Physics Friday 68

Maxwell Speed Distribution

Previously, we described the Boltzmann factor. Multiplying a normalization constant to give an actual probability, we obtain the Boltzmann distribution:
.
Now, suppose that we have a system of tiny non-interacting particles (an ideal gas), with the energy being purely kinetic. Thus, the probability that a particle of mass m has a speed v is proportional to .
However, in 3-dimensional velocity space, the velocity vectors that give the same speed form a sphere, with radius v; the higher the speed v, the larger the number of possible velocity vectors there are. Thus, the distribution of speeds is proportional to , the surface area of the sphere in velocity space. Combining, we have
. Normalizing, we have
,
(where we’ve used the fact that )
so
.
This is the Maxwell speed distribution. Now, we can find three important speeds with this:
I. Most probable speed:
This is vp>0 where . The derivative is
,
and this is zero for vp>0 when
.

II. Mean speed
.

III. Root-mean-square speed:
.

The last of these gives us the average kinetic energy of ideal gas molecules:

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3 Responses to “Physics Friday 68”

  1. Physics Friday 69 « Twisted One 151’s Weblog Says:

    […] speed. This means that the pressure is the same on all sides of the box, and . Now, the Maxwell speed distribution tells us that for ideal gas molecule, the average kinetic energy is given by . This allows us to […]

  2. Physics Friday 70 « Twisted One 151’s Weblog Says:

    […] Friday 70 By twistedone151 We noted previously that for an ideal gas, the molecule speeds should follow the Maxwell speed distribution; from this, […]

  3. Physics Friday 72 « Twisted One 151’s Weblog Says:

    […] we previously derived the ideal gas law PV=NkT=nRT from a particle bouncing in a box, we used the Maxwell speed distribution, which was in turn derived using the Boltzmann factor with regards to the kinetic energy of the […]

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