If the collision frequency of hydrogen molecules in a closed chamber at $27^{\circ} \mathrm{C}$ is $\mathrm{Z}$, then the collision frequency of the same system at $127^{\circ} \mathrm{C}$ is :

<p>The collision frequency ($Z$) of gas molecules is proportional to the square root of the absolute temperature ($T$) of the system. Mathematically, it can be represented as $Z \propto \sqrt{T}$. This implies that when the temperature changes, the collision frequency changes as well according to the formula:</p> <p>$Z_1 = Z_0 \sqrt{\frac{T_1}{T_0}}$</p> <p>where:</p> <ul> <li>$Z_1$ is the collision frequency at temperature $T_1$,</li> <li>$Z_0$ is the initial collision frequency at temperature $T_0$,</li> <li>$T_1$ is the final absolute temperature, and</li> <li>$T_0$ is the initial absolute temperature.</li> </ul> <p>To solve the given problem, we first need to convert the provided temperatures from Celsius to Kelvin (since absolute temperature in Kelvin should be used):</p> <ul> <li>$T_0 = 27^{\circ}C = 27 + 273 = 300$ K</li> <li>$T_1 = 127^{\circ}C = 127 + 273 = 400$ K</li> </ul> <p>Using the formula for collision frequency change and substituting the given values:</p> <p>$Z_1 = Z_0 \sqrt{\frac{400}{300}} = Z_0 \sqrt{\frac{4}{3}}$</p> <p>This can be simplified to:</p> <p>$Z_1 = Z_0 \times \frac{2}{\sqrt{3}}$</p> <p>Thus, the collision frequency of the system at $127^{\circ}C$ is $\frac{2}{\sqrt{3}}$ times the collision frequency at $27^{\circ}C$, making Option B the correct answer:</p> <p>$\frac{2}{\sqrt{3}} \mathrm{Z}$</p>