# Thermodynamics, rubber band

1. Jun 30, 2013

### fluidistic

1. The problem statement, all variables and given/known data
Hi guys, I'm trying to solve questions from past exams; here comes one.
There are 2 possible fundamental equations for the rubber bands:
$S=L_0\gamma \left ( \frac{aU}{L_0} \right ) ^{1/2} -L_0 \gamma \left [ \frac{1}{2} \left ( \frac{L}{L_0} \right ) ^2 + \frac{L_0}{L} -\frac{3}{2} \right ]$ and $S=L_0\gamma \exp \left ( \frac{aUn}{L_0} \right ) -L_0 \gamma \left [ \frac{1}{2} \left ( \frac{L}{L_0} \right ) ^2 + \frac{L_0}{L} -\frac{3}{2} \right ]$ where $L_0=nl_0$, a and gamma are constants. L is the length of the rubber band and U is the internal energy while $l_0$ is the molar length of the rubber band when the tension vanishes.
1)Which one of these equation is acceptable? Why?
2)From the right fundamental equation, determine the state equation $f(T,L)$ that relates the tension f with T and L.
3)We put 2 rubber bands in contact with the parameters $T_0^i$, $l_0^i$,$n_0^i$ where i=1, 2. Assume that there's no flow of matter between the 2 bands nor heat flux around the system. Calculate T in function of the initial parameters.

2. Relevant equations
$\left ( \frac{\partial S}{\partial L} \right ) _{U,n}=\frac{f(T,L)}{T}$ (I believe).

3. The attempt at a solution
1)The second expression for S isn't an extensive function because of the argument in the exponential. Basically if I double the system (n'=2n), I wouldn't get S'=2S, which isn't acceptable. On the other hand the first expression for S is an extensive function. So my answer would be the first expression is acceptable because the 2nd one isn't due to a lack of extensivity.
2)I used the expression given in "Relevant equations" and I reached that $f(T,L)=T \gamma \left ( \frac{L_0^2}{L^2} - \frac{L}{L_0} \right )$. So when L=L_0 I get that the tension is worth 0 which seems good. When $L>L_0$ I get that the tension is negative... I have no idea if this is good. Intuitively I know that if I stretch the rubber band the tension would be like a restore force, but I don't know if a negative force means a force that goes against the stretching, by convention.
So I don't know here if my answer is correct. What do you think?

Edit:3) I know that the total entropy will be the sum of the entropy of each one of the rubber bands. Same for the internal energy. I was thinking about using the definition of temperature to get it. Namely that $\left ( \frac{\partial S}{\partial U} \right ) _{L,n} =1/T$.
But this soon becomes a mess. I reached that $1/T=\frac{\partial S ^{(1)}}{\partial U}+ \frac{\partial S ^{(2)}}{\partial U}$ where $U=U^{(1)}+U^{(2)}$. But I don't have $S^{i}$ in terms of U.
So I don't really know how to tackle this part. Any tip is appreciated!

Last edited: Jun 30, 2013
2. Jul 1, 2013

### Staff: Mentor

Here's what I think: I think you analyzed parts 1 and 2 correctly, including your assessment of the sign of f. Apparently, in this formulation, positive f represents compression, and negative f represents tension. You know this from dU = TdS - f dL. In this version, fdL is the same as pdV in the ordinary form of the equation, and, as we know, p is isotropic compressive stress.

As far as part 3 is concerned, I need to think about it some more. But I'm pretty sure that the overall entropy change is not zero.

Chet

3. Jul 1, 2013

### fluidistic

Thank you very much for the feedback.
I think you are right about the entropy for the part 3), I'll have to reconsider this too.
As a side note, in another problem I must show a relation and I must assume that the change in entropy is 0 but I don't understand why it would be 0.
The problem is in Callen's book (1ed. page 114) and reads as follows:
I have solved the problem and reached the good answer but I had to assume that the entropy of the total system before and after opening the hole remains constant.
I don't understand how on Earth would the entropy stay constant before and after the small hole is drilled. Is it because the substance is the same in each containers? Even so, I still don't understand...

4. Jul 1, 2013

### Staff: Mentor

Regarding part 3: Can it be assumed that the lengths of each of the two rubber bands does not change after they are brought into thermal contact?

I'm thinking that one needs to look at each of the two rubber bands individually. If the lengths of each of the two rubber bands does not change, then the change in internal energy of the system does not change. That would mean that the change in internal energy for one rubber band must equal minus the change in internal energy of the other rubber band. Therefore, one would need to determine the change in the internal energy of a rubber band as a function of temperature at constant length. This would be analogous to determining the change in internal energy of a gas as a function of temperature at constant volume.

Chet

5. Jul 7, 2013

### fluidistic

Sorry for the late reply. Hmm I don't think we're meant to assume non stated assumptions.
For example I assumed, in the final exam worth 100% of my grade that I took a few days ago and that I totally failed, that the internal energy of a magnetic system only depended on the temperature and not volume. It turned out that this was a good assumption but since I didn't justify it mathematically, I got absolutely no consideration for my answer (which was the right result).