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N molecules each of mass m of a gas A and 2N molecules each of mass 2m of gas B are contained in the same vessel which is maintained at temperature T. The mean square velocity of molecules of B type is v² and the mean square rectangular component of the velocity of A type is denoted by ω². Then ω²/v²
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- 2
- 1
- 1/3
- 2/3
Correct Option: D
Mean kinetic energy of the two types of molecules should be equal. The mean square velocity of A type molecules = ω² + ω² + ω² = 3ω²
| Therefore, | m(3ω²) = | (2m)v² | ||
| 2 | 2 |
This gives ω²/v² = 2/3
