Question

If r.m.s. velocity of hydrogen molecules is 4 times that of an oxygen molecule at $47^{\circ}\text{C}$, the temperature of hydrogen molecules is (Molecular weight of Hydrogen and Oxygen are 2 and 32 respectively)}

• A. $23^{\circ}\text{C}$
• B. $47^{\circ}\text{C}$
• C. $80^{\circ}\text{C}$
• D. $114^{\circ}\text{C}$

Hint:

If the ratio of velocities squared equals the inverse ratio of molar masses, the temperatures must be identical.

Answer 1

The Correct Option is B

Step 1: Concept
The root mean square (r.m.s.) velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.

Step 2: Meaning
Let $v_H$ and $v_O$ be the r.m.s. velocities of hydrogen and oxygen respectively. Given $v_H = 4v_O$, $M_H = 2$, $M_O = 32$, and $T_O = 47 + 273 = 320 \text{ K}$.

Step 3: Analysis
$\frac{v_H}{v_O} = \sqrt{\frac{T_H}{M_H} \cdot \frac{M_O}{T_O}} \implies 4 = \sqrt{\frac{T_H}{2} \cdot \frac{32}{320}}$.
$16 = \frac{T_H}{2} \cdot \frac{1}{10} \implies T_H = 16 \times 20 = 320 \text{ K}$.

Step 4: Conclusion
$T_H = 320 - 273 = 47^{\circ}\text{C}$. Final Answer: (B)