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Lead's melting point as a function of pressure (Thermo')

  • #1
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Homework Statement


When lead is melted at atmospheric pressure, the melting point is 600K, the density decreases from 1.101x104 to 1.065x104 kgm-3. The latent heat is 24.5 kJ kg-1

Estimate the melting point of lead at a pressure of 100 atm.

Homework Equations



I haven't been able to find any equations for how the melting of a solid varies with pressure. Intuitively I'd say the melting point will go down, but that's really just a guess.

The Attempt at a Solution


If the density decreases, the material expands. This means it is doing work on the surroundings, so some energy that goes in does that. I looked at some data from a paper (not what the question wants) that tells me that the melting point of lead goes up as pressure goes up in the range of 800 to 1200 MPa.
 

Answers and Replies

  • #2
DrClaude
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Look up the Clausius-Clapeyron relation.
 
  • #3
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Okay, yup, I've looked it up

dp/dT = L/T(Vf-Vi)

So by subbing in the values I get the gradient of the coexistance curve, which turns out to be 0.113 at 1 atm.
I'm not sure how I would find the gradient of the curve for 600 atm?
 
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  • #4
DrClaude
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Okay, yup, I've looked it up

dp/dT = L/T(Vf-Vi)

So by subbing in the values I get the gradient of the coexistance curve, which turns out to be 1.13x10-4 at 1 atm.
I'm not sure how I would find the gradient of the curve for 600 atm?
I guess you mean 100 atm. You can simply make the assumption that the gradient is constant. If you look at a phase diagram, you will see that the coexistance curve can be reasonably approximated as a straight line.
 
  • #5
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Okay well in that case the melting point would be the same? If the gradient of the line is independent of pressure then so is the melting point given fixed values of L and deltaV.
 
  • #6
DrClaude
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If the gradient of the line is independent of pressure then so is the melting point given fixed values of L and deltaV.
That is not correct. The only assumption you are making is that the coexistance curve is a straight line.
 
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  • #7
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Okay, so if the coexistance curve is a straight line, then dP/dT is constant. This means L/T(Vf-Vi) is constant. So if L and (Vf -Vi) are constant then doesn't that make T constant?
 
  • #8
DrClaude
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Okay, so if the coexistance curve is a straight line, then dP/dT is constant. This means L/T(Vf-Vi) is constant. So if L and (Vf -Vi) are constant then doesn't that make T constant?
[Edit: I made a mistake, forget about the coexistance curve being a straight line, which it is not.] You have ##\Delta P / \Delta T = \mathrm{const.}##, therefore what happens for ##\Delta P \neq 0##?
 
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  • #9
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Well if delta p doesn't equal zero then delta T can't equal zero either.


Is the trick to integrate both sides? If I do that I get P = ln(L/T(Vf-Vi))+C.
Subbing in the values for 1 atm I get C=3.18 but does this value only stand for 1atm? I don't think I can use it for 100atm...

I'm tempted to just guess and say that if you multiply the pressure by 100 the melting point also gets multiplied by 100.
 
  • #10
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Right, got it. If dp/dT is constant then P=kT + C where k is he constant of proportionality 0.113. Using the initial values in the question, C = -66.8.
The equation now stands as P = 0.113T -66.8
Letting P = 100 (as opposed to the previous 1), we can find T to be 1476K

Thanks a lot for all your help DrClaude!
 
  • #11
DrClaude
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I'm sorry, but I have been leading you in a the wrong direction. I made a mistake with that idea that the coexistance curve was linear. Let me think a bit to get you back in the right direction.
 
  • #12
DrClaude
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Ok, I got my brain back in gear. What is constant is ##L/\Delta V##, where ##\Delta V = V_\mathrm{l} - V_\mathrm{s}##, the difference in volume between the two phases, which we can approximate as constant for a solid-liquid transition.

Is the trick to integrate both sides? If I do that I get P = ln(L/T(Vf-Vi))+C.
Subbing in the values for 1 atm I get C=3.18 but does this value only stand for 1atm? I don't think I can use it for 100atm...
You have to be careful here: it is a definite integral you have to do.

I'm tempted to just guess and say that if you multiply the pressure by 100 the melting point also gets multiplied by 100.
Wouldn't that give you the same result whatever the value of ##L/\Delta V##?
 
  • #13
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So if dP/dT = k/T where k is L/ΔV.
Integrating both sides, we get P2 - P1 = kln(T2/T1) This is the definite integral?
 
  • #14
DrClaude
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So if dP/dT = k/T where k is L/ΔV.
Integrating both sides, we get P2 - P1 = kln(T2/T1) This is the definite integral?
Yes, that looks correct. You now need to rewrite this as an equation for ##T_2##.
 
  • #15
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Okay now I'm getting absolutely collossal numbers.
T2=T1exp(P2-P1/k)
T2=600exp(107/68.06)
That's with the pressure in pascals to match the rest of the units. Surely the melting point shouldn't depend on units like that?
 
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  • #16
DrClaude
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Okay now I'm getting absolutely collossal numbers.
T2=T1exp(P2-P1/k)
T2=600exp(107/68.06)
That's with the pressure in pascals to match the rest of the units. Surely the melting point shouldn't depend on units like that?
Your value for k is incorrect.
 

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