A sample contains mixture of helium and oxygen gas. The ratio of root mean square speed of helium and oxygen in the sample, is :

Solution

Physics sir · 3 steps

Same sample means same temperature, so the rms speed depends only on the molar mass.

1. Same temperature. Helium and oxygen are mixed in one sample, so they are at the same temperature T. Only the molar masses differ.

   vᵣₘₛ = √(3RT/M) ⇒ v_(He)/v_(O₂) = √(M_(O₂)/M_(He))

2. Molar masses. Helium is monatomic, so M_(He) = 4 g mol⁻¹. Oxygen gas is diatomic O₂, so M_(O₂) = 32 g mol⁻¹.
3. So. v_(He)/v_(O₂) = √(32/4) = √8 = 2√2. The lighter helium moves faster, and the answer is (c).

The idea to keep. At the same temperature, vᵣₘₛ ∝ 1/√M. The lighter gas is always faster, so check that your ratio comes out greater than 1 when the lighter gas is on top.

Answer from the NTA final answer key. Solution by PrepWiser.