Question bank — Real gas effects — dissociation, recombination

True ::: Breaking a chemical bond costs energy, so the reaction absorbs heat from the gas — that is exactly why $T_c$ ends up lower than a "perfect combustion" estimate.

True ::: Recombination is the reverse (exothermic) reaction; as the gas cools it re-forms molecules and dumps the bond energy back into the flow — if there is time for it to happen.

False ::: "Calorically perfect" means fixed composition and constant $c_p,\gamma$ by definition; dissociation changes composition, so it is precisely the thing that breaks the perfect-gas assumption.

False ::: Dissociation increases the mole count (e.g. ${\text{A}}_2\to 2\text{A}$), so by Le Chatelier's Principle higher pressure pushes the balance back toward molecules — $\alpha$ goes down, not up.

True ::: Equilibrium flow recovers recombination energy and keeps $M$ low as pieces re-glue; frozen flow leaves that energy locked up, so equilibrium is the higher ideal bound on $v_e$.

False ::: It hurts through a lower $T_c$ but helps through a lower $M$ (since $v_e\propto 1/\sqrt{M}$); in equilibrium flow the stolen energy comes back too, so the net effect can even be favourable.

True ::: $Da={\tau }_{flow}/{\tau }_{chem}$; large $Da$ means chemistry keeps up (equilibrium), small $Da$ means the flow outruns the reactions (frozen).

$\alpha =0$ the reacting-gas model and the calorically-perfect model give the same answer. True ::: With nothing dissociated the composition is fixed and no bond energy is diverted, so both models collapse to the same $T_c$, $M$, and $v_e$ — this is the degenerate limiting case.

$v_e\propto \sqrt{T_c}$, we should push $T_c$ as high as possible for maximum thrust." Error ::: Past a point, extra heat just breaks more bonds instead of raising $T$, and $v_e$ actually depends on $T_c/M$, not $T_c$ alone — so blindly maximising $T_c$ hits diminishing returns.

$v_e$ because the composition is fixed and simple." Error ::: Frozen flow loses the recombination energy that equilibrium flow recovers; it is the lower bound on $v_e$, not the higher one.

$M$, dissociation must always raise $v_e$." Error ::: This ignores the other lever — dissociation also lowers $T_c$. It is a tug-of-war between $T_c\downarrow$ and $M\downarrow$; the winner depends on how much energy recombination returns downstream.

$K_p$ tells us how fast the reaction goes, so a large $K_p$ means the gas reaches equilibrium quickly." Error ::: $K_p$ is a thermodynamic balance point (from $\ln K_p=-\Delta G^{\circ }/R_{u}T$), not a rate; how fast equilibrium is reached is a kinetics question captured by the Damköhler Number, not by $K_p$.

Error ::: Real flow sits between frozen and equilibrium, usually a few percent below equilibrium, because recombination is fast but not infinitely fast — finite $Da$.

$T_c$." Error ::: Ignoring it overpredicts $T_c$ (you never pay the bond-breaking cost), so it is optimistic, not conservative.

Because ::: each joule spent snapping a bond is a joule not available to raise the gas temperature — energy is conserved, so the split is temperature-heating versus bond-breaking.

Because ::: higher pressure suppresses dissociation (Le Chatelier, since dissociation adds moles), keeping more energy as heat and giving a hotter, more complete burn.

$M$ helpful even when $T_c$ stays fixed? Because ::: $v_e\propto \sqrt{T_c/M}$, so lighter species leave the nozzle faster for the same thermal energy — small pieces "run out" faster than big ones.

Because ::: recombination is favoured by cooling; the chamber is hottest (drives dissociation), while the nozzle cools the gas as it accelerates, shifting the balance toward re-forming molecules.

Because ::: short residence time ${\tau }_{flow}$ makes $Da={\tau }_{flow}/{\tau }_{chem}$ small, so the gas exits before the recombination chemistry can catch up.

$\alpha \to 0$? Nothing to trade off ::: no bonds broken means $T_c$ isn't lowered and $M$ isn't lowered — the reacting gas behaves exactly like a calorically perfect gas.

$\alpha \to 1$ (fully dissociated)? Maximum energy diverted and minimum $M$ ::: $T_c$ is depressed the most and the mixture is all light fragments; whether this is a net gain hinges entirely on how much recombination the nozzle can recover.

$Da\gg 1$ limit, which idealization applies and why? Equilibrium flow ::: the chemistry is effectively infinitely fast relative to the flow, so the gas stays in chemical equilibrium at every point and recovers the full recombination energy.

$Da\ll 1$ limit, which idealization applies and why? Frozen flow ::: the flow is far faster than the chemistry, so composition is locked at its chamber value and the bond energy leaves the nozzle unrecovered.

$\Delta H_d$ (the bond energy) were zero, what would dissociation do to $T_c$? Nothing ::: with no energy cost to break the "bond," splitting molecules would neither steal nor return heat — $T_c$ would be untouched and only the $M$ effect would remain.