Equipartition of Energy
Most recent answer: 06/22/2011
Q:
If heavier gas molecules move more slowly than light gas molecules, why is the average kinetic
energy independent of the mass
- Susovon (age 17)
Suri.West Bengal ,India
- Susovon (age 17)
Suri.West Bengal ,India
A:
You can see how it might work out, since for slow-moving particles the kinetic energy is mv2/2, if particles with bigger m tend to have smaller v, then each particle could have the same average mv2/2. That is in fact exactly how it turns out in thermal equilibrium. This is part of what's known as the equipartition theorem.
So why is it true that <v2> (meaning the average of v2 over time) is just proportional to 1/m at some fixed absolute temperature T? Here's the outline of the core of the argument.
In equilibrium, the probability that any state will be occupied is proportional to e-(E/kT), where E is the state energy and k is a constant (Boltzmann's). The proof of this statement is easy, but we will skip it here. Then one needs that for particles moving much slower than the speed of light the number of states near any particular velocity v is independent of v. Let's call the number of states per little volume in v-space f. So the average of the kinetic energy can be obtained by integrating (mv2/2)e^-(mv2/2kT)*f over all v, then dividing by the integral of e^-(mv2/2kT) *f over all v. That's obviously independent of f and it's easy to check that that is also independent of m. In 3-D, it's also easy to check that the answer comes out to be 3kT/2.
Here's a way to look at it. Draw a 3-D space to represent the possible v's. The probabilities of states with small v's are highest. The probability starts to fall off when v gets big enough for E to be around kT. So right away you can see that the average energy is around kT, and that m has nothing to do with it. The integrals are just to get the exact number right.
Mike W.
please follow up if there's some particular aspect here which needs to be clarified.
So why is it true that <v2> (meaning the average of v2 over time) is just proportional to 1/m at some fixed absolute temperature T? Here's the outline of the core of the argument.
In equilibrium, the probability that any state will be occupied is proportional to e-(E/kT), where E is the state energy and k is a constant (Boltzmann's). The proof of this statement is easy, but we will skip it here. Then one needs that for particles moving much slower than the speed of light the number of states near any particular velocity v is independent of v. Let's call the number of states per little volume in v-space f. So the average of the kinetic energy can be obtained by integrating (mv2/2)e^-(mv2/2kT)*f over all v, then dividing by the integral of e^-(mv2/2kT) *f over all v. That's obviously independent of f and it's easy to check that that is also independent of m. In 3-D, it's also easy to check that the answer comes out to be 3kT/2.
Here's a way to look at it. Draw a 3-D space to represent the possible v's. The probabilities of states with small v's are highest. The probability starts to fall off when v gets big enough for E to be around kT. So right away you can see that the average energy is around kT, and that m has nothing to do with it. The integrals are just to get the exact number right.
Mike W.
please follow up if there's some particular aspect here which needs to be clarified.
(published on 06/22/2011)