# Thermodynamics of Rubber Band

1. Apr 30, 2014

### decerto

1. The problem statement, all variables and given/known data

The equation of state for a rubber band with temperature T is $\mathcal{F}=aT\left[\frac{L}{L_0}-\left(\frac{L_0}{L}\right)^2\right]$

Where $\mathcal{F}$ is the tension, L is the stretched length and L_0 is the unstretched length

a) Write the Central Equation for the rubber band

b) Derive the energy equation for the rubber band $\left(\frac{\partial U}{\partial L}\right)_T$

c) Show that U is a function of T only

2. Relevant equations

$dU=dQ+dW$

$dU=\left(\frac{\partial U}{\partial T}\right)_LdT +\left(\frac{\partial U}{\partial L}\right)_TdL$

3. The attempt at a solution

a) For the central equation some variant of $dU=TdS+\mathcal{F}dL$ I pressume?

b) Comparing the two relevant equations $\left(\frac{\partial U}{\partial L}\right)_T=\mathcal{F}$ ?

c)No real idea how to show this

2. May 1, 2014

### TSny

OK

This is incorrect. Note that from $dU=TdS+\mathcal{F}dL$ you can get $\left(\frac{\partial U}{\partial L}\right)_S=\mathcal{F}$. But you need to get an expression for $\left(\frac{\partial U}{\partial L}\right)_T$.

One approach is to rearrange your central equation for dS and then use one of your relevant equations to substitute for dU. That should express dS in terms of dT and dL. Then try to see what to do.

Once you get the answer for (b) this will be easy.

3. May 2, 2014

### decerto

With $dS=\frac{1}{T}dU-\frac{F}{T}dL$ I'm slightly confused on how to incorporate dT $dU=\left(\frac{\partial U}{\partial T}\right)_LdT +\left(\frac{\partial U}{\partial L}\right)_TdL$ is wrong yes? I think I just made that up.

4. May 2, 2014

### decerto

$dU=TdS +\mathcal{F}dL$

$\left(\frac{\partial U}{\partial L}\right)_T=T\left(\frac{\partial S}{\partial L}\right)_T +\mathcal{F}\left(\frac{\partial L}{\partial L}\right)_T$

$\left(\frac{\partial U}{\partial L}\right)_T=T\left(\frac{\partial S}{\partial L}\right)_T +\mathcal{F}$

Then using the maxwell relation $\left(\frac{\partial \mathcal{F}}{\partial T}\right)_L=\left(\frac{\partial S}{\partial L}\right)_T$

We have

$\left(\frac{\partial U}{\partial L}\right)_T=T\left(\frac{\partial \mathcal{F}}{\partial T}\right)_L +\mathcal{F}$

Using the equation of state we then have $\left(\frac{\partial U}{\partial L}\right)_T=2\mathcal{F}$

5. May 2, 2014

### TSny

OK. This is good. But, check your signs in the Maxwell relation. Note $\mathcal{F}$ for the rubber band corresponds to $-P$ for a usual thermodynamic system.

6. May 2, 2014

### decerto

So its zero which makes the next part trivial, thanks for the help.