Average time between collisions $\tau$ is given by $\tau = \frac{\lambda}{v_{rms}}$, where $\lambda$ is the mean free path.

1. $\lambda \propto \frac{1}{n} \propto V$ (where $n$ is number density).
2. $v_{rms} \propto \sqrt{T}$.
3. For an adiabatic process, $TV^{\gamma-1} = \text{constant}$, so $T \propto V^{1-\gamma}$ and $\sqrt{T} \propto V^{\frac{1-\gamma}{2}}$.

Therefore, $\tau \propto \frac{V}{V^{\frac{1-\gamma}{2}}} = V^{1 - \frac{1-\gamma}{2}} = V^{\frac{2-1+\gamma}{2}} = V^{\frac{\gamma+1}{2}}$.
Hence, $q = \frac{\gamma+1}{2}$.

Mole fraction in liquid: $x_A = 0.40,\ x_B = 0.60$

By Raoult's Law:

$P_A = x_A \cdot P_A^0 = 0.40 \times 7000 = 2800\ \text{Pa}$

$P_B = x_B \cdot P_B^0 = 0.60 \times 12000 = 7200\ \text{Pa}$

$P_{\text{total}} = 2800 + 7200 = 10000\ \text{Pa}$

In vapour phase: $y_A = \dfrac{P_A}{P_{\text{total}}} = \dfrac{2800}{10000} = \mathbf{0.28}$

$y_B = 1 - 0.28 = \mathbf{0.72}$

• \(\text{Ca}^{2+}\): Z=20
• \(\text{Ar}\): Z=18
• \(\text{K}^+\): Z=19
• \(\text{Cl}^-\): Z=17
• \(\text{S}^{2-}\): Z=16

Doppler effect: source towards listener → frequency increases.

Volume $V = \frac{\pi d^2 h}{4} = \frac{\pi \times 12.6^2 \times 34.2}{4} \approx 4262.2 \text{ cm}^3$.

Since inputs have 3 significant figures, $V \approx 4260 \text{ cm}^3$.

Error calculation: $\frac{\Delta V}{V} = 2\frac{\Delta d}{d} + \frac{\Delta h}{h}$

$\Delta V = 4260 \left( 2 \frac{0.1}{12.6} + \frac{0.1}{34.2} \right) \approx 4260 (0.01587 + 0.00292) \approx 80 \text{ cm}^3$.

Final value: $4260 \pm 80 \text{ cm}^3$.