NBPhO 2026

Overview

A real fire extinguisher releases CO₂ through a converging–diverging (de Laval) nozzle. Three interlocking ideas drive the whole problem:

1. Reversible + adiabatic = isentropic. No matter what tortured path the fluid takes through the nozzle (including transiting the local speed of sound at the throat), the specific entropy $s$ is constant along every streamline. Phase fractions at the exit are therefore fixed by the lever rule on the $T$–$s$ diagram alone — pressures, velocities and Mach numbers along the channel never appear in the bookkeeping.

2. The exit pressure sits below the triple point. $p_\text{atm} = 0.10\,\text{MPa}$ is well below $p_t = 0.518\,\text{MPa}$. At the exit, liquid CO₂ cannot exist — only solid (dry ice) and vapour are stable, and they coexist on the sublimation curve.

3. Two starting states with very different entropies. The cylinder upside down feeds saturated liquid into the throat ($s_\ell$, low entropy); the cylinder upright feeds saturated vapour ($s_g$, higher entropy). Because both end up at the same $p$, $T$ but with different $s$, the lever rule says the orientations differ in how much of the exhaust is solid.

The numerical work in parts (i)–(ii) is essentially careful diagram reading. The physics lives in part (iii), which derives — and answers — the question hidden behind part (ii)‘s scenario (b): why does adiabatic expansion of pure saturated vapour produce solid CO₂ at all?

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Part (i) — Phase(s) and temperature at the exit

Why only solid and vapour can leave the nozzle

The CO₂ triple point at $(T_t, p_t) = (216.6\,\text{K},\,0.518\,\text{MPa})$ is the unique point where solid, liquid and vapour coexist. The horizontal slice $p = p_\text{atm} = 0.10\,\text{MPa}$ in the $p$–$T$ diagram lies entirely below this triple point — about a factor of five below in pressure. It crosses only the sublimation curve. Hence at any temperature whatsoever, a CO₂ system held at $p_\text{atm}$ admits two stable phases at most: solid and vapour, possibly in coexistence.

Why the exit state lies inside the solid-vapour dome

Tracing each starting state vertically downward on the $T$–$s$ diagram (constant $s$, the isentropic process) until it intersects the $0.10\,\text{MPa}$ isobar shows the intersection happens inside the solid-vapour two-phase region — the constant-$s$ line at $s = s_\ell(T_0)$ (scenario a) or $s = s_g(T_0)$ (scenario b) does not exit through the saturated-vapour boundary before reaching the $0.1$ MPa isobar. The exit is therefore a coexisting solid-vapour mixture.

Inside the solid-vapour dome, $T$ and $p$ are no longer independent: the temperature is locked to the sublimation temperature $T_\text{sub}(p_\text{atm})$. From the $T$–$s$ diagram the $0.10\,\text{MPa}$ isobar runs flat at

$$\boxed{\,T_\text{exit} \;=\; T_\text{sub}(p_\text{atm}) \;\approx\; 195\,\text{K} \;\approx\; -78\,°\text{C}.\,}$$

The exhaust is a fog of solid CO₂ (“dry ice”) suspended in CO₂ vapour, both phases at the same temperature $T_\text{exit}$ — that is the defining property of being on the sublimation curve.

Physics remark — why a de Laval shape. A purely converging nozzle can accelerate a subsonic fluid only up to the local speed of sound at the throat ($M = 1$). To reach supersonic exhaust the channel must re-widen: in a steady compressible flow with $M > 1$, increasing cross-section accelerates the fluid (the area-velocity relation flips sign across $M = 1$). The de Laval geometry is what turns a high-pressure CO₂ reservoir into a fast, cold jet — and it is what makes the practical CO₂ extinguisher into a snow-maker rather than a sputtering vent.

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Part (ii) — Mass fraction of solid for the two scenarios

Master equation: lever rule on the $T$–$s$ diagram

Steady reversible adiabatic flow forces $\Delta s = 0$ along every streamline, so the exit and inlet share a specific entropy. The relevant inlet states sit at the two edges of the liquid-vapour dome at $T_0 = 298\,\text{K}$:

Scenario	Inlet phase	Entropy from the $T$–$s$ diagram
(a) cylinder inverted	saturated liquid at $T_0$	$s_a \approx 1.30\,\text{kJ\,kg}^{-1}\,\text{K}^{-1}$
(b) cylinder upright	saturated vapour at $T_0$	$s_b \approx 1.58\,\text{kJ\,kg}^{-1}\,\text{K}^{-1}$

(Both values are read off the dome at $T_0 = 25\,°\text{C}$, where the dome is narrow because we are only six degrees below the critical point.)

The exit state lies on the $0.10\,\text{MPa}$ isobar within the solid-vapour dome at $T_\text{exit}\approx 195\,\text{K}$. From the diagram,

$$s_s \;\approx\; 0.70\,\text{kJ\,kg}^{-1}\,\text{K}^{-1}, \qquad s_v \;\approx\; 3.60\,\text{kJ\,kg}^{-1}\,\text{K}^{-1}.$$

Consistency check via Clausius–Clapeyron

The gap between these two readings ought to equal $L_\text{sub}/T_\text{sub}$ by the Clausius–Clapeyron relation across the sublimation curve. With $L_\text{sub}\approx 5.7\times 10^{5}\,\text{J\,kg}^{-1}$ for CO₂ at $-78\,°\text{C}$,

$$\frac{L_\text{sub}}{T_\text{sub}} \;\approx\; \frac{5.7\times 10^{5}}{195} \;\approx\; 2.93\,\text{kJ\,kg}^{-1}\,\text{K}^{-1},$$

which matches $s_v - s_s \approx 2.90\,\text{kJ\,kg}^{-1}\,\text{K}^{-1}$ from the diagram to within reading precision. The numbers are self-consistent.

Lever rule

A two-phase state on the horizontal $0.10\,\text{MPa}$ segment in the solid-vapour dome with specific entropy $s$ is split by mass as

$$s \;=\; (1-x)\,s_v \;+\; x\,s_s \quad\Longrightarrow\quad x \;=\; \frac{s_v - s}{s_v - s_s},$$

with $x$ the mass fraction of solid (dry ice) and $1-x$ the mass fraction of vapour.

Scenario (a) — saturated liquid feed

$$x_a \;=\; \frac{s_v - s_a}{s_v - s_s} \;\approx\; \frac{3.60 - 1.30}{3.60 - 0.70} \;=\; \frac{2.30}{2.90}.$$ $$\boxed{\,x_a \;\approx\; 0.79\,\text{ — about 80\,\% of the exhaust mass is dry ice.}\,}$$

Scenario (b) — saturated vapour feed

$$x_b \;=\; \frac{s_v - s_b}{s_v - s_s} \;\approx\; \frac{3.60 - 1.58}{3.60 - 0.70} \;=\; \frac{2.02}{2.90}.$$ $$\boxed{\,x_b \;\approx\; 0.70\,\text{ — still about 70\,\% dry ice, even with a pure-vapour feed.}\,}$$

Educational remark — both orientations make snow. A naive expectation is that scenario (b) — vapour into the nozzle — should leave a dry vapour jet at the exit, while only scenario (a) makes snow. The result $x_b \approx 0.7$ contradicts this. Physically, CO₂’s saturated-vapour curve in $T$–$s$ bends back to the left near the critical point: at $25\,°\text{C}$ the saturated vapour entropy ($1.58$) is far less than the saturated-vapour entropy at the eventual exit temperature ($3.60$), so the constant-$s$ line plunges deep into the dome. Part (iii) makes this quantitative.

Why scenario (a) gives more solid than (b). Scenario (a)‘s starting point has lower entropy (liquid CO₂ at $25\,°\text{C}$ has $\sim 0.28\,\text{kJ\,kg}^{-1}\,\text{K}^{-1}$ less entropy than its saturated vapour, by $L_\text{vap}/T = 83/298 \approx 0.28$). Lower entropy at fixed exit pressure means the lever-rule split sits closer to the low-entropy end (the solid), so a larger mass fraction is solid.

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Part (iii) — Condition for condensation upon vanishingly small expansion

Reformulating the question on the $T$–$s$ diagram

A pure substance is in equilibrium with its saturated vapour at $T$. Only the vapour escapes through the nozzle; it expands reversibly and adiabatically. The starting state on the $T$–$s$ diagram is $(T,\,s_g(T))$ on the right boundary of the two-phase dome.

Reversible-adiabatic expansion is isentropic, i.e. moves the state vertically downward in $T$–$s$: $T$ falls, $s$ stays at $s_g(T)$. After an infinitesimal pressure drop the state sits at $(T - dT,\,s_g(T))$.

The dome’s right boundary at the new temperature is at entropy $s_g(T - dT) = s_g(T) - (ds_g/dT)\,dT$, where the slope $ds_g/dT$ is taken along the saturation curve (with $p$ varying as $p_\text{sat}(T)$, not held fixed).

• If $ds_g/dT > 0$, then $s_g(T - dT) < s_g(T)$: the boundary recedes leftward as we cool, our point lies outside (to the right of) it, and the vapour ends superheated. No condensation.
• If $ds_g/dT < 0$, then $s_g(T - dT) > s_g(T)$: the boundary sweeps past our point as we cool, our state ends inside the dome, and a small mass of liquid appears. Condensation.

The criterion for condensation is therefore $\dfrac{ds_g}{dT}\bigg|_\text{sat curve} < 0$.

Computing the slope of the saturated-vapour curve

The chain rule, with the vapour as a single-phase fluid travelling along $p = p_\text{sat}(T)$, gives

$$\frac{ds_g}{dT}\bigg|_\text{sat} \;=\; \left(\frac{\partial s}{\partial T}\right)_p \;+\; \left(\frac{\partial s}{\partial p}\right)_T \frac{dp_\text{sat}}{dT}.$$

Each ingredient is standard:

• $(\partial s/\partial T)_p = c_p/T$, from $dh = T\,ds + v\,dp$ at $dp = 0$.
• $(\partial s/\partial p)_T = -(\partial v/\partial T)_p$ — the Maxwell relation derived from $dG = -s\,dT + v\,dp$.
• For the vapour modelled as an ideal gas, $v = RT/(M p)$, so $(\partial v/\partial T)_p = v/T$ and hence $(\partial s/\partial p)_T = -v/T$.
• $dp_\text{sat}/dT = L/[T(v_g - v_\ell)] \approx L/(T\,v_g)$ — the Clausius–Clapeyron equation with the standard “the liquid takes negligible volume” approximation.

Substituting:

$$\frac{ds_g}{dT}\bigg|_\text{sat} \;=\; \frac{c_p}{T} \;-\; \frac{v}{T}\cdot\frac{L}{T\,v} \;=\; \frac{c_p}{T} - \frac{L}{T^2} \;=\; \frac{c_p T - L}{T^2}.$$

The vapour-side specific volume $v$ cancels exactly between the Maxwell term and the Clausius–Clapeyron term — a clean simplification that signals the result is robust to the detailed equation of state of the vapour, as long as it is approximately ideal.

The condensation criterion $ds_g/dT < 0$ therefore reduces to

$$\boxed{\;\;L \;>\; c_p\,T.\;\;}$$

A complementary derivation that exposes the physical content

Along an isentrope of the (ideal-gas) vapour, the same Maxwell-relation algebra gives $\left(\partial T/\partial p\right)_s = v/c_p$. Along the saturation curve, the inverted Clausius–Clapeyron gives $dT/dp\big|_\text{sat} = T\,v_g / L$. Condensation upon expansion means the isentropic cooling per unit pressure drop is greater than the saturation-curve cooling — i.e. the vapour cools below the saturation temperature corresponding to its current pressure and falls into the dome:

$$\left(\frac{\partial T}{\partial p}\right)_s \;>\; \frac{dT}{dp}\bigg|_\text{sat} \quad\Longleftrightarrow\quad \frac{v}{c_p} \;>\; \frac{T\,v_g}{L}.$$

With $v = v_g$ on the saturation boundary, this is exactly $L > c_p T$. The condition has a single, transparent reading: the latent heat must exceed the thermal energy $c_p T$ available in the gas to soak up the cooling.

Consistency checks

• $L \to 0$ limit. A substance with no condensation enthalpy can never condense from expansion. The criterion reduces to $0 > c_p T$, never satisfied. ✓
• $T \to T_c$ (approach to the critical point). Real $L$ vanishes as $\sim (T_c - T)^{1/2}$, while $c_p$ diverges. So $L < c_p T$ near $T_c$, no condensation upon expansion. Caveat: the formula’s ideal-gas premise breaks down here — exactly the regime relevant to CO₂ at $25\,°\text{C}$, only six degrees below $T_c$. The qualitative prediction “no condensation if $L < c_p T$” does not survive into the strongly non-ideal critical region; CO₂ in part (ii) does condense en route from $25\,°\text{C}$ to $-78\,°\text{C}$ in spite of $c_p$ diverging at the start. The criterion is stated for “an arbitrary pure substance” with the implicit ideal-gas $c_p$, and it is the infinitesimal expansion at the start of the saturation curve that the criterion describes — finite expansions far from the starting point can produce condensation by any of several mechanisms.
• Dimensions. $[c_p T] = \text{J\,kg}^{-1}$, same as $[L]$. ✓

Application: water vapour at $T = 373\,\text{K}$

Plugging in the data ($L \approx 2260\,\text{kJ\,kg}^{-1}$, $c_p \approx 2.0\,\text{kJ\,kg}^{-1}\,\text{K}^{-1}$, $T = 373\,\text{K}$):

$$c_p T \;=\; 2.0 \times 373 \;\approx\; 7.5\times 10^{2}\,\text{kJ\,kg}^{-1},$$ $$\frac{L}{c_p T} \;\approx\; \frac{2260}{746} \;\approx\; 3.0.$$

Since $L$ exceeds $c_p T$ by a factor of three, water vapour in equilibrium with boiling water at $100\,°\text{C}$ does condense into droplets upon adiabatic expansion. The visible white “steam” issuing from a kettle spout or a teakettle whistle is, in fact, a fog of micrometre-scale water droplets produced by precisely this mechanism (compounded by mixing with cool ambient air, but the adiabatic-expansion contribution alone suffices).

Educational remark — why dry hot air doesn’t fog when you let it expand. Air at $373\,\text{K}$ has no liquid phase to condense into (it is far above the boiling points of N₂ and O₂ at any pressure achievable by expansion in normal devices). The $L > c_p T$ criterion presupposes a saturation curve to sit on. For superheated vapour, undersaturated gas, or gases in their non-condensable regime, isentropic expansion just cools the gas — that is how a refrigerator’s expansion valve, a CO₂ cartridge, or a bicycle pump operated in reverse all work, without producing droplets.

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Summary of results

Part	Result
(i) — phases and temperature	Solid + vapour mixture at the sublimation temperature $T_\text{exit} \approx 195\,\text{K}$ ($-78\,°\text{C}$).
(ii) — scenario (a), liquid feed	$x_a \approx 0.79$ (≈ 80,% dry ice by mass)
(ii) — scenario (b), vapour feed	$x_b \approx 0.70$ (≈ 70,% dry ice by mass)
(iii) — condensation criterion	$L > c_p T$
(iii) — water at $373\,\text{K}$	$L/c_p T \approx 3$, so condensation does occur.