# Homework Help: Thermodynamics - Paramagnetism proof

1. Nov 5, 2006

### ultimateguy

Show that the entropy of a two-state paramagnet, expressed as a function of temperature, is S=Nk[ln(2coshx)-xtanhx], where x=μB/kT. Check that this formula has the expected behavior as T->0 and T->infinity.

Not even sure where to start...

2. Nov 5, 2006

### Physics Monkey

Hi ultimateguy,

Welcome to Physics Forums. I guess this question basically has two parts. The first part is mathematical, you should just take the two limits x -> 0 and x -> infinity and see what you get. The other bit is more physical, you have to ask yourself what you expect to happen? You might start by taking the limit and seeing if anything looks familiar.

Last edited: Nov 5, 2006
3. Nov 5, 2006

### ultimateguy

But the main part of the question is to show the steps to get to that equation. They want me to derive it.

4. Nov 5, 2006

### Physics Monkey

What theoretical tools are available to you? Can you compute the partition function of the system? Or are you meant to tackle the problem with more elementary means?

5. Nov 5, 2006

### ultimateguy

I don't know what a partition function is. Is it the same as a piecewise defined function?

I'm guessing that what I'm supposed to do is use 1/T=ds/dU and U=-NμBtanh(μB/kT).

I have an idea, can I go from 1/T=ds/dU to dU/T=ds and then integrate?

6. Nov 5, 2006

### Physics Monkey

Excellent idea, try it out and see what you get.

Edit: The partition function is an important quantity you will probably meet later on which enables you calculate many thermodynamic quantities very simply. The log of the partition function is proportional to the free energy F = E - TS of your system.

Last edited: Nov 5, 2006
7. Nov 5, 2006

### ultimateguy

I think I might be on the right track now.

C = dU/dT = Nk((μB/kT)^2/cosh^2(μB/kT))
1/T=dS/dU

so C/T=(dS/dU)(dU/dT)=dS/dT

and dS/dT=(1/T)(Nk((μB/kT)^2/cosh^2(μB/kT)))

and if this is right, how on earth do I integrate that to get S?

EDIT: when i try the first idea, I get S = -NμB*integral(tanh(μB/kT)/T), how do i integrate that?

Last edited: Nov 5, 2006
8. Nov 5, 2006

### quasar987

While our friend is calculating, do you have an idea why E-TS is called the free energy?

9. Nov 5, 2006

### Physics Monkey

You probably want to start by changing variables from 1/T to the x variable defined in your first post. Then integration by parts is appropropriate. However, your final expression involving tanh isn't right, I would recommend going back to the expression involving sech^2 and change variables to x, then integrating by parts.

10. Nov 5, 2006

### Physics Monkey

Basically because under certain circumstances it represents the energy available to do "useful work." Structurally, this is fairly clear since F = E - TS = - PV is basically the total energy minus the "heat", again with the remainder being though of as available for "useful work." But what constitutes the free energy depends on the environment, so Helmholtz is appropriate when you control the volume, but Gibbs is more appropriate when you control the pressure.

Edit: But I should warn you against taking such things too literally.

Last edited: Nov 5, 2006
11. Nov 5, 2006

### ultimateguy

So if I change the variables to x, I would get this?

dS/dT = (Nk^2/uB)(x^3)(sech^2(x))

When I integrate, do I just integrate by parts with respect to x and then change x back to uB/kT?

12. Nov 5, 2006

### Physics Monkey

You're on the right track now. What you want to do next is multiply both sides by dT and then write dT in terms of dx. Youll have then something like dS = f(x) dx which you can integrate by parts directly to find S(x).

13. Nov 5, 2006

### ultimateguy

So what I need to integrate is dS=Nk(T^2)(x^3)sech^2(x)dx?

14. Nov 5, 2006

### Physics Monkey

Yes, except that your units aren't quite right and you should write the T in terms of x, but you're almost there!

15. Nov 5, 2006

### ultimateguy

I figured out the units problem, and integrated dS=-Nkx(sech^2(x))dx and got Nk(ln(coshx)-xtanhx). That's the solution except there's supposed to be a 2 in front of the coshx..... I don't know where the 2 comes in.

16. Nov 5, 2006

### Physics Monkey

Remember that whenever you integrate there is always a constant of integration. The Nk ln 2 term is this constant (use the properties of log to see this). You can determine this constant by making sure the Third Law is satisfied in the limit T -> 0. This is part of making sure your expression has the correct limits.

17. Nov 5, 2006

### ultimateguy

I still don't understand, if T->0 then S->0 and if T->infinity then S->-infinity.

So if the result of the integration is Nk(ln(coshx)-xtanhx + ln2). I can see how it works, I just don't understand how I'm supposed to know to assume that the constant has that exact value.

Edit: I think I see it now. The lowest value the entropy can have is Nkln2, so when T->0 it must be that value. Is that right?

Last edited: Nov 5, 2006
18. Nov 5, 2006

### Physics Monkey

You do want S -> 0 as T -> 0, but you need that 2 to make things work. To see this note that as T -> 0, x -> infinity. When x is very large you can approximate tanh x as 1 and cosh x is e^x/2. So if you want the expression S(x) = Nk(ln(cosh x) - x tanh x + C) to go to zero as x gets large, you must choose C in a very special way. This is the origin of the choice of constant. Work it out for yourself to see.

Note that you can also figure the constant out by looking at the high temperature limit. As T goes to infinity, you expect the entropy to become simply N k ln 2 which again works out only if you choose C just right.

Last edited: Nov 5, 2006
19. Nov 5, 2006

### ultimateguy

I got it! Thank you very much for the help.

Usually I just do these problems, get frustrated and quit. This one was actually fun to work through and figure out. I have to stop by here more often.

Thanks again!

20. Nov 5, 2006

### Physics Monkey

My pleasure, be sure to come back if you need help again.