A rigid vessel of volume 2.0 × 10⁻³ m³ is kept at 300 K…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A rigid vessel of volume $2.0 \times 10^{-3}\ \text{m}^3$ is kept at $300\ \text{K}$ . It holds helium (molar mass $4.0 \times 10^{-3}\ \text{kg/mol}$ ) and neon (molar mass $2.020 \times 10^{-2}\ \text{kg/mol}$ ), each at a partial pressure of $4.0\ \text{Pa}$ . At $t = 0$ a pinhole of area $1.0 \times 10^{-8}\ \text{m}^2$ is opened in the wall, as shown in the figure, and the gas leaks out into a vacuum. The gas is so thin that each atom leaves on its own, without colliding near the hole. The number of atoms of one gas that hit unit area of the wall each second is $\tfrac14 n\bar v$ , where $n$ is that gas's number density and $\bar v = \sqrt{8RT/(\pi M)}$ its mean speed. Take $R = 8.314\ \text{J/(mol K)}$ and $k_B = 1.38 \times 10^{-23}\ \text{J/K}$ .

A closed box of gas with small blue dots for helium atoms and larger orange dots for neon atoms, mixed together. The box is 2.0 times 10 to the minus 3 cubic metres at 300 K. A tiny hole in the right wall, of area 1.0 times 10 to the minus 8 square metres, opens into a vacuum.

How many helium atoms leave each second just after the hole is opened?
After how long has the partial pressure of helium fallen to half its first value?
What is the ratio of the partial pressure of helium to that of neon at $t = 600\ \text{s}$ ?
At first the escaping stream holds more helium than neon. At what time does it hold equal numbers of the two?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Write the leak as an equation for each gas on its own, solve it, then compare the two gases.

Step 1. Mean speeds. $\bar v_{\text{He}} = \sqrt{\dfrac{8 \times 8.314 \times 300}{\pi \times 4.0 \times 10^{-3}}}\ \text{m/s} = 1260\ \text{m/s}$ and likewise $\bar v_{\text{Ne}} = 561\ \text{m/s}$ .

Step 2. First rate (a). Each gas starts at $n = \dfrac{p}{k_BT} = 9.662 \times 10^{20}\ \text{m}^{-3}$ . Helium atoms leave at

\displaystyle \tfrac14 n\bar v A = \tfrac14 \times 9.662 \times 10^{20}\ \text{m}^{-3} \times 1260\ \text{m/s} \times 1.0 \times 10^{-8}\ \text{m}^2 = 3.04 \times 10^{15}\ \text{s}^{-1}

Step 3. The law of the leak. $V\dfrac{dn}{dt} = -\tfrac14 n\bar v A$ , so $n = n_0e^{-t/\tau}$ with $\tau = \dfrac{4V}{A\bar v}$ . For helium $\tau = 635\ \text{s}$ ; for neon $\tau = 1427\ \text{s}$ . The pressure of each gas follows its $n$ , since the temperature is fixed.

Step 4. Half the helium (b). $t = \tau_{\text{He}}\ln 2 = 635\ \text{s} \times 0.693 = 440\ \text{s}$ .

Step 5. The ratio (c). Both started equal, so

\displaystyle \frac{p_{\text{He}}}{p_{\text{Ne}}} = \exp\left(-\frac{t}{\tau_{\text{He}}} + \frac{t}{\tau_{\text{Ne}}}\right) = \exp\left(-\frac{600}{635} + \frac{600}{1427}\right) = 0.592

Step 6. An even stream (d). Equal escape rates need $\dfrac{n_{\text{He}}}{n_{\text{Ne}}} = \dfrac{\bar v_{\text{Ne}}}{\bar v_{\text{He}}} = 0.445$ . So $e^{-kt} = 0.445$ , with $k = \dfrac{1}{\tau_{\text{He}}} - \dfrac{1}{\tau_{\text{Ne}}} = 8.742 \times 10^{-4}\ \text{s}^{-1}$ :

\displaystyle t = \frac{0.810}{8.742 \times 10^{-4}\ \text{s}^{-1}} = 926\ \text{s}

After that, neon atoms leave faster than helium atoms.

Answer: (a) $3.04 \times 10^{15}\ \text{s}^{-1}$ . (b) $440\ \text{s}$ . (c) $0.592$ . (d) $926\ \text{s}$ , about $15.4$ minutes.

Check. At the start helium leaves $\sqrt{2.020 \times 10^{-2}/4.0 \times 10^{-3}} = 2.25$ times as fast as neon (Graham's law). Its share inside the vessel falls, and the two rates meet when its density is $0.445$ of neon's.

Watch out. The rate is not fixed: it falls as the vessel empties. Dividing the number of atoms by the first rate gives a time that is too short and has no half-life in it.

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