NBPhO 2026, Теория и експеримент — Задача 9. CO₂ fire extinguisher

i) (1,5 т.)

At atmospheric pressure $p_{\mathrm{atm}} = 1.0 \times 10^5\ \mathrm{Pa}$, which is below the triple-point pressure $p_t = 5.18 \times 10^5\ \mathrm{Pa}$, no liquid CO$_2$ 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 atm, $T_{\mathrm{sub}} \approx 195\ \mathrm{K}$ ($\approx -78\ {}^{\circ}\mathrm{C}$).

Tracing the isentrope from the initial state to $p = 1\ \mathrm{bar}$ 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 stream is a solid–vapour mixture at $T \approx 195\ \mathrm{K}$ in both cases.

Grading.

• Recognising that liquid is impossible at $p_{\mathrm{atm}} < p_t$, 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 $T = T_{\mathrm{sub}}(1\ \mathrm{atm}) \approx 195\ \mathrm{K}$ ($\approx -78\ {}^{\circ}\mathrm{C}$), within $\pm 10\ \mathrm{K}$ 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

ii) (3,5 т.)

The expansion through the nozzle is reversible and adiabatic, so it is isentropic: the entropy per unit mass $s$ 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 $p_{\mathrm{atm}} = 1\ \mathrm{bar}$, well below the triple-point pressure $p_t = 5.18\ \mathrm{bar}$. From part (i), the exit state is in the two-phase (solid+vapour) region at $T_{\mathrm{sub}}(1\ \mathrm{bar}) \approx 195\ \mathrm{K}$ ($-78\ {}^{\circ}\mathrm{C}$).

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

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

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

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

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

Mass fractions. Applying the lever rule to each:

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 CO$_2$ per unit mass discharged — which is why fire extinguishers are designed to draw liquid.

Grading.

• Recognising that “reversible adiabatic” means isentropic ($s = \mathrm{const}$), so the endpoint is found by tracing vertically downward on the $T$–$s$ diagram 0.6 pts
• Deriving the lever rule for mass fraction $x$ in the two-phase region: $x = (s_{\mathrm{vap}} - s_{\mathrm{end}})/(s_{\mathrm{vap}} - s_{\mathrm{sol}})$ 0.7 pts
• Reading the initial entropy $s_a$ from the saturated-liquid line at $T_0 = 298\ \mathrm{K}$ (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 $s_{\mathrm{sol}}$ at the 1-bar solid boundary 0.3 pts
• Reading $s_{\mathrm{vap}}$ 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

iii) (3 т.)

Consider an infinitesimal isentropic (reversible adiabatic) expansion of the saturated vapour from $(T, p_{\mathrm{sat}}(T))$. Two trajectories meet at this point: the saturation curve $p_{\mathrm{sat}}(T)$, 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 $n_{\mathrm{vap}}$ 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

Treating $n_{\mathrm{liq}}$ as $T$-independent and using $p_{\mathrm{sat}} = n_{\mathrm{vap}} k_B T \approx (\mathrm{const}) \cdot T \cdot \exp(-\ell/k_B T)$, 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 $\ln p_{\mathrm{sat}}$:

Equivalently,

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

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

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

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.

iv) (? points)

Consider an infinitesimal isentropic (reversible adiabatic) expansion of the saturated vapour from $(T, p_{\mathrm{sat}}(T))$. Two trajectories meet at this point: the saturation curve $p_{\mathrm{sat}}(T)$, 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 $n_{\mathrm{vap}}$ 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

Treating $n_{\mathrm{liq}}$ as $T$-independent and using $p_{\mathrm{sat}} = n_{\mathrm{vap}} k_B T \approx (\mathrm{const}) \cdot T \cdot \exp(-\ell/k_B T)$, 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 $\ln p_{\mathrm{sat}}$:

Equivalently,

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

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

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

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: $n_{\mathrm{vap}}/n_{\mathrm{liq}} \propto \exp(-\ell/k_B T)$ (those who know and use the Clausius–Clapeyron equation do not need to derive it) 0.6 pts
• Obtaining the Clausius–Clapeyron form $\frac{\mathrm{d}p_{\mathrm{sat}}}{\mathrm{d}T} = pL/(R_s T^2)$ 0.6 pts
• Adiabat slope $\frac{\mathrm{d}T}{\mathrm{d}p} = R_s T/(c_p p)$ from ideal-gas adiabatic relation 0.4 pts
• Comparing slopes to obtain the condition $L > c_p T$ 0.6 pts
• Numerical check for water giving $L/(c_p T) \approx 3 > 1$, hence condensation 0.3 pts