# Degrees of freedom and RMS-speed of CO2?

1. Feb 14, 2010

### Thymo

Are the RMS-speed of gas particles related to degrees of freedom?

I give up! I must consult help no matter how embarrassing it is!
Any help is greeted with a big smile!

Does the formula for root mean square speed of particles in a gas (below) apply for all particles?

$$v_{rms}=sqrt{frac{3k_{b}T}{m}}$$

I understand that it's derived from the kinetic energy of monatomic gases:

1) $$E_{k}=\frac{3}{2}k_{b}T$$

2) $$\frac{1}{2}mv_{rms}^2=\frac{3}{2}k_{b}T$$

3) $$v_{rms}=\sqrt{\frac{3k_{b}T}{m}}$$

However, the formula of the kinetic energy of diatomic gases is (from Cappelen's "Rom Stoff Tid: Fysikk 1"):
$$E_{k}=\frac{5}{2}k_{b}T$$

Thus the RMS speed must be:

$$v_{rms}=\sqrt{\frac{5k_{b}T}{m}}$$

No?

Monatomic particles have three translational degrees of freedom, diatomic particles have three translational and two rotational (as they are linear and the rotation around the axis that pierces both particles are "freezed out"), oui?
Is this what is reflected in their formulas for kinetic energy?

The research and lack of sleep resulted in this conclusion:

$$v_{rms}=\sqrt{\frac{D_{f}k_{b}T}{m}}$$

Where $$D_{f}$$ stand for degrees of freedom

So the RMS speed for carbon dioxide in 23$$\circ$$C must be:
$$v_{rms}=\sqrt{\frac{5 \times\ 1.38\ \times\ 10^{-23} \ J \ K^{-1} \times\ 296K}{44 \times\ 1.66 \times\ 10^{-27} \ kg}}$$

5 degrees of freedom comprising three translational, two rotational (linear molecule) and 0 vibrational as they are negliguble at room temperature.

$$v_{rms}=530\frac{m}{s}=1900\frac{km}{h}$$

Now, I will be very happy for any feedback (link to a site or anything) on whether my reasoning or calculation is correct or wrong.

~~~~ Thymo

PS: This is not homework, I'm in first grade physics.

Last edited: Feb 14, 2010
2. Feb 14, 2010

### Thymo

Well, thanks a lot! :grumpy:

I posted the same question in norwegian at a Norwegian site, Realisten.com, and got two replies before this garbagee could produce ONE!

Hehe...

Anyways, I'm here to update about what I know now.

The speed of particles are always given by:
$$v_{rms}=\sqrt{\frac{3k_{b}T}{m}}$$

The "3" on the right side reflects the three translational degrees of freedom. It does not reflect the total degrees of freedom because speed is translational movement. The kinetic energy in the form of vibrational and rotational does not result in speed.

I still have a slight puzzle though.
If:
$$E_{k}=\frac{5}{2}k_{b}T$$
For CO2 particles.

Doesn't that imply that:
$$\frac{1}{2}mv^2=\frac{5}{2}k_{b}T$$
too?

But if we solve for $$v$$, we get:
$$v=\sqrt{\frac{5k_{b}T}{m}}$$

Which is wrong. So, is the $$E_{k}$$ for CO2 different from the normal formula for $$E_{k}$$?

Any help is much appriciated!

3. Feb 14, 2010

### conway

Norway must be some country.

Your formula with the 5/2 is giving you the kinetic plus the rotational. I think you know that. The formula with the 3/2 will give you the "center-of-mass" kinetic energy, not counting anything else the molecule is doing.