NEET Physics Kinetic Interpretation of Temperature MCQ Question
If the root mean square (rms) speed of nitrogen molecules at 300 K is 516 m/s, what is the rms speed of oxygen molecules (O2) at the same temperature, given that the molecular mass of nitrogen (N2) is 28 u and oxygen (O2) is 32 u?
482 m/s
516 m/s
550 m/s
600 m/s
Correct Answer
Detailed Explanation
The rms speed is inversely proportional to the square root of the molecular mass. Therefore, v_rms(O2) = v_rms(N2) * sqrt(M(N2) / M(O2)) = 516 * sqrt(28/32) = 482 m/s.
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