Question:
At what temperature is the root mean square (rms) speed of oxygen molecules ($O_2$) equal to that of helium molecules ($He$) at $27^\circ\text{C}$? (Given molar mass of $O_2 = 32\text{ g/mol}$, $He = 4\text{ g/mol}$)}
At what temperature is the root mean square (rms) speed of oxygen molecules ($O_2$) equal to that of helium molecules ($He$) at $27^\circ\text{C}$? (Given molar mass of $O_2 = 32\text{ g/mol}$, $He = 4\text{ g/mol}$)}
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Since oxygen is 8 times heavier than helium ($32 / 4 = 8$), its absolute temperature must be exactly 8 times higher to maintain the same rms speed: $8 \times 300\text{ K} = 2400\text{ K}$.
Updated On: May 31, 2026
- $2400\text{ K}$
- $2127\text{ K}$
- $2427^\circ\text{C}$
- $2127^\circ\text{C}$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The root mean square (rms) velocity of gas molecules is given by $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$, where $T$ is the temperature in Kelvin and $M$ is the molar mass.
Step 2: Meaning
For the rms velocities of two gases to be equal, their ratio of absolute temperature to molar mass must be equal: $\frac{T_1}{M_1} = \frac{T_2}{M_2}$.
Step 3: Analysis
Let the temperature of oxygen be $T_1$, with molar mass $M_1 = 32\text{ g/mol}$. The temperature of helium is $T_2 = 27^\circ\text{C} = 27 + 273 = 300\text{ K}$, with molar mass $M_2 = 4\text{ g/mol}$. Equating the ratios: \[ \frac{T_1}{32} = \frac{300}{4} \implies T_1 = \left(\frac{32}{4}\right) \times 300 = 8 \times 300 = 2400\text{ K}\]
Step 4: Conclusion
The rms speed of oxygen molecules is equal to that of helium at $2400\text{ K}$. Final Answer: (A)
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