Kinetic energy change during phase change

  • Thread starter rht
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  • #1
rht
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Main Question or Discussion Point

hi all,
its been hours that i could not find a decent answer for a 'simple' Q:

during a phase change (say, boiling) the temp' doesnt change, as we all know.
we also know that the temp' is a measure of the system kinetic energy (KE).

im interested to know how the average KE AND its distribution (follow the Maxwell-Boltzmann dist', right?) looks like, both in the liquid and in the vapor.

how can we explain the fact that once mol' go to the vapor, with higher KE, the temp' stays the same?

do u have a plot of the KE dist' of both phases?

same goes for melting...

tnx,
roy
 

Answers and Replies

  • #2
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we also know that the temp' is a measure of the system kinetic energy (KE).
Temperature is related to the unordered kinetic energy, but the relation depends on the medium and its degrees of freedom (=> it can change during melting/boiling).

The Maxwell-Boltzmann distribution applies to gases only.
 
  • #3
rht
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tnx, but this doesnt really answer the question....

let me put it in another way - doest the gas molecules have the same avg' KE or higher than the liquid molecules?
 
  • #4
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The molecules in the liquid are held together by attractive forces (potential energy), and when you add heat to cause evaporation, most of the heat goes into overcoming the attractive forces.
 
  • #5
rht
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yes, this i know..
can u illustarte the avg' KE of the vapor during the boiling process (until all liquid was vaporized)?
 
  • #6
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The average kinetic energy should stay similar. This is another way to express what Chestermiller wrote.
 
  • #7
rht
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sorry for the nagging, but i feel this is exactly the gap i have -


im still trying to figure out what is going under boiling and melting - where u introduce heat to the system all the time, so the liquid keep its temp' (100C for water for example).
if the hottest mol' leaves the liquid first, their KE should be higher in the vapor than in the liquid (which holds a lot of slower mol').
in that case, the temp' of the vapor should be higher than in the liquid BUT during boiling, the temp' should be const' ?!?!?!? - and this is my hole in understanding.

can u ellaborate how the liquid and the vapor KE distribution will look like during boiling (or melting)?

by the way, u r right about the dist' - in liquid is just the Boltzmann (although, doesnt make much of a difference for this discussion :smile:).

tnx again
 
  • #8
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Understand that KE can be the same numerically but still quite different. Think about something vibrating, and something thrusting forward.
 

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