NBPhO 2026

Part i) (1.5 points)

At atmospheric pressure $p_\text{atm} = 1.0\times 10^5\,\text{Pa}$, which is below the triple-point pressure $p_t = 5.18\times 10^5\,\text{Pa}$, no liquid CO₂ can exist. The exit stream is therefore either pure vapour, pure solid, or a solid–vapour mixture. Pure solid is impossible: a rigid (incompressible) solid cannot expand or accelerate through a converging–diverging nozzle. So the exit is either pure vapour, or a solid–vapour mixture; in the latter case the temperature must be the sublimation temperature at $1\,\text{atm}$, which is $T_\text{sub} \approx 195\,\text{K}$ ($\approx -78\,^\circ\text{C}$).

Tracing the isentrope from the initial state to $p = 1\,\text{bar}$ on the $T$-$s$ diagram (provided): for both scenarios (a) and (b), the trajectory crosses the saturation boundary into the two-phase region. Hence the exit stream is a solid–vapour mixture at $T \approx 195\,\text{K}$ in both scenarios.

Grading

• Recognising that liquid is impossible at $p_\text{atm} < p_t$, so the stream is solid, vapour, or a mixture: 0.5 pts
• Excluding pure solid (cannot adiabatically expand through nozzle): 0.2 pts
• If solid–vapour coexistence, identifying $T = T_\text{sub}(1\,\text{atm}) \approx 195\,\text{K}$ ($\approx -78\,^\circ\text{C}$), within $\pm 10\,\text{K}$: 0.3 pts
• Tracing the isentrope on the $T$-$s$ diagram and concluding the stream is in the two-phase region for one scenario: 0.3 pts
• Two-phase for both scenarios: 0.2 pts

Part ii) (3.5 points)

The expansion through the nozzle is reversible and adiabatic, so it is isentropic: the entropy per unit mass is conserved between the inlet and the exit. On the $T$-$s$ diagram, the trajectory is therefore vertical (constant $s$) until it crosses a phase boundary or a chosen isobar.

The exit pressure is $p_\text{atm} = 1\,\text{bar}$, well below the triple-point pressure $p_t = 5.18\,\text{bar}$. From part (i), the exit state is in the two-phase (solid+vapour) region at $T_\text{sub}(1\,\text{bar}) \approx 195\,\text{K}$ ($-78\,^\circ\text{C}$).

Lever rule. In the two-phase region at fixed $p$, $T$, the total entropy per unit mass for a mixture with mass fraction $x$ of solid (and $1-x$ vapour) is

$$s_\text{end} = x\,s_\text{sol} + (1-x)\,s_\text{vap},$$ $$x = \frac{s_\text{vap} - s_\text{end}}{s_\text{vap} - s_\text{sol}},$$

where $s_\text{sol}$ and $s_\text{vap}$ are the entropies of pure solid and pure vapour at the boundaries of the two-phase region at $p = 1\,\text{bar}$.

Reading entropies from the $T$-$s$ diagram. At $p = 1\,\text{bar}$ (sublimation isobar at $T \approx -78\,^\circ\text{C}$), the boundaries are

$$s_\text{sol} \approx 400(50)\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}, \quad s_\text{vap} \approx 3250(50)\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}.$$

The initial entropies for the two scenarios are read at $T_0 = 298\,\text{K}$ ($25\,^\circ\text{C}$) on the saturation dome:

• Scenario (a), saturated liquid at $T_0$: $s_a \approx 2100(50)\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$.
• Scenario (b), saturated vapour at $T_0$: $s_b \approx 2550(50)\,\tfrac{\text{J}}{\text{kg}\cdot\text{K}}$.

Mass fractions. Applying the lever rule to each:

$$x_a = \frac{3250 - 2100}{3250 - 400} = \frac{1150}{2850} \approx 0.40,$$ $$x_b = \frac{3250 - 2550}{3250 - 400} = \frac{700}{2850} \approx 0.25.$$

So the liquid-fed extinguisher (a) produces a stream that is $\sim 40\,\%$ dry ice (and $60\,\%$ vapour) by mass, while the vapour-fed extinguisher (b) produces a stream that is only $\sim 25\,\%$ dry ice. The liquid scenario yields about $1.6$ times more solid CO₂ per unit mass discharged — which is why fire extinguishers are designed to draw liquid.

Grading

• Recognising that “reversible adiabatic” means isentropic ($s = \text{const}$), so the endpoint is found by going vertically downward on the $T$-$s$ diagram: 0.6 pts
• Deriving the lever rule for mass fraction $x$ in the two-phase region: $x = (s_\text{vap} - s_\text{end})/(s_\text{vap} - s_\text{sol})$: 0.7 pts
• Reading the initial entropy $s_a$ from the saturated-liquid line at $T_0 = 298\,\text{K}$ (scenario a): 0.5 pts
• Reading the initial entropy $s_b$ from the saturated-vapour line at $T_0$ (scenario b): 0.5 pts
• Reading $s_\text{sol}$ at the $1$-bar solid boundary: 0.3 pts
• Reading $s_\text{vap}$ at the $1$-bar vapour boundary: 0.3 pts
• Computing $x_a \approx 0.40 \pm 0.05$ for scenario (a): 0.3 pts
• Computing $x_b \approx 0.25 \pm 0.05$ for scenario (b): 0.3 pts

Part iii) (3 points)

Consider an infinitesimal isentropic (reversible adiabatic) expansion of the saturated vapour from $(T, p_\text{sat}(T))$. Two trajectories meet at this point: the saturation curve $p_\text{sat}(T)$, and the adiabat (constant entropy). The expansion follows the adiabat. If the adiabat falls below the saturation curve in $T$-$p$ space (i.e., the vapour cools more slowly than the saturation temperature drops with pressure), then a finite fraction of the vapour condenses into droplets. Otherwise the expansion remains in the pure-vapour region.

Saturation slope (Clausius–Clapeyron). The saturated vapour above a liquid has number density $n_\text{vap}$ in thermal equilibrium with the liquid. By the Boltzmann distribution, the probability of a molecule occupying the vapour state (energy $\ell$ per molecule above the liquid) versus the liquid state is

$$\frac{n_\text{vap}}{n_\text{liq}} \propto \exp\left(-\frac{\ell}{k_BT}\right).$$

Treating $n_\text{liq}$ as $T$-independent and using $p_\text{sat} = n_\text{vap}k_BT \approx (\text{const})\cdot T\cdot\exp(-\ell/k_BT)$, the dominant temperature dependence is exponential. Per unit mass, $\ell/k_B = L/R_s$ where $R_s$ is the specific gas constant of the vapour. Differentiating $\ln p_\text{sat}$:

$$\frac{1}{p_\text{sat}}\frac{\mathrm{d}p_\text{sat}}{\mathrm{d}T} = \frac{L}{R_sT^2}.$$

Equivalently,

$$\left.\frac{\mathrm{d}T}{\mathrm{d}p}\right|_\text{sat} = \frac{R_sT^2}{pL}.$$

Adiabat slope. For the ideal-gas adiabat $Tp^{(1-\gamma)/\gamma} = \text{const}$, with $R_s = c_p - c_v$ and $\gamma = c_p/c_v$:

$$\left.\frac{\mathrm{d}T}{\mathrm{d}p}\right|_\text{adiab} = \frac{R_sT}{c_pp}.$$

Condensation condition. The vapour condenses on expansion iff the adiabat cools less than the saturation curve, i.e. $\left.\frac{\mathrm{d}T}{\mathrm{d}p}\right|_\text{adiab} < \left.\frac{\mathrm{d}T}{\mathrm{d}p}\right|_\text{sat}$:

$$\frac{R_sT}{c_pp} < \frac{R_sT^2}{pL},$$ $$\boxed{\,L > c_pT.\,}$$

Numerical check (water at $T = 373\,\text{K}$). $L = 2260\,\tfrac{\text{kJ}}{\text{kg}}$, $c_p = 2.0\,\tfrac{\text{kJ}}{\text{kg}\cdot\text{K}}$:

$$\frac{L}{c_pT} = \frac{2260}{2.0\cdot 373} \approx 3.0 > 1.$$

So water vapour does condense on isentropic expansion. This is precisely why steam from a kettle appears as a visible white plume of fine droplets, rather than as invisible pure water vapour.

Grading

• Setting up the comparison: condensation occurs iff the adiabat lies below the saturation curve in $T$-$p$ space: 0.5 pts
• Deriving the saturation slope via Boltzmann: $n_\text{vap}/n_\text{liq} \propto \exp(-\ell/k_BT)$ (those who know and use the Clausius–Clapeyron equation do not need to derive it): 0.6 pts
• Obtaining the Clausius–Clapeyron form $\mathrm{d}p_\text{sat}/\mathrm{d}T = pL/(R_sT^2)$: 0.6 pts
• Adiabat slope $\mathrm{d}T/\mathrm{d}p = R_sT/(c_pp)$ from ideal-gas adiabatic law: 0.4 pts
• Comparing slopes to obtain the condition $L > c_pT$: 0.6 pts
• Numerical check for water giving $L/(c_pT) \approx 3 > 1$, hence condensation: 0.3 pts