Home/
Competitive Exams/
JEE Main & Advanced/
Physics/
Kinetic Theory and Thermodynamics/
Question 31064
JEE Physics · Medium

Kinetic Theory and Thermodynamics: RMS Speed MCQ

Solve this quality-checked JEE multiple-choice question, then review the correct answer and explanation.

Timer-based practiceDetailed answer reviewMobile-friendly flow
RMS SpeedMediumQuestion 31064

Question

At what temperature is RMS speed of nitrogen molecules equal to RMS speed of oxygen at 300 K? (M_N₂=28, M_O₂=32)
  1. A
    262.5 K
    Correct
  2. B
    300 K
  3. C
    343 K
  4. D
    400 K

Correct answer

262.5 K

Explanation

v_rms ∝ √(T/M). √(T_N₂/28) = √(300/32). T_N₂ = 300×28/32 = 262.5 K.

Review Kinetic Theory and Thermodynamics notesStart timed chapter practice