Problem 15.6: Speed of particles in ideal gas

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In this animation N = nR (i.e., kB = 1). This, then, gives the ideal gas law as PV = NT. The average values shown, < >, are calculated over intervals of one time unit so that the average rate change in momentum is equal to the pressure times the area, A (where A = 1).  Restart.  This animation shows the distribution of speeds in an ideal gas based on the Maxwell-Boltzmann distribution as shown by the smooth black curve on the graph for a given temperature:

n(v) dv/N = (2/π)1/2  (m/kBT)3/2 v2 exp(−mv2/2kBT) .

What happens to the distribution as you increase the energy (temperature)?  Since there is a speed distribution, when we talk about a characteristic speed of a gas particle at a particular temperature, we use one of three characteristic speeds:

  • Average speed (<v> = ∫ vn(v)dv).
  • Most probable speed (find maximum of n(v)).
  • Root-mean-square (rms) speed (<v2>1/2 = [ ∫v2n(v)dv ]1/2).

Find an expression for each (in terms of m and kBT).  Identify which peak is which characteristic speed.