Confined in a box, what is the average distance between a particle hitting a side?
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Answered by
3
Heya....
Let's prove....
Average kinetic energy of gas molecule hitting a container surface is 2kBT instead of entire gas energy that is, 32kBT....
Where, KB is Boltzmann constant and T is thermodynamics temperature...
So,,,
d¢(v, ¢)= 12vnf~(v)sin(¢)cos(¢)dvd¢
where,, f~is maxwells speed distribution...
{£} we get....
(£) fπ¢f00mv34vsin(¢)cos(¢)
where,,
(£) = 42-√kBT 3/23m= √ not equals to 2kBT...
-- Be Brainly....
Let's prove....
Average kinetic energy of gas molecule hitting a container surface is 2kBT instead of entire gas energy that is, 32kBT....
Where, KB is Boltzmann constant and T is thermodynamics temperature...
So,,,
d¢(v, ¢)= 12vnf~(v)sin(¢)cos(¢)dvd¢
where,, f~is maxwells speed distribution...
{£} we get....
(£) fπ¢f00mv34vsin(¢)cos(¢)
where,,
(£) = 42-√kBT 3/23m= √ not equals to 2kBT...
-- Be Brainly....
Answered by
7
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