Exercises — Real gas effects — dissociation, recombination

What $\alpha =0$ means: zero moles have broken apart, so every molecule is still whole. The mixture is pure ${\text{A}}_2$, no free $\text{A}$ atoms. Which way it leans: entirely toward molecules (the left side). This is the picture at low temperature or very high pressure, where bonds survive.

(a) Recombination — small pieces glue into a bigger molecule (releases heat). (b) Dissociation — a molecule is torn into fragments (absorbs heat). (c) Recombination — two atoms form one molecule (releases heat). Rule of thumb: more molecules on the right = recombination; more fragments on the right = dissociation.

Step — energy diverted. Each dissociated mole eats $\Delta H_d$; the fraction dissociated is $\alpha$, so per mole of product: $E_{\text{lost}}=\alpha \Delta H_d=0.15\times 6.0=0.90\text{MJ/mol}.$ Step — as a percent of the intended heating: $\frac{E_{\text{lost}}}{Q}=\frac{0.90}{4.0}=0.225=22.5\%.$ Why it matters: nearly a quarter of the heat that should raise $T_c$ is instead locked in broken bonds — a large hit to flame temperature if it is never recovered.

Why this tool: the parent gives $v_e\propto 1/\sqrt{M}$ at fixed $T_c$ and $\gamma$, so the ratio of exhaust velocities is a ratio of $1/\sqrt{M}$. $\frac{v_{e,2}}{v_{e,1}}=\frac{1/\sqrt{M_2}}{1/\sqrt{M_1}}=\sqrt{\frac{M_1}{M_2}}=\sqrt{\frac{20}{16}}=\sqrt{1.25}\approx \mathrm{1.118.}$ Result: about a $+11.8\%$ increase in $v_e$ from lighter species alone — the "good side" of the tug-of-war.

Step — simplify the denominator. The two brackets $(1-\alpha )(1+\alpha )$ are a difference of squares: multiplying out, the cross terms $-\alpha$ and $+\alpha$ cancel, leaving $1-{\alpha }^2$. And $(2\alpha {)}^2=4{\alpha }^2$. So the expression collapses to $K_p=\frac{4{\alpha }^2}{1-{\alpha }^2}\cdot \frac{p}{p^{\circ }}.$ Step — put in $p=p^{\circ }$ (so $p/p^{\circ }=1$) and $K_p=1$: $\frac{4{\alpha }^2}{1-{\alpha }^2}=\mathrm{1.0.}$ Step — clear the fraction: $4{\alpha }^2=1-{\alpha }^2\Rightarrow 5{\alpha }^2=1.$ Step — solve: ${\alpha }^2=0.2\Rightarrow \alpha =\sqrt{0.2}\approx \mathrm{0.447.}$ Reading it: about $45\%$ of the ${\text{A}}_2$ has torn apart at this temperature and pressure — heavy dissociation. (We keep the positive root only; a negative $\alpha$ is physically meaningless.)

Why $K_p$ stays fixed: $K_p$ depends only on temperature, through $\ln K_p=-\Delta G^{\circ }/(R_{u}T)$ — where $\Delta G^{\circ }$ is the reaction's standard Gibbs score and $R_u$ the universal gas constant (both in the crib sheet). Change pressure, not temperature → $K_p$ is untouched. What responds is $\alpha$. Step — general form with pressure back in: $K_p=\frac{4{\alpha }^2}{1-{\alpha }^2}\cdot \frac{p}{p^{\circ }}.$ With $p/p^{\circ }=2$ and $K_p=1$: $1=\frac{4{\alpha }^2}{1-{\alpha }^2}\cdot 2=\frac{8{\alpha }^2}{1-{\alpha }^2}.$ Step — solve: $8{\alpha }^2=1-{\alpha }^2\Rightarrow 9{\alpha }^2=1\Rightarrow \alpha =\sqrt{\frac{1}{9}}=\frac{1}{3}\approx \mathrm{0.333.}$ Reading it: dissociation dropped from $0.447$ to $0.333$ — Le Chatelier in numbers. Squeezing the gas pushes the equilibrium back toward whole molecules.

The figure below plots $\alpha$ against the pressure ratio $p/p^{\circ }$ for $K_p=1$: the curve falls as you move right, and the two black dots mark exactly the points we just computed ($0.447$ at $p^{\circ }$, $0.333$ at $2p^{\circ }$). Trace the red curve downhill — that downhill slope is Le Chatelier.

Baseline yardstick: $v_e\propto \sqrt{T_c/M}=\sqrt{3600/22}$. We compare each scenario as a ratio.

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 came back and $M$ is lighter → 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%: the lighter gas still wins, but the lost heat nearly cancels it.

The lesson: dissociation is not automatically bad. Even frozen, the $M$-benefit can edge out the $T_c$-loss. Equilibrium is clearly best. This is exactly the tug-of-war the parent warned about, now with numbers.

Step — definition: $Da={\tau }_{flow}/{\tau }_{chem}=0.20/2.0=\mathrm{0.10.}$ Step — interpret: $Da=0.1\ll 1$. The gas leaves the nozzle in one-tenth of the time a reaction needs, so recombination has no time to happen. Conclusion: frozen flow — the bond energy stays locked up and is lost. To recover it you would need a longer nozzle (bigger ${\tau }_{flow}$) or faster chemistry (smaller ${\tau }_{chem}$), both pushing $Da$ upward.

(a) Dissociation at each pressure. Using $K_p=\frac{4{\alpha }^2}{1-{\alpha }^2}\cdot \frac{p}{p^{\circ }}=1$:

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

(b) Flame temperature (frozen, using the given linear cost).

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

(c) Verdict. Tripling pressure cut $\alpha$ from $0.447$ to $0.277$ and raised $T_c$ from $3100$ K to about $3290$ K. Higher pressure → less dissociation → hotter, more complete combustion, exactly as Le Chatelier predicts (fewer moles favoured under squeeze). This is why real engines run high chamber pressure.

The figure below shows both trends at once: the black curve is $\alpha$ falling as pressure rises from $p^{\circ }$ to $3p^{\circ }$, and the red curve is $T_c$ rising over the same range — the two computed points are marked on each. Reading the two curves together makes the "less dissociation → hotter chamber" chain visible in one glance.

Step — compute each $\sqrt{T_c/M}$ (proportional to $v_e$):

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

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

Reading it: here the higher-pressure case is about $1.0\%$ lower in frozen $v_e$! The $T_c$ gain from suppressing dissociation was outweighed by the heavier $M$. Lesson: raising pressure is not a free lunch for $v_e$ in frozen flow — it trades a hotter chamber for a heavier gas. The real payoff of high $p_c$ comes through equilibrium recovery and higher mass flow, not this simple frozen ratio. Mastery means seeing that no single lever wins in isolation.