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An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is v̄, m is its mass and k_B is Boltzmann constant, then its temperature will be:
- mv̄²/6k_B
- mv̄²/3k_B
- mv̄²/7k_B
- mv̄²/5k_B
Correct answer: mv̄²/3k_B
Solution
The rms speed is set by translational motion: (1/2) m vbar^2 = (3/2) k_B T, so T = m vbar^2/(3 k_B). Rotational and vibrational modes do not enter the rms velocity.
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