Determing triple point temperature

In summary: You don't need to calculate three different points, you can use the equation dP/dT= ΔH/TΔV to calculate the slope of the line at any given temperature.
  • #1
ebunny91
13
0

Homework Statement



The metal Eborium (Eb) has three solid phases: a, b, c. At a pressure of 0.45 atm, the a and b phase coexist at 70 K (temperature). The molar volume of Eb(a) is 1.23 liter/mole and that of Eb(b) is 1.47 liter/mole. The heats of transformation are as follows:

Eb(a) ---> Eb(c) delta H (a --> c) = 375.8 J/mole

Eb(b) ---> Eb(c) delta H (b --> c)= 24.3 J/mole

At a pressure of 0.59 atm, three phases may be present. Determine the temperature at which the system must be brought to insure that three phases are present (Assume this is close to 77 K (temperature)).


Homework Equations



Clausius-Clapeyron: 1)dp/dT= deltaH/(T deltaV) or 2)ln(p2/p1)= - (deltaH/R)(1/T2 - 1/T1)


The Attempt at a Solution



I integrated eqn 1) on both sides and got p= (deltaH/deltaV) lnT2/T1 and got 2 equations after that by plugging eveything in :

p= (375.8/1.23) ln T2/77 and p= (24.3/1.47) ln T2/77. When I equate these two equations I get T2=77 K which doesn't seem right. What am I doing wrong? What's confusing me is that the question says "three solid phases a, b and c" so am I using the right equations?
 
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  • #2
Welcome to PF, ebunny91! :smile:

I'm afraid your integration of equation 1 and your subsequent substitution are both not quite right.
Do you have a reason to assume that deltaV is independent of temperature?
Furthermore your LHS should become p2-p1 instead of just p (there is an integration constant).
Then you substituted the volumes instead of the delta-volume.
And you substituted a T1 of 77 K, while your problem states that the measurement is at 70 K (or is there a typo involved?)

I believe your second equation should not be used, since it assumes a transition to a gas phase that behaves like an ideal gas.


However, there is third way to approach this problem.
For small changes, you can apply ΔT≈(dT/dp) Δp...
 
  • #3
It was a typo yea. It's supposed to be 77 K.

I know that at the triple point all three solid phases should be in equilibrium at the same temperature and pressure. The problem already gives us the pressure (0.59 atm) so now we need to find the temperature at the triple point. (I'm just typing this out to help me understand).

We can't use the 2nd equation because there aren't any gases involved, right?

The amount of given info is confusing me. I think I am overlooking a really simple concept, but I'm not sure what it is.
 
  • #4
There are 3 p-T phase lines involved that intersect.

You have a point (p1,T1) on one of the three lines, and you can calculate the slope dp/dT of that specific line.
Furthermore you have the pressure p2 of the triple point...
 
  • #5
Oh ok, so dP/dT= ΔP/ΔT = P2-P1/T2-T1 = 0.59-0.45 atm/T2 - 77K

and then maybe I can use dP/dT= ΔH/TΔV to get the slope of the same line? But I don't have ΔH (a-->b)...
 
  • #6
Can you perhaps think up a process that starts with a and ends in b?
 
  • #7
Well maybe I can use the two equations to get the ΔH (a-->b)

First eqn:
Eb(a) ---> Eb(c) ΔH (a --> c) = 375.8 J/mole

and flip/negate the second eqn:

Eb(c) ---> Eb(b) ΔH (b --> c) = - 24.3 J/mole

So Eb(a) ---> Eb(b) ΔH = ΔH (a --> c) + ΔH (b --> c)
= 375.8 - 24.3 = 351.5 J/mol

so dP/dT = ΔH/TΔV
= ΔH/T2 (V(b) - V(a))
= 351.5/T2 (1.47 - 1.23)

So by equating the two equations:

351.5/T2 (1.47 - 1.23) = 0.59-0.45 atm/T2 - 77K

Then solving for T2, I get 77.00736116 K. Which seems about right since the question says to assume the temperature at the triple point is close to 77 K (temperature)). Is this the right method and is my answer correct?
 
  • #8
Looks good!

However, you're not supposed to "solve" T2.
You're just supposed to calculate dp/dT at the measured temperature T1.
The formula dp/dT is for one specific point (p1,T1) where the volumes were measured.
At a different T, the volumes will be different, so the formula wouldn't hold.
 
  • #9
But then I would only have the slope of the line. I have to determine the temperature at which the system must be brought to insure that the three phases are present. Do I have to do something more to get the temperature of the triple point?
 
  • #10
ebunny91 said:
But then I would only have the slope of the line. I have to determine the temperature at which the system must be brought to insure that the three phases are present. Do I have to do something more to get the temperature of the triple point?

Huh? :confused:

You have a point on the line, the slope of the line at that point, and you have the final pressure.

Just don't fill in a variable T to find the slope at that point.
 
  • #11
I have two equations for the slope of a line:

351.5/T2 (1.47 - 1.23) and 0.59-0.45 atm/T2 - 77K

Since it is the same line, the two equations (slopes) should be equal. So I equated them to solve for final temperature.

The question tells me to assume that the final temperature is close to 77K so can't we assume that the molar volumes are the same?
 
  • #12
Your first equation for the slope of the line is wrong.
It's not at T2.

And no, you can't assume the molar volumes are the same at another temperature.
Molar volumes typically change with temperature.
 
  • #13
So in equation 1, it's at T1 = 77K. Then by equating the equations I only have one unknown which is T2 and that turns out to be 77.00736044 K
 
  • #14
ebunny91 said:
So in equation 1, it's at T1 = 77K. Then by equating the equations I only have one unknown which is T2 and that turns out to be 77.00736044 K

Yes, that's what I meant.

One more thing though.
I think you have forgotten to convert your units to SI units.
Did you calculate with "atmosphere" and "liter"?
 
  • #15
Oh I forgot to convert the atm to joules/liter. Thanks!
 
  • #16
Cheers! :smile:

(I hope you meant atm to Pascal, and liter to m3.)
 

1. What is a triple point?

A triple point is the temperature and pressure at which a substance exists in equilibrium as a solid, liquid, and gas. This means that at the triple point, all three phases of the substance can coexist in perfect balance.

2. Why is determining the triple point temperature important?

Determining the triple point temperature is important because it is a unique property of a substance and can be used to accurately measure temperature and pressure. It also provides valuable information about the behavior and characteristics of a substance.

3. How is the triple point temperature determined?

The triple point temperature is determined experimentally by subjecting a sample of the substance to varying pressures and temperatures until all three phases are in equilibrium. The temperature and pressure at this point are then recorded as the triple point.

4. Can the triple point temperature change?

Yes, the triple point temperature can change with a change in the purity of the substance or a change in the atmospheric pressure. It is important to note the conditions under which the triple point temperature was determined for accurate measurements.

5. What are some common examples of substances with a triple point?

Some common examples of substances with a triple point include water, carbon dioxide, and sulfur dioxide. These substances have well-defined triple points that are used in various industries for accurate temperature and pressure measurements.

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