda }_{He}}{{\lambda }_{Ar}}=\frac{d_{Ar}^2}{d_{He}^2}={\left(\frac{3.6}{2.1}\right)}^2=(1.714{)}^2\approx 2.94$ Helium ka $\lambda$ lagbhag 2.9× lamba hai. Kyun: jitna chota molecule ka diameter, utna chota uska collision cross-section $\sigma =\pi d^2$, toh woh hits ke beech bahut aage nikal jaata hai. Mass kabhi $\lambda$ mein enter nahi hoti — sirf size aur crowding.

(a) $\bar{v}=\sqrt{\frac{8RT}{\pi M}}=\sqrt{\frac{8(8.314)(273)}{\pi (0.028)}}=\sqrt{2.063\times 10^5}\approx 454\text{m/s}$.

(b) $\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}=\frac{(1.38\times 10^{-23})(273)}{\sqrt{2}\pi (3.7\times 10^{-10}{)}^2(1.01\times 10^5)}$. Numerator $=3.767\times 10^{-21}$. Denominator $=\sqrt{2}(3.1416)(1.369\times 10^{-19})(1.01\times 10^5)=6.143\times 10^{-14}$. $\lambda \approx 6.13\times 10^{-8}\text{m}\approx 61\text{nm}$

(c) $Z$ nikalne ka sabse saaf, drop-free tarika yeh hai ki ek molecule actually per second kitni distance travel karta hai use per collision ke beech ki average distance se divide karo: $Z=\frac{\text{distance travelled per second}}{\text{distance per collision}}=\frac{\bar{v}}{\lambda }$ Yahan plain $\bar{v}$ kyun, koi $\sqrt{2}$ nahi? $\sqrt{2}$ already $\lambda$ mein baked in hai (usne free path ko chota kiya kyunki targets move karte hain). Agar tum top ko bhi $\sqrt{2}$ se multiply karo toh tum usi effect ko double-count kar rahe ho. Doosre tarike se: collision rate hai (relative speed)$\times n\sigma =\sqrt{2}\bar{v}n\sigma$, aur $\lambda =1/(\sqrt{2}n\sigma )$ toh $1/\lambda =\sqrt{2}n\sigma$; $Z={\bar{v}}_{rel}n\sigma =\sqrt{2}\bar{v}n\sigma$ form karne se equals $\bar{v}/\lambda$ — do $\sqrt{2}$ cancel karke $\bar{v}/\lambda$ reh jaata hai. $Z=\frac{454}{6.13\times 10^{-8}}\approx 7.4\times 10^9{\text{s}}^{-1}$ Ek molecule lagbhag 7 billion baar per second collide karta hai — swarm itna busy hota hai. Collisions ke beech average time $1/Z\approx 1.35\times 10^{-10}$ s hai.

Effusion rate $\propto \bar{v}\propto \sqrt{T}$. Toh $\frac{{\text{rate}}_2}{{\text{rate}}_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{4}=2$ Rate double ho jaata hai. Kyun chaar guna nahi? Speed $\sqrt{T}$ ke saath badhti hai, $T$ ke saath nahi, kyunki kinetic energy $\frac{1}{2}m{\bar{v}}^2\propto T$ ⇒ $\bar{v}\propto \sqrt{T}$. Dekho Diffusion and effusion.

$d$ ke liye boxed formula ko invert karo: $\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}P}\Rightarrow d^2=\frac{k_{B}T}{\sqrt{2}\pi P\lambda }$ $d^2=\frac{(1.38\times 10^{-23})(300)}{\sqrt{2}\pi (1.0\times 10^5)(7.0\times 10^{-8})}=\frac{4.14\times 10^{-21}}{3.111\times 10^{-2}}=1.331\times 10^{-19}{\text{m}}^2$ $d=\sqrt{1.331\times 10^{-19}}\approx 3.65\times 10^{-10}\text{m}=0.365\text{nm}$ Yeh kyun matter karta hai: ek bulk quantity measure karna ($\lambda$, viscosity ya diffusion experiments se) humein invisible atoms ko weigh aur size karne deta hai — kinetic theory ki ek mahaan triumph. Dekho Viscosity and thermal conductivity of gases.

(a) $v_{rms}=\sqrt{3k_{B}T/m}$ se, dono sides square karo, phir per-molecule (single-particle) mass $m$ ke liye solve karo: $m=\frac{3k_{B}T}{v_{rms}^2}=\frac{3(1.38\times 10^{-23})(273)}{485^2}=\frac{1.1305\times 10^{-20}}{2.3522\times 10^5}=4.806\times 10^{-26}\text{kg}$

(b) Gas identify karne ke liye tumhe molar mass chahiye, jo per-molecule mass times Avogadro's number hai: $M=m{N}_A=(4.806\times 10^{-26})(6.022\times 10^{23})=0.02894\text{kg/mol}\approx 28.9\text{g/mol}$. Yeh essentially air / N₂ hai ($\approx 28$–$29$ g/mol).

(c) L5.1 jaisa hi inversion: $d^2=\frac{k_{B}T}{\sqrt{2}\pi P\lambda }=\frac{(1.38\times 10^{-23})(273)}{\sqrt{2}\pi (1.01\times 10^5)(9.0\times 10^{-8})}=\frac{3.767\times 10^{-21}}{4.038\times 10^{-2}}=9.33\times 10^{-20}{\text{m}}^2$ $d\approx 3.05\times 10^{-10}\text{m}\approx 0.31\text{nm}$ Moral: ek measured speed aur ek measured length se, kinetic theory tumhe us atom ki mass, identity, aur size de deta hai jo tum kabhi dekh nahi sakte.

$v_{rms}$ diya hai, per-molecule mass $m$ nikalo (molar mass NAHI) ::: $m=3k_{B}T/v_{rms}^2$; phir molar mass $M=m{N}_A$ $\lambda$, $T$, $P$ diya hai, $d$ nikalo ::: $d=\sqrt{k_{B}T/(\sqrt{2}\pi P\lambda )}$ Universal ratio $v_{rms}/\bar{v}$ ::: $\sqrt{3\pi /8}\approx 1.085$ Collision frequency ::: $Z=\bar{v}/\lambda$ ($\sqrt{2}$ already $\lambda$ ke andar hai)