Question 234: Thermodynamics & KTG

Question

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 :

Accepted Answer

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:
$$Z_1 = Z_0 	imes \frac{2}{\sqrt{3}}$$
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:
$$\frac{2}{\sqrt{3}} \mathrm{Z}$$

Answer

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

Answer

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

Answer

$$\frac{3}{4} \mathrm{Z}$$

Answer

$$\frac{4}{3} \mathrm{Z}$$

Thermodynamics & KTG

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 :

Choose the correct answer: