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Effect of pressure on Melting point

by emailanmol
Tags: effect, melting, point, pressure
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emailanmol
#1
Mar19-12, 06:56 AM
P: 297
Hey, I have read that INCREASING the external pressure on solids INCREASES their melting point.(except for ice)

WHY DOES THIS HAPPEN?

(This is similar to the effect on Boiling point which rises when external pressure is increased.
I know this happens because boiling takes place at temp where vapour pressure of liquid state is equal to external pressure and since external pressure is higher , the liquid will boil at a higher vapour pressure.And since vapour pressure is proportional to temp , a higher temp is needed to boil the liquid.)
CAN YOU EXPLAIN WHY THIS HAPPENS FOR MELTING POINT USING SIMILAR ARGUMENTS?

2)

Also WHAT IS THE DEFINITION OF MELTING POINT.(Lol.I dont want answers saying temp where solid converts to liquid.)
I want it in terms of pressure(like the way its defined for boiling point, tenp at which vapour pressure of liquid equals external pressure)

I am guessing MELTING POINT is TEMP at which SOLID VAPOUR PRESSURE is equal to LIQUID VAPOUR PRESSURE
AM I RIGHT?
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moogull
#2
Mar19-12, 08:51 AM
P: 82
The most simple interpretation I think is like this: if you are pushing down harder on a substance (i.e. more pressure) then a molecule must overcome a greater force to break free from the forces holding it together as a solid and become a liquid or gas. Therefore you have to have a greater temperature in order for it to have enough energy to break solid bonds? There is not physics to this answer really, but I'll start looking in a thermo book I have and come up with a better answer in terms of pressure.
emailanmol
#3
Mar19-12, 09:16 AM
P: 297
Hey, Moogull.

Thank you so much for your reply (and effort)
You make a great point.

Incidently even i was thinking that molecules would need more kinetic energy to escape than before and thus a higher temp.

It would be really cool if someone comes up with something in terms of vapour pressure :-)

moogull
#4
Mar19-12, 09:18 AM
P: 82
Effect of pressure on Melting point

I'm in class right now, but I'll find something for you to look at, don't worry!
emailanmol
#5
Mar19-12, 09:24 AM
P: 297
Sure :-)
Thanks a lot
moogull
#6
Mar19-12, 11:08 AM
P: 82
I'm not so sure that vapor pressure would be a good way to think about a melting phase transition. I wouldn't go so far as to say that solid vapor pressure and liquid vapor pressure are equal. Here is a link about vapor pressures for solids and liquids: http://www.chem.purdue.edu/gchelp/liquids/vpress.html

If you were to have a mixed state (solid and liquid), or just a solid or a liquid of a substance in a closed container then there would only be one vapor pressure that you could measure. I don't know if solid vapor pressure vs. liquid vapor pressure is conventional. In fact if you have a solid in a closed container then at certain pressures (for certain substances) it will sublimate/go directly to a gas and that is where you have solid vapor pressure, but if you have a substance that is melting, I don't know too much about the vapor pressure in that container. I guess there would be some contribution to vapor pressure from the liquid and maybe from the solid as well, but if the liquid is less dense then it will be sitting on top of the solid and the liquid will be the one contributing to vapor pressure (in a continuum like model).

Here is another link that talks about the thermodynamics of melting in crystals (its a pdf)
Thermodynamics of Crystal-Melt Phase Change

That resource talks about phase changes in terms of enthalpy. I haven't done too many searches, but it seems like vapor pressure is not talked about too much in melting phase change.
DrDu
#7
Mar19-12, 11:23 AM
Sci Advisor
P: 3,596
Quote Quote by moogull View Post
I'm not so sure that vapor pressure would be a good way to think about a melting phase transition. I wouldn't go so far as to say that solid vapor pressure and liquid vapor pressure are equal.
No, emailanmol is right. Both liquid and solid are in equilibrium and obviously are in equilibrium with the vapour. Hence vapour pressure has to be the same. If not, e.g. liquid would evaporate and condense on the solid.

The change of the phase equilibrium line with pressure is given by the Clausium Clapeyron equation
dT/dp=T Delta V_m /Q_m, where Delta V_m is the difference of molar Volumes of the two phases and Q_m the heat of melting. To prove this relation, you need a good deal of thermodynamics:

http://en.wikipedia.org/wiki/Clausiu...eyron_relation
emailanmol
#8
Mar19-12, 11:25 AM
P: 297
Hey, the links you posted are amazing. This is all that I needed.Now I have better understanding.
Thanks a lot
moogull
#9
Mar19-12, 11:30 AM
P: 82
Thanks DrDu,

As he said, there is a lot of thermo behind the euqation, and I am only in my first semester of it so I'm kind of shooting in the dark. Question to DrDu, so solid vapor pressure would be those molecules that boil off of the solid and liquid vp is off of the liquid, so would the total vapor pressure in a closed container be equal to rate of condensation plus rate of deposition?
emailanmol
#10
Mar19-12, 11:36 AM
P: 297
Hey DrDu,

Thanks for clearing that up.Could you explain in laymens' terms on how solid vapour pressure depends on temp and external presure.
Does it increase or does it decrease or remains constant.


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