Derivation of Average Speed of Gaseous Molecules

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

P=NVmv2‾3=NV2Ekinetic3=NVkT=NkTVP=\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. NVmv2‾3=NVkT\frac{N}{V}\frac{m\overline{v^2}}{3}=\frac{N}{V}kT

  2. Canceling out NV\frac{N}{V} from both sides we get:

    mv2‾3=kT⇒v2‾=3kTm\frac{m\overline{v^2}}{3}=kT\Rightarrow{\overline{v^2}=\frac{3kT}{m}}

  3. Since R = Nak:

    v2‾=3(R/Na)(T)m=3RTNam\overline{v^2}=\frac{3(R/N_a)(T)}{m}=\frac{3RT}{N_{a}m}

  4. Finally, to calculate the average speed we find vrms (Root Mean Square of Speed):

    vrms=v2‾=3RTNam=3RTMv_{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 convert it to (J)/(mol)(K).

R=8.314JmolKR=8.314\frac{J}{molK}