IPhO 2025, Теория — Задача 3. Champagne!

Q3-1

SOLUTION:

A.1. Laplace's law: $P_b = P_0 + \dfrac{2\sigma}{a}$

Q3-2

SOLUTION:

A.2.1. $a_c$ is so $c_\ell = c_b$

A.2.2. $c_b = k_H P_b = k_H(P_0 + \frac{2\sigma}{a})$ and $c_0 = k_H P_0$ so $a_c = \dfrac{2\sigma}{P_0(c_\ell/c_0 - 1)}$

A.2.3. $a_c = 0.3\,\mu\mathrm{m}$

Q3-3

SOLUTION:

A.3.1. The number of moles of CO$_2$ (ideal gas) inside the bubble is $n_c = \frac{4}{3}\pi a^3 \dfrac{P_0}{RT_0}$

A.3.2. Equation : balance of CO$_2$ in the bubble

A.3.3 $\dfrac{dn_c}{dt} = 4\pi a^2 \dfrac{da}{dt}\dfrac{P_0}{RT} = j 4\pi a^2 \Rightarrow \dfrac{da}{dt} = j\dfrac{RT}{P_0}$

A.3.4. Model 1: $\dfrac{da}{dt} = \dfrac{DRT}{aP_0}(c_\ell - c_0)$ so $a^2 = a_0^2 + \dfrac{2DRT_0}{P_0}(c_\ell - c_0)t$

A.3.5. Model 2: $\dfrac{da}{dt} = \dfrac{KRT_0}{P_0}(c_\ell - c_0)$ so $a = a_0 + \dfrac{KRT_0}{P_0}(c_\ell - c_0)t$

A.3.6. Experimental data : $\dfrac{da}{dt}$ is constant: model 2

A.3.7 Slope of the experimental data : $\dot a \approx 150/0.62 \approx 0.24\,\mathrm{mm}\cdot\mathrm{s}^{-1}$

A.3.8 $K = 1.0 \times 10^{-4}\,\mathrm{m}\cdot\mathrm{s}^{-1}$

Q3-4

SOLUTION:

A.4.1. Main forces: buoyancy $\frac{4}{3}\pi a^3 \rho_\ell g_0$, drag force $6\pi\eta a v$, weight is negligible: $\frac{\rho_{\mathrm{CO_2}}}{\rho_\ell} = \frac{P_e M_{\mathrm{CO_2}}}{RT\rho_\ell} \approx 10^{-3}$ : $m_b \ll m_\ell$

A.4.2. Simplified equation is a balance between buoyancy and drag force $\frac{4}{3}\pi a^3 \rho_\ell g_0 = 6\pi\eta a v$ so $v = \frac{2}{9\eta}a^2\rho_\ell g_0$.

A.4.3. Time between two bubbles: $\Delta t = 1/f_b$

A.4.4. Using $\eta = \dfrac{2\rho_\ell g_0}{9} \times \dfrac{a^2}{v}$ for the penultimate bubble $( n - 1 )$ with $a_{n-1} \approx 0.19\,\mathrm{mm}$

A.4.5. $v(t_{n-1}) = \dfrac{z(t_n) - z(t_{n-2})}{2 \times f_\mathrm{b}^{-1}} = 4.5\,\mathrm{cm}\cdot\mathrm{s}^{-1}$

A.4.6. $\eta \approx 2 \times 10^{-3}\,\mathrm{Pa}\cdot\mathrm{s}$

Q3-5

SOLUTION:

A.5.1. $v = \frac{dz}{d}\,t = \frac{2\rho_\ell g_0}{9\eta}a^2$ and $\frac{da}{d}\,t = q_a$ so $\frac{dz}{d}\,a = \frac{2\rho_\ell g_0}{9q_a\eta}a^2$. Neglecting $a(z = 0)$, $z = \dfrac{2\rho_\ell g_0}{27q_a\eta}a^3$ so $a_{H_\ell} = \left(\dfrac{27q_a\eta H_\ell}{2\rho_\ell g_0}\right)^{1/3}$

Source typography note: The active blue replacement in the original key displays bare $d$ denominators followed by baseline $t$, $t$ and $a$. This revision artifact is retained literally above; it is not asserted to be an intended new equation. The struck red extra $dz/dt$ equation is excluded from active content.

A.5.2. $a_{H_\ell} = 3.9 \times 10^{-4}\,\mathrm{m}$ for $\eta = 2.0 \times 10^{-3}\,\mathrm{Pa}\cdot\mathrm{s}$

Q3-6

SOLUTION:

A.6.1 The rate of bubbles reaching the free surface by unit time is $N_\mathrm{b}f_\mathrm{b}$

A.6.2. So the volume of CO$_2$ released per unit time at the free surface is: $\dfrac{\mathrm{d}V}{\mathrm{d}t} = \frac{4}{3}\pi a_{H_\ell}^3 N_\mathrm{b}f_\mathrm{b}$

A.6.3. According to A.5, $\dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{18\pi N_\mathrm{b}f_\mathrm{b}\eta H_\ell}{\rho_\ell g_0}q_a$

A.6.4. With $q_a = \dfrac{da}{dt} = \dfrac{RT_0}{P_0}K(c_\ell - c_0)$ according to A3.

A.6.5. In the bubble, $c_\mathrm{b} \approx c_0$. Using the ideal gas law, the total number $n$ of CO$_2$ moles in $V_\ell$ verifies: $\dfrac{dn}{dt} = -\dfrac{P_0}{RT_0}\dfrac{\mathrm{d}V}{\mathrm{d}t} = -\dfrac{18\pi N_\mathrm{b}f_\mathrm{b}\eta K H_\ell}{\rho_\ell g_0}(c_\ell - c_0)$

With $c_\ell = \dfrac{n}{V_\ell}$, we get a first order linear ODE $\dfrac{\mathrm{d}c_\ell}{\mathrm{d}t} = \dfrac{1}{V_\ell}\dfrac{\mathrm{d}n}{\mathrm{d}t} = -\dfrac{18\pi N_\mathrm{b}f_\mathrm{b}\eta K H_\ell}{\rho_\ell g V_\ell}(c_\ell - c_0)$

A.6.6. Exponential decay with characteristic time: $\tau = \dfrac{\rho_\ell g V_\ell}{18\pi N_\mathrm{b}f_\mathrm{b}\eta K H_\ell}$

Q3-7

SOLUTION:

B.1.1. and 1.2. Variation of kinetic energy: system : the rim (perimeter $\ell = 2\pi r$) and the volume $\delta V = h\ell dt = h2\pi r v_\mathrm{f}dt$; during $dt$ the volume $\delta V$ get a kinetic energy $dE_c = \frac{1}{2}\rho_\ell \delta V\, v_\mathrm{f}^2 = \frac{1}{2}\rho_\ell h \ell v_\mathrm{f}dt = \pi r \rho_\ell h v_\mathrm{f}dt$.

B.1.3. surface tension energy: $E_s = \sigma S$ for a surface $S$

B.1.4. $\delta E_s = -2\sigma \ell v_\mathrm{f}dt = -4\sigma\pi r v_\mathrm{f}dt$.

B.1.5. Kinetic energy theorem: the lost energy is $\delta E_s/2 < 0$ so $dE_c + \delta E_s = \delta E_s/2$

B.1.6. $v_f = \sqrt{2\sigma/\rho_\ell h}$

If partial answer: $v_f = \sqrt{\sigma/\rho_\ell h}$ obtained only by dimensional analysis: 0.2 pt to the question

Q3-8

SOLUTION:

B.2.1. Pressure forces on $m_p$: $F_z = P(t)S - P_0S$

B.2.2. Volume $V(t) = V_0 + Sz$

B.2.3. Adiabatic and reversible compression for an ideal gas: $PV(t)^\gamma = P_0 V_0^\gamma$ so $P(t) = P_0\left(\dfrac{V_0}{V_0 + Sz}\right)^\gamma = P_0\left(\dfrac{1}{1 + Sz/V_0}\right)^\gamma$

B.2.4. Approximation: $P(t) \approx P_0(1 - \gamma\frac{Sz}{V_0})$

B.2.5. Pressure force: $F_z = -\gamma S^2 P_0 \dfrac{z}{V_0}$

B.2.6. Newton's 2nd law: $m_p \ddot z = -\gamma S^2 P_0 \dfrac{z}{V_0}$ so $m_p \ddot z + \gamma S^2 P_0 \frac{z}{V_0} = 0$

B.2.7. Harmonic oscillator of angular frequency $\omega_0^2 = S^2 \dfrac{P_0\gamma}{m_p V_0}$

B.2.8. $f_0 = \dfrac{1}{2\pi}\sqrt{\dfrac{S^2 P_0\gamma}{m_p V_0}}$

Q3-9

SOLUTION:

B.3.1. The maximal value of $f_0$ is $f_0 = 40\,\mathrm{kHz}$ is obtained for $r = r_c$

B.3.2. Exact expression of $f_0$ with $m = \frac{8r^3}{3}\rho_g$ and $S = \pi r_c^2$: $f_0 = \dfrac{1}{2\pi}\sqrt{\dfrac{3r_c\pi^2 P_0\gamma}{8\rho_g V_0}}$ so $f_0 = \dfrac{1}{2\pi}\sqrt{\dfrac{\gamma P_0}{\rho_g}}\sqrt{\dfrac{3\sqrt{3}\pi}{16a}}\sqrt{\dfrac{\rho_\ell g_0}{\sigma}}$ or $a = \dfrac{3\sqrt{3}}{64\pi}\dfrac{\gamma P_0}{\rho_g f_0^2}\sqrt{\dfrac{\rho_\ell g_0}{\sigma}}$

B.3.3. $a = 0.53\,\mathrm{mm}$

Determination of $h$

B.3.4. $r_c = \dfrac{2}{\sqrt{3}}a^2\sqrt{\dfrac{\rho_\ell g_0}{\sigma}}$ and $r_c = 0.15\,\mathrm{mm}$ so $v_f = \dfrac{r_c}{t_b} = 5.0\,\mathrm{m}\cdot\mathrm{s}^{-1}$

B.3.5. $h = \dfrac{2\sigma}{\rho_\ell v_f^2} = \dfrac{3t_\mathrm{b}^2}{2a^4}\sqrt{\dfrac{\sigma^3}{\rho_\ell^3 g_0}}$

B.3.6. Numerical value $h = 3.7\,\mu\mathrm{m}$

Q3-10

SOLUTION:

C.1.1. Conservation of CO$_2$ molecules: $n_T = n_V + n_L = n_V + k_H(T_0)P_i V_L$

C.1.2. Ideal gas law: $n_V = \dfrac{P_i V_G}{RT_0}$

C.1.3. For $T_0 = 6\,{}^\circ\mathrm{C}$: $P_i = 4.81\,\mathrm{bar}$

C.1.4. For $T_0 = 20\,{}^\circ\mathrm{C}$: $P_i = 7.76\,\mathrm{bar}$

Q3-11

The saturated vapor pressure $P_{\mathrm{sat}}^{\mathrm{CO_2}}$ for the CO$_2$ solid/gas transition follows : $\log_{10}\left(\dfrac{P_{\mathrm{sat}}^{\mathrm{CO_2}}}{P_0}\right) = A - \dfrac{B}{T+C}$ with $T$ in K, $A = 6.81$, $B = 1.30\times 10^3\ \mathrm{K}$ and $C = -3.49\ \mathrm{K}$.

C.2 Give the numerical value $T_f$ of the CO$_2$ gas at the end of the expansion, after opening a bottle, if $T_0 = 6\,{}^\circ\mathrm{C}$ and if $T_0 = 20\,{}^\circ\mathrm{C}$, if no phase transition occured. Choose which statements are true (several statements possible):

1. At $T_0 = 6\,{}^\circ\mathrm{C}$ a grey-white fog appears while opening the bottle.
2. At $T_0 = 6\,{}^\circ\mathrm{C}$ a blue fog appears while opening the bottle.
3. At $T_0 = 20\,{}^\circ\mathrm{C}$ a grey-white fog appears while opening the bottle.
4. At $T_0 = 20\,{}^\circ\mathrm{C}$ a blue fog appears while opening the bottle.

SOLUTION:

C.2.1. The adiabatic reversible expansion goes from $P_i$ to $P_0$.

C.2.2. $T_f = T_0\left(\dfrac{P_i}{P_0}\right)^{(1/\gamma)-1}$

C.2.3. For $T_0 = 6\,{}^\circ\mathrm{C}$: $P_i = 4.69\,\mathrm{bar}$ and $T_f = 195.3\,\mathrm{K} = -77.8\,{}^\circ\mathrm{C}$.

C.2.4. For $T_0 = 20\,{}^\circ\mathrm{C}$: $P_i = 7.45\,\mathrm{bar}$ and $T_f = 184.3\,\mathrm{K} = -88.8\,{}^\circ\mathrm{C}$.

C.2.5. First method: comparison $P_{\mathrm{sat}}(T_f)$ and $P_f = P_0$.
Second method: evaluation of the transition temperature at $P_0$ and comparison with $T_f$.

C.2.6. First method: $P_{\mathrm{sat}}^{\mathrm{CO_2}}(T_f = 6\,{}^\circ\mathrm{C}) = 1.07\,\mathrm{bar} > P_0$ . As the solid-liquid frontier has a positive slope in $P, T$ state-diagram, the final state of CO$_2$ is gaseous. $P_{\mathrm{sat}}^{\mathrm{CO_2}}(T_f = 20\,{}^\circ\mathrm{C}) = 0.41\,\mathrm{bar} < P_0$ . As the solid-gas frontier has a positive slope in $P, T$ state-diagram, the final gaseous state hypothesis is inconsistent and a phase transition has occured in the latter case.

Second method: $T_{trans} = \dfrac{B}{A - \log_{10}\left(\dfrac{P_0}{P_0}\right)} - C$. $T_{trans} = 194.4\,\mathrm{K} = -78.8\,{}^\circ\mathrm{C}$. For $T_0 = 6\,{}^\circ\mathrm{C}$: $T_f = 195.3\,\mathrm{K} > T_{trans}$; the final state of CO$_2$ is gaseous. For $T_0 = 20\,{}^\circ\mathrm{C}$: $T_f = 184.3\,\mathrm{K} < T_{trans}$; the final gaseous state hypothesis is inconsistent and a phase transition has occured.

C.2.7. The true statements are: 1 and 4.

During bottle opening, the cork stopper pops out. We now determine the maximum height $H_c$ it reaches.

Q3-12

Assume that the friction force $F$ due to the bottleneck on the cork stopper is $F = \alpha A$ where $A$ is the area of contact and $\alpha$ is a constant to determine. Initially, the pressure force slightly overcomes the friction force. The cork's mass is $m = 10\,\mathrm{g}$, its diameter $d = 1.8\,\mathrm{cm}$ and the length of the cylindrical part initially stuck in the bottleneck is $\ell_0 = 2.5\,\mathrm{cm}$. Once the cork has left the bottleneck, you can neglect the net pressure force.

C.3 Give the numerical value of $H_c$ if the external temperature is $T_0 = 6\,{}^\circ\mathrm{C}$. (1.3pt)

SOLUTION:

C.3.1. Let us evaluate the work of the friction force. $\vec{F} = -\alpha\,\pi d\,(\ell_0 - z)\vec{u}_z$. Initially, this force slightly compensates the pressure force: $F = \pi\alpha d \ell_0 = \pi\dfrac{d^2}{4}(P_i - P_0)$ so $\alpha = (P_i - P_0)\dfrac{d}{4\ell_0}$

C.3.2. $\vec{F} = -(P_i - P_0)\pi d^2\dfrac{(\ell_0 - z)}{4\ell_0}\vec{u}_z$ The total work is therefore: $W_f = -\alpha\pi d\dfrac{\ell_0^2}{2} = -\dfrac{(P_i - P_0)\pi d^2}{8}\ell_0$

C.3.3. and C.3.4. Work of the internal pressure force:

First method: the variation of internal energy of the gas is:

As its expansion is adiabatic: $\Delta U_g = W_{\mathrm{cork}\to\mathrm{CO2}} = -W_{\mathrm{CO2}\to\mathrm{cork}}$ The cork stopper receives therefore a work from this gas equals to $-\Delta U_g$.

Second method: let us write $P$ the internal pressure during the expansion. The work received by the cork is:

$W_{\mathrm{CO2}\to\mathrm{cork}} = \int_{V_G}^{V_F} P\,dV$, where $V_F = V_G + \dfrac{\pi d^2 \ell_0}{4}$ and $P_0 V_F^{\gamma} = P_i V_G^{\gamma}$.

The integration leads to the same result.

C.3.5. The work due to the external pressure $P_0$ is: $W_e = -P_0\cdot\dfrac{\pi d^2}{4}\ell_0$

C.3.6. Energy balance. The cork stopper has an initial kinetic energy: $E_c = -\Delta U_g + W_f + W_e$

(The work of the weight is negligible and should not be taken into account).

At $T_0 = 6\,{}^\circ\mathrm{C}$: $P_i = 4.69\,\mathrm{bar}$ . $W_f = -1.17\,\mathrm{J}$; $W_e = -0.64\,\mathrm{J}$; $\Delta U_g = -2.57\,\mathrm{J}$; $E_c = 0.76\,\mathrm{J}$

C.3.7. The maximum height reached by the cork stopper is therefore: $H_c = \dfrac{E_c}{mg_0} = \dfrac{-\Delta U_g + W_f + W_e}{mg_0}$

C.3.8. $H_c = 7.7\,\mathrm{m}$

If the candidates assumed a constant pressure $P_i$ for the gaseous CO$_2$ during its expansion, they would find a work done by CO$_2$ on the cork equal to: $P_i(\pi \ell_0 d^2/4) = 3\,\mathrm{J}$ instead of $2.56\,\mathrm{J}$ and finally $H_c = 12\,\mathrm{m}$. The difference is not negligible!

Q3-13

[1] Liger-Belair et al, Am. J. Enol. Vitic., Vol. 50, No. 3 (1999).

[2] Liger-Belair et al, Sc. Reports 7, 10938 (2017).