Question

The rms velocity of molecules of oxygen gas is given by \( v = \sqrt{\frac{3kT}{m}} \) at some temperature \( T \). The molecules of another gas have the same rms velocity at temperature \( T' = \frac{T}{16} \). The second gas is:

• A. Hydrogen
• B. Helium
• C. Nitrogen
• D. Neon

Hint:

For equal rms velocities, \( \frac{T}{M} \) must be constant; lighter gases achieve same speed at lower temperature.

Answer 1

The Correct Option is A

Step 1: Use rms velocity relation.
\[ v = \sqrt{\frac{3kT}{m}}. \] For same rms velocity, \[ \sqrt{\frac{T}{m_{\text{O}_2}}} = \sqrt{\frac{T/16}{m_{\text{gas}}}}. \] Step 2: Simplify.
\[ m_{\text{gas}} = \frac{m_{\text{O}_2}}{16}. \] Since molecular mass of \( \text{O}_2 = 32 \), \[ m_{\text{gas}} = 2. \] This corresponds to hydrogen (\( \text{H}_2 \)).
Step 3: Final Answer.
Hence, the gas is hydrogen.