Degree of Freedom Formula for Molecules with n Atoms

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Homework Help Overview

The discussion revolves around determining the formula for the degrees of freedom of molecules with n atoms, specifically focusing on translation, rotation, and vibration. Participants are exploring the distinctions between monatomic and diatomic gases in this context.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the degrees of freedom associated with different types of molecules, questioning whether the values of 3n for monatomic and 5n for diatomic gases pertain to translation, rotation, or vibration.

Discussion Status

The conversation is ongoing, with some participants providing insights into the degrees of freedom for monatomic and diatomic gases. There is a request for further clarification on the relationship between translational and rotational degrees of freedom, indicating a productive exploration of the topic.

Contextual Notes

Participants are discussing specific types of gases (monatomic and diatomic) and their properties, which may influence the degrees of freedom. There is an emphasis on understanding the underlying physics rather than simply applying formulas.

kidia
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Please I need help.

Can anybody can give me the formula for determine the number of degree of freedom of molecules undergoing translation,rotation and vibration of n number of atom.
 
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it depends on what the molecules are. monatomic? (it's 3n.) diatomic? (it's 5n). more than that, i dunno.
 
rbj can u clariyfy more if that 3n for monotomic and 5n for diatomic is for translation,rotation or vibration
 
kidia said:
rbj can u clariyfy more if that 3n for monotomic and 5n for diatomic is for translation,rotation or vibration

sorry, kidia. they moved the thread but they didn't leave a little "moved" arrow behind.

for monatomic gases (He Ne Ar, the inert gasses), each molecule is a single atom and virtually all of the mass of the molecule is concentrated at the nucleus. they're a simple ball with the mass all concentrated at the center. even if you were to spin the ball, there would be very little rotational kinetic energy in that spin because the mass is all at the center. no moment of inertia. so these molecules have 3 degrees of freedom of translation, and only those three. up-down (z-axis), left-right (x-axis), and forward-backward (y-axis). no rotation or vibration.

diatomic gasses (O2 N2) have two identical atoms bonded together. for each atom, the mass is concetrated at the nucleus. so this structure is like a dumbell structure. besides the 3 translational motions (x, y, z-axis) that the monatomic gasses have, there are 2 more rotational degrees of freedom. imagine the dumbell lined up on the z-axis. there would be a non-zero moment of inertia along the x-axis and along the y-axis, but not along the z-axis.

it's obvious (due to symmetry) why the 3 translational degrees of freedom should be the same (contain the same average kinetic energy) and why the 2 identical rotational degrees of freedom contain the same average kinetic energy, but someone else will have to explain why the 2 rotational degrees of freedom contain the same amount of average kinetic energy per degree of freedom as the 3 translational degrees of freedom. can a real physicist explain that?
 
thanx very much rbj I catch u.
 

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