Kinetic theory — pressure derivation, temperature as mean KE: Exercises

1.7.9 · D4 · Physics › Thermodynamics › Kinetic theory — pressure derivation, temperature as mean KE

Poore note mein yahi constants use hote hain:

Core toolbox, saari cheezein parent note mein build ki gayi hain:

Level 1 — Recognition

Goal: sahi formula dhundho aur plug in karo. Abhi koi trap nahi.

L1.1 — Pressure ke liye kaun si speed?

Ek container mein $N$ molecules hain, har ek ka mass $m$ hai, volume $V$ mein. Tumhe bataya gaya hai $\overline{v^{2}} = 2.4 \times 1 0^{5} m^{2} / s^{2}$ , saath mein $N = 3.0 \times 1 0^{23}$ , $m = 4.7 \times 1 0^{- 26}$ kg, $V = 2.0 \times 1 0^{- 3} m^{3}$ . Pressure nikalo.

Recall Solution

WHAT: seedha $P = \frac{1}{3} N m \overline{v^{2}} / V$ ka use karo. WHY: hume $\overline{v^{2}}$ directly diya gaya hai, isliye koi square-rooting ki zaroorat nahi. $P = \frac{1}{3} \cdot \frac{( 3.0 \times 1 0 ^{23} ) ( 4.7 \times 1 0 ^{- 26} ) ( 2.4 \times 1 0 ^{5} )}{2.0 \times 1 0 ^{- 3}}$ Andar ka numerator: $(3.0 \times 1 0^{23}) (4.7 \times 1 0^{- 26}) = 1.41 \times 1 0^{- 2}$ . Times $2.4 \times 1 0^{5}$ gives $3.384 \times 1 0^{3}$ . $P = \frac{1}{3} \cdot \frac{3.384 \times 1 0 ^{3}}{2.0 \times 1 0 ^{- 3}} = \frac{1}{3} (1.692 \times 1 0^{6}) \approx 5.64 \times 1 0^{5} Pa$

L1.2 — Temperature se mean KE nikalna

$T = 350$ K par ek molecule ki average translational kinetic energy nikalo.

Recall Solution

WHAT: $\overline{E_{k}} = \frac{3}{2} k_{B} T$ use karo. WHY: mean KE sirf temperature par depend karti hai, is baat par nahi ki gas kaun si hai. $\overline{E_{k}} = \frac{3}{2} (1.38 \times 1 0^{- 23}) (350) \approx 7.25 \times 1 0^{- 21} J$

Level 2 — Application

Goal: do steps chain karo, units convert karo, molar vs per-molecule form choose karo.

L2.1 — Oxygen ka rms speed

Oxygen gas $O_{2}$ ( $M = 32 g/mol = 0.032 kg/mol$ ) $T = 400$ K par rakha hai. $v_{r m s}$ nikalo.

Recall Solution

WHAT: molar form $v_{r m s} = 3 R T / M$ . WHY: hume single-molecule mass $m$ nahi, molar mass $M$ pata hai, isliye $R$ aur $M$ waala form use karo. $v_{r m s} = \frac{3 ( 8.314 ) ( 400 )}{0.032} = \frac{9976.8}{0.032} = 3.118 \times 1 0^{5} \approx 558 m/s$

L2.2 — rms speed se waapis temperature nikalna

Helium ( $M = 0.004 kg/mol$ ) ka $v_{r m s} = 1360$ m/s hai. Uska temperature kya hai?

Recall Solution

WHAT: $v_{r m s} = 3 R T / M$ ko invert karke $T$ nikalo. WHY: root hatane ke liye dono sides square karo, phir $T$ isolate karo. $v_{r m s}^{2} = \frac{3 R T}{M} \Rightarrow T = \frac{M v _{r m s}^{2}}{3 R}$ $T = \frac{( 0.004 ) ( 1360 ) ^{2}}{3 ( 8.314 )} = \frac{( 0.004 ) ( 1.8496 \times 1 0 ^{6} )}{24.942} = \frac{7398.4}{24.942} \approx 297 K$

Level 3 — Analysis

Goal: reason karo ki ek quantity kaise respond karti hai jab doosri badle — ratios, sirf numbers nahi.

L3.1 — Temperature double karna

Ek gas ko constant volume par $300$ K se $600$ K tak garam kiya jaata hai. Kitne factor se (a) $v_{r m s}$ badlega, (b) pressure badlega, (c) mean KE badlegi?

Recall Solution

WHAT: full numbers ki jagah proportionalities use karo. WHY: ratios mein saare constants cancel ho jaate hain, toh hum sirf track karte hain ki har quantity $T$ ke saath kaise scale karti hai. (a) $v_{r m s} \propto T$ . $300 \to 600$ jaane par $T$ $2$ se multiply hota hai, toh $v_{r m s}$ $2 \approx 1.414$ se scale hota hai. (b) Constant $V$ aur $N$ par, $P V = N k_{B} T \Rightarrow P \propto T$ . Toh pressure double hoga (factor $2$ ). (c) $\overline{E_{k}} = \frac{3}{2} k_{B} T \propto T$ , toh yeh bhi double hogi (factor $2$ ). Dekho Maxwell-Boltzmann Speed Distribution: poora speed curve $2$ se right ki taraf stretch hota hai — neeche figure dekho.

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

L3.2 — Do gases, same temperature

Hydrogen ( $M_{H_{2}} = 0.002$ kg/mol) aur oxygen ( $M_{O_{2}} = 0.032$ kg/mol) same temperature share karte hain. Ratio $v_{r m s} (H_{2}) / v_{r m s} (O_{2})$ nikalo.

Recall Solution

WHAT: equal $T$ par do rms speeds ka ratio. WHY: equal $T$ par $3 R T$ factor identical hota hai, toh sirf masses alag hain. $\frac{v _{r m s} ( H _{2} )}{v _{r m s} ( O _{2} )} = \frac{3 R T / M _{H_{2}}}{3 R T / M _{O_{2}}} = \frac{M _{O_{2}}}{M _{H_{2}}} = \frac{0.032}{0.002} = 16 = 4$ Hydrogen molecules theek 4× tez chalti hain — phir bhi dono gases ki mean KE identical hai. Figure dekho.

Level 4 — Synthesis

Goal: pressure derivation, gas law, aur energy ko ek problem mein combine karo.

L4.1 — Pressure do tareekon se agree karni chahiye

$0.5$ mol gas $V = 0.01 m^{3}$ mein $T = 290$ K par rakhi hai. (a) $P V = n R T$ use karke pressure nikalo. (b) Agar har molecule ka mass $m = 6.6 \times 1 0^{- 26}$ kg hai, toh $v_{r m s}$ nikalo aur confirm karo ki microscopic pressure $P = \frac{1}{3} N m v_{r m s}^{2} / V$ part (a) se match karta hai.

Recall Solution

(a) Macroscopic route. $P = \frac{n R T}{V} = \frac{( 0.5 ) ( 8.314 ) ( 290 )}{0.01} = \frac{1205.5}{0.01} = 1.2055 \times 1 0^{5} Pa$ (b) Microscopic route. Pehle $N = n N_{A} = (0.5) (6.02 \times 1 0^{23}) = 3.01 \times 1 0^{23}$ . Molar mass $M = m N_{A} = (6.6 \times 1 0^{- 26}) (6.02 \times 1 0^{23}) = 0.03973 kg/mol$ . $v_{r m s} = \frac{3 R T}{M} = \frac{3 ( 8.314 ) ( 290 )}{0.03973} = \frac{7233.2}{0.03973} = 1.821 \times 1 0^{5} \approx 426.7 m/s$ Ab microscopic pressure: $P = \frac{1}{3} \frac{N m v _{r m s}^{2}}{V} = \frac{1}{3} \frac{( 3.01 \times 1 0 ^{23} ) ( 6.6 \times 1 0 ^{- 26} ) ( 426. 7 ^{2} )}{0.01}$ $(3.01 \times 1 0^{23}) (6.6 \times 1 0^{- 26}) = 0.019866$ . Times $v_{r m s}^{2} = 1.821 \times 1 0^{5}$ : $= 3617.6$ . $P = \frac{1}{3} \cdot \frac{3617.6}{0.01} = \frac{1}{3} (3.6176 \times 1 0^{5}) \approx 1.206 \times 1 0^{5} Pa ✓$ Dono routes agree karte hain — yahi kinetic theory ka poora point hai: microscopic motion experimental law ko reproduce karta hai.

L4.2 — Ek poore mole ki kinetic energy

L4.1 wali gas ke liye, saare molecules ki total translational kinetic energy nikalo.

Recall Solution

WHAT: total KE $= N \cdot \overline{E_{k}} = \frac{3}{2} N k_{B} T$ . Equivalently $\frac{3}{2} n R T$ . WHY: mole form zyada clean hai kyunki $N k_{B} = n R$ . $E_{t o t} = \frac{3}{2} n R T = \frac{3}{2} (0.5) (8.314) (290) = \frac{3}{2} (1205.5) \approx 1808 J$

Level 5 — Mastery

Goal: aisi problems jahan obvious pehla move galat ho, ya jahan koi limiting case handle karna pade.

L5.1 — rms aur mean speed alag hote hain

Chaar molecules ki speeds hain $300, 400, 500, 600 m/s$ . Mean speed $\overset{v}{ˉ}$ aur rms speed $v_{r m s}$ compute karo, aur dikhao ki $v_{r m s} > \overset{v}{ˉ}$ .

Recall Solution

WHAT: same numbers ke do alag averages. WHY: pressure aur energy ko $\overline{v^{2}}$ chahiye (phir uska root), plain mean nahi — squaring fast molecules ko zyada weight deta hai. Mean speed: $\overset{v}{ˉ} = \frac{300 + 400 + 500 + 600}{4} = \frac{1800}{4} = 450 m/s$ Squares ka mean: $\overline{v^{2}} = \frac{30 0 ^{2} + 40 0 ^{2} + 50 0 ^{2} + 60 0 ^{2}}{4} = \frac{90000 + 160000 + 250000 + 360000}{4} = \frac{860000}{4} = 215000$ $v_{r m s} = 215000 \approx 463.7 m/s$ Wakai $463.7 > 450$ . Yeh gap isliye hai kyunki squaring average karne se pehle badi speeds ko inflate kar deta hai — yeh ek real, general inequality hai, koi coincidence nahi.

L5.2 — Degenerate case: ek akela molecule

Ek akela molecule jiska mass $m = 5 \times 1 0^{- 26}$ kg hai, $x$ -axis ke saath $v_{x} = 500$ m/s par $L = 0.1$ m ke cube mein bounce kar raha hai. Right wall par woh kauna sa time-averaged force lagata hai, aur tum kaisa "pressure" assign karoge? Yeh discuss karo ki smooth-pressure picture yahan kyon toot jaati hai.

Recall Solution

WHAT: parent derivation Step 3 se single-molecule result $F_{1} = m v_{x}^{2} / L$ use karo. WHY: sirf ek molecule hone se average karne ke liye koi bheed nahi, lekin uske baar baar maarne ka time-average phir bhi well defined hai. $F_{1} = \frac{m v _{x}^{2}}{L} = \frac{( 5 \times 1 0 ^{- 26} ) ( 500 ) ^{2}}{0.1} = \frac{( 5 \times 1 0 ^{- 26} ) ( 2.5 \times 1 0 ^{5} )}{0.1} = \frac{1.25 \times 1 0 ^{- 20}}{0.1} = 1.25 \times 1 0^{- 19} N$ Wall area $L^{2} = 0.01 m^{2}$ par assign kiya gaya pressure: $P = \frac{F _{1}}{L ^{2}} = \frac{1.25 \times 1 0 ^{- 19}}{0.01} = 1.25 \times 1 0^{- 17} Pa$ Physics point: yeh "pressure" aslaan $Δ t = 2 L / v_{x} = 4 \times 1 0^{- 4}$ s ke interval par alag-alag maarne wale whacks ka time average hai. Ek real gauge essentially zero padhega sivaay individual pings ke. Parent note ka smooth, steady pressure tab hi emerge karta hai jab $N$ astronomically bada ho taaki kicks overlap hokar ek continuous push ban sakein. Ek akela molecule woh limit dikhata hai jahan macroscopic concept ghul jaata hai.

Active recall

Recall Cover-and-check summary

Pressure formula mein seedha kaun si speed jaati hai? ::: $\overline{v^{2}}$ (squares ka mean), na $v_{r m s}$ aur na $\overset{v}{ˉ}$ . Agar $T$ constant $V$ par double ho, toh $P$ aur $v_{r m s}$ ka kya hoga? ::: $P$ double hoga; $v_{r m s}$ $2$ se badhega. Equal $T$ par do gases ke rms speeds ka ratio? ::: $M_{2} / M_{1}$ (bhaari wala slow hoga). Kaun sa constant $N$ ke saath pair karta hai, aur kaun sa $n$ ke saath? ::: $k_{B}$ $N$ ke saath; $R$ $n$ ke saath; aur $R = N_{A} k_{B}$ . Kya $v_{r m s}$ kabhi $\overset{v}{ˉ}$ ke barabar hota hai? ::: Sirf tab jab saare molecules ki speed ek jaise ho.