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 vx. 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 = Nak. 

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

Finally, to calculate the avarage speed we find vrms(Root Meean Square of Speed).

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

M = Nam 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. \(R=8.314\frac{J}{molK}\)

One Response to “ Derivation of Average Speed of Gaseous Molecules ”

  1. 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.

Leave a Reply

You can use these XHTML tags: <a href="" title=""> <abbr title=""> <acronym title=""> <blockquote cite=""> <code> <em> <strong>