QJEEM24Phcf0f: JEE Main 2024 Physics - Class 11 JEEM24Phcf0f Solution: If the collision frequency of hydrogen molecules in a closed…

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:

$$Z_1=Z_0\sqrt{\frac{T_1}{T_0}}$$

where:

• $Z_1$ is the collision frequency at temperature $T_1$ ,
• $Z_0$ is the initial collision frequency at temperature $T_0$ ,
• $T_1$ is the final absolute temperature, and
• $T_0$ is the initial absolute temperature.

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):

• $T_0 = 27^{\circ}C = 27 + 273 = 300$ K
• $T_1 = 127^{\circ}C = 127 + 273 = 400$ K

Using the formula for collision frequency change and substituting the given values:

$$Z_1=Z_0\sqrt{\frac{400}{300}}=Z_0\sqrt{\frac{4}{3}}$$

This can be simplified to:

$$Z_1=Z_0\times \frac{2}{\sqrt{3}}$$

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:

$$\frac{2}{\sqrt{3}}\mathrm{Z}$$