Question

The temperature at which oxygen molecules will have same r.m.s. speed as helium molecules at 57$^\circ$C is (molecular masses of oxygen and helium are 32 and 4 respectively.) ______.

• A. 1320 K
• B. 2240 K
• C. 2640 K
• D. 3230 K

Hint:

Heavier molecules are sluggish! To make a massive Oxygen molecule ($32$ g/mol) move as fast as a tiny Helium atom ($4$ g/mol), you must heat it up by the exact ratio of their masses ($32/4 = 8$ times hotter in Kelvin). $330 \text{ K} \times 8 = 2640 \text{ K}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The root mean square speed ($v_{rms}$) is given by $\sqrt{\frac{3RT}{M}}$. For the speeds to be equal, $\frac{T_1}{M_1} = \frac{T_2}{M_2}$.

Step 2: Formula Application:
Temperature of Helium ($T_{He}$) = $57 + 273 = 330$ K. $M_{He} = 4$, $M_{O_2} = 32$. $\frac{330}{4} = \frac{T_{O_2}}{32}$.

Step 3: Explanation:
$T_{O_2} = \frac{330 \times 32}{4}$ $T_{O_2} = 330 \times 8 = 2640$ K.

Step 4: Final Answer:
The temperature is 2640 K.