Conservation of energy of an oxygen molecule

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Homework Help Overview

The discussion revolves around a problem in the context of energy conservation, specifically regarding the kinetic energy of an oxygen molecule falling from a height in a vacuum. The original poster attempts to determine the height required for the kinetic energy at the bottom to match the average energy of an oxygen molecule at a specified temperature.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the conservation of energy principle, questioning the initial calculations and assumptions regarding the average energy of the oxygen molecule. There is also a suggestion to reconsider the formula used for Eavg based on the molecular characteristics of oxygen.

Discussion Status

The discussion includes attempts to clarify the correct approach to calculating the average energy of the oxygen molecule, with some participants suggesting adjustments to the formula based on its diatomic nature. There is no explicit consensus on the correct answer, but guidance has been offered regarding the formula to use.

Contextual Notes

Participants note constraints related to the timing of the homework deadline and the limitations of their computational tools, which may affect their ability to verify their answers before submission.

jincy34
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[SOLVED] conservation of energy

Homework Statement



From what height must an oxygen molecule fall in a vacuum so that its kinetic energy at the bottom equals the average energy of an oxygen molecule at 200 K?

Homework Equations



conservation of energy
Eavg=3/2KbT
Epot=mgh

The Attempt at a Solution



I used conservation of energy and set
Epot.=Eavg. and got 7855.8 m. But it is wrong.
 
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What is the correct ans? This one seems to be right. Or my computer's not working.
 
I get to see the answer only after it is due. It is due thursday. when i tried this answer, the computer says it is wrong.
 
Eavg=3/2KbT
Oxygen molecule is a diatomic molecule. Therefore Eavg = 5/2kbT. Try this.
 
Thank you.
 
rl.bhat said:
Eavg=3/2KbT
Oxygen molecule is a diatomic molecule. Therefore Eavg = 5/2kbT. Try this.

Oy. Forgot about that altogether. Thanks from me too.
 

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