Calculating thermal kinetic energy of a liquid.

In summary, the formula for calculating the speed of atoms in a liquid due to thermal motion is (3/2)kT, as it is for gases. This is known as the equipartition theorem in classical statistical mechanics. However, this rule does not hold true in quantum analysis, as shown by the example of a quantum harmonic oscillator. Additionally, in liquids, the interactions between particles must also be taken into account when calculating average kinetic energy, making the classical method less accurate.
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I want to know, given a liquid at a particular temperature and pressure, how fast the atoms in it are moving due to thermal motion. I know that you can do this by calculating the thermal kinetic energy of the atoms, and then figure out the speed from there.

I also know that for gases the formula is (3/2)kT, but is the same true for liquids? If not, then what is the formula?
 
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  • #2
Yes, the same is true for liquids.

It's because the kinetic energy term is quadratic in momentum and can always be factored out of the total partition function- Q=Q(KE) * Q(PE)

Any quadratic degree of freedom has an energy of 1/2 k T, a result generally referred to as the equipartition theorem.

However, the 1/2 kT rule is only true in classical statistical mechanics. It is not true in a quantum analysis.

For example, a quantum harmonic oscillator in its ground state has energy E=1/2 hbar w0, with K.E.=1/4 hbar w0, which is not equal to 1/2 kT.
 
  • #3
I think you could easily show that the average kinetic energy is proportional to T, but in liquids particularly, you can't neglect the interactions of the various liquid particles. Again, as christian said, this is for the classical method.
 

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