Maxwell's distribution
f = e^(-E/kT) where E = (1/2) mv^2
can be looked at in different ways. It is a Chi Square distribution with
respect to velocity v, and exponential with respect to kinetic energy E.
The _most likely (mode)_ kinetic energy is zero but the _mean_ kinetic
energy is not zero . The distribution decays exponentially with higher
energies.
George
On 11/20/2014 6:13 PM, meekerdb wrote:
If it were the momentum or velocity the mean would be zero, but it
wouldn't be exponential. If you just considered the speed (absolute
magnitude of velocity) in a particular direction you get an
exponential distribution. Is that what the graph represents?
Brent
On 11/20/2014 5:03 PM, LizR wrote:
The average kinetic energy of an air molecule is zero, I imagine,
because they're all travelling in different directions and cancel
out? Or doesn't it work like that?
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