Kinetic molecular theory — derivation of P = (1 - 3)ρv²_rms

2.4.5 · Chemistry › States of Matter (Quantitative)

Assumptions (KYA use karne ki permission hai)

YEH KYUN MATTER KARTA HAI: elastic collisions se hum velocity ko cleanly reverse kar sakte hain; koi forces nahi matlab molecules hits ke beech straight lines mein travel karti hain; negligible size matlab woh ek doosre ko block nahi karte. Yeh simplifications maths ko exact banate hain ek ideal gas ke liye.

Problem setup (KAISE shuru karte hain)

$N$ molecules lo, har ek ka mass $m$ hai, ek cube mein jiska side $L$ hai (toh volume $V = L^{3}$ ).

Figure — Kinetic molecular theory — derivation of P = (1 - 3)ρv²_rms

Ek molecule lo jiske velocity components $v_{x}, v_{y}, v_{z}$ hain. Hum pehle woh force nikalenge jo yeh ek wall pe exert karta hai ( $x$ -axis ke perpendicular wall), phir generalise karenge.

Step 1 — Ek collision mein momentum change

Molecule right wall se takrata hai $x$ -velocity $+ v_{x}$ ke saath aur $- v_{x}$ ke saath wapas bounce karta hai (elastic, sirf $x$ -component reverse hoti hai).

$Δ p = m (- v_{x}) - m (v_{x}) = - 2 m v_{x}$

Yeh step kyun? Sirf $x$ -component matter karta hai ek wall ke liye jo $x$ -direction face kar rahi hai — $y, z$ motion wall ke along slide karti hai aur koi push deliver nahi karti. Wall ko $+ 2 m v_{x}$ milta hai (Newton's third law).

Step 2 — Yeh kitni baar hit karta hai?

Usi wall pe do hits ke beech, molecule opposite wall tak jaata hai aur wapas aata hai: distance $2 L$ at speed $v_{x}$ .

$collisions ke beech time Δ t = \frac{2 L}{v _{x}}$

Yeh step kyun? Collisions ka rate = 1 per round-trip. Zyada hits per second → zyada force.

Step 3 — Ek molecule se force

Force = momentum transfer ka rate:

$f_{x} = \frac{momentum per hit}{time per hit} = \frac{2 m v _{x}}{2 L / v _{x}} = \frac{m v _{x}^{2}}{L}$

Yeh step kyun? Force yeh hai ki kitni tezi se momentum deliver hota hai. Notice karo $v_{x}^{2}$ appear hota hai: tez molecules zyada strong hit karte hain aur zyada baar — $v_{x}$ ke do factors.

Step 4 — Saare $N$ molecules is wall pe

$F_{x} = \frac{m}{L} (v_{x 1}^{2} + v_{x 2}^{2} + \dots + v_{x N}^{2}) = \frac{m N}{L} \overline{v_{x}^{2}}$

jahan $\overline{v_{x}^{2}}$ squared $x$ -velocities ka mean hai.

Yeh step kyun? Total force sirf saare individual pushes ka sum hai; $N$ factor out karne se sum $N \times$ average ban jaata hai.

Step 5 — Us wall pe pressure

$P = \frac{F _{x}}{area} = \frac{F _{x}}{L ^{2}} = \frac{m N v _{x}^{2}}{L ^{3}} = \frac{m N v _{x}^{2}}{V}$

Step 6 — Isotropy use karo (random directions) $x$ hatane ke liye

Motion random hai, toh koi direction special nahi:

$\overline{v_{x}^{2}} = \overline{v_{y}^{2}} = \overline{v_{z}^{2}}$

Kyunki $v^{2} = v_{x}^{2} + v_{y}^{2} + v_{z}^{2}$ hai, averages lete hain:

$\overline{v^{2}} = 3 \overline{v_{x}^{2}} \Rightarrow \overline{v_{x}^{2}} = \frac{1}{3} \overline{v^{2}}$

Yeh step kyun? Yeh key trick hai. Humne sirf $x$ -wall measure kiya, lekin pressure saari walls pe same hoti hai kyunki motion ka koi preferred direction nahi hai. Toh $\overline{v_{x}^{2}}$ ko full 3-D speed se replace karte hain.

Step 7 — Final result

$P V = \frac{1}{3} N m \overline{v^{2}} ⟺ P = \frac{1}{3} ρ v_{r m s}^{2}$

jahan $ρ = \frac{N m}{V}$ density hai aur $v_{r m s} = \overline{v^{2}}$ .

Worked examples

Common mistakes

Active recall

Recall Cloze check (answer karne ke baad reveal karo)

Har wall collision mein momentum change = == $2 m v_{x}$ ==.
Usi wall pe hits ke beech time = == $2 L / v_{x}$ ==.
Ek molecule se ek wall pe force = == $m v_{x}^{2} / L$ ==.
Key isotropy relation: $\overline{v_{x}^{2}} =$ == $\frac{1}{3} \overline{v^{2}}$ ==.
Final: $P =$ == $\frac{1}{3} ρ v_{r m s}^{2}$ ==, aur $v_{r m s} =$ == $3 R T / M$ ==.

Recall Feynman: 12-saal ke bachche ko explain karo

Socho ek box hai jisme super-bouncy tennis balls har jagah udd rahi hain. Jab bhi koi ball wall se takraati hai toh wapas bounce karti hai aur wall ko ek chhota sa dhakka deti hai. Itni zyada balls itni baar hit karti hain ki woh saare chhote dhakke ek steady push jaise feel hote hain wall pe — wahi steady push pressure hai. Agar balls tez chalein (hotter box), toh woh zyada strong dhakka deti hain aur zyada baar, toh pressure badhta hai. Kyunki box 3-D hai aur balls equally har direction mein udti hain, har wall sirf ek third bouncing feel karti hai — wahin se $\frac{1}{3}$ aata hai!

Connections

Ideal Gas Equation PV = nRT — combine karo taaki $K E_{a v g} = \frac{3}{2} k_{B} T$ mile.
Maxwell-Boltzmann Speed Distribution — $v_{m p}, \overline{v}, v_{r m s}$ deta hai.
Graham's Law of Diffusion — rate $\propto v_{r m s} \propto 1/ M$ .
Average Kinetic Energy and Temperature
Real Gases and van der Waals Equation — kya toot jaata hai jab postulates fail hote hain.
Newton's Laws — Impulse and Momentum — is derivation ka physics engine.

Ek elastic collision mein molecule ka momentum change kya hota hai wall (x-facing) se?

2 m v_{x}

(velocity

+ v_{x}

- v_{x}

reverse hoti hai).

Ek molecule ke usi wall pe successive collisions ke beech time kya hota hai?

Δ t = 2 L / v_{x}

(length

2 L

ka round trip speed

v_{x}

pe).

Ek molecule dwara ek wall pe exert ki gayi force?

f = m v_{x}^{2} / L

Force mein

v_{x}

nahi

v_{x}^{2}

kyun appear karta hai?

Tez molecules zyada strong hit karte hain (

\propto v_{x}

) aur zyada baar hit karte hain (

\propto v_{x}

), do factors dete hain.

Derivation mein use ki gayi isotropy relation state karo.

\overline{v_{x}^{2}} = \overline{v_{y}^{2}} = \overline{v_{z}^{2}} = \frac{1}{3} \overline{v^{2}}

Density ke terms mein final KMT pressure equation?

P = \frac{1}{3} ρ v_{r m s}^{2}

P V

N, m, \overline{v^{2}}

ke terms mein express karo.

P V = \frac{1}{3} N m \overline{v^{2}}

P V = \frac{1}{3} N m \overline{v^{2}} = N k_{B} T

v_{r m s}

formula derive karo.

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

v_{r m s}

kyun use karte hain average speed kyun nahi?

Pressure

\overline{v^{2}}

pe depend karta hai;

v_{r m s} = \overline{v^{2}} \neq = \overline{v}

Ek common error jo pressure 2× chhoti deta hai?

Δ p = m v_{x}

use karna

2 m v_{x}

ki jagah (bounce momentum reverse karta hai yeh bhool jaana).

Same T pe, H₂ aur O₂ ke

v_{r m s}

ka ratio?

32/2 = 4

(halka = tez).

Physically factor 1/3 kahan se aata hai?

3 spatial dimensions speed ko equally share karte hain; har wall ek third feel karti hai.