Thermodynamics of an elastic band

[bon]

Homework Statement

The Gibbs free energy of an elastic band satisfied dG = -SdT - LdF where F is the tension and L the string's length.

At fixed temp the entropy is given by S = So - a(L-Lo)^2 + b(L-Lo)^2

Where Lo is the length of the elastic at zero tension and a and b are positive constants.

Show that the thermal expansion coefficient at zero tension is negative.

Homework Equations

The Attempt at a Solution

So I'm trying to show that partial L wrt partial T at constant F < 0.

Using a maxwell-type relation we can say dL/Dt)f = dS/dF)T

But I'm not sure where to go from here..the expression for S doesn't contain F and I can't see how to solve..
thanks!

 

[Mapes]

At fixed temp the entropy is given by S = So - a(L-Lo)^2 + b(L-Lo)^2

Is this part correct? It seems like an odd way of writing it; I suspect there's a typo or another condition in there somewhere.

 

[bon]

It is correct yes! And no typo..this was on a past exam paper.

 

[Mapes]

It is correct yes! And no typo..this was on a past exam paper.

It doesn't seem odd that the entropy expression could be simplified to S = So + c(L-Lo)^2, where c = b-a? I don't see a way to solve it unless it's also assumed that a > b.

 

[rwisz]

I would approach this problem by first understanding the concepts of thermodynamics and elasticity. Thermodynamics deals with the relationships between different forms of energy and their ability to do work, while elasticity deals with the ability of a material to return to its original shape after being deformed. In this case, the elastic band can be seen as a system with both thermal and elastic energy.

The given equation for Gibbs free energy, dG = -SdT - LdF, shows that the change in free energy is dependent on both temperature and tension. This makes sense, as changing the temperature or tension of the elastic band will result in a change in its shape and therefore its free energy.

To show that the thermal expansion coefficient at zero tension is negative, we need to find the partial derivative of length with respect to temperature at constant tension, or dL/dT|F. Using the Maxwell relation, we can rewrite this as dL/dT|F = dS/dF|T.

Substituting the given expression for entropy, S = So - a(L-Lo)^2 + b(L-Lo)^2, we can find the partial derivative of S with respect to F at constant T, which is simply the coefficient of (L-Lo)^2. This equals 2b(L-Lo), which is always positive since b is a positive constant.

Therefore, we can conclude that dL/dT|F is always negative, meaning that the thermal expansion coefficient at zero tension is negative. This makes sense intuitively, as increasing the temperature of the elastic band will cause it to expand, but at zero tension, the elastic band will not be able to expand and will instead contract as the temperature increases.