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

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

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

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 $k = \dfrac{1}{\tau_{\text{He}}} - \dfrac{1}{\tau_{\text{Ne}}} = 8.742 \times 10^{-4}\ \text{s}^{-1}$:

After that, neon atoms leave faster than helium atoms.