Negative temperature

Let's think about the concept of temperature for a moment. What is a temperature, in terms of how you measure it in real life? When we measure temperature, we put a small system (a thermometer) in contact with a larger system. The thermometer comes to equilibrium with the larger system by exchanging energy until the total entropy of both systems is maximized. What we read off as temperature has to do with how the fluid inside the thermometer expands or contracts.

Systems at negative temperature come to equilibrium by the same means. Recall that, when we put two systems together that are isolated, the total entropy is maximized, S = S1+S2 while the total energy is held constant, E=E1+E2. Thus the sign of dS must be positive. But dS is just related to the temperature, when the energy changes.

A thermometer put in contact with a system at negative temperature will come to equilibrium in the same way it would as if it were put into contact with a system at positive temperature. The two systems will exchange energy until their temperatures are the same. Whatever temperature the thermometer is at in the end will depend on how much energy it exchanged with the system.

Not all systems can have negative temperatures. Only systems that have entropy functions that contain regions of negative slope do. Some systems have entropy functions that increase monotonically with energy for all energies, and thus, these systems cannot have negative slopes. Fluids have the latter kind of entropy function. Magnets can have the former. The basic difference between the two is that fluids can have an infinite energy (because the momenta are unbounded), but magnets can be saturated at a maximum energy.

TA office hours

Sophia's office hours this week will be 1pm tomorrow (Wed) in ESB 3413.

Correction to problem 6d on PS6

It turns out that neither bottle can be frozen the way mentioned. Here is a new problem statement for 6d:

An alternative way to freeze the solutions is to change the pressure. At 0 °C for each bottle, what pressure must be reached in order for freezing to begin? Assume the water freezes first, and assume constant molar densities. Why are these numbers unrealistic? What is special about water freezing here, relative to other liquids?

Hints for problem 5 on PS6

If you are having trouble with problem 5, go back and take a look at the lecture "Stability" to see a related derivation of tangent line properties, for the liquid-vapor transition.

Also, don't forget that the conditions for the critical point for the liquid-vapor transition were:
d^2 a /dv^2 = 0 and d^3 a / dv^3 = 0
A similar set of two equations exists for the mixture.

PS6 problems 3 and 4

Two corrections:

On problem 3, the radius of a cell is 10 um (microns).

On problem 4, the unsubscripted x's are just the same x's as those that precede them.

PS7-8 and online notes

Per the class vote, PS7 and PS8 will be due the Friday following their original Wednesday due dates. Note that this is not the case for PS6.

It was pointed out to me that some of the graphs are inverted in the lecture notes. I've tried to fix this for the lectures going back to "Solutions", so there are updated versions on the website. Let me know if you're still having problems.

Comments on midterm grades

Just to clarify:

The number in blue ink at the top of your midterm exam called "midterm average" is your average in the class for the midterm exam plus PS1-4. Both a percentage and letter grade are listed. Note that it is not the average of everyone in the class on the midterm.

Also, the green dot on everyone's exam indicates that you turned in your survey.

Scott

PS5 correction

It just came to my attention that I inverted the derivative in problem 4 on PS5. The left hand side in both expressions should be (dT/dP).

Class Tuesday

In order to make up one of the lectures from this past week, this coming Tuesday's discussion class (Nov 13) will consist of a normal lecture (1 hour) followed by time as needed to answer questions about the problem set. I have arranged to keep the classroom for additional time.

Info on the midterm

Here is a summary of information about the midterm:

You can pick up your exam at the front chemical engineering office no earlier than 9am Monday morning, Nov. 5.

The exam is due back at the office no later than 11am Tuesday morning, Nov. 6.

Please write neatly on separate paper and staple your work together.

Happy studying!

Problem 2 on the midterm practice problems

I was told there may be some confusion about part 2b. Here is how to approach that problem.

For any phase, we can write the entropy as a function of the pressure and temperature:

S = S(T,P)

The total derivative is:

dS = (dS/dT) dT + (dS/dP) dP

= (Cp/T) dT - (alpha V) dP

Where Cp is the constant pressure heat capacity and alpha is the thermal expansion coefficient. In the second line, the (dS/dP) term was converted with a Maxwell relation to get -(dV/dT).

If we have a line along which the entropies are equal between two phases:

S1 = S2 (along Kauzmann locus)

The full differential must also be equal:

dS1 = dS2 (along Kauzmann locus)

Substituting the above relation:

(Cp1/T) dT - (alpha1 V1) dP = (Cp2/T) dT - (alpha2 V1) dP
(along Kauzmann locus)

Now simplifying:

(dP/dT) = Delta(Cp) / [T Delta(alpha V)] (along Kauzmann locus)

In general, the Kauzmann locus will extend into regions of (T,P) space where one is comparing a metastable state to a stable one. In the helium example given, the Kauzmann locus occurs in some parts of the phase diagram where the liquid is supercooled and the crystal is stable. It also occurs in places where the crystal is superheated and the liquid is stable. Where the Kauzmann locus intersects the phase boundary between the liquid and crystal, there is a Kauzmann point.