NTA Abhyas JEE Main2020PhysicsKinetic Theory of GasesPractice
The temperature at which a molecule of nitrogen will have the same rms velocity as a molecule of oxygen at 127 ° C is
Options
- A457   ℃
- B273   ℃
- C350   ℃
- D77   ℃
Correct answer
D. 77   ℃
Step-by-step solution
Root mean square velocity v r m s = 3 R T M where R is gas constant, T the temperature and M molecular weight. Given, M N 2 = 28 ,         M O 2 = 32 ,       T O 2 = 127   ℃ = 127 + 273 = 400 K ∴                                 v O 2 v N 2 = T O 2 M O 2 × M N 2 T N 2 = 400 32 × 28 T N 2 = 1 ⇒                       &