Question

Equal number of gas molecules A (mass 𝑚 and radius 𝑟) and B (mass 2𝑚 and radius 2𝑟) are placed in two separate containers of equal volume. At a given temperature, the ratio of the collision frequency of B to that of A is
(Assume the gas molecules as hard spheres)

• A. \(\sqrt2:1\)
• B. 2\(\sqrt2:1\)
• C. 1:\(\sqrt2\)
• D. 1:2\(\sqrt2\)

Answer 1

The Correct Option is B

Equal numbers of gas molecules A (mass $m$, radius $r$) and B (mass $2m$, radius $2r$) are placed in two separate containers of equal volume. At a given temperature, find the ratio of the collision frequency of B to that of A. (Assume hard-sphere molecules.)

Working

Collision frequency for a gas is proportional to $$Z \propto n \, \sigma \, \bar{v}$$ where • $n$ = number density (same for A and B since equal number & equal volume) • $\sigma$ = collision cross section • $\bar{v}$ = mean speed of molecules.

Step 1: Collision cross-section

For hard spheres: $$\sigma = \pi d^2 = \pi (2r)^2 = 4\pi r^2$$ For B, radius is $2r$, so $$\sigma_B = \pi (4r)^2 = 16\pi r^2$$ Ratio: $$\frac{\sigma_B}{\sigma_A} = \frac{16\pi r^2}{4\pi r^2} = 4$$

Step 2: Mean speed

Mean speed varies as $$\bar{v} \propto \frac{1}{\sqrt{m}}$$ For A: mass = $m$ For B: mass = $2m$ $$\frac{\bar{v}_B}{\bar{v}_A} = \frac{1/\sqrt{2m}}{1/\sqrt{m}} = \frac{1}{\sqrt{2}}$$

Step 3: Collision frequency ratio

$$\frac{Z_B}{Z_A} = \frac{\sigma_B}{\sigma_A} \cdot \frac{\bar{v}_B}{\bar{v}_A} = 4 \cdot \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$

Answer

Ratio of collision frequencies = $2\sqrt{2} : 1$
✔ Correct option: Option 2