## Derivation of Average Speed of Gaseous Molecules

We can start by analyzing equation #15 from the Derivation of Ideal Gas Law below:

\(P=\frac{N}{V}\frac{m\overline{v^2}}{3}=\frac{N}{V}\frac{2E_{kinetic}}{3}=\frac{N}{V}kT=\frac{NkT}{V}\)

Because we want to derive the equation to the speed of gaseous molecules, the most important factor come to mind is the speed v_{x}. Therefore, the equation will be

1. \(\frac{N}{V}\frac{m\overline{v^2}}{3}=\frac{N}{V}kT\)

Canceling out \(\frac{N}{V}\) from both side we get:

2. \(\frac{m\overline{v^2}}{3}=kT\Rightarrow{\overline{v^2}=\frac{3kT}{m}}\)

Since R = N_{a}k.

3. \(\overline{v^2}=\frac{3(R/N_a)(T)}{m}=\frac{3RT}{N_{a}m}\)

Finally, to calculate the avarage speed we find v_{rms}(Root Meean Square of Speed).

**4.** \(v_{rms}=\sqrt{\overline{v^2}}=\sqrt{\frac{3RT}{N_{a}m}}=\sqrt{\frac{3RT}{M}}\)

M = N_{a}m in the equation above is the mass of one mole of molecules (the molecular mass).

* The gas constant R must be expressed in correct units for the situation in which it is being used. In the ideal gas equation where pv=nRT, it is logical to use units (L)(atm)/(mol)(K).

*In regard to speed, however, energy unit must be taken into account. Therefore, it is more appropriate to to convert it to (J)/(mol)(K).

**A. **\(R=8.314\frac{J}{molK}\)

Please be careful with this. The RMS speed IS NOT equivalent to the average speed. In fact, the RMS speed will always be greater than the average speed.