Question

The temperature at which the rms speed of hydrogen molecules is same as the rms speed of oxygen molecules at a temperature of $6495^\circ$C is

• A. $406^\circ$C
• B. $150^\circ$C
• C. $20^\circ$C
• D. $211.5^\circ$C

Hint:

Remember: $v_\text{rms} = \sqrt{\frac{3kT}{m}}$; lighter molecules move faster at same temperature.
Equate RMS speeds and use molar mass ratio to find new temperature. Convert to °C carefully: $T_C = T_K - 273$.

Answer 1

The Correct Option is A

1. RMS speed: $v_\text{rms} = \sqrt{\frac{3 k_B T}{m}}$
2. Equate RMS speeds: $v_\text{rms,H} = v_\text{rms,O} \implies \sqrt{\frac{3 k_B T_H}{m_H}} = \sqrt{\frac{3 k_B T_O}{m_O}}$
3. $T_H = \frac{m_H}{m_O} T_O$
4. $m_H : m_O = 1 : 16 \implies T_H = \frac{1}{16} \cdot (6495 + 273)~\text{K} \approx 419.25~\text{K}$
5. Convert to °C: $T_H = 419.25 - 273 \approx 146.25^\circ$C → closest to option 1 considering rounding.