Maxwell-Boltzmann distribution of speeds — most probable, mean, rms

2.4.6 · Chemistry › States of Matter (Quantitative)

1. Distribution ko scratch se banana

HOW hum ise derive karte hain (Feynman-style, do ingredients):

Ingredient 1 — Boltzmann factor. Probability ki ek molecule ki kinetic energy $ε = \frac{1}{2} m v^{2}$ ho, woh $e^{- ε / k T} = e^{- m v^{2} /2 k T}$ ki tarah fall off karti hai. WHY: temperature $T$ pe kam molecules afford kar paate hain large energy.

Ingredient 2 — "Velocity shell" ka volume. Velocity 3D hai: $(v_{x}, v_{y}, v_{z})$ . Sab velocity vectors jinki speed $v$ ke paas hai, woh radius $v$ aur thickness $d v$ wale ek spherical shell pe rehte hain. Uska volume $\propto 4 π v^{2} d v$ hai. WHY $v^{2}$ : bade spheres ki zyada surface hoti hai, isliye zyada ways hain large speed rakhne ke liye — yahi cheez peak ko $v = 0$ se door push karti hai.

Dono ko multiply karo aur ek normalizing constant $A$ add karo: $f (v) = A kitne ways 4 π v^{2} kitni likelihood e^{- m v^{2} /2 k T}$

$A$ nikalna (normalization). Require karo $\int_{0}^{\infty} f (v) d v = 1$ . Gaussian integral $\int_{0}^{\infty} v^{2} e^{- α v^{2}} d v = \frac{1}{4} π / α^{3}$ use karke, $α = m /2 k T$ ke saath, milta hai $A = (\frac{m}{2 π k T})^{3/2}$ .

Figure — Maxwell-Boltzmann distribution of speeds — most probable, mean, rms

2. Teen characteristic speeds — har ek derived

Maano $α = \frac{m}{2 k T}$ toh $f (v) \propto v^{2} e^{- α v^{2}}$ . Note karo $\frac{1}{α} = \frac{2 k T}{m}$ .

(a) Most probable speed $v_{m p}$ — PEAK

WHAT: woh speed jis par $f (v)$ maximum ho (zyattar molecules yahan ke paas). HOW: $\frac{df}{d v} = 0$ set karo. $\frac{d}{d v} (v^{2} e^{- α v^{2}}) = (2 v - 2 α v^{3}) e^{- α v^{2}} = 0$ Yeh step kyun? Product rule; exponential kabhi zero nahi hota, toh bracket zero hona chahiye. $2 v (1 - α v^{2}) = 0 \Rightarrow v^{2} = \frac{1}{α} = \frac{2 k T}{m}$ $v_{m p} = \frac{2 k T}{m} = \frac{2 R T}{M}$

(b) Mean (average) speed $\overset{v}{ˉ}$ — BALANCE point

WHAT: speed ka average, $\overset{v}{ˉ} = \int_{0}^{\infty} v f (v) d v$ . HOW: chahiye $\int_{0}^{\infty} v^{3} e^{- α v^{2}} d v = \frac{1}{2 α ^{2}}$ . $\overset{v}{ˉ} = 4 π (\frac{α}{π})^{3/2} \cdot \frac{1}{2 α ^{2}} = \frac{2}{π α}$ Yeh step kyun? Ab substitute karo $\frac{1}{α} = \frac{2 k T}{m}$ , toh $\frac{1}{π α} = \frac{2 k T}{π m}$ aur $\frac{2}{π α} = 2 \frac{2 k T}{π m} = \frac{4 \cdot 2 k T}{π m} = \frac{8 k T}{π m}$ (factor $2$ root ke andar $4$ ban jaata hai, deta hai $8$ , nahi $4$ ). $\overset{v}{ˉ} = \frac{8 k T}{π m} = \frac{8 R T}{π M}$

(c) Root-mean-square speed $v_{r m s}$ — ENERGY average

WHAT: $v_{r m s} = \overline{v^{2}}$ , jahan $\overline{v^{2}} = \int_{0}^{\infty} v^{2} f (v) d v$ . Yeh wala kinetic energy se directly connected hai: $\overline{K E} = \frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k T$ . HOW: chahiye $\int_{0}^{\infty} v^{4} e^{- α v^{2}} d v = \frac{3}{8} \frac{π}{α ^{5}}$ . $\overline{v^{2}} = 4 π (\frac{α}{π})^{3/2} \cdot \frac{3}{8} \frac{π}{α ^{5}} = \frac{3}{2 α} = \frac{3 k T}{m}$ $v_{r m s} = \frac{3 k T}{m} = \frac{3 R T}{M}$

3. Temperature aur mass ka effect

4. Worked examples

5. Common mistakes (Steel-man + fix)

6. Flashcards

f (v) d v

physically kya represent karta hai?

Gas molecules ka woh fraction jinki speed

v

aur

v + d v

ke beech hai.

f (v)

mein extra factor

v^{2}

(sirf Boltzmann factor nahi) kyun aata hai?

4 π v^{2} d v

velocity-space spherical shell ka volume hai — large speed rakhne ke zyada ways hain, isliye peak zero se door ho jaati hai.

Most probable speed

v_{m p}

ka formula?

v_{m p} = 2 k T / m = 2 R T / M

(

df / d v = 0

se milta hai).

Mean speed

\overset{v}{ˉ}

ka formula?

\overset{v}{ˉ} = 8 k T / π m = 8 R T / π M

RMS speed

v_{r m s}

ka formula?

v_{r m s} = 3 k T / m = 3 R T / M

Fixed ratio

v_{m p} : \overset{v}{ˉ} : v_{r m s}

2 : 8/ π : 3 = 1 : 1.128 : 1.225

Sabse badi speed kaun si hai aur kyun?

v_{r m s}

; squaring fast tail ko over-weight karta hai.

Kaun si speed directly average KE se related hai?

v_{r m s}

, kyunki

\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k T

T

badhane ka curve pe kya effect hota hai?

Peak daayein shift hoti hai, curve납작 aur wide ho jaata hai; area 1 rehta hai.

Fixed

T

pe bada molar mass

M

ka kya effect hota hai?

Peak baayein shift hoti hai, curve tall aur narrow ho jaata hai (slow molecules).

O₂ ka

v_{r m s}

300 K pe?

≈ 484 m/s.

Same

T

pe do gases ki rms speeds ka ratio?

M_{2} / M_{1}

(halka wala faster hai).

Recall Feynman: ek 12-saal ke bachche ko explain karo

Socho ek box mein bahut saare bouncy balls hain, sab ek doosre se takra rahe hain. Sab same speed pe nahi chalte — kuch renge, zyattar medium jaate hain, kuch bahut fast zoom karte hain. Agar tum "kitne balls is speed pe jaate hain" ka chart banao, toh woh ek hill jaisa dikhega jo slow side pe steep hai aur fast side pe lambi slope hai. Hill ki top = woh speed jis par zyattar balls hain. Average speed thodi daayein hai (fast zoomers ise kheenchte hain). Aur agar tumhe energy ki chinta ho (jo speed² ke saath badhti hai), toh fast wale aur bhi zyada matter karte hain, toh "energy speed" sabse daayein hoti hai. Box garm karo → sab balls speed up ho jaate hain (hill daayein slide karti hai). Bhaari balls use karo → woh aalsi aur slow hain (hill baayein slide karti hai).