Relation between kinetic energy and temperature

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The term 3/2 in the equation \(\frac{1}{2} mv^2 = \frac{3}{2} kT\) arises from the three translational degrees of freedom of a particle, each contributing \(\frac{1}{2} kT\) to the total kinetic energy. This relationship is explained by the equipartition theorem, which states that energy is equally distributed among all degrees of freedom. The derivation involves equating two ideal gas equations: one derived experimentally (PV=RT) and the other from kinetic theory (PV=Nmc²/3). The Hyperphysics site is recommended for further exploration of these concepts. Understanding this relationship is crucial for grasping the fundamentals of kinetic theory and thermodynamics.
johnathon
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Where does the 3/2 come from?
\frac{1}{2} mv^2 = \frac{3}{2} kT
 
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A particle can move in any of three directions (that's where the 3 comes from), with kt/2 being the kinetic energy carried by motion on each the x,y or z dimensions.
This link gives a short and sweet bit of book work. That Hyperphysics site is good for many things, actually.
 
johnathon said:
Where does the 3/2 come from?
\frac{1}{2} mv^2 = \frac{3}{2} kT
There are three translational degrees of freedom, each contributing 1/2kt to the total energy. This from the equipartition theorem.
 
Taking it back a step the 3kT/2 can be found by equating the two ideal gas equations,one being obtained experimentally(PV=RT) the other being obtained theoretically using kinetic theory(PV=Nmc bar squared/3)
 

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