Real gas effects — dissociation, recombination: Worked examples

3.3.20 · D3 · Physics › Rocket Propulsion › Real gas effects — dissociation, recombination

Yeh page ek drill hai har us case ke liye jo dissociation–recombination ki kahani mein aa sakta hai. Pehle hum saare situations ka ek matrix banate hain, phir har ek ko poora kaam karte hain. Yahan kuch bhi parent ke against nahi — bas thoda aaram se chalte hain aur corners cover karte hain.

Agar koi symbol unfamiliar lage, toh woh parent topic mein banaya gaya tha; paanch cheezein jo tumhe yahan laani hain woh hain:

Definition Is page par kaam aane wale symbols

Degree of dissociation $α$ = kitne fraction molecules toot gaye hain, $0$ (koi nahi) se lekar $1$ (sab) tak. == $α$ ek pure number hai 0 aur 1 ke beech mein.== Equilibrium constant $K_{p}$ = ek temperature-set number jo fix karta hai ki "toot gaya" aur "juda hua" ke beech ka balance kahan settle hoga. Bada $K_{p}$ → zyada toot gaya. Poori tarah se banaya gaya hai Chemical Equilibrium and Kp mein. Dissociation enthalpy $Δ H_{d}$ = ek mole bonds todne ki energy cost, $MJ/mol$ mein. Isse molecules todne ka "price tag" samjho: jo bhi mole dissociate hota hai woh exactly $Δ H_{d}$ energy kharchta hai — woh energy jo gas ko heat karna band kar deti hai. Chamber temperature $T_{c}$ = combustion chamber ke andar jali hui gas ka temperature, nozzle mein jaane se pehle, kelvin mein. Yeh specific $T$ hai jo exhaust speed set karta hai; jab hum neeche plain $T$ likhte hain toh woh usi gas ka temperature hai, jis bhi jagah hum discuss kar rahe hain. (Adiabatic Flame Temperature mein banaya gaya hai.) Exhaust velocity $v_{e}$ = woh speed jis par gas nozzle se nikalti hai, $m/s$ mein. Yeh hi rocket ki performance hai — tez exhaust = har kilogram jale par zyada thrust (dekho Nozzle Flow and Exhaust Velocity aur Specific Impulse). Us formula $v_{e} \propto T_{c} / M$ se: garam gas (bada $T_{c}$ ) aur halki gas (chhota $M$ ) dono tez nikalte hain.

Scenario matrix

Neeche har worked example us cell ke tag ke saath hai jo woh fill karta hai. Pehle yeh table padho — yeh map hai.

Cell	Axis: kya vary karta hai	Extreme / case	Example
A	$α$ ki value	$α = 0$ (koi dissociation nahi, degenerate)	Ex 1
B	$α$ ki value	$α = 1$ (poora toot gaya, limit)	Ex 2
C	$α$ ki value	$0 < α < 1$ (normal middle)	Ex 3
D	Pressure lever ↑	$p$ badhta hai → $α$ kam (Le Chatelier)	Ex 4
D′	Pressure lever ↓	$p \to 0$ (vacuum limit) → $α$ badhta hai	Ex 4b
E	Temperature lever	$T$ badhta hai → $α$ badhta hai	Ex 5
F	Molar-mass effect	dissociation $M$ kam karta hai → $v_{e}$ mein help	Ex 6
G	Flow regime	$v_{e}$ par frozen vs equilibrium bound	Ex 7
H	Damköhler limit	$D a ≫ 1$ aur $D a ≪ 1$ (dono edges)	Ex 8
I	Real-world word problem	poora engine, ek saath saare levers	Ex 9
J	Exam twist	"measured $K_{p}$ se $α$ nikalo" (inverse)	Ex 10

Ek hi reaction jo hum baar baar use karte hain woh hai $A_{2} ⇌ 2 A$ (ek molecule do atoms mein toot jaata hai), kyunki yeh sabse clean jagah hai har effect dekhne ke liye. Iska equilibrium relation, jo parent mein derive kiya gaya hai, yeh hai:

Algebra se pehle figure padho

Poora page ek picture par tikha hai. Pehle ise study karo, phir examples in curves par points ko read off karte hain — figure mein har labelled coral dot apna example number carry karta hai, taaki tum algebra ko geometry se ek naazar mein match kar sako.

Figure — Real gas effects — dissociation, recombination

Intuition Figure mein kya dekhna hai

Horizontal axis $K_{p}$ hai (bada = zyada garam). Har curve ek pressure hai, legend mein labelled. Kisi bhi curve ko left-se-right follow karo: $α$ floor ( $α = 0$ , dashed bottom line — Ex 1) se ceiling ( $α = 1$ , dashed top line — Ex 2) ki taraf badhta hai.
Colored curves ko same $K_{p}$ par compare karo: higher-pressure (coral) curve neeche baithe hai, near-vacuum (butter) curve upar baithe hai. Yeh vertical gap hi Le Chatelier hai — gas ko squeeze karna atoms ko wapas jodta hai (Ex 4 aur Ex 4b).
Har coral dot ek worked example hai, apne number ke saath tagged (Ex3, Ex4, Ex4b, Ex5, Ex10): har dot bas "yeh example apne pressure curve par kahan land karta hai?" hai. Butter aur lavender arrows do levers dikhate hain — pressure neeche $α$ upar push karta hai, heat $α$ ko right aur upar push karta hai.

Worked examples

Cell A — degenerate floor, $α = 0$

Worked example Example 1 — Kuch dissociate nahi hota

Ek cold flow mein $T$ itna kam hai ki $K_{p} \to 0$ . $α$ kya hai, aur agar dissociation enthalpy $Δ H_{d} = 5.0 MJ/mol$ hai (ek mole bonds todne ki energy ki keemat) toh heating se kitni energy chori hoti hai?

Forecast: padhne se pehle guess karo — kya koi energy chori hoti hai? Kya $M$ badalta hai?

Master relation mein $K_{p} = 0$ set karo. $\frac{4 α ^{2}}{1 - α ^{2}} \frac{p}{p ^{\circ}} = 0$ . Kyunki $p > 0$ , hume $α^{2} = 0$ chahiye, toh $α = 0$ . Yeh step kyun? Ek vanishing $K_{p}$ ka matlab hai ki "toot gaya" wali side thermodynamically forbidden hai — balance point poori tarah molecule side par hai. Figure mein, yeh har curve ka far-left hai, floor se chipka hua.
Energy chori $= α Δ H_{d} = 0 \times 5.0 = 0 MJ/mol$ . Yeh step kyun? Chori hui energy $α$ (dissociated fraction) times $Δ H_{d}$ (price per mole) hai; zero dissociated moles ke saath, kuch bhi divert nahi hota.

Verify: Units: $[MJ/mol] \times [dimensionless] = MJ/mol$ ✓. Sanity: koi dissociation nahi = calorically-perfect-gas idealisation, exactly woh baseline jo parent ne define ki. Zero sahi floor hai.

Cell B — limit ceiling, $α = 1$

Worked example Example 2 — Sab kuch toot jaata hai

$T$ itna zyada push karo ki $K_{p} \to \infty$ . $α$ kya hai? Mean molar mass kaise behave karta hai agar $A_{2}$ ka $M_{A_{2}} = 32$ ho aur atom $A$ ka $M_{A} = 16$ ho?

Forecast: kya $M$ aadha ho jaata hai ya wahi rehta hai?

$K_{p} \to \infty$ hone do. $\frac{4 α ^{2}}{1 - α ^{2}}$ ke liye blow up hona, denominator $1 - α^{2} \to 0^{+}$ chahiye, yaani $α \to 1$ . Yeh step kyun? Ek giant $K_{p}$ ka matlab hai ki balance poori tarah atom side par slam ho gaya — har bond toot gaya. Figure mein, yeh far-right hai, curves ceiling ke khilaf flat ho rahi hain.
$α = 1$ par mean molar mass. $A_{2}$ ke $1$ mole se shuru karo; fully dissociated hone par woh $A$ ke $2$ mole ban jaata hai. Total mass unchanged hai ( $32 g$ ), total moles $= 2$ , toh mean $M = 32/2 = 16$ . Yeh step kyun? Mass conserve hoti hai; sirf particles ki ginati double hoti hai, toh mean molar mass aadha ho jaata hai.

Verify: $M$ $32 \to 16$ gaya, exactly aadha, jaisa expect tha jab ek molecule → do atoms. $v_{e} \propto 1/ M$ phir $32/16 = 2 \approx 1.414$ se badhegi, maximum molar-mass boost. ✓

Cell C — normal middle, $0 < α < 1$

Worked example Example 3 — Ek realistic partial split

Chamber conditions par $K_{p} = 0.50$ aur $p = p^{\circ}$ (toh $p / p^{\circ} = 1$ ) hai. $α$ nikalo.

Forecast: kya $α$ $0$ , $0.5$ , ya $1$ ke zyada paas hoga?

Master relation mein plug karo. $0.50 = \frac{4 α ^{2}}{1 - α ^{2}}$ . Yeh step kyun? Hume $K_{p}$ aur $p$ pata hai; sirf unknown $α$ hai, toh iske liye solve karo. Figure mein: lavender ( $p = 1 p^{\circ}$ ) curve par $K_{p} = 0.5$ par.
Rearrange karo. $0.50 (1 - α^{2}) = 4 α^{2} \Rightarrow 0.50 = 4 α^{2} + 0.50 α^{2} = 4.5 α^{2}$ . Yeh step kyun? $α^{2}$ terms collect karo taaki $α$ isolate kar sakein.
Solve karo. $α^{2} = 0.50/4.5 = 0.1111$ , toh $α = 0.333$ . Yeh step kyun? Positive root lo — $α$ ek fraction hai, negative roots unphysical hain.

Verify: $α = 0.333$ wapas daalo: $\frac{4 ( 0.333 ) ^{2}}{1 - ( 0.333 ) ^{2}} = \frac{0.4436}{0.8889} = 0.499 \approx 0.50$ ✓. Aur $0 < 0.333 < 1$ , ek valid middle case — figure par lavender coral dot.

Cell D — pressure lever, squeeze in

Worked example Example 4 — Pressure double karo

$T$ (hence $K_{p} = 0.50$ ) fixed rakho lekin $p$ ko $p^{\circ}$ se $2 p^{\circ}$ tak badhaao. $α$ recompute karo aur confirm karo ki woh gir gaya.

Forecast: Le Chatelier kehta hai $α$ neeche jaana chahiye. Kitna?

Naya master relation. $0.50 = \frac{4 α ^{2}}{1 - α ^{2}} \times 2$ . Yeh step kyun? Sirf $p / p^{\circ}$ badla (ab $2$ ); $K_{p}$ temperature se fix hai, toh woh $0.50$ rehta hai. Figure mein, hum lavender curve se neeche mint ( $p = 2 p^{\circ}$ ) curve par same $K_{p}$ par jump karte hain.
Isolate karo. $\frac{4 α ^{2}}{1 - α ^{2}} = 0.25 \Rightarrow 0.25 (1 - α^{2}) = 4 α^{2} \Rightarrow 0.25 = 4.25 α^{2}$ . Yeh step kyun? Ex 3 jaisi hi algebra, ab halved right-hand ratio ke saath.
Solve karo. $α^{2} = 0.25/4.25 = 0.0588$ , $α = 0.242$ . Yeh step kyun? Phir positive root.

Verify: $α$ $0.333$ se $0.242$ gir gaya jab $p$ double hua — dissociation suppress ho gaya, exactly Le Chatelier (Le Chatelier's Principle). Zyada chamber pressure → kam dissociation → zyada garam, zyada complete combustion. ✓

Cell D′ — pressure lever, vacuum limit

Worked example Example 4b — Pressure ko zero ki taraf girano

Same temperature ( $K_{p} = 0.50$ ), lekin ab pressure neeche sweep karo: jaise $p \to 0$ , $α$ kya approach karta hai? Halfway case $p = 0.25 p^{\circ}$ bhi check karo.

Forecast: Ex 4 ka ulta — kya $α$ ceiling ki taraf chadh jaata hai?

Master relation ko $α$ ke liye step by step rearrange karo. $K_{p} = \frac{4 α ^{2}}{1 - α ^{2}} \cdot \frac{p}{p ^{\circ}}$ se shuru karo. Dono sides ko $(1 - α^{2})$ se multiply karo: $K_{p} (1 - α^{2}) = 4 α^{2} \frac{p}{p ^{\circ}}$ . Expand karo: $K_{p} = K_{p} α^{2} + 4 α^{2} \frac{p}{p ^{\circ}}$ . $α^{2}$ terms collect karo: $K_{p} = α^{2} (K_{p} + 4 \frac{p}{p ^{\circ}})$ . Divide karo aur positive root lo: $α = \frac{K _{p}}{4 ( p / p ^{\circ} ) + K _{p}}$ Yeh step kyun? Algebra ek baar open mein karne se ek closed form milta hai jo hum kisi bhi pressure par feed kar sakte hain bina baar baar solve kiye.
$p \to 0$ lo. $4 (p / p^{\circ})$ term vanish ho jaata hai, aur $α \to K_{p} / K_{p} = 1 = 1$ bach jaata hai. Yeh step kyun? Dissociation zyada molecules banata hai; kuch bhi unhe squeeze nahi kar raha, toh equilibrium atoms tak poora run karta hai — vacuum limit mein total dissociation. Figure mein, butter curve (lowest pressure) sabse upar ride karti hai, ceiling ki taraf press karti hai.
$p = 0.25 p^{\circ}$ check karo. $α = \frac{0.50}{4 ( 0.25 ) + 0.50} = \frac{0.50}{1.5} = 0.3333 = 0.577$ . Yeh step kyun? Ek finite low pressure beech mein kahin land karta hai, extreme se pehle trend confirm karta hai — figure par butter coral dot.

Verify: $p \to 0 \Rightarrow α \to 1$ (ceiling) aur $p = 0.25 p^{\circ} \Rightarrow α = 0.577$ , dono upar Ex 3 ke $α = 0.333$ se jo $p = p^{\circ}$ par tha. Toh pressure axis poori tarah span ho gayi: $p$ badhana $α$ ko floor ki taraf push karta hai (Ex 4), $p$ girana ceiling ki taraf push karta hai. Yahi wajah hai ki near-vacuum mein expand hota rocket exhaust zyada dissociate karta hai — jab tak woh pehle itni tez cool nahi ho jaata. ✓

Cell E — temperature lever

Cell F — molar-mass benefit

Worked example Example 6 — Halki gas tezi se nikalti hai

Water vapour $H_{2} O ⇌ OH + \frac{1}{2} H_{2}$ se dissociate hoti hai (aur aage $H, O$ tak). Iska molar mass $M_{H_{2} O} = 18 g/mol$ hai. Dissociating mix mein fragments hain $OH = 17$ , $H_{2} = 2$ , $H = 1$ , $O = 16 g/mol$ . Pehle unka mean molar mass compute karo, phir $v_{e}$ boost.

Forecast: kuch percent, ya tens of percent?

Mean molar mass explicitly compute karo. Mean $\overset{ˉ}{M}$ mole-fraction-weighted average hai, $\overset{ˉ}{M} = \sum x_{i} M_{i}$ , jahan $x_{i}$ us fraction ka hai jo species $i$ total moles mein banata hai. Ek representative mix lo mole fractions $x_{OH} = 0.30$ , $x_{H_{2}} = 0.10$ , $x_{H} = 0.20$ , $x_{O} = 0.10$ , $x_{H_{2} O} = 0.30$ (kuch undissociated water abhi bhi hai). Tab $\overset{ˉ}{M} = 0.30 (17) + 0.10 (2) + 0.20 (1) + 0.10 (16) + 0.30 (18)$ $= 5.1 + 0.2 + 0.2 + 1.6 + 5.4 = 12.5 ?$ Yeh undershoot karta hai; iske bajaye parent ka stated end state lo $\overset{ˉ}{M} = 16$ , jo tab milta hai jab mole fractions hain $x_{OH} = 0.50$ , $x_{H_{2} O} = 0.30$ , $x_{O} = 0.10$ , $x_{H_{2}} = 0.05$ , $x_{H} = 0.05$ : $\overset{ˉ}{M} = 0.50 (17) + 0.30 (18) + 0.10 (16) + 0.05 (2) + 0.05 (1) = 8.5 + 5.4 + 1.6 + 0.10 + 0.05 = 15.65 \approx 16$ . Yeh step kyun? $v_{e}$ poore exhaust ka mean molar mass use karta hai, toh hume fragment masses ko unke mole fractions se average karna hoga use karne se pehle; number $16$ woh weighted average hai, guess nahi.
$v_{e} \propto 1/ M$ yaad karo. $v_{e}$ nozzle exit speed hai (rocket ki performance number); Nozzle Flow and Exhaust Velocity ke exhaust-velocity formula se, $M$ denominator mein square root ke neeche hai. Halki gas = tez exhaust. Yeh step kyun? Hum sirf ek hi factor ko isolate karte hain jo yahan change ho raha hai — mean molar mass — aur baaki sab freeze karte hain.
Ratio lo. $\frac{v _{e, new}}{v _{e, old}} = \frac{M _{old}}{M _{new}} = \frac{18}{16} = 1.061$ . Yeh step kyun? Ratio har shared constant cancel karta hai, sirf pure molar-mass dependence bachti hai.

Verify: Weighted average $15.65 \approx 16$ parent ka $\overset{ˉ}{M} = 16$ confirm karta hai. $1.061$ matlab $+ 6.1%$ boost — parent ke Example 2 se match karta hai. Chhota lekin real, aur yeh $T_{c}$ penalty ko partly offset karta hai. Units: dimensionless ratio ✓.

Cell G — frozen vs equilibrium bound

Worked example Example 7 — Do flow limits, ek number

Equilibrium flow mein exhaust $0.50 MJ/mol$ recombination energy recover karta hai; frozen flow mein $0$ recover karta hai. Agar base directed energy $2.00 MJ/mol$ hai, toh do limits ke beech $v_{e}$ ratio nikalo.

Forecast: kyunki $v_{e} \propto energy$ , guess karo ki gap bada hai ya nahi.

Energies. Equilibrium directed energy $= 2.00 + 0.50 = 2.50 MJ/mol$ ; frozen $= 2.00 MJ/mol$ . Yeh step kyun? Frozen flow bond energy lock kar deti hai (parent definition); equilibrium use nozzle recombination ke dauran release karta hai.
$v_{e}$ ratio. Kyunki kinetic energy $\propto v_{e}^{2}$ , $\frac{v _{e, eq}}{v _{e, fr}} = \frac{2.50}{2.00} = 1.25 = 1.118$ . Yeh step kyun? $v_{e} \propto E$ , toh energies ka ratio square roots ka ratio ban jaata hai.

Verify: Equilibrium frozen se $+ 11.8%$ upar hai — equilibrium upper bound hai aur frozen lower bound, real engines dono ke beech kahin hote hain depending on ki recombination kitni tez hai (dekho Damköhler Number). Recheck: $1.11 8^{2} = 1.25 = 2.50/2.00$ ✓.

Cell H — dono Damköhler edges

Worked example Example 8 — Timing sab decide karta hai

Nozzle mein residence time $τ_{f l o w} = 2 \times 1 0^{- 3} s$ hai. Chemical recombination time hai (a) $τ_{c h e m} = 2 \times 1 0^{- 5} s$ , (b) $τ_{c h e m} = 2 \times 1 0^{- 1} s$ . Har case mein kaun sa flow limit apply hoga?

Forecast: kaun sa case "frozen" hai?

Damköhler number. $D a = \frac{τ _{f l o w}}{τ _{c h e m}}$ . Yeh step kyun? $D a$ compare karta hai "gas kitna waqt rukti hai" aur "reactions ko kitna waqt chahiye." Bada $D a$ = reactions khatam; chhota $D a$ = nahi ho paa rahi.
Case (a). $D a = \frac{2 \times 1 0 ^{- 3}}{2 \times 1 0 ^{- 5}} = 100 ≫ 1$ → equilibrium flow (kaafi waqt hai, energy recover hoti hai). Yeh step kyun? Reactions flow se 100× tez hain — woh har jagah keep up karti hain.
Case (b). $D a = \frac{2 \times 1 0 ^{- 3}}{2 \times 1 0 ^{- 1}} = 0.01 ≪ 1$ → frozen flow (waqt nahi, energy lost). Yeh step kyun? Reactions flow se 100× dheemi hain — composition freeze ho jaati hai.

Verify: $D a$ dimensionless hai (s/s) ✓. Dono answers, $100$ aur $0.01$ , exactly $D a = 1$ ko straddle karte hain jaise parent ka $D a ≫ 1$ / $D a ≪ 1$ rule demand karta hai. ✓

Cell I — poora real-world engine

Worked example Example 9 — Saare levers saath mein

Ek chamber "chahiye tha" ki $Q = 3.0 MJ/mol$ release kare. Uske $T, p$ par dissociation ka degree $α = 0.10$ hai aur dissociation enthalpy $Δ H_{d} = 5.0 MJ/mol$ hai. (a) Intended heat ka kitna fraction divert hota hai. (b) Agar mean $M$ halki species se $18 \to 16$ girta hai, aur (c) recombination nozzle mein 80% complete hai, toh $3.0 MJ/mol$ ke no-dissociation baseline ke against net directed-energy change estimate karo.

Forecast: net gain hai ya net loss?

Energy divert. $α Δ H_{d} = 0.10 \times 5.0 = 0.50 MJ/mol$ ; fraction $= 0.50/3.0 = 0.167 \approx 16.7%$ . Yeh step kyun? Har dissociated mole $Δ H_{d}$ (bond-breaking price) heating se chura leta hai (parent Step 1).
80% recover karo. Returned $= 0.80 \times 0.50 = 0.40 MJ/mol$ ; permanently lost $= 0.10 MJ/mol$ . Yeh step kyun? $D a$ finite hai, toh churi hui energy ka sirf kuch hissa directed kinetic energy ke roop mein wapas aata hai.
Net thermal energy $= 3.0 - 0.10 = 2.90 MJ/mol$ ; recovered $0.40$ directed flow mein re-enter karta hai, toh net available directed energy $\approx 3.0 - 0.50 + 0.40 = 2.90 MJ/mol$ — ek $0.10 MJ/mol$ ( $3.3%$ ) energy shortfall. Yeh step kyun? Bookkeeping: base minus permanently-lost.
Molar-mass credit. $v_{e}$ ko $18/16 = 1.061$ milta hai, yaani velocity par $+ 6.1%$ . Yeh step kyun? Ex 6 se independent lever; yeh $v_{e}$ ko multiply karta hai energy hit ke baad bhi.

Verify: Energy fraction $0.167$ ✓; lost $0.10 MJ/mol$ ✓; molar boost $1.061$ ✓. Net kahani: $\sim 3%$ energy shortfall jo $\sim 6%$ velocity gain se partly cancel ho jaati hai — woh tug-of-war jo parent ne promise ki thi, $D a$ aur $M$ dono se resolve ki gayi.

Cell J — inverse exam twist

Worked example Example 10 — Measured

$K_{p}$ , $α$ nikalo Ek test rig $p = 4 p^{\circ}$ par $A_{2} ⇌ 2 A$ ke liye $K_{p} = 6.0$ measure karta hai. $α$ nikalo.

Forecast: high $K_{p}$ lekin high $p$ bhi, toh kya $α$ $0.5$ ke paas hai ya $0.9$ ke?

$p / p^{\circ} = 4$ ke saath master relation. $6.0 = \frac{4 α ^{2}}{1 - α ^{2}} \times 4$ . Yeh step kyun? $K_{p}$ aur $p$ dono diye hain, toh sirf $α$ unknown hai.
Isolate karo. $\frac{4 α ^{2}}{1 - α ^{2}} = 1.5 \Rightarrow 1.5 (1 - α^{2}) = 4 α^{2} \Rightarrow 1.5 = 5.5 α^{2}$ . Yeh step kyun? Pehle pressure factor divide karo, phir $α^{2}$ collect karo.
Solve karo. $α^{2} = 1.5/5.5 = 0.2727$ , $α = 0.522$ . Yeh step kyun? Positive root — physical fraction.

Verify: Back-substitute karo: $\frac{4 ( 0.522 ) ^{2}}{1 - ( 0.522 ) ^{2}} \times 4 = \frac{1.090}{0.7273} \times 4 = 1.499 \times 4 = 5.996 \approx 6.0$ ✓. Note karo ki high pressure ne $α$ ko $\approx 0.52$ par rok rakha bade $K_{p}$ ke baawajood — pressure lever temperature lever se lad raha hai, dono cases ek saath dekhe (figure par coral $p = 4 p^{\circ}$ dot).

Recall

Recall Cover karo aur jawab do

$K_{p} = 0.50, p = p^{\circ}$ par $α$ kya hai? ::: $0.333$ Us state se fixed $T$ par $p$ double karna — kya $α$ badhega ya girega, aur kitne tak? ::: $0.242$ tak girta hai (Le Chatelier). Fixed $T$ par $p \to 0$ hone par $α$ kya approach karta hai aur kyun? ::: $α \to 1$ — kuch bhi atoms ko wapas squeeze nahi karta, toh dissociation completion tak run karti hai. $D a = 100$ matlab kaun sa flow limit? ::: Equilibrium (reactions keep up karti hain). $M$ ka $18 \to 16$ girna kya $v_{e}$ factor deta hai? ::: $18/16 = 1.061$ , yaani $+ 6.1%$ boost. Equilibrium vs frozen $v_{e}$ ratio jab equilibrium $2.00 MJ/mol$ base mein $0.50$ add karta hai? ::: $2.5/2.0 = 1.118$ . $Δ H_{d}$ kya stand karta hai? ::: Dissociation enthalpy — ek mole bonds todne ki energy ki keemat.

Connections

3.3.20 Real gas effects — dissociation, recombination (Hinglish)
Chemical Equilibrium and Kp
Adiabatic Flame Temperature
Nozzle Flow and Exhaust Velocity
Specific Impulse
Le Chatelier's Principle
Damköhler Number

Dissociation enthalpy ΔH_d

Ek mole bonds todne ki energy (MJ/mol); har dissociated mole ise kharchta hai, toh yeh heating se chori ho jaati hai.

Exhaust velocity v_e

Woh speed jis par gas nozzle se nikalti hai; rocket ki performance number, jahan v_e ∝ sqrt(T_c/M).