O2 molecues speed in air, and their limit Temperature is the proportional measure of kinetic energy of the random motion of the constituent micro particles in a system as per wikipedia.
Now I understand that O2 molecules are randomly moving in air at normal temperatures.
I could not find if we could define the speed of these molecules at a certain temperature.
Obviously these particles with rest mass cannot move as fast as light. So they must cause a max limit to temperature.
I wanted to know if there was a limit to the temperature of air (gases) due to this speed limit of their constituent gas molecules? Is this temperature the absolute max (like absolute min 0 K)? Have we ever seen anything close to this in the universe?
Question:


*

*What is the speed of the random motion of O2 molecules in air at normal temperature (20Celsius)

*Is there a function or equation that would describe the connection directly between this speed of the O2 molecules and the temperature of the air they are in? Does this speed (must be smaller then c) give a limit to the absolute max temperature?
 A: The quantity you're looking for is the rms (root-mean-square) speed.
The relationship is given by $$v_{rms} = \sqrt{\frac{3RT}{M}}$$ where R is the gas constant, T is the temperature and M is the molar mass. 


*

*Using the equation above and the values $$R = 8.3145 \frac{kg\ m^2}{s^2\ K\ mol},\ \ \ T = 293.15\ K,\ \ \ M = 3.2\times10^{-2}\ kg$$ you can plug in and evaluate for the quantity you're looking for. Note that the molar mass is 32 g/mol because oxygen is a diatomic molecule.

*The equation above is a description of a statistical relationship between temperature and speed. As I understand, there is a theoretical maximum possible temperature according to the laws of physics as we know them, I think posited by Max Planck.
A: "I could not find if we could define the speed of these molecules at a certain temperature." 
Here is the non-relativistic answer. Temperature is related to the energy per degree of freedom by $E=\frac{1}{2}k_BT$. $k_B$ is Boltzmann's constant. The speed of an oxygen molecule represents 3 degrees of freedom, one for each dimension or direction. Thus the total kinetic energy is $E_k=\frac{3}{2}k_BT = \frac{1}{2} m(v-\left<v\right>)^2$ from which you find that $v-v\left<v\right>=\sqrt{3k_BT/m}$. At 20'C this gives 477 m/s. It is important to consider the deviation from average $v-\left<v\right>$ because the average velocity does not affect the temperature. 
Your question triggered me to check out the relativistic generalization. The definition of T now becomes $3 k_B T   =   \left< ( v − \left<v\right> ) . ( p − \left<p\right> ) \right>$ according to this site http://numericana.com/answer/heat.htm#temperature. If the average speed is zero, in the rest frame of the gas, $3 k_B T   =   \left< m\gamma v^2 \right>$. Clearly there is no upper limit to T, as $\gamma$ is unbounded.
