Exercises — Real gas effects — dissociation, recombination

$\alpha =0$ ka matlab: zero moles tute hain, isliye har molecule abhi bhi poora hai. Mixture pure ${\text{A}}_2$ hai, koi free $\text{A}$ atoms nahi. Yeh kis taraf lean karta hai: bilkul molecules ki taraf (left side). Yeh picture low temperature ya bahut high pressure par hoti hai, jahan bonds survive karte hain.

(a) Recombination — chhote pieces ek bade molecule mein jud jaate hain (heat release hoti hai). (b) Dissociation — ek molecule fragments mein toot jaata hai (heat absorb hoti hai). (c) Recombination — do atoms ek molecule banate hain (heat release hoti hai). Rule of thumb: right side par zyada molecules = recombination; right side par zyada fragments = dissociation.

Step — energy diverted. Har dissociated mole $\Delta H_d$ consume karta hai; dissociated fraction $\alpha$ hai, isliye product ke har mole ke liye: $E_{\text{lost}}=\alpha \Delta H_d=0.15\times 6.0=0.90\text{MJ/mol}.$ Step — intended heating ka percent ke roop mein: $\frac{E_{\text{lost}}}{Q}=\frac{0.90}{4.0}=0.225=22.5\%.$ Kyun matter karta hai: almost ek-chauthaayi heat jo $T_c$ badhani chahiye thi wo broken bonds mein lock ho jaati hai — agar yeh kabhi recover na ho to flame temperature ko ek bada nuksaan hai.

Yeh tool kyun: parent deta hai $v_e\propto 1/\sqrt{M}$ fixed $T_c$ aur $\gamma$ par, isliye exhaust velocities ka ratio $1/\sqrt{M}$ ka ratio hai. $\frac{v_{e,2}}{v_{e,1}}=\frac{1/\sqrt{M_2}}{1/\sqrt{M_1}}=\sqrt{\frac{M_1}{M_2}}=\sqrt{\frac{20}{16}}=\sqrt{1.25}\approx \mathrm{1.118.}$ Result: sirf lighter species se $v_e$ mein lagbhag $+11.8\%$ increase — tug-of-war ka "good side."

Step — denominator simplify karo. Do brackets $(1-\alpha )(1+\alpha )$ difference of squares hain: multiply out karne par, cross terms $-\alpha$ aur $+\alpha$ cancel ho jaate hain, $1-{\alpha }^2$ bachta hai. Aur $(2\alpha {)}^2=4{\alpha }^2$. Isliye expression collapse hoke $K_p=\frac{4{\alpha }^2}{1-{\alpha }^2}\cdot \frac{p}{p^{\circ }}$ ho jaata hai. Step — $p=p^{\circ }$ daalo (isliye $p/p^{\circ }=1$) aur $K_p=1$: $\frac{4{\alpha }^2}{1-{\alpha }^2}=\mathrm{1.0.}$ Step — fraction clear karo: $4{\alpha }^2=1-{\alpha }^2\Rightarrow 5{\alpha }^2=1.$ Step — solve karo: ${\alpha }^2=0.2\Rightarrow \alpha =\sqrt{0.2}\approx \mathrm{0.447.}$ Reading it: is temperature aur pressure par ${\text{A}}_2$ ka lagbhag $45\%$ toot chuka hai — heavy dissociation. (Sirf positive root rakhte hain; negative $\alpha$ physically meaningless hai.)

$K_p$ kyun fixed rehta hai: $K_p$ sirf temperature par depend karta hai, $\ln K_p=-\Delta G^{\circ }/(R_{u}T)$ ke through — jahan $\Delta G^{\circ }$ reaction ka standard Gibbs score hai aur $R_u$ universal gas constant (dono crib sheet mein hain). Pressure change karo, temperature nahi → $K_p$ unchanged. Jo respond karta hai wo $\alpha$ hai. Step — pressure ke saath general form: $K_p=\frac{4{\alpha }^2}{1-{\alpha }^2}\cdot \frac{p}{p^{\circ }}.$ $p/p^{\circ }=2$ aur $K_p=1$ ke saath: $1=\frac{4{\alpha }^2}{1-{\alpha }^2}\cdot 2=\frac{8{\alpha }^2}{1-{\alpha }^2}.$ Step — solve karo: $8{\alpha }^2=1-{\alpha }^2\Rightarrow 9{\alpha }^2=1\Rightarrow \alpha =\sqrt{\frac{1}{9}}=\frac{1}{3}\approx \mathrm{0.333.}$ Reading it: dissociation $0.447$ se gir ke $0.333$ ho gayi — Le Chatelier in numbers. Gas ko squeeze karna equilibrium ko wapas whole molecules ki taraf push karta hai.

Neeche ka figure $\alpha$ ko pressure ratio $p/p^{\circ }$ ke against plot karta hai $K_p=1$ ke liye: curve girta hai jab aap right move karte ho, aur do black dots exactly wahi points mark karte hain jo humne abhi compute kiye ($p^{\circ }$ par $0.447$, $2p^{\circ }$ par $0.333$). Red curve ko downhill trace karo — woh downhill slope hi Le Chatelier hai.

Baseline yardstick: $v_e\propto \sqrt{T_c/M}=\sqrt{3600/22}$. Hum har scenario ko ratio ke roop mein compare karte hain.

Equilibrium case ($T_c=3600$, $M=18$): $\frac{v_e^{\text{eq}}}{v_e^{\text{base}}}=\sqrt{\frac{3600/18}{3600/22}}=\sqrt{\frac{22}{18}}=\sqrt{1.222}\approx \mathrm{1.106.}$ → +10.6%: energy wapas aayi aur $M$ lighter hai → pure win.

Frozen case ($T_c=3100$, $M=18$): $\frac{v_e^{\text{fr}}}{v_e^{\text{base}}}=\sqrt{\frac{3100/18}{3600/22}}=\sqrt{\frac{3100\times 22}{3600\times 18}}=\sqrt{\frac{68200}{64800}}=\sqrt{1.0525}\approx \mathrm{1.026.}$ → +2.6%: lighter gas abhi bhi jeet rahi hai, lekin lost heat almost ise cancel kar deti hai.

Sabak: dissociation automatically bura nahi hai. Even frozen mein, $M$-benefit $T_c$-loss ko edge out kar sakta hai. Equilibrium clearly best hai. Yahi exactly woh tug-of-war hai jo parent ne warn kiya tha, ab numbers ke saath.

Step — definition: $Da={\tau }_{flow}/{\tau }_{chem}=0.20/2.0=\mathrm{0.10.}$ Step — interpret karo: $Da=0.1\ll 1$. Gas nozzle ek reaction ke liye zaruri time ka ek-dashmaansh mein chhod deta hai, isliye recombination ke paas koi time nahi hota. Conclusion: frozen flow — bond energy locked up rehti hai aur lost ho jaati hai. Ise recover karne ke liye ya to longer nozzle chahiye (bada ${\tau }_{flow}$) ya faster chemistry (chhota ${\tau }_{chem}$), dono $Da$ ko upar push karte hain.

(a) Har pressure par dissociation. $K_p=\frac{4{\alpha }^2}{1-{\alpha }^2}\cdot \frac{p}{p^{\circ }}=1$ use karte hue:

• $p=p^{\circ }$ par: $4{\alpha }^2=1-{\alpha }^2\Rightarrow 5{\alpha }^2=1\Rightarrow \alpha =\sqrt{0.2}\approx \mathrm{0.447.}$
• $p=3p^{\circ }$ par: $12{\alpha }^2=1-{\alpha }^2\Rightarrow 13{\alpha }^2=1\Rightarrow \alpha =\sqrt{1/13}\approx \mathrm{0.277.}$

(b) Flame temperature (frozen, given linear cost use karke).

• $p^{\circ }$ par: $T_c=3600-500\cdot (0.447/0.447)=3600-500=3100\text{K}.$
• $3p^{\circ }$ par: $T_c=3600-500\cdot (0.277/0.447)=3600-500\cdot 0.6197=3600-309.9\approx 3290\text{K}.$

(c) Verdict. Pressure triple karne se $\alpha$ $0.447$ se $0.277$ tak cut hua aur $T_c$ $3100$ K se lagbhag $3290$ K tak raise hua. Higher pressure → less dissociation → hotter, more complete combustion, exactly jaisa Le Chatelier predict karta hai (squeeze mein fewer moles favoured hote hain). Yahi wajah hai ki real engines high chamber pressure par run karte hain.

Neeche ka figure dono trends ek saath dikhata hai: black curve $\alpha$ ko girte hue dikhati hai jab pressure $p^{\circ }$ se $3p^{\circ }$ tak badhta hai, aur red curve $T_c$ ko badhte hue same range mein — dono computed points har curve par marked hain. Dono curves ko ek saath padhne se "less dissociation → hotter chamber" chain ek nazar mein visible ho jaata hai.

Step — har ek $\sqrt{T_c/M}$ compute karo ($v_e$ ke proportional):

• $p^{\circ }$: $\sqrt{3100/18}=\sqrt{172.22}\approx \mathrm{13.12.}$
• $3p^{\circ }$: $\sqrt{3290/19.5}=\sqrt{168.72}\approx \mathrm{12.99.}$

Step — ratio: $\frac{v_e(3p^{\circ })}{v_e(p^{\circ })}=\frac{12.99}{13.12}\approx \mathrm{0.990.}$

Reading it: yahan higher-pressure case frozen $v_e$ mein lagbhag $1.0\%$ lower hai! Dissociation suppress karne se $T_c$ gain, heavier $M$ se outweigh ho gaya. Sabak: frozen flow mein $v_e$ ke liye high $p_c$ koi free lunch nahi hai — yeh ek hotter chamber ko heavier gas ke saath trade karta hai. High $p_c$ ka asli faayda equilibrium recovery aur higher mass flow ke through aata hai, is simple frozen ratio se nahi. Mastery ka matlab hai yeh dekhna ki koi single lever akele nahi jeetta.