Derivation of the average translational kinetic energy of a molecule

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SUMMARY

The discussion focuses on the derivation of the average translational kinetic energy of a molecule, specifically addressing the relationship between the components of velocity. The user expresses confusion regarding the addition of vector components, particularly how the average of the squared velocities in three dimensions leads to the equation (v^2)av = 3(vx^2)av. The clarification provided emphasizes that the average squared velocity components are equal, allowing for the simplification of the equation.

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  • Understanding of basic physics concepts, particularly kinetic energy.
  • Familiarity with vector mathematics and operations.
  • Knowledge of statistical mechanics principles.
  • Proficiency in manipulating equations involving averages and squares.
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Students in physics, particularly those studying thermodynamics and statistical mechanics, as well as educators seeking to clarify concepts related to kinetic energy and vector mathematics.

NeuronalMan
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Homework Statement



Hello, this is not actually a homework problem. I just can't seem to understand the derivation of the average translational kinetic energy of a molecule. I am startled by the way the velocities are added.


Homework Equations




My undergraduate level textbook says that (vx^2)av = (vy^2)av = (vz^2)av, but then it says (v^2)av = (vx^2)av + (vy^2)av + (vz^2)av = 3(vx^2)av

How can velocities being vectors be added this way? I am surely missing something here.


The Attempt at a Solution

 
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I thinking you are missing the importance of subscripts. It is certainly true that the square of the magnitude of the 3-d velocity vector is always

v2=vx2+vy2+vz2

what does this become when you replace the components with their averages which are all equal to each other?
 

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