Question

A flask of volume 0.1 m\(^3\) contains He (monatomic) and O\(_2\) (diatomic) gases in the ratio 4:1. The flask is maintained at a temperature 27 °C. The ratio of the rms speeds of the He-atoms and O\(_2\)-molecules is:
(Atomic mass of He = 4 amu, and O\(_2\) = 32 amu)

• A. 2 : 1
• B. 4 : 1
• C. \(2 \sqrt{2} : 1\)
• D. \(\sqrt{2} : 1\)

Hint:

RMS speed depends on particle mass: \(v_\text{rms} \propto 1/\sqrt{m}\). For gases at the same temperature, lighter particles move faster. Multiply by factor if multiple types of particles exist in mixture.

Answer 1

The Correct Option is C

Step 1: Recalling the rms speed expression.
v_rms = √(3k_B T/m), where m denotes the mass of a single gas particle.

Step 2: Particle count and masses.
The flask holds He and O₂ in a 4:1 numerical ratio. Helium atomic mass = 4 amu, oxygen molecular mass = 32 amu.

Step 3: Ratio computation.
v_rms(He)/v_rms(O₂) = √(m_O₂/m_He) = √(32/4) = √8 = 2√2.

Step 4: Final result.
Hence, the rms speed ratio of He atoms to O₂ molecules is 2√2 : 1.