Question

What is the rms speed of molecules of water vapor in air at 0°C?
My answer is : v=√（6kT/m)=869.84m/s,wherw mis is the mass of a single H2O molecule. But the answer in my book is 615m/s,why didn't the book add the rotational energy of the molecule? At what temperature the molecule gets the rotational energy?

Answer 1

I believe the idea here is that temperature is the measure of the translational kinetic energy of molecules. You are correct that based on the equipartition theorem that you would get a total thermal energy of 6kT. But since you only consider translational contribution, you get 3kT

Answer 2

@michele_laino

Answer 3

I think that it is necessary to apply this formula:
\[{v_{rms}} = \sqrt {\frac{{3kT}}{m}} = \sqrt {\frac{{3 \cdot 1.38 \cdot {{10}^{ - 16}} \cdot 273}}{{18 \cdot 1.67 \cdot {{10}^{ - 24}}}}} = ...?\]

Answer 4

Why don't we imclude the rotational degrees of freedom?

Answer 5

I think,maybe the 273.15K temperature is not high enough to let the water molecules attain the rotational energy because the energy is discrete according to the quantum theory.

Answer 6

@michele_laino can you pls help?

Answer 7

substantially, the oscillatory and rotatory motions of the molecule of water, can be neglected at low temperatures. Furthermore the internal degrees of freedom of a molecule are like "frozen" at low temperatures, and they contribute to the thermal capacity of a gas, only at sufficiently high temperatures.
Of course such effects, related to the internal degrees of freedom of a molecule, can be described using the Quantum Theory only