Kinetic theory — pressure derivation, temperature as mean KE: Worked examples

1.7.9 · D3 · Physics › Thermodynamics › Kinetic theory — pressure derivation, temperature as mean KE

Yeh sab kuch parent note ke teen boxed facts pe tikaa hai, jo plain words mein dobara likhe gaye hain taaki kuch bhi assume na ho:

Ideal Gas Law PV=nRT dekho macroscopic law ke liye jisse hum compare karte rehte hain, aur Maxwell-Boltzmann Speed Distribution dekho jahan $v_{r m s}$ , $\overset{v}{ˉ}$ aur most-probable speed alag ho jaati hain.

Scenario matrix

Neeche har cell ek type ka question hai. "Example" column batata hai kaun sa worked example prove karta hai ki tum isse handle kar sakte ho.

Cell	Kya khaas hai isme	Example
A. T → speed	Temperature diya hai, $v_{r m s}$ nikalo	Ex 1
B. speed → T	Speed diya hai, temperature nikalo	Ex 2
C. micro → P	$N, m, V, v$ diya hai, pressure nikalo	Ex 3
D. Do gases, same T	Masses ke across speeds/KE compare karo	Ex 4
E. Ratio / scaling	$T$ ya $m$ ek factor se change ho — koi numbers plug nahi	Ex 5
F. Degenerate $T \to 0$	Limiting/zero input, kya toot ta hai	Ex 6
G. Real-world word problem	Tyre / balloon style, translate karke solve karo	Ex 7
H. Exam twist	$v_{r m s}$ vs $\overset{v}{ˉ}$ trap, ya heavier≠higher-P	Ex 8

Pehle poore landscape ka figure, taaki dekh sako cells aapas mein kaise related hain:

Figure — Kinetic theory — pressure derivation, temperature as mean KE

Example 1 — Cell A: temperature se speed

Step 1. Molar form $v_{r m s} = 3 R T / M$ choose karo. Yeh step kyun? Humein $M$ (per mole) diya gaya hai, $m$ (per molecule) nahi. Jis form mein $M$ already hai usse use karne se $N_{A}$ se divide karna bachta hai.

Step 2. Plug in karo: $v_{r m s} = \frac{3 ( 8.314 ) ( 300 )}{0.032} = \frac{7482.6}{0.032} = 233831$

Step 3. Root lo: $v_{r m s} \approx 483.6 m/s$ Yeh step kyun? Square root "mean-square" ko undo karta hai; yeh hume kuch aisa deta hai jiska unit speed hai.

Verify: Nitrogen ki $517 m/s$ se thoda neeche, kyunki oxygen bhaari hai ( $32$ vs $28$ ), aur $v \propto 1/ M$ toh ratio hona chahiye $28/32 = 0.935$ , jo deta hai $517 \times 0.935 \approx 483$ . ✓ Units: $J/mol \div kg/mol = J/kg = m^{2} / s^{2} = m/s$ . ✓

Example 2 — Cell B: speed se wapas temperature

Step 1. $\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k_{B} T$ se shuru karo aur $T$ ke liye solve karo: $T = \frac{m v ^{2}}{3 k _{B}} = \frac{m v _{r m s}^{2}}{3 k _{B}}$ Yeh step kyun? Hum mean-KE relation ko invert kar rahe hain kyunki hume speed pata hai aur temperature chahiye — Example 1 ka bilkul ulta.

Step 2. Numbers plug karo ( $v_{r m s}^{2} = 137 0^{2} = 1.8769 \times 1 0^{6}$ ): $T = \frac{( 6.64 \times 1 0 ^{- 27} ) ( 1.8769 \times 1 0 ^{6} )}{3 ( 1.38 \times 1 0 ^{- 23} )}$

Step 3. Upar aur neeche compute karo: $T = \frac{1.2463 \times 1 0 ^{- 20}}{4.14 \times 1 0 ^{- 23}} \approx 301 K$

Verify: Lagbhag room temperature — $300 K$ par helium ke consistent. Reverse-check: $301 K$ par, $v_{r m s} = 3 (1.38 \times 1 0^{- 23}) (301) /6.64 \times 1 0^{- 27} \approx 1370 m/s$ . ✓

Example 3 — Cell C: microscopic numbers se pressure

Step 1. $P = \frac{1}{3} \frac{N m v _{r m s}^{2}}{V}$ use karo (kyunki $\overline{v^{2}} = v_{r m s}^{2}$ ). Yeh step kyun? Yeh direct microscopic pressure formula hai — isme har quantity di gayi hai.

Step 2. Numerator ke pieces: $N m = (3.0 \times 1 0^{24}) (5.3 \times 1 0^{- 26}) = 0.159 kg$ , aur $v_{r m s}^{2} = 48 0^{2} = 230400 m^{2} / s^{2}$ .

Step 3. Combine karo: $P = \frac{1}{3} \cdot \frac{( 0.159 ) ( 230 400 )}{0.010} = \frac{1}{3} \cdot \frac{36633.6}{0.010} = \frac{1}{3} (3.66336 \times 1 0^{6})$ $P \approx 1.22 \times 1 0^{6} Pa \approx 12 atm$

Verify: Units: $\frac{kg \cdot m ^{2} / s ^{2}}{m ^{3}} = \frac{N \cdot m}{m ^{3}} = \frac{N}{m ^{2}} = Pa$ . ✓ ( $1 atm = 1.013 \times 1 0^{5} Pa$ , toh $\approx 12 atm$ — ek dense box ke liye reasonable.)

Example 4 — Cell D: ek temperature par do gases

Step 1 (a). $\overline{E_{k}} = \frac{3}{2} k_{B} T$ sirf $T$ par depend karta hai, mass par nahi: $\frac{E _{k, H_{2}}}{E _{k, O_{2}}} = \frac{\frac{3}{2} k _{B} T}{\frac{3}{2} k _{B} T} = 1$ Yeh step kyun? Yahi "temperature = mean KE" ka gehra content hai — ek shared $T$ par, har species identical average translational energy carry karta hai (dekho Boltzmann Constant and Equipartition).

Step 2 (b). Equal KE lekin different mass, iska matlab different speed hoga. Kyunki $\frac{1}{2} m v_{r m s}^{2}$ equal hai: $\frac{v _{r m s, H_{2}}}{v _{r m s, O_{2}}} = \frac{M _{2}}{M _{1}} = \frac{0.032}{0.002} = 16 = 4$ Yeh step kyun? Energy fix karke mass vary karne par speed $1/ M$ ke scale hoti hai; halka gas faster hota hai.

Verify: Hydrogen $16 \times$ halka hai → $16 = 4 \times$ faster, energies equal. Cross-check energy: $\frac{1}{2} M_{1} v_{1}^{2}$ vs $\frac{1}{2} M_{2} v_{2}^{2}$ : $(0.002) (4^{2}) = 0.032 = (0.032) (1^{2})$ . ✓

Example 5 — Cell E: numbers plug kiye bina scaling

Step 1 (a) pressure. Fixed $N, V$ par: $P V = N k_{B} T \Rightarrow P \propto T$ . Toh $P \to 4 P$ (factor $4$ ). Yeh step kyun? Rigid box = constant $V$ ; sealed = constant $N$ . Sirf $T$ move kar raha hai, aur $P$ use linearly track karta hai.

Step 2 (b) rms speed. $v_{r m s} = 3 k_{B} T / m \propto T$ . Toh $v_{r m s} \to 4 v_{r m s} = 2 v_{r m s}$ (factor $2$ ). Yeh step kyun? Speed $T$ ke square root ke andar hai, isliye $T$ mein $\times 4$ sirf speed mein $\times 2$ deta hai.

Step 3 (c) mean KE. $\overline{E_{k}} = \frac{3}{2} k_{B} T \propto T$ . Toh $\overline{E_{k}} \to 4 \overline{E_{k}}$ (factor $4$ ).

Verify: Consistency: KE $\propto v_{r m s}^{2}$ , aur wakai $(2)^{2} = 4$ . ✓ Pressure aur energy $T$ ki tarah scale karte hain (factor 4); speed $T$ ki tarah (factor 2). Koi number kabhi aaya hi nahi — pure scaling.

Example 6 — Cell F: degenerate limit $T \to 0$

Step 1. Jab $T \to 0$ : $v_{r m s} = 3 k_{B} T / m \to 0 = 0$ , aur $\overline{E_{k}} = \frac{3}{2} k_{B} T \to 0$ , aur $P = \frac{1}{3} ρ \overline{v^{2}} \to 0$ . Yeh step kyun? Har quantity mein $T$ ka factor hai (directly, ya root ke neeche), toh sab absolute zero par saath milkar vanish ho jaate hain.

Step 2. Limit ki sanity: zero speed matlab molecules frozen still, zero pressure matlab wall hits nahi — internally consistent (dekho Internal Energy of Ideal Gas).

Step 3. Yeh toot ta kahan hai: ideal model assume karta hai point particles sirf translational motion ke saath aur koi quantum effects nahi. Real gases $0 K$ se pehle liquefy/solidify ho jaate hain, aur quantum "zero-point" motion ek molecule ko truly rest mein hone se rokta hai. Yeh step kyun? Ek achha scientist validity ka domain batata hai: yeh ek degenerate input hai jahan formula ab bhi clean $0$ return karta hai, lekin physics ideal-gas territory chod chuki hai.

Verify: Exactly $T = 0$ par plug karne par literally $v_{r m s} = 0$ , $\overline{E_{k}} = 0$ , $P = 0$ milta hai — ek self-consistent (agar unphysical bhi) prediction. ✓

Example 7 — Cell G: ek real-world word problem

Step 1. Words ko translate karo. "Rigid, no leaks" $\Rightarrow V$ aur $N$ constant. Toh $P V = N k_{B} T$ reduce ho jaata hai $P \propto T$ mein, matlab $\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}$ . Yeh step kyun? Yeh identify karna ki kya constant hai, ek paragraph ko ek proportionality mein badal deta hai — microscopic reason: zyada garm molecules wall se zyada baar aur zyada force se takraate hain, aur $P / T$ constant rehta hai.

Step 2. $P_{2}$ ke liye solve karo: $P_{2} = P_{1} \cdot \frac{T _{2}}{T _{1}} = 2.20 \times 1 0^{5} \cdot \frac{330}{290}$

Step 3. Compute karo: $P_{2} = 2.20 \times 1 0^{5} \times 1.1379 \approx 2.50 \times 1 0^{5} Pa$

Verify: Rise of $\frac{330 - 290}{290} = 13.8%$ ; pressure $\frac{2.50 - 2.20}{2.20} = 13.6%$ badha (rounding). ✓ Isliye tyre pressure cold check karna chahiye.

Example 8 — Cell H: exam trap ( $v_{r m s}$ vs $\overset{v}{ˉ}$ , aur heavier ≠ more pressure)

Step 1 (a). Ordinary average — jodo aur divide karo: $\overset{v}{ˉ} = \frac{200 + 400 + 400 + 600 + 800}{5} = \frac{2400}{5} = 480 m/s$ Yeh step kyun? $\overset{v}{ˉ}$ plain arithmetic mean hai — average flight distance jaise cheez ke liye useful hai, pressure ke liye nahi.

Step 2 (b). rms — square karo, average karo, root lo: $\overline{v^{2}} = \frac{20 0 ^{2} + 40 0 ^{2} + 40 0 ^{2} + 60 0 ^{2} + 80 0 ^{2}}{5} = \frac{40000 + 160000 + 160000 + 360000 + 640000}{5}$ $= \frac{1 360 000}{5} = 272000 \Rightarrow v_{r m s} = 272000 \approx 521.5 m/s$ Yeh step kyun? Pressure aaya tha momentum-transfer $\propto v_{x}^{2}$ se, toh mean of the squares physically relevant average hai.

Step 3 (c). $P = \frac{1}{3} ρ \overline{v^{2}}$ mein $v_{r m s}$ use karo. Aur $v_{r m s} \geq \overset{v}{ˉ}$ hamesha hota hai, kyunki squaring fast molecules ko extra weight deta hai; equality sirf tab jab har molecule ki speed identical ho. Yeh step kyun? Yahi classic trap hai jiske baare mein parent note ne warning di thi — do "averages" alag jobs ke liye alag tools hain. Bonus trap: equal $T$ par ek bhaari gas itni slower hoti hai ki pressure unchanged rehta hai, kyunki $\frac{1}{2} m \overline{v^{2}} = \frac{3}{2} k_{B} T$ mein final $P = N k_{B} T / V$ mein koi mass nahi hai.

Verify: $v_{r m s} = 521.5 > \overset{v}{ˉ} = 480$ , jaisa required tha. ✓ Gap isliye hai kyunki speeds ka spread nonzero hai.

Active recall

Recall Answers cover karo aur self-test karo

Pressure formula mein kaun si speed jaati hai, $\overset{v}{ˉ}$ ya $v_{r m s}$ ? ::: $v_{r m s}$ , kyunki pressure $\overline{v^{2}}$ (mean of squares) se aata hai, speeds ke mean se nahi. Same temperature par, do gases ke liye kya bada hoga mean KE — ya equal hoga? ::: Equal; $\overline{E_{k}} = \frac{3}{2} k_{B} T$ sirf $T$ par depend karta hai. Agar $T$ fixed $V, N$ par quadruple ho jaaye, toh $P$ aur $v_{r m s}$ ka kya hoga? ::: $P \to 4 P$ ( $T$ mein linear); $v_{r m s} \to 2 v_{r m s}$ ( $T$ ka square root). Constant $V, N$ par, kaun sa ratio constant hota hai? ::: $P / T$ constant hota hai. $T = 0$ par $v_{r m s} = 0$ model ki limit kyun signal karta hai, fact kyun nahi? ::: Real gases condense ho jaati hain aur quantum zero-point motion true rest forbid karta hai; ideal formula ab bhi $0$ return karta hai lekin physics ab apply nahi hoti.

Parent topic par wapas jao · related: Degrees of Freedom and Molar Heat Capacity, Pressure as Force per Area, Elastic Collisions and Momentum.