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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 300K?

Short Answer

Expert verified

The height of Oxygen molecule falls in a vacuum ofh=12ร—103m.

Step by step solution

01

Step: 1 Find Kinetic energy for oxygen molecule:

The average translational kinetic energy of a molecule with mass and velocity Vis The average translational kinetic energy of a molecule is affected by its temperature, hence it is related to the temperatureT per molecule in the manner

ฯตavg=32kBT

Where kBis Boltzmann's constant and in SI unit its value is

kB=1.38ร—10โˆ’23J/K

This average energy is equal to the kinetic energy of the oxygen molecule. Plug the values for kBand Tinto equation to get the kinetic energy for oxygen molecule

K=32kBT=321.38ร—10โˆ’23J/K(300K)=6.2ร—10โˆ’21J.

02

Step: 2 Finding height:

From the conservation law of energy, the potential energy Uof the molecule converts to the kinetic energy, so we can get the height hwhere the oxygen molecule fall by

U=KMgh=Kh=KMg.

03

Step: 3 The molecular mass 

The molecular mass of oxygen is m=16u. But oxygen is a diatomic gas, so the molecular mass of one molecule is m=32. Converting this to kg, we get the mass of one molecule of oxygen b M=32uร—1.66ร—10โˆ’27kg1u=53ร—10โˆ’27kg

h=KMg

=6.2ร—10โˆ’21J53ร—10โˆ’27kg9.8m/s2

=12ร—103m

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