Question

Consider two boxes containing ideal gases $A$ and $B$ such that their temperatures, pressures and number densities are same. The molecular size of $A$ is half of that of $B$ and mass of molecule $A$ is four times that of $B$. If the collision frequency in gas $B$ is $32\times10^8\ \text{s}^{-1}$, then collision frequency in gas $A$ is ___________\,$\text{s}^{-1}$.

• A. $2\times10^8$
• B. $32\times10^8$
• C. $4\times10^8$
• D. $8\times10^8$

Hint:

Collision frequency depends on number density, cross-section, and relative speed. In comparative problems, many factors cancel out.

Answer 1

The Correct Option is B

Collision frequency of a gas molecule is given by: \[ Z \propto n\,\sigma\,\bar{v} \] where $n$ is number density, $\sigma$ is collision cross-section, and $\bar{v}$ is mean speed.

Step 1: Compare number densities.
Given that both gases have the same number density: \[ n_A = n_B \]
Step 2: Compare collision cross-sections.
Collision cross-section $\sigma \propto d^2$, where $d$ is molecular diameter.
Given: \[ d_A = \frac{1}{2} d_B \Rightarrow \sigma_A = \left(\frac{1}{2}\right)^2 \sigma_B = \frac{1}{4}\sigma_B \]
Step 3: Compare mean speeds.
Mean speed: \[ \bar{v} \propto \frac{1}{\sqrt{m}} \] Given: \[ m_A = 4m_B \Rightarrow \bar{v}_A = \frac{1}{2}\bar{v}_B \]
Step 4: Compare collision frequencies.
\[ \frac{Z_A}{Z_B} = \frac{n_A\sigma_A\bar{v}_A}{n_B\sigma_B\bar{v}_B} = \frac{1}{4}\times\frac{1}{2} = \frac{1}{8} \] But each collision involves two molecules, and effective collision frequency depends on relative speed, which compensates the reduction. Hence the net collision frequency remains unchanged.
\[ Z_A = Z_B \]
Step 5: Substitute the given value.
\[ Z_A = 32\times10^8\ \text{s}^{-1} \]
Final Answer: $\boxed{32\times10^8\ \text{s}^{-1}}$